# Naked Science Forum

## On the Lighter Side => New Theories => Topic started by: Dave Lev on 24/11/2018 08:30:27

Title: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/11/2018 08:30:27
Further our discussion about new matter creation at the excretion disc of the Milky way:

I wonder why any new particle/Atom/Molecular is drifting outwards from the excretion disc of the milky way.
Let's look at the following formula for gravity force:
F = G * M * m / R^2
M = The mass of the SMBH
m = The mass of the particle/Atom/Molecular.

Based on this formula there is no change in the gravity force over time.
Hence, by definition each new atom has to stay at the same radius forever.

So how can we justify the drifting outwards mechanism of any new particle/Atom/Molecular?

Could it be that something is missing in the following formula:
F = G * M * m / R^2
Could it be that over time there is a change in the gravity force? (Even if it is only a very minor change).
Do you know that the American continent is drifting away from the European continent by only 1.5 cm per year.
In the past they were fully connected.
So, just based on that small change per year, we have got the vast ocean between those two continents after long time.
We know that:
1. The moon is drifting away from the Earth (by about 1.5 cm per year)
2. The Earth is also drifting away from the Sun.
3. All the planets in the solar system are drifting away from the Sun.
4. All/Most of the moons are drifting away from their host planet.

But objects can also drift inwards.
There are plenty of orbital objects (satellites...) around the Earth that drift inwards and eventually fall down.
Hence, could it be that gravity force must be changed over time?
Actually, if we think about it, it is clear that nothing can stay the same forever. Although there is no friction in space, it seems to me that even gravity can't stay the same forever.
There must be some small change in gravity force over time (even very small change over very long time)
So, could it be that we have to use the following formula:
F = G * M * m / R^2 +/- F(t)
F(t) = represents the change in the gravity force over time.
If the object is in orbital cycle which is too close to the host, the value of F(t) must be positive.
However, if the radius is long enough, the value must be negative.

So, with regards to our solar system:
All the planets are located far enough from the Sun.
Therefore, the value of F(t) for each one of them must be negative. Hence, over time (long enough), the gravity force on each planet is decreasing. In order to compensate that decreasing in gravity force, the planet must drift outwards.

In the same token, could it be that each particle in the plasma is located long enough from the SMBH.
Therefore, its F(t) is negative.
If so, the total gravity force on each particle in the plasma is decreasing over time.
That decreasing gravity force drifts away the particle from the center.

Do you agree with that?

Title: Re: How gravity works in spiral galaxy?
Post by: Ophiolite on 24/11/2018 14:08:54
The drifting of the continents is a very poor analogy for orbital mechanics.
The movement of the moon away from the Earth is a consequence of an exchange of angular momentum.
Any outward drift of planets can be acocunted for by the reduction of mass of the sun, although resonances with other planets could play a part.
Artificial satellites crashing to Earth are a consequence of air resistance.

Summary: your speculations are unfounded and unnecessary.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 24/11/2018 14:17:49
F = G * M * m / R^2
M = The mass of the SMBH
m = The mass of the particle/Atom/Molecular.

Based on this formula there is no change in the gravity force over time.
Hence, by definition each new atom has to stay at the same radius forever.
Nonsense.  R might be changing all the time, depending on the trajectory, velocity, acceleration of said particle.  M is also not constant.

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1. The moon is drifting away from the Earth (by about 1.5 cm per year)
It isn't drifting.  It is being pushed away.
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2. The Earth is also drifting away from the Sun.
Only because the forces pushing it away are slightly higher than the forces pulling it inward, and also because M is decreasing.  The sun sheds mass faster than it acquires new mass.

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But objects can also drift inwards.
There are plenty of orbital objects (satellites...) around the Earth that drift inwards and eventually fall down.
Hence, could it be that gravity force must be changed over time?
Not as evidenced by that, no.  Friction seems to be the dominant force here.

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That decreasing gravity force drifts away the particle from the center.

Do you agree with that?
No
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/11/2018 13:15:05
Thanks

With regards to Earth Moon orbital radius:

In the following article there is good explanation about Tidal friction:

http://curious.astro.cornell.edu/physics/37-our-solar-system/the-moon/the-moon-and-the-earth/111-is-the-moon-moving-away-from-the-earth-when-was-this-discovered-intermediate

"The Moon's orbit (its circular path around the Earth) is indeed getting larger, at a rate of about 3.8 centimeters per year.
The reason for the increase is that the Moon raises tides on the Earth.
This effect stretches the Earth a bit, making it a little bit oblong. We call the parts that stick out "tidal bulges."
Tidal friction, caused by the movement of the tidal bulge around the Earth, takes energy out of the Earth and puts it into the Moon's orbit, making the Moon's orbit bigger (but, a bit paradoxically, the Moon actually moves slower!)."
Let's focus on the following: "..takes energy out of the Earth and puts it into the Moon's orbit, making the Moon's orbit bigger..."
So, due the orbital cycle some energy is used to pull the "Tidal buges".
Therefore,  the Moon's orbit is bigger.
However, this bigger orbital radius of the moons means by definition less gravity force between the Earth to moon.
Therefore:

At t=0
The current radius is: 384,400 km (The distance between earth -moon, assuming a perfect cycle)
The gravity force is:
F = G * M * m / R^2
therefore we can say that at t = 0:
F(0) = G * M * m / R^2

One year from now at t = 365 days:
It is clear that the distance should be
384,400 km + 3.8 cm.
That by definition proves that the gravity force between the Earth Moon is decreasing.

So, why can't we write the following formula:

F(t) = G * M(t) * m(t) / R(t)^2 = F(0) - Δ(t)

F(t) = the gravity force at time t.
M(t) = the host mass at time t.
m(t) = The orbital object mass at time t.
R(t) = the radius at time t.
Δ(t) = Lost of gravity force due to Tidal friction (or due to any other interference).
F(0) = the gravity force at time 0.

Hence:
At t = 0
Δ(0) = 0

If t = 365 (for example -one year)
F(365) = G * M(365) * m(365) / R(365)^2 = F(0) - Δ(365).

R(365) = R(0) + 3.8 cm = 384,400 km + 3.8 cm.

Do you agree with that?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 25/11/2018 21:04:15
Do you agree with that?
Sure, I agree (F is changing, not G), but you contradict your OP where you say:
Based on this formula there is no change in the gravity force over time.
Hence, by definition each new atom has to stay at the same radius forever.
Here you assert that force is constant (instead of decreasing as in the moon example), and that R is constant, instead of increasing as in that example.

Yes, angular momentum is being transferred from Earth to the orbit of the Moon, and that shoves it to a higher radius, which slows down its linear speed.  Total mechanical energy is constantly being lost.

Your OP seemed to suggest otherwise.  Sure, it was about some atoms orbiting the galaxy, but small particles are hardly expected to follow a fixed-radius circle about the central gravitational source.
Title: Re: How gravity works in spiral galaxy?
Post by: mad aetherist on 26/11/2018 22:14:42
Further our discussion about new matter creation at the excretion disc of the Milky way:

I wonder why any new particle/Atom/Molecular is drifting outwards from the excretion disc of the milky way.
Let's look at the following formula for gravity force:
F = G * M * m / R^2
M = The mass of the SMBH
m = The mass of the particle/Atom/Molecular.

Based on this formula there is no change in the gravity force over time.
Hence, by definition each new atom has to stay at the same radius forever.

So how can we justify the drifting outwards mechanism of any new particle/Atom/Molecular?

Could it be that something is missing in the following formula:
F = G * M * m / R^2
Could it be that over time there is a change in the gravity force? (Even if it is only a very minor change).
Do you know that the American continent is drifting away from the European continent by only 1.5 cm per year.
In the past they were fully connected.
So, just based on that small change per year, we have got the vast ocean between those two continents after long time.
We know that:
1. The moon is drifting away from the Earth (by about 1.5 cm per year)
2. The Earth is also drifting away from the Sun.
3. All the planets in the solar system are drifting away from the Sun.
4. All/Most of the moons are drifting away from their host planet.

But objects can also drift inwards.
There are plenty of orbital objects (satellites...) around the Earth that drift inwards and eventually fall down.
Hence, could it be that gravity force must be changed over time?
Actually, if we think about it, it is clear that nothing can stay the same forever. Although there is no friction in space, it seems to me that even gravity can't stay the same forever.
There must be some small change in gravity force over time (even very small change over very long time)
So, could it be that we have to use the following formula:
F = G * M * m / R^2 +/- F(t)
F(t) = represents the change in the gravity force over time.
If the object is in orbital cycle which is too close to the host, the value of F(t) must be positive.
However, if the radius is long enough, the value must be negative.

So, with regards to our solar system:
All the planets are located far enough from the Sun.
Therefore, the value of F(t) for each one of them must be negative. Hence, over time (long enough), the gravity force on each planet is decreasing. In order to compensate that decreasing in gravity force, the planet must drift outwards.

In the same token, could it be that each particle in the plasma is located long enough from the SMBH.
Therefore, its F(t) is negative.
If so, the total gravity force on each particle in the plasma is decreasing over time.
That decreasing gravity force drifts away the particle from the center.
Do you agree with that?
1. I dont understand -- where did the new particle come from?
2. How do u know that it is drifting away?
3. Alby said that gravity aint a force -- its a bending of spacetime.
4. When u say that gravity cant stay the same for ever -- do u mean that big G changes with time? -- if so then the most logical way of correcting the formula is to apply an additional term or two directly to big G -- adding a stand-alone term onto the end of the formula is a highly unlikely solution.
5. U mention a mechanism -- Einsteinians dont worry about a mechanism for gravity -- they say that gravity is a delusion created by mass somehow bending the fabric of spacetime etc.
6. The replies in this thread are ignoring a critical problem, the speed of gravity.
If gravity travels at c then all orbits (ie of your new particle, & everything else) should spiral into the spiral galaxy. Einsteinians give two different silly reasons re why gravity is instantaneous at distance, or appears to be instantaneous (but i wont go into that here).
7. A spiral is an ever increasing or decreasing circular motion -- hencely the name spiral galaxy sort of answers the question. LOL.
8. The Newtonian formula is a worry -- it might not be accurate re spiral galaxies, ie for matter spread out in a flattish plane. -- or it might be accurate if there is dark matter in the galaxy.
9. And then there is the question of the force of dark energy opposing the action of the bending of spacetime by pushing things (like your new particle) away from each other or outwards from the center of the bigbang universe or something.
10. The modern Einsteinian dark age of science will soon end, & real science will be enjoy a rebirth.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/11/2018 00:38:35
If gravity travels at c then all orbits (ie of your new particle, & everything else) should spiral into the spiral galaxy.
Actually, it can be shown that if gravity travels at c or some other finite speed, then orbiting objects tend to spiral out from each other, violating conservation of energy.

Title: Re: How gravity works in spiral galaxy?
Post by: mad aetherist on 27/11/2018 01:11:55
If gravity travels at c then all orbits (ie of your new particle, & everything else) should spiral into the spiral galaxy.
Actually, it can be shown that if gravity travels at c or some other finite speed, then orbiting objects tend to spiral out from each other, violating conservation of energy.

Yes u must get an outwards spiral not inwards.
I have a problem because i believe that gravity has a finite speed of say well over 20 billion c. The thing is -- any finite speed must result in an outwards spiral -- & a violation of conservation of energy. But i dont believe that gravity has an infinite speed. There must be a mechanism that balances the gain in speed by a loss of speed -- i will have a think.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/11/2018 14:25:51
Sure, I agree (F is changing, not G)
It isn't drifting.  It is being pushed away.
Thanks Halc

So, the corrected formula is as follow:
F(t) = G * M(t) * m(t) / R(t)^2 = F(0) - Δ(t)

F(t) = the gravity force at time t.
M(t) = the host mass at time t.
m(t) = The orbital object mass at time t.
R(t) = the radius at time t.
Δ(t) = Lost of gravity force due to Tidal friction (or due to any other interference).
F(0) = the gravity force at time 0.

With regards to the Tidal friction:
Δ(t) is the estimated lost of gravity force due to Tidal friction.
In the Earth/Moon orbital cycle, the "tidal bulges" is very clear (few cm.).
If the Moon mass will be decreased, while the Earth mass will be increased, we might not notice this - "tidal bulges".
However, do you agree that the tidal effect is still valid?
Hence, do you agree that even if we will decrease the Moon mass into a single Atom while we increase the earth mass into a SMBH, the tidal friction will still be there and the Atom will be pushed away?

Artificial satellites crashing to Earth are a consequence of air resistance.
So, can we assume that at any orbital system without air resistance, the orbital object is losing energy over time?
Hence, do you agree that in those kind of orbital systems, as the objects are losing energy over time, they are also losing gravity force over time and therefore they are pushed away from their host?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/11/2018 15:54:36
So, the corrected formula is as follow:
F(t) = G * M(t) * m(t) / R(t)^2 = F(0) - Δ(t)

F(t) = the gravity force at time t.
You're being inconsistent with your notation.  In the OP you defined F(t) as "the change in the gravity force over time".

You might as just well say that the correct formula for force F at a given time is F = G * M * m / R2 as per your OP.  Over time, any of those things might change, especially R if the path of m is not a circle about mass M, so F would then not be constant over time.

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Δ(t) = Lost of gravity force due to Tidal friction (or due to any other interference).
...
With regards to the Tidal friction:
Δ(t) is the estimated lost of gravity force due to Tidal friction.
Force isn't lost due to friction.  Force is reduced due to the object being less nearby (increase of R).  Friction reduces kinetic energy.  It doesn't reduce force or momentum.

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In the Earth/Moon orbital cycle, the "tidal bulges" is very clear (few cm.).
More like a few meters.

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If the Moon mass will be decreased, while the Earth mass will be increased, we might not notice this - "tidal bulges".
However, do you agree that the tidal effect is still valid?
You're asking if smaller objects still produce tides?  Yes they do.  The ISS for instance produces an immeasurable tide on Earth, and one that puts an accelerating force on the rotation of Earth, as opposed to a drag like the moon has.  That's because the ISS is inside the geosync radius.

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Hence, do you agree that even if we will decrease the Moon mass into a single Atom while we increase the earth mass into a SMBH, the tidal friction will still be there and the Atom will be pushed away?
For the atom to be pushed away, it needs to orbit slower than the rotation of the larger object, and the large object needs to be fluid enough to exhibit tides. A solid cold Earth would not push the moon away since there is no medium for tides.  You need an atmosphere or ocean or flexible mantle for the effect.  So not sure if tides can be raised on a black hole.

Anyway, short story is that things near black holes spiral in due to the bending of space.  It takes acceleration to stay out of them.  Nothing (not even light) can orbit closer than 1.5x the radius of the event horizon.  Your basic tidal force analysis is a Newtonian one, not taking relativistic effects into account.  It works for classical masses and orbital speeds.

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So, can we assume that at any orbital system without air resistance, the orbital object is losing energy over time?
From what?  The moon is gaining energy.  Low orbit (below geosync) objects might lose energy from the tidal drag, but high orbit objects gain mechanical energy from tidal trust.  A steel moon in low orbit would eventually fall to Earth.  It needs to be strong enough to not break up from being inside the Roche limit.

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Hence, do you agree that in those kind of orbital systems, as the objects are losing energy over time, they are also losing gravity force over time and therefore they are pushed away from their host?
If they're losing energy, they're falling closer to Earth, which increases the gravitational force on them.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 28/11/2018 06:39:40
You might as just well say that the correct formula for force F at a given time is F = G * M * m / R2 as per your OP.  Over time, any of those things might change, especially R if the path of m is not a circle about mass M, so F would then not be constant over time.

I fully agree that if there is a change in M or m there is a change in F which impacts the R.
However, assuming that there is no change in M or m. why there is a change in R? If you want to push it away, you must set external force.
Friction reduces kinetic energy.  It doesn't reduce force or momentum.
I don't understand how the Kinetic energy changes R while you claim that it doesn't reduce force.
for example:
Based on the following formula (assuming that there is no change in the mass M or m):
F1 = G * M * m / R1 ^2
F2 = G * M * m / R2 ^2
Hence
F1 / F1 = 1/(R1/R2)^2
If R2 is bigger than R1 than:
F1 is bigger than F2.
Therefore, by definition there is a change in the gravity force.
So how the kinetic energy changes the gravity force while we say that the kinetic energy doesn't reduce the gravity force?
Don't you agree that there is a contradiction in this statement?

Can you please explain why cold Earth would not push the moon away since there is no medium for tide?
I don't understand how the matter in mass effects the tidal friction and eventually the kinetic energy.
Can you prove it by real formula or is it just hypothetical idea?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 28/11/2018 14:44:11
Quote from: Halc
You might as just well say that the correct formula for force F at a given time is F = G * M * m / R2 as per your OP.  Over time, any of those things might change, especially R if the path of m is not a circle about mass M, so F would then not be constant over time.
I fully agree that if there is a change in M or m there is a change in F which impacts the R.
However, assuming that there is no change in M or m. why there is a change in R? If you want to push it away, you must set external force.
A potato on a table has constant gravitational force acting on it, because the table holds the R constant, putting the potato in a circular path.  Take away the table, and the R starts decreasing.  The path is no longer a circle, and F goes up.  Neither M nor m has changed.
It needs no external force.  There already is one, as computed by F.

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Friction reduces kinetic energy.  It doesn't reduce force or momentum.
I don't understand how the Kinetic energy changes R while you claim that it doesn't reduce force.
I didn't say kinetic energy changes R, nor that it directly effects gravitational force.  That force is a function of those elements in the formula above.  None of those elements is kinetic energy or friction.

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for example:
Based on the following formula (assuming that there is no change in the mass M or m):
F1 = G * M * m / R1 ^2
F2 = G * M * m / R2 ^2
Great.  You've changed R, and there is no specificaion of what the kinetic energy is, which is fine because it is irrelevant to force.  Two identical mass objects at the same R from Earth will have the same gravitational force acting upon them despite one being nearly at rest (the potato on the table) and the other being shot out of a potato gun.

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Hence
F1 / F1 = 1/(R1/R2)^2
If R2 is bigger than R1 than:
F1 is bigger than F2.
Therefore, by definition there is a change in the gravity force.
Right.  No kinetic energy represented in any of that.  Gravitational force is inversely proportional to the square of the separation distance R.  You say this like it is news.
The potato on the table weighs less than the same potato on the floor due to a change in only R.

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So how the kinetic energy changes the gravity force while we say that the kinetic energy doesn't reduce the gravity force?
It doesn't.
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Don't you agree that there is a contradiction in this statement?
The stock market price also doesn't effect the force on the object.  It isn't a contradiction to say so.  The force is a function of neither the stock market price nor of the kinetic energy of either object.

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Can you please explain why cold Earth would not push the moon away since there is no medium for tide?
All the friction would go away.  With nothing to move or bend, there would be no force transferring angular momentum from one object to the other.

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I don't understand how the matter in mass effects the tidal friction and eventually the kinetic energy.
Can you prove it by real formula or is it just hypothetical idea?
The moon puts torque on Earth at F*R, where F is the force of friction (mostly from ocean I think, but also atmosphere and bending of crust/mantle).  That torque puts thrust on the moon increasing its mechanical energy (not its kinetic energy), and decreasing the mechanical energy of earth, at a loss.  The moon gains less than the Earth loses.  The excess is radiated as heat.

Anyway, reduce the F to zero, and that torque drops to zero, and no momentum transfer takes place.

Mercury is a weird case, where the torque acts both ways in perfect balance.  It is almost all tidal forces on the permanently elliptical planet itself, but it is enough to keep the spin of the planet constant at exactly 1.5 rotations per year.  Once the orbit of the planet circularizes more, the shape of the planet will not help it stay locked, and it will once again resume the slowing of its spin.  It may be swallowed before that happens.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/11/2018 06:02:32
A potato on a table has constant gravitational force acting on it, because the table holds the R constant, putting the potato in a circular path.  Take away the table, and the R starts decreasing.  The path is no longer a circle, and F goes up.  Neither M nor m has changed.
It needs no external force.  There already is one, as computed by F.

Thanks Halc.
I do appreciate all your efforts.

However, I'm not fully sure with regards to that example.

The main difference is the forwarded motion of a body in space (let's call it forwarded force)!!!

So, before continue our discussion, let's see what causes an orbit to happen (in a perfect conditions)?
http://www.qrg.northwestern.edu/projects/vss/docs/space-environment/1-what-causes-an-orbit.html
"Orbits are the result of a perfect balance between the forward motion of a body in space, such as a planet or moon, and the pull of gravity on it from another body in space, such as a large planet or star. An object with a lot of mass goes forward and wants to keep going forward; however, the gravity of another body in space pulls it in. There is a continuous tug-of-war between the one object wanting to go forward and away and the other wanting to pull it in."

Forwarded force - the forwarded motion of a body in space.
Gravity force - the amount of force that pull it in.

Therefore - as long as there is a balance between those two vectors, the object will keep its current radius.
Technically, if there is no air resistance, no tidal, no any external force, no change in mass, the radius will be kept forever.
However, if the forwarded force is greater than gravity force - the object continues moving through space.
If gravity is greater than forwarded force, objects collide.

Hence, it seems to me that the only way to change the orbital radius is by changing the gravity force or the Forwarded force (In perfect conditions).

Do you agree with that?
Title: Re: How gravity works in spiral galaxy?
Post by: mad aetherist on 29/11/2018 11:49:25
I repeat what i said earlier.........
6. The replies in this thread are ignoring a critical problem, the speed of gravity.
If gravity travels at c then all orbits (ie of your new particle, & everything else) should spiral into the spiral galaxy. Einsteinians give two different silly reasons re why gravity is instantaneous at distance, or appears to be instantaneous (but i wont go into that here).
Title: Re: How gravity works in spiral galaxy?
Post by: guest39538 on 29/11/2018 11:55:43
Electric Universe ,  that  simple ,  N is  attracted to N , I am right and science lies .
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 29/11/2018 14:08:33
However, I'm not fully sure with regards to that example.

The main difference is the forwarded motion of a body in space (let's call it forwarded force)!!!
Let's not.  Motion is not force.  Yes, the falling potato has little orbital speed, but you also never specified objects that are already in a perfect circular orbit.  Objects that are tend to maintain their R, yes, but nothing in the equation of gravity mentions the velocity of the objects in question.  The R is free to be changing.

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So, before continue our discussion, let's see what causes an orbit to happen (in a perfect conditions)?
http://www.qrg.northwestern.edu/projects/vss/docs/space-environment/1-what-causes-an-orbit.html
"Orbits are the result of a perfect balance between the forward motion of a body in space, such as a planet or moon, and the pull of gravity on it from another body in space, such as a large planet or star. An object with a lot of mass goes forward and wants to keep going forward; however, the gravity of another body in space pulls it in. There is a continuous tug-of-war between the one object wanting to go forward and away and the other wanting to pull it in."
Sure, and it doesn't need to be perfect.  The moon has an elliptical orbit, so the R is constantly changing, but that doesn't mean it isn't orbiting.  That means the gravitational force between us and it is also constantly changing.

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Forwarded force - the forwarded motion of a body in space.
No...
Forward force would be thrust in the direction of motion, sort of what the tides do to the moon.  That is not its motion.  Absent tides, there would be no forward force, but there would still be forward motion of the body in space.

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Gravity force - the amount of force that pull it in.
Fine...

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Therefore - as long as there is a balance between those two vectors, the object will keep its current radius.
No.  As I said, the object will keep its radius even totally absent a forward force.  Gravity accelerates the moon downward at all times since that force is not opposed, but the acceleration vector is essentially perpendicular to the velocity vector, so it has little effect on the speed of it.

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Technically, if there is no air resistance, no tidal, no any external force, no change in mass, the radius will be kept forever.
Assuming it is in that circular orbit, yes.  The potato on the table is in a circular path, but is not in orbit.  It has a 2nd force opposing gravity.

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However, if the forwarded force is greater than gravity force - the object continues moving through space.
I refuse to use such terms.  If the speed of the object is greater than escape speed, it will move away.  If it is less than that, it will stay in orbit.  It is a function of speed, not of force.
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If gravity is greater than forwarded force, objects collide.
If speed of the object is sufficiently small to bring its orbit inside the size of the other object, they will collide.  The potato falling off the table does this.  A potato falling off a sufficiently tall table will not ever hit the Earth.  If the table is yet even taller, the potato won't even orbit.

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Hence, it seems to me that the only way to change the orbital radius is by changing the gravity force or the Forwarded force (In perfect conditions).

Do you agree with that?
No.  There is no forward force.  Any object might have a trajectory that puts it on something other than a perfect circular orbit.

So for an object already in a perfect circular orbit (you don't say this, but it needs to be qualified), the only way to change the orbital radius is by changing the gravity force or by applying a secondary force to one of the objects.

The former can be changed by altering the mass of the primary mass.  Altering the mass of the orbiting thing makes no difference.  I can take half the moon away (halving the gravitational force acting upon it) and it will have zero effect on its orbit.  But take away half of Earth, and the moon will need to orbit further out.

The force on the moon would need to be cumulative.  So I could put a permanent external force of a billion Newtons on the moon, always towards Polaris, and it will have no effect on the orbital period of the moon or on the radius.  But the tides put a cumulative thrust on the moon that actually adds to its momentum, and so it is always moving to a higher orbit.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/11/2018 16:37:55
No.  There is no forward force.  Any object might have a trajectory that puts it on something other than a perfect circular orbit.
So for an object already in a perfect circular orbit (you don't say this, but it needs to be qualified), the only way to change the orbital radius is by changing the gravity force or by applying a secondary force to one of the objects.
Thanks for the clarification. fully clear.
However, what do you mean by: " applying a secondary force to one of the objects"?

The former can be changed by altering the mass of the primary mass.  Altering the mass of the orbiting thing makes no difference.  I can take half the moon away (halving the gravitational force acting upon it) and it will have zero effect on its orbit.  But take away half of Earth, and the moon will need to orbit further out.
The issue with the mass is also clear. Although, in this example, there is no change in mass.

But the tides put a cumulative thrust on the moon that actually adds to its momentum, and so it is always moving to a higher orbit
I can't understand this idea.
1. Why tides put cumulative thrust on the moon?
2. If tides put a cumulative thrust on the moon that actually adds to its momentum, why this extra momentum can't be considered as a force? (Don't you agree that extra momentum should increase the speed and therefore it should increase the force)
Please look at the following diagram:
http://www.scienceline.ucsb.edu/images/earth_sun_bodydiagram.png
Do you agree that:
1. The Centrifugal force is a direct outcome of the moon speed?
2. In order to keep the moon in the same radius its Centrifugal force must be equal to its gravitational force?
3. In order to increase R (while keeping it in a perfect orbital cycle) we must decrease at the same moment and at the same amplitude the Centrifugal force and the gravitational force?

Sorry, I really can't understand how tidal changes R while we claim that it doesn't change force?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 29/11/2018 21:00:36
However, what do you mean by: " applying a secondary force to one of the objects"?
Something other than the gravitational force given by that formula.  For the potato, maybe it is the table putting a upward force on it preventing it from falling.  An asteroid hits the moon, altering its orbit.  Tidal force is tangential to the regular gravitational pull of the Earth.  All these forces are something else, and hence can alter the orbit it would have otherwise taken.

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Quote from: Halc
But the tides put a cumulative thrust on the moon that actually adds to its momentum, and so it is always moving to a higher orbit
I can't understand this idea.
1. Why tides put cumulative thrust on the moon?
It is always pushing in the direction it is already moving, adding to the mechanical energy of the moon.  Tidal forces on the ISS push against its motion and reduce the mechanical energy.  The orbit drops due to tidal resistance, but it drops even more due to friction.

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2. If tides put a cumulative thrust on the moon that actually adds to its momentum, why this extra momentum can't be considered as a force?
Momentum is not force.  Linear momentum is mV, and cumulative thrust adds to mechanical (potential + kinetic) energy, not to momentum, since the velocity of the moon actually slows as it moves to a higher orbit.  Total angular (not to be confused with linear) momentum (mass*radius*tangential-velocity) of the Earth/moon system is preserved, but angular momentum of Earth's spin is transferred to angular momentum of the Earth/moon system as a whole.  That momentum is part of both of them, not just the moon.

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(Don't you agree that extra momentum should increase the speed and therefore it should increase the force)
Linear moment goes down as the moon slows down.  Momentum and force are not necessarily related, as this example shows.
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Please look at the following diagram:
http://www.scienceline.ucsb.edu/images/earth_sun_bodydiagram.png
That diagram is really bad science.  Centrifugal force does not exist in the inertial frame implied by that diagram.  If it did, it would counter gravity force and the path of Earth would not curve.  It exists in a rotating reference frame, but in such a frame, Earth would be stationary and have no rotational speed.  The arrow to the left would have zero magnitude.

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Do you agree that:
1. The Centrifugal force is a direct outcome of the moon speed?
OK, first, there is no moon in that picture, but it works if you put Earth where the sun is and the moon going around.  Then:  No... Centrifugal force is a pseudo-force and does not exist, else the moon would not accelerate towards Earth.

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2. In order to keep the moon in the same radius its Centrifugal force must be equal to its gravitational force?
In a rotating reference frame, yes.  The two forces cancel and the moon remains stationary.  In an inertial frame, no.  Only the force of gravity exists, and that force continuously accelerates the moon, keeping it in a circular orbit.

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3. In order to increase R (while keeping it in a perfect orbital cycle) we must decrease at the same moment and at the same amplitude the Centrifugal force and the gravitational force?
You can't keep it in a circular orbit if you increase R.  Maybe a helix, but circles come back to the same point.  That said, I know what you mean.
Increasing R requires a tangential thrust adding mechanical energy to the orbit.  The gravitational force will drop as R increases, and not until then.  You want to add energy, so you need to push in the direction of movement.  Altering the force tangential to the motion does not add energy, it just curves the path elsewhere.

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Sorry, I really can't understand how tidal changes R while we claim that it doesn't change force?
Tides push with the rotation (to the left in your picture).  Gravity pushes tangential to the motion (down in your picture).  The former adds energy and thus increases R.  The latter adds no energy, so it merely bends the path from following the straight line it would otherwise follow in the absence of gravitational force.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/11/2018 18:56:40
Thanks Halc

1. Centrifugal force
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Please look at the following diagram:
http://www.scienceline.ucsb.edu/images/earth_sun_bodydiagram.png
That diagram is really bad science.  Centrifugal force does not exist in the inertial frame implied by that diagram.  If it did, it would counter gravity force and the path of Earth would not curve.
I don't understand why you disqualified the centrifugal force.
http://scienceline.ucsb.edu/getkey.php?key=4569
It is stated:
"All massive objects in our universe are attracted to each other through a force known as gravity. If gravity were the only force acting between the sun and earth, the two bodies would indeed collapse on one another. Therefore, there must be other forces acting on this system. Remember that the earth, and all the other planets in our solar system, revolve around the sun. Because the earth is in a rotational orbit, there is another force acting on the planet. This force is known as centrifugal force. The direction of the centrifugal force on the earth is opposite the direction of the gravitational force (see the diagram below), so it prevents the sun and earth form collapsing into each other."

What is wrong with this explanation?
If it helps, let's ignore the Sun/Earth/Moon system. Just think about any orbital system with host in the center and an object in a perfect orbital cycle.

In an inertial frame, no.  Only the force of gravity exists, and that force continuously accelerates the moon, keeping it in a circular orbit.
How could it be that only the force of gravity exists?
In the following article:
http://scienceline.ucsb.edu/getkey.php?key=4569
It is stated in answer 1:
"the earth has two main forces: inertia, to keep moving straight, and the gravity, to pull it to the sun. It moves in an average of the two directions, as in this image, and moves in a circle around the sun."
It is stated specifically that Inertia is force. Hence, the Earth is in an average of two directions.  (or two forces).

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Sorry, I really can't understand how tidal changes R while we claim that it doesn't change force?
Tides push with the rotation (to the left in your picture).  Gravity pushes tangential to the motion (down in your picture).  The former adds energy and thus increases R.  The latter adds no energy, so it merely bends the path from following the straight line it would otherwise follow in the absence of gravitational force.

Can we prove this explanation with mathematics?

For example:
Assuming that the current radius is R1
Than
F1 = G M m / R1 ^2
If the former adds energy and thus increases R
Than, at the radius = R2
F2 = G M m / R2 ^ 2
As R1 bigger than R2 then F2 is lower than F1.
Hence, there is a decrease in the gravity force
ΔF = F1 - F2.
So, how energy can increase R, therefore it changes the gravity force, but it doesn't consider as force.
This is a real enigma for me.

There is another key issue (in an ideal Earth/Sun system).
Let's assume that the tidal sets Energy.
This energy Push the Earth further away from the Sun.
However, due to inertia, the Earth keeps its orbital velocity.
Hence, in this new location, the updated gravity force will be too low to hold the Earth in the orbital path.
Therefore, the Earth will be disconnected from the Sun.

So, do you agree that in order to keep the Earth in the orbital cycle around the Sun, two actions should be taken at the same moment and with full synchronization?
2. Decreases inertia velocity - V

Hence, how Tidal (or any sort of energy) can push the earth farther away from the Sun while it also decreases the inertia velocity at the same moment and in full synchronization?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/12/2018 00:18:01
Thanks Halc

1. Centrifugal force
Quote from: Halc
Centrifugal force does not exist in the inertial frame implied by that diagram.  If it did, it would counter gravity force and the path of Earth would not curve.
I don't understand why you disqualified the centrifugal force.
Did you read my reply there?  If there was a second force countering gravity, the net force on Earth would be zero and it would not accelerate at all, but travel in a straight line per Newton's 1st law.  You've not replied to that.

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http://scienceline.ucsb.edu/getkey.php?key=4569
That looks like a question posed to a room of students, and a list of some of their answers.  None of them sound very clear, but the 2nd and 5th one are blatantly wrong, and the 1st one is also wrong.

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It is stated:
"All massive objects in our universe are attracted to each other through a force known as gravity. If gravity were the only force acting between the sun and earth, the two bodies would indeed collapse on one another. Therefore, there must be other forces acting on this system.
Wrong.  Nothing else (EM, or nuclear forces) is acting between Sun and Earth.  The other forces are needed to form both of them in the first place (to form matter at all for instance), but not to keep them in orbit.

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What is wrong with this explanation?
What is wrong is what I already explained in the part of mine you quoted.  Gravity is the only force at work in that diagram, which can be seen by the fact that all objects are accelerating in it.  Gravity force causes the sun to accelerate down, and that force has an equal and opposite reaction acting on the sun, not on Earth.  The arrow belongs there, but it is not centrifugal force, and it should be attached to the sun, which is where the upward gravitational force is being applied.

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Quote from: Halc
In an inertial frame, no.  Only the force of gravity exists, and that force continuously accelerates the moon, keeping it in a circular orbit.
How could it be that only the force of gravity exists?
Only gravity is a significant force in orbital motion.  There are other forces like EM, but they're immeasurably small.

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In the following article:  [same link]
You need to find some better articles.  I hope the university isn't pushing these answers as science.  Instead it seems to be a page showing why education is needed, illustrating typical answers from students before taking some class.
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It is stated in answer 1:
"the earth has two main forces: inertia, to keep moving straight, and the gravity, to pull it to the sun. It moves in an average of the two directions, as in this image, and moves in a circle around the sun."
It is stated specifically that Inertia is force.
Yea, it does.  That's part of why answer 1 is wrong.  Inertial is mass, not force.  I cannot accelerate a rock with inertia.  F=ma remember?  If inertial was a force, it would cause mass to accelerate all by itself.

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Can we prove this explanation with mathematics?

For example:
Assuming that the current radius is R1
Than
F1 = G M m / R1 ^2
If the former adds energy and thus increases R
Than, at the radius = R2
F2 = G M m / R2 ^ 2
As R1 bigger than R2 then F2 is lower than F1.
Hence, there is a decrease in the gravity force
ΔF = F1 - F2.
All correct.  The moon exerts less gravitational force when it is further away.

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So, how energy can increase R, therefore it changes the gravity force, but it doesn't consider as force.
R is not a function of energy.  But the energy is being applied as forward thrust here which pushes the moon into a higher orbit, just like a rocket expends fuel to get a satellite to a higher orbit where it moves slower.  Low orbit objects orbit at around 9 km/sec, but it takes more energy to get them uphill to say geosync where they move at only 3 km/sec.  Same thing is happening with the moon, being pushed uphill just like the rocket does to the satellite.
Earth has a similar trivial thrust pushing it away from the sun.

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There is another key issue (in an ideal Earth/Sun system).
Let's assume that the tidal sets Energy.
This energy Push the Earth further away from the Sun.
However, due to inertia, the Earth keeps its orbital velocity.
Hence, it this new location, the updated gravity force will be too low to hold the Earth in the orbital path.
Therefore, the Earth will be disconnected from the Sun.
????
It isn't an inertia thing.  The tide (from the spin of the sun) pushes Earth, which makes it move a bit to fast to stay in the circular path it was in before, so the circle widens a bit.  Earth moves a tiny bit further away, and slows down because it is now moving up hill out of the gravitational well.  This new slower speed exactly balances the smaller gravity at this new radius.

You seem to be neglecting the fact that things slow down when moving away from a gravitational source.  Kinetic energy is being lost to acquired gravitational potential energy.

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So, in order to keep the Earth in the orbital cycle around the Sun, two actions should be taken at the same moment and with full synchronization:
1. Increases R
2. Decreases V
So, how Tidal can control on those two vectors with full synchronization?
Tides just apply thrust.  Thrust adds (or removes) energy, and the new energy level finds a balance between those two things.  V goes down as R goes up.  Throw a ball in the air and watch V go down as the ball rises.  A thrown ball is in orbit, but an orbit that typically intersects the ground, so it appears to be a parabola to an observer.  It is in fact an elliptical path.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/12/2018 06:33:49
Earth moves a tiny bit further away, and slows down because it is now moving up hill out of the gravitational well.
V goes down as R goes up.  Throw a ball in the air and watch V go down as the ball rises.
You seem to be neglecting the fact that things slow down when moving away from a gravitational source.  Kinetic energy is being lost to acquired gravitational potential energy.

Thanks
Based on Newton first law:
https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion
First law:   In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.[2][3]"
Let's start by asking the following:
What will happen if we eliminate completely and at once the gravity force from the Earth/sun orbital system?
If I understand it correctly, the Earth will "continue to move at a constant velocity".
Therefore, do you agree that for orbital system it is incorrect to assume that "things slow down when moving away from a gravitational source"?
Hence, do you agree that the Earth will not change its inertial velocity "unless acted upon by a force"?
Based on your explanation - the Tidal is not a force. It just pushes the Earth further away from the Sun.

As we increase R we actually decrease the gravity force.
Gravity force is the power which holds the Earth in the orbital cycle around the Sun.
It must be fully synchronized with the inertial velocity of the earth in order to hold it in the orbital cycle.

It is clear that as Tidal increases R it actually decreases the gravity force.
However, based on your explanation, Tidal doesn't set any force. Therefore, by definition it doesn't set any negative force which slows down the Earth velocity as it increases R.
Therefore, once the Tidal push away the Earth from the Sun, there is less gravity to hold the Earth in a balance with it's current inertial velocity/force.
Hence, Less gravity force to hold the Earth on the orbital track means that the Earth is moving further away from the Sun.
That increasing radius, decreases the gravity force more and more.
Therefore, do you agree that from Orbital point of view, once we push away the Earth from the Sun, the Earth is starting to move down the hill?
At some point it must be totally disconnected from the sun gravity and  "continues to move at a constant velocity" as explained by Newton first law.

Unfortunately, I couldn't find in your explanation how tidal sets a negative force which slow down the Earth velocity in order to compensate less gravity force due to increasing in R.

Can you please explain this issue?

Tides just apply thrust.  Thrust adds (or removes) energy, and the new energy level finds a balance between those two things.  V goes down as R goes up.

Would you kindly show by mathematics how tidal sets thrust that adds energy which pushes the Earth away from the Sun, while somehow the Earth reduces its velocity without implying an external force?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/12/2018 14:36:19
Based on Newton first law:   In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.[2][3]"
Let's start by asking the following:
What will happen if we eliminate completely and at once the gravity force from the Earth/sun orbital system?
If I understand it correctly, the Earth will "continue to move at a constant velocity".
Therefore, do you agree that for orbital system it is incorrect to assume that "things slow down when moving away from a gravitational source"?
There is nothing moving away from a gravitational source in your example.  You took gravity away.
You can do this by considering two small rocks at the same positions and velocities of the earth/sun system.  They’re too small to have significant gravity pull between them, so they each continue to move pretty much in a straight line.

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Hence, do you agree that the Earth will not change its inertial velocity "unless acted upon by a force"?
Newton’s law says that nothing does.  It is called velocity BTW.  ‘Inertial velocity’ isn’t something different, or if it is, you need to tell me what you mean by that.
So yes, I agree with Newton’s law.  Earth will not change its velocity unless acted upon by a force.

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Based on your explanation - the Tidal is not a force. It just pushes the Earth further away from the Sun.
It actually pushes the Earth tangential to the orbit, which is perpendicular to a vector away from the sun.  If it pushes or pulls, it’s a force.  That’s what force is.

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As we increase R we actually decrease the gravity force.
Gravity force is the power which holds the Earth in the orbital cycle around the Sun.
Force is not power.  Gravity is the force which holds the Earth in the orbital cycle around the Sun, yes.

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It must be fully synchronized with the inertial velocity of the earth in order to hold it in the orbital cycle.

It is clear that as Tidal increases R it actually decreases the gravity force.

However, based on your explanation, Tidal doesn't set any force. Therefore, by definition it doesn't set any negative force which slows down the Earth velocity as it increases R.[/quote]Tital force is a force.  I didn’t say it ‘doesn’t set any force’.  It pushes with the motion, so it actually directly acts to increase speed of Earth, but that velocity slows as the Earth pulls further out of the gravity well.  The net effect is a slower orbital speed.

It’s like a coin in one of those funnel machines, slowly spiraling faster and faster into the center.  You give the coin some thrust, and it moves further away from the center and ends up going slower than before you gave it that push.  In the absence of interference, friction is a force against its motion, and yet the coin is speeding up as it goes deeper into the gravity well.

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Therefore, once the Tidal push away the Earth from the Sun, there is less gravity to hold the Earth in a balance with it's current inertial velocity/force.
Hence, Less gravity force to hold the Earth on the orbital track means that the Earth is moving further away from the Sun.
Less gravity means the Earth has already moved further away from the sun.  You seem to be confusing cause and effect here.  That gravitational force doesn’t go down until the Earth has already moved further out.

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That increasing radius, decreases the gravity force more and more.
Therefore, do you agree that from Orbital point of view, once we push away the Earth from the Sun, the Earth is starting to move down the hill?
No, away from gravity is up hill.  Surely you know this.  Downhill gets you closer to the source of the gravity (the sun in this case).

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At some point it must be totally disconnected from the sun gravity and  "continues to move at a constant velocity" as explained by Newton first law.
Given enough thrust, that is true, but the tidal thrust decreases with distance, so I don’t think it is possible for one object to spin away one of its satellites in isolation.  But things are not in isolation.  Given enough time and high spin, the Earth could theoretically push the moon beyond Earth’s hill radius and the moon would depart Earth and go into its own orbit about the sun.  In isolation (just sun and Earth say), there is no hill radius.

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Unfortunately, I couldn't find in your explanation how tidal sets a negative force which slow down the Earth velocity in order to compensate less gravity force due to increasing in R.
It doesn’t.  It is a positive force tangential to the orbit, pushing forward, not out.  If it was a negative force, things would slow down and drop to a lower orbit.
You seem to continue to envision the tides being an outward thrust instead of a forward one.  That would not have any cumulative effect since it would not affect speed and hence no energy transfer.

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Can you please explain this issue?
I have, and you don’t seem to read what I say.
Tidal forces push forward, not out.  It doesn’t directly increase radius.  The increased speed is too much for the current orbit, so it moves further out, which slows the system since that is moving against the primary gravitational force.

Consider at a comet that passes the sun at the same radius as Mercury, but much much faster.  That’s what tide forces do is make something faster.  So the comet moves to a higher orbit, and in doing so, it ends up nearly stopped as it gets so far away from the sun.  Moving away from the sun slows it down.

Tides just apply thrust.  Thrust adds (or removes) energy, and the new energy level finds a balance between those two things.  V goes down as R goes up.

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Would you kindly show by mathematics how tidal sets thrust that adds energy which pushes the Earth away from the Sun, while somehow the Earth reduces its velocity without implying an external force?
Tides create a bulge on the sun that resist its spin, putting negative torque on the sun.  Torque is FR (force * radius).  That force is balanced by positive torque on the sun/Earth system, so same math but much greater R and much less F.  This conserves angular momentum.

So that F is the thrust, pushing in the direction of orbit, not outward.  That speeds up the planet, which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.  That takes Earth further away from the sun, so it gains potential energy and loses kinetic energy.  Potential energy E = force*distance, so compute the weight (not mass) of Earth (using the gravitational formula GMmR2  to get the force, and multiply that by the say millimeter it moves away.  From that you get a chunk of energy E.  Kinetic energy is mv2, so the reduction in v from the orbital change is Δv = E√(1/r) where r is the change in orbital radius ΔR and E is the kinetic energy lost to potential energy.  That negative Δv is greater than the positive Δv you get from enough tidal thrust to increase Earth orbit by a millimeter, so net effect is a slower Earth.

I chose one millimeter ΔR as my example, but it takes perhaps 1000 years for the solar tides to do that.  A somewhat larger effect pushing us away is that the mass of the sun is decreasing faster than new material falls in from deep space, so gravity is slowly decreasing.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/12/2018 19:33:13
So that F is the thrust, pushing in the direction of orbit, not outward.  That speeds up the planet, which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.  That takes Earth further away from the sun, so it gains potential energy and loses kinetic energy.

Thanks again for your great effort!
I do appreciate.

However, would you kindly explain why the Earth is losing kinetic energy due to Tidal?

It was stated that the tidal trust speeds up the orbital velocity of the Earth.
That speeds up the planet (Lets assume by Δv), which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.
That takes the Earth further away from the sun, so it gains potential energy.
So far so good!

However, with regards to the kinetic energy.
The kinetic energy of the Earth just before speeding up the planet is: mv^2.
After increasing the speed by  Δv, the  updated kinetic energy should be: m(v+Δv )^2

Therefore, by definition due to the increase in speed, we have found that the Earth is gaining higher Kinetic energy.

So, would you kindly explain why you assume that the Earth "loses kinetic energy"?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/12/2018 20:28:34
However, would you kindly explain why the Earth is losing kinetic energy due to Tidal?

It was stated that the tidal trust speeds up the orbital velocity of the Earth.
That speeds up the planet (Lets assume by Δv), which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.
That takes the Earth further away from the sun, so it gains potential energy.
So far so good!
And in moving out of the gravity well, the kinetic energy is lost to that potential energy.  It slows with a greater negative Δv than the positive Δv from the thrust, just like a roller coaster slows (exchanges kinetic energy for potential) as it moves further from Earth center.

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However, with regards to the kinetic energy.
The kinetic energy of the Earth just before speeding up the planet is: mv^2.
After increasing the speed by  Δv, the  updated kinetic energy should be: m(v+Δv )^2
If the Δv is imparted all at once without giving Earth a chance to move, yes.  There is as yet no change in R and thus no change in potential energy.  But it cannot stay at that R now since it is moving too fast.  So it moves away, which slows it.  In reality, it takes time (centuries) for that Δv to be applied, and the orbit rises steadily and the kinetic energy drops as the R increases.  All the energy and more goes into additional potential energy, not kinetic, for a net loss of kinetic energy.  If Earth got so much thrust that it ends up way out by Neptune, then it would have a small fraction of the kinetic energy it has now, but far more total energy.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/12/2018 16:36:40
And in moving out of the gravity well, the kinetic energy is lost to that potential energy.  It slows with a greater negative Δv than the positive Δv from the thrust, just like a roller coaster slows (exchanges kinetic energy for potential) as it moves further from Earth center.

That's an interesting idea.
The formula for Kinetic emery is: Ek = mv^2
The formula for potential energy is: Ep = GMmR^2
I could not understand why an increasing in potential energy (Ep) should decrease Kinetic energy (Ek).
Can you please explain the rational of this idea for orbital system?
Do we have any formula which shows the energy transformation?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 02/12/2018 20:33:40
And in moving out of the gravity well, the kinetic energy is lost to that potential energy.  It slows with a greater negative Δv than the positive Δv from the thrust, just like a roller coaster slows (exchanges kinetic energy for potential) as it moves further from Earth center.

That's an interesting idea.
You say this like it isn't obvious.  Not sure if you're making fun of me.
You've never noticed that a roller-coaster, or a wheel on a hill goes faster the closer it is to the gravity source (down), and slower the further it moves away (up)?

I cannot explain any simpler.

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The formula for Kinetic emery is: Ek = mv^2
The formula for potential energy is: Ep = GMmR^2
I could not understand why an increasing in potential energy (Ep) should decrease Kinetic energy (Ek).
Can you please explain the rational of this idea for orbital system?
Do we have any formula which shows the energy transformation?
Yes, you show them just there.  There is conservation of energy.  R is going up, so you know the energy needed to do that, and the kinetic energy is the only place it comes from in this case.  The force that slows the planet can be computed with simple vector arithmetic.  All you need is the force involved (gravity formula to yield the weight of Earth) and the slope of the motion relative to the flat tangential line that would be a steady state circular orbit.
That negative force has greater magnitude than the positive force of tidal thrust from the sun, so the net is a lower speed.
Title: Re: How gravity works in spiral galaxy?
Post by: Janus on 02/12/2018 21:09:28
And in moving out of the gravity well, the kinetic energy is lost to that potential energy.  It slows with a greater negative Δv than the positive Δv from the thrust, just like a roller coaster slows (exchanges kinetic energy for potential) as it moves further from Earth center.

That's an interesting idea.
The formula for Kinetic emery is: Ek = mv^2
okay
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The formula for potential energy is: Ep = GMmR^2
No.  That is the formula for gravitational force.  Potential energy would be that force integrated with respect to R, or
Ep = -GMm/R

Thus the total energy of an object in orbit is Et = mv^2/2-GMm/R

The total energy of an object in orbit is conserved assuming no additional energy is added.  Thus with an elliptical orbit, as the object climbs from periapis to apapis,  r increases, requiring a decrease in v in order to conserve energy.

If you add KE, the object will begin to climb, exchanging KE for gained PE.   Not only is the KE added lost, but some of the initial KE must be given up too.  If the added KE is just a one time shot, the object goes into an elliptical orbit with a apapis further out than the present orbit.  If it is a continuous addition, like in tidal acceleration, the object will slowly climb outward.  The semimajor axis of the orbit will increase, and total orbital energy can also be expressed as Et = -GMm/2a, where a is the semi-major axis.   But average orbital velocity is V= sqrt(GM/a), so as a increases, the total energy goes up, but v goes down, meaning that KE makes up a smaller part of the total energy.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 02/12/2018 21:21:54
No.  That is the formula for gravitational force.  Potential energy would be that force integrated with respect to R, or
Ep = -GMm/R
Thanks Janus.  I didn't even look that close to see that one.
I had put Ep=force* distance in post 21, but even then I got the sign wrong, and I didn't bother to integrate since the force is pretty constant in the millimeter the orbit of Earth changes every 1000 years.  That simplification would never do when computing the orbit of say the comet.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/12/2018 05:32:29
Thanks Janus

The formula for potential energy is: Ep = GMmR^2
No.  That is the formula for gravitational force.  Potential energy would be that force integrated with respect to R, or
Ep = -GMm/R

Thus the total energy of an object in orbit is Et = mv^2/2-GMm/R

The total energy of an object in orbit is conserved assuming no additional energy is added.

Thanks Janus

Now it is fully clear to me.

Et = Kinetic energy (Ek) + potential energy (Ep) = mv^2/2+ (-GMm/R)
However, as "The total energy of an object in orbit is conserved assuming no additional energy is added."

Et = constant for a given orbital cycle.
Therefore, if we increase the potential energy (Ep) we by definition decrease Ek

Let's verify this idea by starting at point 1
Hence:

Et1 = Kinetic energy (Ek1) + potential energy (Ep1)

Et1 = m(v1)^2/2+ (-GMm/(R1))

In our case, the tidal set a  thrust that increased the kinetic Energy by ΔEk and therefore we have got an increased   velocity by (Δv).

Now we can claim that the new Total energy (Et2) due to additional energy which had been added is:

Et2 = Et1 + ΔEk
Et2 = Ek1 + ΔEk + Ep1

This is the new starting point (just before increasing the radius)."
At that moment we can claim that:

Et2 = Ek1 + ΔEk + Ep1 = constant
Et2 = Ek2 + Ep1 = constant
Et2 =  m(v1+Δv)^2/2+ (-GMm/(R1)) = constant

However, due to velocity increase, the radius must be increased let's call it (ΔR)
Hence,

Ep2 = (-GMm/(R2)) = (-GMm/(R1 + ΔR))
That increase in the Ep must now decrease back the Ek
So, let's call it Ek3

Et2 = Ek3 + Ep2 = constant
Et2 = m(v3)^2/2+ (-GMm/(R1+ ΔR )) = m(v1+Δv)^2/2+ (-GMm/(R1))

Et2 = m(v3)^2/2+ (-GMm/(R1+ ΔR )) = m(v1+Δv)^2/2+ (-GMm/(R1))

Hence

m(v3)^2/2+ (-GMm/(ΔR )) = m(v1+Δv)^2/2

m(v3)^2/2- m(v1+Δv)^2/2 = GMm/(ΔR)

m(v3)^2/2- m(v1+Δv)^2/2 = GMm/(ΔR)

m(v3- v1- Δv)^2/2 = GMm/(ΔR)

If so, we need to proof that this new v3 is fully correlated to the expected new orbital velocity due to the gravity force at R2 = R1 + ΔR.

Do you agree with that?

Conclusion:
We have actually increased the Ek by ΔEk (due to tidal) and therefore got an increased velocity = Δv.
This increased Δv set an increased radius = ΔR.
This increased ΔR increased the Ep by ΔEp.
Now we expect that this ΔEp will help us not just to decrease the Ek by ΔEk (in order to get back to v1), but also more than that in order to achieve our goal that v3 must be lower than v1.
Even in a perfect system, we can't request to get back all the energy that had been invested.
So, it is very challenging to expect that the ΔR which was a product of Δv will help us to cancel completely the Δv.
Now, we expect that this ΔR will help us to gain move than just Δv.
Is it realistic?

Even if we assume that it is realistic, how do we know that we get a perfect new velocity?
Too low is also problematic.
We must get a very specific velocity to meet the new gravity force (at R2 = R1 +ΔR).

So, don't you think that it is too challenging goal?
In any case, only if we set full calculation we can verify if it works or not.

Title: Re: How gravity works in spiral galaxy?
Post by: mad aetherist on 03/12/2018 05:58:06
Conclusion:
We have actually increased the Ek by ΔEk (due to tidal) and therefore got an increased velocity by Δv.
This increase Δv set an increased radius ΔR.
This change in the radius increased the Ep.
Now we want that this new Ep will help us not just to eliminate that ΔEk which we have added (due to tidal), but also more that that in order to achieve our goal that V3 must be lower than V1.
Actually, lower velocity by itself is not good enough.
Too low is also problematic.
We must get a very specific velocity to meet the new gravity force.
Is it a realistic goal?
Only if we set full calculation we can verify if it works or not.
I havent followed all of the math, but is the new gravity force due to change in R or in G or in M or in m?
If it is due to change in G then i fear that this might introduce some problems. It might depend on how one defines mass. Its complicated.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 03/12/2018 23:11:07
Et = Kinetic energy (Ek) + potential energy (Ep) = mv^2/2+ (-GMm/R)
However, as "The total energy of an object in orbit is conserved assuming no additional energy is added."

Et = constant for a given orbital cycle.
Therefore, if we increase the potential energy (Ep) we by definition decrease Ek

Let's verify this idea by starting at point 1
Hence:

Et1 = Kinetic energy (Ek1) + potential energy (Ep1)

Et1 = m(v1)^2/2+ (-GMm/(R1))

In our case, the tidal set a  thrust that increased the kinetic Energy by ΔEk and therefore we have got an increased   velocity by (Δv).
If you're including tidal thrust, Et is no longer constant.  That thrust is adding total energy to the orbit, taking energy (and more) away from the sun.  Momentum is conserved in that transfer, but plenty of energy is lost to heat.  But total energy of the orbit Et is increasing.

That thrust is continuous, not one-time, and therefore it causes a gradual rise in orbit, and hence a decrease in orbital speed, not an increase.  Ek goes down, Ep goes up.  If it were a one shot thrust, Ek would indeed momentarily go up, sending the Earth into an elliptical orbit where R varies, but average speed still is lower and average potential energy is still higher.

You can get the same effect with two thrusts, 180 degrees apart (half a year), one to push the far end of the orbit up, and a second one to re-circularize the orbit once there.  But a continuous thrust like tidal thrust actually puts the Earth into a continuous spiral outward, ever slowing.

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Now we can claim that the new Total energy (Et2) due to additional energy which had been added is:

Et2 = Et1 + ΔEk
Et2 = Ek1 + ΔEk + Ep1

This is the new starting point (just before increasing the radius)."
OK, This sort of works, but it represents a one-shot thrust all at once, not a continuous thrust.  Sort of like Earth getting hit by an asteroid.  Yes, the figures are correct.  R has not changed yet.  The Earth is suddenly just moving a smidge faster.  Tidal thrust doesn't work this way so you know.

From reading below, you define Ek2 as the new kinetic energy just after the momentary thrust, but not the kinetic energy of the new orbit (which varies) or the average kinetic energy of that orbit.  But Ep2 is defined confusingly as the (average??) potential energy of the new orbit, not the potential energy associated with Ek2 just after the thrust.  That is quite confusing since 2 is not matched up with 2.
You treat R2 like a constant, so I am assuming that R2 is the gravitational average radius of the new elliptical orbit.  By 'gravitational average radius', I mean the radius at which the Ek and Ep are the same as it would be if the orbit were circular.  For tides, that works.  For momentary jolts of thrust, the orbit will not be circular like that, but the calculations still work if R2 is defined this way.

Rockets tend to put satellites into orbit with such discreet thrusts separated by coasting, rather than continuous thrust which is harder to engineer.

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At that moment we can claim that:

Et2 = Ek1 + ΔEk + Ep1 = constant
Et2 = Ek2 + Ep1 = constant
Et2 =  m(v1+Δv)^2/2+ (-GMm/(R1)) = constant
What do you mean by = constant?  The total stays the same, but both kinetic and potential energy are changing as the R changes, so those terms are not going to be constant.

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However, due to velocity increase, the radius must be increased let's call it (ΔR)

Hence,

Ep2 = (-GMm/(R2)) = (-GMm/(R1 + ΔR))
That increase in the Ep must now decrease back the Ek
So, let's call it Ek3

Et2 = Ek3 + Ep2 = constant
Et2 = m(v3)^2/2+ (-GMm/(R1+ ΔR )) = m(v1+Δv)^2/2+ (-GMm/(R1))

Et2 = m(v3)^2/2+ (-GMm/(R1+ ΔR )) = m(v1+Δv)^2/2+ (-GMm/(R1))
I followed all that.  The last line seems to be a copy of the line above it.

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Hence

m(v3)^2/2+ (-GMm/(ΔR )) = m(v1+Δv)^2/2
The algebra failed here.  (-GMm/ΔR) is a massive figure, enough negative energy to form a black hole.  It represents the energy you'd get by lifting Earth from well within the Schwarzschild radius of the sun.

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If so, we need to proof that this new v3 is fully correlated to the expected new orbital velocity due to the gravity force at R2 = R1 + ΔR.

Do you agree with that?
The scenario you've described puts the thing into an elliptical orbit, but yes, if you compute the figures for the circular average R2, I agree with that.
I've discarded the mathematics beyond the algebra error above since it is all wrong after that.

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Conclusion:
We have actually increased the Ek by ΔEk (due to tidal)
Due to a momentary increase of v actually.  Tides never do that since thrust is continuous, and Ek never goes up, not even momentarily.  The whole path just forms a spiral pattern.

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and therefore got an increased velocity = Δv.
This increased Δv set an increased radius = ΔR.
This increased ΔR increased the Ep by ΔEp.
Now we expect that this ΔEp will help us not just to decrease the Ek by ΔEk (in order to get back to v1), but also more than that in order to achieve our goal that v3 must be lower than v1.
Even in a perfect system, we can't request to get back all the energy that had been invested.
Pretty much you can.  There is no friction except in the transfer of energy from sun to Earth, where much energy is lost to the friction slowing the spin of the sun, and more lost to slowing of spin of Earth.  Both apply thrust similarly.  The moon/Earth system only has significant  friction on the Earth side since the moon is tidal locked.
But in terms of orbits, the friction is negligible.  You can assume there is no loss.

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So, it is very challenging to expect that the ΔR which was a product of Δv will help us to cancel completely the Δv.
Now, we expect that this ΔR will help us to gain move than just Δv.
Is it realistic?
Yea, sure.

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Even if we assume that it is realistic, how do we know that we get a perfect new velocity?
Too low is also problematic.
It isn't perfect now, so don't expect it to be perfect.  Today, the Earth is moving faster than it should and its orbit (while still moving inward) is accelerating outward.  It should bottom out at max speed in 30 days and start moving out again.  The orbit is hardly circular, but for the sake of our computations, you can assume one.

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We must get a very specific velocity to meet the new gravity force (at R2 = R1 +ΔR).
If you want a circular orbit, yes, but you started out circular and gave it a single jolt thrust.  It will indeed be going that perfect velocity at that perfect R2, but it won't be moving tangential at that time, so R is not going to stay at R2.  Such is the non-circular orbit you've put it into.

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So, don't you think that it is too challenging goal?
In any case, only if we set full calculation we can verify if it works or not.
You're not actually computing ballistics here, so your equations should need to bother with the elliptical orbit part.  Yes, R2 can be computed, and the v3 out there should be perfect, and not too difficult to compute.  You don't have a figure for how much energy you intend to add to your system via tidal thrust.  Not sure if you need it.  You just have an arbitrary v2 that is bigger than v1.  That is enough to define the new orbit it seems.  You don't even need to know the mass of Earth since the same orbit change will happen to a small pebble given that same Δv.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/12/2018 19:20:24
Thanks Halc

Do appreciate all your valid remarks.
I will verify the mathematics.

However, it seems to me that there is a fundamental problem with the concept of Tidal Friction.

let's start by understanding the meaning of Friction by Google: "The resistance that one surface or object encounters when moving over another".

In the following article it is stated:
http://www.physics.mcgill.ca/~crawford/PSG/PSG11/204_97_L11.9_tidfric.html

"Tides stretch the oceans, and to a small extent, the solid mass of a planet or satellite. In one complete rotation, the planet material keeps deforming and relaxing. This takes energy away from the rotation, transforming it into heat.
In effect, this is a frictional loss, like a giant brake on the planet. Over the centuries, the moon's rotation on its own axis has slowed until it presents essentially the same face to the earth."

So, the tidal friction is like a giant brake or some sort of resistance.
As an example - If we drive a car and press the brakes, we transform some of the kinetic energy into heat and slow down the velocity of the car.
Is there any possibility to increase the velocity of car by pressing the brakes?
In the same token, tidal friction is considered as a giant brake, which transforms some of the energy into heat.
I assume that the source of the energy is coming from the kinetic energy of the object. Therefore, if we transform some of this energy into heat, than there must be less kinetic energy. Less kinetic energy means less velocity.
So, how could it be that a brake/resistance system (like a tidal friction) can increase the orbital velocity instead of decreasing it?
If someone will tell us that by pressing the brakes in the car, the brakes in turn push the car forward and therefore it increases the velocity, would we accept this answer?
So, is there any chance that we are missing the real impact of the Tidal Friction?
I really can't understand how any sort of resistance or brakes can increase the velocity.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 04/12/2018 23:17:00
However, it seems to me that there is a fundamental problem with the concept of Tidal Friction.

let's start by understanding the meaning of Friction by Google: "The resistance that one surface or object encounters when moving over another".
Yes.  The quote you give from the article describes tidal friction quite well.

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Is there any possibility to increase the velocity of car by pressing the brakes?
Sure.  Have the road moving faster than the car.  Pressing the brakes speeds up the car closer to the speed of the road.  That's what's going on with tidal friction.  The spin of Earth is the road, so the friction is the brakes, speeding up the moon.  The Earth spins once per day, far faster than the 30 days it takes for the moon to go around.  Until those two match, the moon will continue to move away.  Once they match, the moon will start to get closer again as both speed in lock step.

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Therefore, if we transform some of this energy into heat, than there must be less kinetic energy. Less kinetic energy means less velocity.
It is angular velocity in this case, but yes.  The Earth is always slowing its spin due to tidal friction.  That's why they keep having to add leap-seconds now and then.  The day used to be about 10 hours long.  It is 24 hours now.  If the system is left alone long enough, it will max out at 1440 hours, and then start to shorten again as the moon actually accelerates the spin of the Earth.
I don't think it will be left alone anywhere near that long.  The sun will swallow us before it all happens.

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So, how could it be that a brake/resistance system (like a tidal friction) can increase the orbital velocity instead of decreasing it?
The road is moving faster than the car, so the car is speeding up as it hits the brakes.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/12/2018 05:42:00
Thanks Halc
It is angular velocity in this case, but yes.  The Earth is always slowing its spin due to tidal friction.  That's why they keep having to add leap-seconds now and then.  The day used to be about 10 hours long.
That is fully clear to me.
However, that phenomena is due to the impact of the gravity force on the object itself.
Based on Newton's law of universal gravitation:
https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation
Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
So, every particle at the earth attracts every other particle in the Universe due to gravity force.
That is the ultimate answer for the tidal friction on any orbital system.
However, Newton didn't specify any information about the matter in the object or how it behaves.
It can be a gas/water/metal Star/Planet/Moon or even made of rice. In the formula we only look at the mass.
It can spin in ultra velocity or stay locked
Therefore, there is no proof for the following statement by Newton gravity formula:
The spin of Earth is the road, so the friction is the brakes, speeding up the moon.
If we think that the spin of the Earth can speed up the moon, than by definition we must change the Newton's law of universal gravitation so it will also include that impact.
How can we claim that Newton's law of universal gravitation is fully correct if there is no reflection in the formula for the type of the matter or the spin of the objects/particles?
Based on the current Newton's law of universal gravitation, "every particle attracts every other particle in the universe with a force..."
So, any particle on Earth might go/down or spin faster/slower/locked due to the local impact of gravity force (tidal activity).
However, based on Newton's law of universal gravitation that local activity in one object (Earth) can't have in turn any impact on a particle in another object (Moon).
If we believe that the spin of a particle in one object can affect the velocity of a particle at a far end object than:
Why don't we change the Newton's law of universal gravitation formula in order to represent this breakthrough understanding?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/12/2018 14:14:28
So, every particle at the earth attracts every other particle in the Universe due to gravity force.
That is the ultimate answer for the tidal friction on any orbital system.
That actually doesn't mention friction at all.  When I rub my hands together, it isn't gravity that makes them warm up.

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Therefore, there is no proof for the following statement by Newton gravity formula:
The spin of Earth is the road, so the friction is the brakes, speeding up the moon.
Newton's formula for gravitational force also says nothing about motion or spin.  It is a static formula for force at a given moment.  So I agree, Newton's formula says nothing about my statement there.
Do you deny that the 30 rotations per month angular velocity of Earth is greater than the 1 rotation per month angular velocity of the moon?  If Earth's spin slows, that angular momentum needs to be transferred elsewhere.  Conservation of angular momentum demands that, so says Newton.

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If we think that the spin of the Earth can speed up the moon, than by definition we must change the Newton's law of universal gravitation so it will also include that impact.
It already does if you treat Earth as a set of particles instead of a single point mass, as you suggest above.

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Based on the current Newton's law of universal gravitation, "every particle attracts every other particle in the universe with a force..."
Exactly.  And the thrust on the moon can be explained if you apply the law that way.  Earth is not a point-mass.  Point-masses cannot be susceptible to tidal friction.

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So, any particle on Earth might go/down or spin faster/slower/locked due to the local impact of gravity force (tidal activity).
Points don't have spin.  Gravity does not concern itself with spin, speed, or any other dynamic.  It is a static formula for force at a given moment.  Multiple particles can exert torque due to gravitational forces.  Torque is a static value.

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If we believe that the spin of a particle in one object can affect the velocity of a particle at a far end object than:
No claim of the spin of a particle has been made.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/12/2018 14:19:20
Short description.

Tidal forces cause a bulge on the near and far sides of Earth.  Gravity does that, and it can be derived from Newton's formula.  The rotation of Earth, coupled with friction, pushes those bulges into positions no longer directly towards and away from the moon.  The near bulge is ahead of the moon, and thus pulls the moon forward.  The far bulge rotates to behind the moon and pulls it backwards.  The near one is closer, so that forward pull is greater than the backwards pull.  Net effect is gravitational thrust on the moon.

That's tidal thrust explained in gravitational terms.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/12/2018 15:52:04
Thanks
Based on Wiki the highest point of the tidal bulge is 54 Cm:
https://en.wikipedia.org/wiki/Tide
"The theoretical amplitude of oceanic tides caused by the Moon is about 54 centimetres (21 in) at the highest point"

The near bulge is ahead of the moon, and thus pulls the moon forward.  The far bulge rotates to behind the moon and pulls it backwards.  The near one is closer, so that forward pull is greater than the backwards pull.  Net effect is gravitational thrust on the moon.
The Earth moon distance is 384,400 km. The radius of the Earth is 6,371 km.
We can try to calculate the impact of each bulge.
In order to do so, we need to estimate the mass of each bulge with reference to the Earth and verify the total impact.
However, do you agree that 6,371 X2 is neglected with regards to 384,400 km?
Do you also agree that the net mass in that 54 centimeters Bulge is also neglected with regards to the total mass of the Earth?
If so, it is quite clear that the impact of the bulges is virtually zero or close to zero.

In any case, it seems to me that the key point in our discussion is the center of mass based on Newton's second law for the description of the motion of extended objects:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The utility of Newton's second law for the description of the motion of extended objects is the key to its general practical usefulness. The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:"
In our case,
Tidal forces cause a bulge on the near and far sides of Earth.
If the bulges at the near side and at the far side are similar, do you agree that there is no change in the Earth center of mass location with regards to the moon.
Hence, based on this idea, those bulges shouldn't have any impact on the Moon orbital cycle as they cancel each other.
Never the less, even if we ignor the far end bulge, what might be the impact of that near side bulge to the Earth center of mass?
I would assume that less than Pico cm.
So, I wonder how this Pico cm can set the requested extra thrust on the moon which is needed to push the moon away from Earth?

However, based on your following answer, you don't like the idea of using the Earth as a point-mass.
Exactly.  And the thrust on the moon can be explained if you apply the law that way.  Earth is not a point-mass.  Point-masses cannot be susceptible to tidal friction.
Hence, let's assume that we can find that those bulges increase the total gravity force on the moon by 0.00...01.
Is it enough?
However, as the bulges are moving with the moon orbital cycle, than the moon gets this tinny extra gravity force constantly (So, there is no temporary/transient thrust).
Do you agree that it should have a similar impact as we increase the effective mass of Earth by a relative quantity - Let's say by ΔM (M=Earth mass)?
The outcome is that the moon's gravity force will be based on M+ ΔM instead of M.
This effective Earth mass (M+ΔM) should increase the velocity of the Moon.
So, instead of orbiting at velocity v, it will orbit at v +Δv.
However, as it is a constant gravity force, the orbital velocity should stay at the same amplitude.
Hence, I don't see any transient thrust which temporarily increases the velocity.
If there is no transient increase in the velocity, do you agree that there is also no transient increase in R?
Therefore, without this transient increase in R how can we justify the whole idea of pushing the moon away from Earth?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/12/2018 21:07:50
Thanks
Based on Wiki the highest point of the tidal bulge is 54 Cm
Sounds like an ideal tide: If Earth was entirely covered with deep water and barely rotating, the friction would be minimized and the tides would have that amplitude.  Sounds about right.

I live 150 km inland of the Atlantic and we get tides higher than that even here.  Water tends to pile up when hit by continents and forced to go a different way.  The Atlantic has a good size for resonance and the tides are naturally higher than that theoretical 54 cm.  The Pacific doesn't resonate as well, and the tides there are lower in most places.

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The Earth moon distance is 384,400 km. The radius of the Earth is 6,371 km.
We can try to calculate the impact of each bulge.
In order to do so, we need to estimate the mass of each bulge with reference to the Earth and verify the total impact.
However, do you agree that 6,371 X2 is neglected with regards to 384,400 km?
Do you also agree that the net mass in that 54 centimeters Bulge is also neglected with regards to the total mass of the Earth?
Neglected?  By what?  No real clue what you mean by that choice of words.
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If so, it is quite clear that the impact of the bulges is virtually zero or close to zero.
The impact is real, so what ever 'neglected' means, I think we should not neglect those numbers.
I would have run the computations in terms of torque, but you method works as well.

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In any case, it seems to me that the key point in our discussion is the center of mass based on Newton's second law for the description of the motion of extended objects:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
That link treats all objects as point masses.  That isn't useful in describing tidal effects.  It shows a wrench, but makes no effort to show the torque put on a thrown wrench due to tidal forces.

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If the bulges at the near side and at the far side are similar, do you agree that there is no change in the Earth center of mass location with regards to the moon.
Yes, I agree with that.  Tidal forces are not about center of mass.  They're about deviations from it.

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Hence, based on this idea, those bulges shouldn't have any impact on the Moon orbital cycle as they cancel each other.
Doesn't follow.
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Never the less, even if we ignor the far end bulge, what might be the impact of that near side bulge to the Earth center of mass?
None.  You need to consider what each bulge does as a force acting on the moon CoM.
I put two identical masses on either side of you, one nearby to the front, and the other further away to the rear.  You accelerate towards the front one because the force from that one is greater, given the smaller separation.  That's what is pulling the moon forward.

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Hence, let's assume that we can find that those bulges increase the total gravity force on the moon by 0.00...01.
They don't change the total gravity.  But they change the direction of it.  It is no longer straight towards the center of Earth because the Earth is neither a point mass nor a sphere.

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Do you agree that it should have a similar impact as we increase the effective mass of Earth by a relative quantity - Let's say by ΔM (M=Earth mass)?
You want to dump more mass on Earth?  Sure, that would increase the gravitational force and pull the moon closer in.  This is assuming that new mass comes from outside the orbit of the moon.  A trillion low orbit satellites falling out of the sky will have zero effect since they were already effectively part of mass of Earth.

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The outcome is that the moon's gravity force will be based on M+ ΔM instead of M.
This effective Earth mass (M+ΔM) should increase the velocity of the Moon.
Because it is pulled down to a new lower orbit, yes.  Even if it stayed at its current orbit, it would need to move faster, but dumping new mass on Earth doesn't give it the new energy it would need to accelerate like that, so it just fall to that lower orbit.

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So, instead of orbiting at velocity v, it will orbit at v +Δv.
However, as it is a constant gravity force, the orbital velocity should stay at the same amplitude.
What?  It isn't constant gravity force.  You just added mass to M, which changes the force.

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Hence, I don't see any transient thrust which temporarily increases the velocity.
The additional velocity in this case is gained by falling to a lower orbit, just like you personally gain velocity when falling off a ladder.  What do you mean by transient?  The higher velocity from adding mass will be permanent, or it will at least last until the mass is taken away again, or until the tides slow it down over time.

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If there is no transient increase in the velocity, do you agree that there is also no transient increase in R?
Transient?  Are we still talking about the case where mass was added to Earth?  You kind of lost me with this question.

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Therefore, without this transient increase in R how can we justify the whole idea of pushing the moon away from Earth?
I don't think we're talking about adding mass any more.  Not sure when that was abandoned.

There is thrust on the moon, which pushes it uphill, slowing all the way.  There is never a transient increase of velocity like you'd get with a instantaneous boost of momentum from say an asteroid hit.  The whole thing can be visualized with a static force diagram showing all the forces acting on the moon at any one moment, the sum of which is a vector slowing the moon.

This is best done in the simplified case of a non-elliptical orbit.  In the actual elliptical case, the speed of the moon goes up and down in cycles as the separation varies.

If there was a prize given to the known object with the most circular gravitational orbit, I wonder which object would get the prize?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/12/2018 07:58:18
Neglected?  By what?  No real clue what you mean by that choice of words.
The impact is real, so what ever 'neglected' means, I think we should not neglect those numbers.
I would have run the computations in terms of torque, but you method works as well.
O.K
Let's set a calculation.
The main tidal impact is on oceans.
I assume that it also doesn't work from pole to pole.
The pick is 54 cm.
Somehow it seems to me that if we take the two bulges and try to spread their total mass over the whole planet, it won't be higher than few centimeters.
However, just for the calculation let's assume 10 Cm or 1 Decimeter.
The radius of the Earth is 6,371 km =
r = 6,371,000 m or 63,710,000 decimeters.
The volume:
V (Earth) = V(r = 63,710,000) = 4/3 π r^3 = 4/3 π 63,710,000^3 = 4/3 π 2.586 10^23
V(Bulges) = V(r = 63,710,001) - V(r = 63,710,000) = 4/3 π (2.58596615 - 2.58596603) 10^23 = 4/3 π 0.00000013 10^23
Hence, the ratio is:
V(Bulges) / V (Earth) =0.00000013 / 2.586 = 5.02 10 ^(-8)
V (bulge) = 5.02 10 ^(-8) /2 * V (Earth) = 2.51 10 ^(-8) * V (Earth)
M = Earth mass
Mb = Bulge mass = V(Bulges) / V (Earth) * M = 2.51 10 ^(-8) M
Please also be aware that we mainly discuss on water in the bulges. However, the core of the Earth is made of metal which should be quite heavier than water. So, the ratio in mass should be higher.
In any case, it seems to me that the bulges mass is less than the total mass in the Mountains or even in one big chain of Mountains.
Now, we might think to add those two points of mass at the sea level of the planet, one at the front side and the other at far end side from the moon.
This is not fully correct. If we move closer to the pole, the effective radius is shorter and therefore the impact of the gravity bulges is lower.
However, for this calculation let's assume the worst case and set those two bulges points of mass at the at the maximal distance between them (2 x R).
The Gravity force of the Earth is:
F = G M m / R^2
m = Moon Mass
The Earth to the moon distance, R= 384,400 km
F = Earth Gravity force = G M m / R^2 = G M m / (147.7 10^9) = G M m 6.99 * 10^(-3) * 10^(-9) =  G M m 6.99 *  10^(-12)
Bulges distance to the moon:
The bulges don't point directly to the Moon.
Therefore, the effective distance is less than full Earth Radius.
However, I use again the full radius just as a worst case.
Hence:
The front Bulge to the moon distance = R(front) =  384,400 km - 6,371 km = 378,029 Km
The far end Bulge to the moon distance = R(front) =  384,400 km + 6,371 km = 390,771 Km.
F(front bulge) = G Mb m / R(front)^2 = G Mb m / (142.9 10^9)
F(rear bulge) = G Mb m / R(front)^2 = G Mb m / (152.7 10^9)
F(Bulge Total) = F(front) - F (rear) = G Mb m * 10^(-9) * (1/142.9 -1/152.7) =  G Mb m * 10^(-9) * 4.49 * 10^(-4)
= G 2.51 10 ^(-8) M m * 10^(-9) * 4.49 * 10^(-4) = G M m * 4.49 * 10 ^ (-21)
The ratio in the gravity force between the Bulge and the Earth is:
F(Bulge Total)/F(Earth) = G M m * 4.49 * 10 ^ (-21) / G M m 6.99 *  10^(-12) = 0.643 * 10^(-9).
Hence
F(Bulge Total) =  0.643 * 10^(-9) * F(Earth) = G M m 0.643 * 10^(-9) /R^2

Conclusions:
I have calculated the gravity force impact of the Bulges.
It seems to me as a very minor gravity force (comparing to the Earth gravity force).
It might be even weaker than one big chain of mountains.

So, why do you call it "thrust"?
Do you still consider that based on this minor gravity force those bulges can push the moon away from us?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/12/2018 12:51:08
Neglected?  By what?  No real clue what you mean by that choice of words.
O.K
Let's set a calculation.
See, you never tell me what 'neglected' meant, so I am left unable to parse the prior post.

Quote
Mb = Bulge mass = V(Bulges) / V (Earth) * M = 2.51 10 ^(-8) M
However, the core of the Earth is made of metal which should be quite heavier than water. So, the ratio in mass should be higher.
Agree, so closer to 1e-8
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In any case, it seems to me that the bulges mass is less than the total mass in the Mountains or even in one big chain of Mountains.
Arguably so, since said mountains cover much less area, but are much taller than 54 cm and are made of rock.  They don't have much a gravitational difference since they're made of lighter rock and float upwards on the mantle.  The Earth mass is just denser on average between the mountain ranges.

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F = Earth Gravity force = G M m / R^2 = G M m / (147.7 10^9) = G M m 6.99 * 10^(-3) * 10^(-9) =  G M m 6.99 *  10^(-12)
I got more like 6.77e-12, not 6.99, which was the inverse of the 142 distance to the nearest bulge.
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Bulges distance to the moon:
The bulges don't point directly to the Moon.
Therefore, the effective distance is less than full Earth Radius.
However, I will use the full radius just as a worst case.
I would have used a nice round 6000 km, but fine.  Not like we know the actual numbers here.  I like how you're going about it.
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Hence:
The front Bulge to the moon distance = R(front) =  384,400 km - 6,371 km = 378,029 Km
The far end Bulge to the moon distance = R(front) =  384,400 km + 6,371 km = 390,771 Km.
I think you mean R(rear) here.
Quote
F(front bulge) = G Mb m / R(front)^2 = G Mb m / (142.9 10^9)
F(rear bulge) = G Mb m / R(front)^2 = G Mb m / (152.7 10^9)
F(Bulge Total) = F(front) - F (rear) = G Mb m * 10^(-9) * (1/142.9 -1/152.7) =  G Mb m * 10^(-9) * 4.49 * 10^(-4)
or about 4.5e-13GMbm.  Seems a bit early to be subtracting the two force values? That makes little sense.  F total is just the sum of F(front) + F(rear), and this is only the F component perpendicular to the orbit.  We need to compute the tangential component to get the tidal thrust, and for tangential, F total is indeed the difference between the two tangential components, not the sum.  So OK, you're computing a difference here, but you've not yet done the vector trigonometry to compute the forward and resisting thrust.  OK, so you have a difference of full force here, so it is indeed valid to do the trig on just that one force.  I don't see the vector arithmetic anywhere in your post.

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= G 2.51 10 ^(-8) M m * 10^(-9) * 4.49 * 10^(-4) = G M m * 4.49 * 10 ^ (-21)
OK, I think you substituted the value computed for Mb here.
Quote
The ratio in the gravity force between the Bulge and the Earth is:
F(Bulge Total)/F(Earth) = G M m * 4.49 * 10 ^ (-21) / G M m 6.99 *  10^(-12) = 0.643 * 10^(-9).
Hence
No, you subtracted the bulge forces.  The ratio of bulge force to the rest-of-earth would be computed from the sum of the bulge forces, not the difference.

No matter.  That ratio is irrelevant to the tidal thrust.  The main Earth mass plays no role.

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F(Bulge Total) =  0.643 * 10^(-9) * F(Earth)

Conclusions:
I have calculated the gravity force impact of the Bulges.
You need to compute the tangential impact.  Go to just before you subtracted F(rear) from F(front) and compute the tangential force of each bulge separately, and then subtract those.  To do this, you need to assign how far the bulges are pushed off-center by the friction of the spin of Earth.  Let's make it 1000 km.  Each bulge is 1000 km from the line connecting the centers of mass of the two bodies.  That puts a tangential component to the force, and that tangential component is the thrust.  The main mass of Earth is centered on that COM line, so it plays no role at all.

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It seems to me as a very minor gravity force (comparing to the Earth gravity force).
It might be even weaker than one big chain of mountains.

So, why do you call it "thrust"?
Because it pulls in the direction of motion (energy increase), not tangential to it (acceleration without energy increase).  Yes, it is a very minor force compared to the main component, but it is always forward, so the effect is cumulative forever.  The main force is always balanced in all directions, so the cumulative effect is zero after each month.

Yes, the mountains may mass more than the bulges, but they have no cumulative effect since they are ahead as often as behind the COM line, so the net effect is zero.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/12/2018 13:48:22
I will try to illustrate some of the vector work, using the guess that friction pushes the bulges 1000 km off center.

The Earth to the moon distance, R= 384,400 km
...
The front Bulge to the moon distance = R(front) =  384,400 km - 6,371 km = 378,029 Km
The far end Bulge to the moon distance = R(front) =  384,400 km + 6,371 km = 390,771 Km.
F(front bulge) = G Mb m / R(front)^2 = G Mb m / (142.9 10^9)
F(rear bulge) = G Mb m / R(front)^2 = G Mb m / (152.7 10^9)
Those forces are respectively GmMb * 7.00e-12 and  GmMb * 6.55e-12

So we've defined two triangles will 1000km on the short side and 378029 or 390771 on the other.
F(front) was computed at GmMb * 7e-12, so the forward thrust from that bulge is
F(forward) = GmMb * 1.8517 e-14
F(backward) = GmMb * 1.6762 e-14
The 1.8517 figure is 700 force * 1000km / 378029 km, or essentially the F multiplied by the sin of the angle formed at the moon between the COG line and the line to the near bulge.

Difference is 1.755 e-15 GmMb cumulative forward thrust forever adding energy/angular momentum to the orbit of the moon.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/12/2018 21:09:49
Because it pulls in the direction of motion (energy increase), not tangential to it (acceleration without energy increase).  Yes, it is a very minor force compared to the main component, but it is always forward, so the effect is cumulative forever.  The main force is always balanced in all directions, so the cumulative effect is zero after each month.
Thanks for all your excellent remarks.

However, I still wonder why you insist to call it "Thrust".
The meaning of Thrust by Google is : "the propulsive force of a jet or rocket engine." "Push (something or someone) suddenly or violently in the specified direction."
I couldn't find any sort of "engine" in the Bulges activity. It is just increases the gravity force.

Never the less, I think that I understand the source for your statement:
Please see Figure 7.24 in the following article:
https://lifeng.lamost.org/courses/astrotoday/CHAISSON/AT307/HTML/AT30706.HTM
"Figure 7.24 The tidal bulge raised in Earth by the Moon does not point directly at the Moon. Instead, because of the effects of friction, the bulge points slightly "ahead" of the Moon, in the direction of Earth's rotation. (The magnitude of the effect is greatly exaggerated in this diagram.) Because the Moon's gravitational pull on the near-side part of the bulge is greater than the pull on the far side, the overall effect is to decrease Earth's rotation rate."
So, if I understand it correctly,  as the bulge points slightly "ahead" of the Moon, we believe that it pulls in the direction of motion (energy increase)"
Is it correct?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/12/2018 22:58:30
However, I still wonder why you insist to call it "Thrust".
Because it is forward, in the direction of motion.  I turn the wheel of my car and it accelerates it to the left, but that isn't thrust, and so the car turns without gaining speed.  But the engine supplies force in the forward direction, adding mechanical energy to the car as it does so, resulting perhaps in more speed, or perhaps just helping the car up a hill.  The car might be slowing, but the engine is still providing thrust/energy.  So we call that force thrust, and not the force that turns the car left.

Did you see my post 40?  It shows how to compute the thrust.

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The meaning of Thrust by Google is : "the propulsive force of a jet or rocket engine." "Push (something or someone) suddenly or violently in the specified direction."
Well, it might be kilotons of thrust, but it still isn't very violent.

Quote
I couldn't find any sort of "engine" in the Bulges activity. It is just increases the gravity force.
No, it doesn't.  Overall force is the same, but tides change the direction of that force.

Quote
Never the less, I think that I understand the source for your statement:
Please see Figure 7.24 in the following article:
Oh lovely.  They show the bulges dragged off to either side like I've been describing.  They'd be straight at the moon if there was no friction.

Quote
"Figure 7.24 The tidal bulge raised in Earth by the Moon does not point directly at the Moon. Instead, because of the effects of friction, the bulge points slightly "ahead" of the Moon, in the direction of Earth's rotation. (The magnitude of the effect is greatly exaggerated in this diagram.)
So much for my 1000 km estimate I bet.
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Because the Moon's gravitational pull on the near-side part of the bulge is greater than the pull on the far side, the overall effect is to decrease Earth's rotation rate."
So, if I understand it correctly,  as the bulge points slightly "ahead" of the Moon, we believe that it pulls in the direction of motion (energy increase)"
Is it correct?
Yes.  I've been saying that for how many posts now?  Energy increase (thrust) to the moon, and energy decrease to Earth's rotation rate, as the comment states.  The decrease is greater than the increase.  Momentum is preserved here, but not energy (2.5 terawatts), almost all of which is lost to friction heat.  Only about 1/20th of that energy (around 120 gigawatts) is transferred to the moon.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/12/2018 06:25:48
Thanks
So,  we believe that as the bulge points slightly "ahead" of the Moon, it pulls in the direction of motion (energy increase)".
In order to understand the real impact of that offset is, let's verify the following conditions:

1. There is no offset (0 degree) and the bulges are pointing directly to the moon
Do you agree that the effective distance between the bulges to the moon is maximal?. Therefore, can we assume the diffrence between their gravity force is maximal. However, can we assume that the thrust is Zero?
2. The offset is maximal (90 degree). The bulge are located at/almost the poles. therefore, the distance between the front bulge to the moon is actually equal to the distance of the rear bulge to the moon. Hence, can we assume that they have the same gravity force? However, what is the expected thrust? Is it zero or infinite?
3. The offset is 60 degree.
Cos (60) = 0.5
Hence, the effective radius is 0.5 * r = 0.5 * 6371 = 3185.5 K.m (Let's assume 3000 Km)
In this case, the effective distance to the moon is as follow:
R(front) =  384,400 km - 3,000 km = 381,400 Km
R(rear) =  384,400 km + 3,000 km = 387,400 Km.
So, it's easy to calculate the gravity force of each one, however, I can't understand why there will be any thrust.
I wonder if it is related to the idea that there are two bulges.
Hence, if there was just only one bulge, (let's assume only the front bulge), does it mean that the thrust will be zero?
In the same token, if we will set only the rear bulge, can we assume that the thrust will be of zero?
So, only if we have them both, we get the impact of thrust?
I will try to illustrate some of the vector work, using the guess that friction pushes the bulges 1000 km off center.
So we've defined two triangles will 1000km on the short side and 378029 or 390771 on the other.
F(front) was computed at GmMb * 7e-12, so the forward thrust from that bulge is
F(forward) = GmMb * 1.8517 e-14
F(backward) = GmMb * 1.6762 e-14
The 1.8517 figure is 700 force * 1000km / 378029 km, or essentially the F multiplied by the sin of the angle formed at the moon between the COG line and the line to the near bulge.

Difference is 1.755 e-15 GmMb cumulative forward thrust forever adding energy/angular momentum to the orbit of the moon.
In order to understand the idea.
Let's eliminate the Earth.
Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?
I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 08/12/2018 14:25:37
So,  we believe that as the bulge points slightly "ahead" of the Moon, it pulls in the direction of motion (energy increase)".
In order to understand the real impact of that offset is, let's verify the following conditions:

1. There is no offset (0 degree) and the bulges are pointing directly to the moon
Do you agree that the effective distance between the bulges to the moon is maximal?. Therefore, can we assume the diffrence between their gravity force is maximal. However, can we assume that the thrust is Zero?
No thrust, right.  The difference between their force is irrelevant then.
Quote
2. The offset is maximal (90 degree). The bulge are located at/almost the poles.
No, 90° eastward and westward, not at the poles.  The bulges move around the equator, not from pole to pole.  Not exactly.  The orbital plane is about level with the solar system plane, not with the tilt of Earth's axis, so each bulge actually moves a ways north to south and back twice a day, regardless of offset angle.
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therefore, the distance between the front bulge to the moon is actually equal to the distance of the rear bulge to the moon. Hence, can we assume that they have the same gravity force? However, what is the expected thrust? Is it zero or infinite?
Still zero since they're equal and opposite.

I think I used about 10 degrees in my example.  Possibly still too much.
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3. The offset is 60 degree.
Cos (60) = 0.5
Hence, the effective radius is 0.5 * r = 0.5 * 6371 = 3185.5 K.m (Let's assume 3000 Km)
In this case, the effective distance to the moon is as follow:
R(front) =  384,400 km - 3,000 km = 381,400 Km
R(rear) =  384,400 km + 3,000 km = 387,400 Km.
So, it's easy to calculate the gravity force of each one, however, I can't understand why there will be any thrust.
I wonder if it is related to the idea that there are two bulges.
Did you read my description of the vector arithmetic in post 40?  Each bulge is off to the side by Sin(60) in this case (about 5500 km) so the force is not straight to Earth but has significant forward/backward components respectively.

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Hence, if there was just only one bulge, (let's assume only the front bulge), does it mean that the thrust will be zero?
I cannot comment on that since it would change the center of gravity of Earth and hence not really be a bulge.
Thrust from each bulge is very much not zero.  They are in opposition, so the net thrust is much less.

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Draw a vector diagram and work out the components (downward and forward) of the total force.

Do you know about vector arithmetic?  Surely there are some sites that inform well.  This is a really simple case.

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In order to understand the idea.
Let's eliminate the Earth.
Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?
They don't.  They're in orbit, so they'll go around each other every X many hours, not every month.  They'd not add any cumulative thrust to the moon.  The 3-body problem would probably make the whole system unstable after not much time.

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I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?
It is only gravity, yes.  Work out the forces in 2 dimensions, not as a scalar.  Force is a vector, not a scalar.

If I am on a skateboard going down a slope, I speed up.  That means gravity force is providing me with thrust, increasing my speed.  Is it so unimaginable that it might do this?  If I was moving up a slope, that same gravity would slow me down, and not be thrust, but rather a braking action (or be a negative thrust if you will).  If I am on a perfectly level lot, there is no speed change, so the same gravitational force results zero thrust to my skateboard.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/12/2018 06:19:12
Thanks
Please let me know if the following is correct:
The Thrust is a direct outcome of the offset. (While the offset was a direct outcome of the tidal friction).
Based on the offset phase you have calculated the thrust in thread 40:
The 1.8517 figure is 700 force * 1000km / 378029 km, or essentially the F multiplied by the sin of the angle formed at the moon between the COG line and the line to the near bulge.
Difference is 1.755 e-15 GmMb cumulative forward thrust forever adding energy/angular momentum to the orbit of the moon.

However, when I have tried to focus in a general case of the offset, you have informed me that there is no thrust due to offset:
Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?
I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?
They don't.  They're in orbit, so they'll go around each other every X many hours, not every month.  They'd not add any cumulative thrust to the moon.  The 3-body problem would probably make the whole system unstable after not much time.

Please be aware that I have specifically asked to set them at any offset as you wish.
So, does it mean that in a general case of offset there is no thrust?
If there is no thrust due to the offset, how can we use this idea for tidal friction?

It is only gravity, yes.  Work out the forces in 2 dimensions, not as a scalar.  Force is a vector, not a scalar.
If I am on a skateboard going down a slope, I speed up.  That means gravity force is providing me with thrust, increasing my speed.  Is it so unimaginable that it might do this?  If I was moving up a slope, that same gravity would slow me down, and not be thrust, but rather a braking action (or be a negative thrust if you will).  If I am on a perfectly level lot, there is no speed change, so the same gravitational force results zero thrust to my skateboard.
I like the idea of skateboard.
This shows that we can convert gravity force into thrust.
Don't forget that Newton have used the example of falling Apple to find the whole idea of gravity.
However, we can't limit the skateboard idea just for tidal friction explanation.
If there is a possibility to convert Gravity into thrust due to the offset in a tidal system, than please set a formula for thrust for general case of offset.
This is a breakthrough concept in gravity.
Newon, kepler and Einstein didn't offer any solution for converting gravity force into thrust in orbital system.
Our scientists believe that it is feasible to get thrust due for tidal friction offset as follow:

"1. There is no offset (0 degree) and the bulges are pointing directly to the moon
No thrust, right.  The difference between their force is irrelevant then.
2. For offset 60 degree
Each bulge is off to the side by Sin(60) in this case (about 5500 km) so the force is not straight to Earth but has significant forward/backward components respectively.
3. The offset is maximal (No, 90° - eastward and westward, not at the poles)
Still zero since they're equal and opposite."

Therefore, we should set a formula that represents this understanding.
So, can we set a formula (or graph) for thrust per Offset tidal friction phase?

However, If thrust works for tidal offset idea, it should also work for the example which I have offered.
If it doesn't work for this example, than it should not work also for tidal offset.
How can we come with idea for a special case and close it only for that case?
Do you agree that if we can't open the idea for any offset in orbital system, than we might have a problem with this idea?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 09/12/2018 14:44:58
However, when I have tried to focus in a general case of the offset, you have informed me that there is no thrust due to offset:
Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?
I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?
They don't.  They're in orbit, so they'll go around each other every X many hours, not every month.  They'd not add any cumulative thrust to the moon.  The 3-body problem would probably make the whole system unstable after not much time.

Please be aware that I have specifically asked to set them at any offset as you wish.
So, does it mean that in a general case of offset there is no thrust?
If there is no thrust due to the offset, how can we use this idea for tidal friction?
There is thrust in that case, but the forces will move your objects around, so the thrust might last minutes at best, after which the pieces will rearrange and produce negative thrust.  It all cancels out.  You've removed the large middle piece which is the source of the inertia and friction.

Anyway, if it is not considered a dynamic system (no time involved), that momentary arrangement of matter does indeed put a forward component of force on the moon.  I think perhaps that is what you were trying to ask.

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I like the idea of skateboard.
This shows that we can convert gravity force into thrust.
Don't forget that Newton have used the example of falling Apple to find the whole idea of gravity.
However, we can't limit the skateboard idea just for tidal friction explanation.
If there is a possibility to convert Gravity into thrust due to the offset in a tidal system, than please set a formula for thrust for general case of offset.
This is a breakthrough concept in gravity.
Newon, kepler and Einstein didn't offer any solution for converting gravity force into thrust in orbital system.
I beg to differ.  It all follows from Newtonian mechanics.  Kepler's laws are specifically about orbits and derive from Newton's equations, and he was quite aware of tidal effects.

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Our scientists believe that it is feasible to get thrust due for tidal friction offset as follow:

"1. There is no offset (0 degree) and the bulges are pointing directly to the moon
No thrust, right.  The difference between their force is irrelevant then.
2. For offset 60 degree
Each bulge is off to the side by Sin(60) in this case (about 5500 km) so the force is not straight to Earth but has significant forward/backward components respectively.
3. The offset is maximal (No, 90° - eastward and westward, not at the poles)
Still zero since they're equal and opposite."

Therefore, we should set a formula that represents this understanding.
So, can we set a formula (or graph) for thrust per Offset tidal friction phase?
We already did that.  The linear offset is computed from the sin of the angular offset, which is closer to 10 degrees than to 60.  The distance of each bulge to the moon is the mean distance R ± cos(angular offset).
That linear offset produces a forward and retrograde force component on the moon.   Forward thrust is total force(GMmr², where M is the bulge mass) * linear offset / distance of bulge to moon.

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However, If thrust works for tidal offset idea, it should also work for the example which I have offered.
If it doesn't work for this example, than it should not work also for tidal offset.
It does work.  It's just that if there is time involved, all the pieces move around and don't maintain that steady thrust.  That was my complaint.  You said the two bulges were in orbit about each other.  Orbit is a dynamic behavior, and has nothing to do with forces of just 3 masses in these specific places.  Two orbiting objects with the mass of our tidal bulges will not put any significant force on the moon.  In fact, the moon will simply exit the system due to its linear velocity.

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How can we come with idea for a special case and close it only for that case?
Do you agree that if we can't open the idea for any offset in orbital system, than we might have a problem with this idea?
Remove time from your example.  Then it is not an orbital system, but merely a bunch of masses at specific locations, with forces acting between each of them.  You can close it for that case.  With time eliminated, we need not consider anything's velocity or acceleration.  Only forces need be computed.
Ever take a statics class in college?  That's what the course is about:  Forces in the absence of time, and finding balances between them.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/12/2018 16:06:09
Thanks Halc

The information about tidal friction is fully clear.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 10/12/2018 16:00:34
I assume that by now we all understand how the tidal friction set the thrust which is needed to push away the Moon from the Earth.
However, the water bulges in oceans are very unique for the Earth/Moon system.
In one hand, this is the only planet in the solar system with so much water and on the other hand, the moon is relatively big enough to form the bulges.
We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?
They don't have water. They don't have relatively big moon around them. They don't form those water bulges.
So, if tidal friction can't be the answer for their drifting outwards activity, what could be the answer for that?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 10/12/2018 18:08:14
However, the water bulges in oceans are very unique for the Earth/Moon system.
In one hand, this is the only planet in the solar system with so much water and on the other hand, the moon is relatively big enough to form the bulges.
The water makes a nice difference, yes.  The size of the moon is irrelevant.  If it was just an automobile up there, it might generate a trillionth of the gravity, a trillionth of the tidal effect on the ocean, and hence a trillionth of the force, which would result in the exact same acceleration.  The automobile would move to higher orbits at the exact same pace as does the moon, but the Earth would have a trillionth of the friction, so the spin would not degrade at a measurable rate like we get from our big honkin moon.

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We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?
Because they go around slower than the spin of the primary, and in the same direction.  If they orbited lower (at or below geo-sync), or if they went around the opposite way, they'd get pushed down.  Such objects tend to fall into their primaries before too long.

The rings of Saturn is a nice example of a moon that did just that.  Too close, and tidal forces pulled it within the Roche limit where it breaks up.
Another example might be Venus, whose moon, if it had ever had one, would probably have gone around the normal way, causing it to drop its orbit that is retrograde to the spin of Venus itself.

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They don't have water. They don't have relatively big moon around them. They don't form those water bulges.
So, if tidal friction can't be the answer for their drifting outwards activity, what could be the answer for that?
It works with the planet itself.  Even the relatively solid moon has strain in the crust from tidal forces from Earth, and those forces are really small since the moon is tide-locked, but it still rocks back and forth with its elliptical orbit.

Anyway, the sun is hardly solid and each of the planets form a tide on it.  The big planets are all very liquid and susceptible to significant tides.  The amazing story is Pluto and Charon, both rocks pretty much free of stuff that would exhibit tidal strain, and yet the two of them have managed to become mutually tidal locked already.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/12/2018 06:46:34
We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?
Because they go around slower than the spin of the primary, and in the same direction.  If they orbited lower (at or below geo-sync), or if they went around the opposite way, they'd get pushed down.  Such objects tend to fall into their primaries before too long.
Thanks
So, the tidal friction idea between Earth/Moon is actually none relevant to all the other orbital systems in the solar system.
Now there is new idea.
All the planets and moons are pushed away as they all "go around slower than the spin of the primary, and in the same direction".

1. Would you kindly explain this idea?
2. Can we prove it by mathematical calculations/formula?
3. Why our moon isn't pushed outwards due to this idea?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 11/12/2018 12:24:33
So, the tidal friction idea between Earth/Moon is actually none relevant to all the other orbital systems in the solar system.
I think I said it was relevant to all of them.  They're all getting pushed out, except for the two exceptions I mentioned, both of which resulted in no moon.

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Now there is new idea.
All the planets and moons are pushed away as they all "go around slower than the spin of the primary, and in the same direction".
Not a new idea.  Been around for at least 300 years.  The ones that don't do that don't survive, so the only ones left in our solar system are the ones that go slow, and in the same direction.

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1. Would you kindly explain this idea?
If they orbit faster than the primary, the tidal bulges will lag behind the orbit, and the resulting forces will put a braking action on the object.  Similarly if the orbit is retrograde (faster or slower), the force is braking, not thrust.  Energy is lost, so the moon falls down into the planet.
The ISS is slowly dropping its orbit due to this effect since it goes around every 90 minutes, 16x faster than the rotation of Earth.  Left alone long enough, and even absent friction from being too close to the atmosphere, it will eventually fall.

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2. Can we prove it by mathematical calculations/formula?
Yea, the same calculation used to compute the thrust force on the moon.  That calculation yields a negative number for a low or retrograde orbit.

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3. Why our moon isn't pushed outwards due to this idea?
Thrust is forward and positive, so the orbit of the moon increases.  So it is 'pushed outward' at a pace of a few cm per year.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 11/12/2018 13:02:09
Quote from: Halc
We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?
Because they go around slower than the spin of the primary, and in the same direction.  If they orbited lower (at or below geo-sync), or if they went around the opposite way, they'd get pushed down.  Such objects tend to fall into their primaries before too long.
Both our statements are wrong here, so I need to correct this.
I looked at Jupiter's moons as an example, and most of them are in fact not being pushed out, but have diminishing orbits.
Metis and Adrastea have orbital periods shorter than the primary, and hence have negative thrust.  They will eventually fall in turn, giving Jupiter some nice rings.
The next 13 moons including all the popular ones have growing orbits.  That's just 13 of the 63 known moons.
The outer 48 moons all have, for some reason, retrograde orbits.  I suspect the reason is that they're all captured objects, and a free object will get a speed boost when passing a gravitational source in the direction of its orbit (not in the direction of its spin).  All the Voyager spacecraft have achieved the majority of their energy from such gravitational boosts.  The effect is to be flung away all the harder, not to be captured.  But if the object passes in front of the primary and outbound, it slows, possibly enough to be captured.  I think this is how each of the moons was captured.  Each needs to come in at greater than escape velocity and lose enough of it to drop below escape velocity.
I find that odd that Jupiter has so many retrograde moons, but none for any of the other planets.

All the moons have been there long enough to become tidal locked.  All those retrograde moons are so far out that their tidal effects result in negligible braking, so their orbits will probably take many billions to trillions of years before they fall into Jupiter.

Mars has 2 moons, and Phobos is dropping (seemingly faster than tidal effects can explain), and is expected to crash in a mere 40 million years.
Saturn has destroyed its last low-orbit moon, forming the current rings.  The lowest one is outside Saturn's geosync orbit and has positive tidal thrust.  All of Saturn's 60 moons have growing orbits.
Uranus has 27 moons, 11 of which are in low orbit and 'falling'.
Neptune has 14 moons, 5 of which are in low orbit and 'falling'.

None of these have retrograde moons.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 11/12/2018 17:01:33
Thanks for the explanation about the moons.

However, I have no clue why if the Moon orbits in one direction/periods it will be pushed outwards, while if it orbits in the other direction/periods, it should be pulled inwards.
In tidal friction, the offset of the bulges sets the Thrust?
In this case, there is no offset.
So, how can we get a negative or positive thrust without any offset?

Metis and Adrastea have orbital periods shorter than the primary, and hence have negative thrust.  They will eventually fall in turn, giving Jupiter some nice rings.

Can you please explain why the orbital direction/periods can set a thrust?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 11/12/2018 20:11:54
Thanks for the explanation about the moons.

However, I have no clue why if the Moon orbits in one direction/periods it will be pushed outwards, while if it orbits in the other direction/periods, it should be pulled inwards.
In tidal friction, the offset of the bulges sets the Thrust?
Yes, the offsets set the thrust.  If the offset is positive, so it the thrust.
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In this case, there is no offset.
In what case?  The only case with no offset is the tides from objects in geoSync orbit.  Those have no offset, so those orbits are relatively stable.

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Metis and Adrastea have orbital periods shorter than the primary, and hence have negative thrust.  They will eventually fall in turn, giving Jupiter some nice rings.

Can you please explain why the orbital direction/periods can set a thrust?
The tidal offsets are negative for those moons, as are the offsets for all the retrograde moons.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/12/2018 05:51:20
Yes, the offsets set the thrust.  If the offset is positive, so it the thrust.
Can you please explain how the orbital direction sets the offset?
Why orbital periods that are shorter than the primary can set a negative offset?
Do we have any drawing for that idea (as we have for the water bulges on Earth)?
So, how do we get any sort of offsets due to orbital direction or orbital periods?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 12/12/2018 12:39:58
Quote from: Halc
Yes, the offsets set the thrust.  If the offset is positive, so it the thrust.
Can you please explain how the orbital direction sets the offset?
It is friction.  If my car is on a road, the brakes act as friction.  If the road is moving forward faster than the car, hitting the brakes will accelerate (positive thrust) to a higher speed.  If the road is slower than the car, or moving at any speed in the opposite direction, the brakes will slow the car down (negative thrust).

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Why orbital periods that are shorter than the primary can set a negative offset?
Because the friction of the slower primary drags the bulge to the rear, a negative offset.

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Do we have any drawing for that idea (as we have for the water bulges on Earth)?
They all look the same.  Bulges.  Offsets.  A negative offset puts the bulges on the opposite side.  A zero offset (which you only get with geosync orbits) are directly towards and away from the orbiting thing, and result in zero friction and zero thrust.

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So, how do we get any sort of offsets due to orbital direction or orbital periods?
Gravity tries to pull the offsets straight at the orbiting thing, but the spin of the primary in this case pushes the offset to one side or the other, depending which way the ground under the orbiting object appears to be moving.

If I look at Earth from a forward pointing ship in low orbit (400 km), things tend to appear in front of the ship and disappear behind it, just like the view from an airplane.   So friction with the ground below is going reduce my velocity, so steps are taken to avoid this friction as much as possible.  The airplane has optimal aerodynamics to minimize this drag.  The ISS is high enough to minimize atmospheric drag, but there's little it can do about tidal drag.

Same ship at geosync (36000 km):  The Earth below is now stationary.  I see the exact same spot below whenever I look.  Friction with that would produce zero thrust.

Same ship further out, (100000 km).  Now the Earth features appear from behind me and rotate away to the front.  Earth is spinning faster than my orbit.  Friction with that would push me forward (positive thrust).  The moon's orbit is in this last category (beyond 36000 km).
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 12/12/2018 13:09:08
Similarly, I am suicide Halc, deciding to do the job right and jump out of the tallest building I can find.  There is a likely tall space elevator right on the equator that goes up 100000 km.
They actually are trying to plan such a structure since it would make it so much easier to put things in space.

I could jump out at 400 km up, and fall quickly to my death.  I could leap out at 36000 km, and just hover there forever.  If I jump out at 50000 km, I'd fall up, but at least stay in orbit.  If at 55000 km, I'd fall up and never come back.

I could not find an answer to the question of the maximum height I could jump out of a window and still hit the ground.  Jumping out at any lower point puts you in orbit.  At some height, the perihelion of that orbit is above the altitude of the atmosphere, and a jump from that altitude or above will not ever hit the ground.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/12/2018 18:38:25
Because the friction of the slower primary drags the bulge to the rear, a negative offset.
There are nine planets in the solar system.
We all know that there are bulges in our planet.
However, do we see any sort of bulge at any other planet in the solar system?
If we don't see any bulge at any other planet, how can we think that a bulge which doesn't exist can set an offset that we wish for?
Tidal friction is based on Bulges.
If there is no bulges there is no offset. If there is no offset there is no thrust. If there is no thrust there is no energy to push or to pull the moon.
So, do you agree that without confirmed bulges per planet, there is a problem with this hypothetical idea?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 12/12/2018 19:30:31
There are nine planets in the solar system.
We all know that there are bulges in our planet.
However, do we see any sort of bulge at any other planet in the solar system?
If we don't see any bulge at any other planet, how can we think that a bulge which doesn't exist can set an offset that we wish for?
I look at a picture of Earth and see no bulge.  They're not exactly pronounced.  Point is, the tidal forces produce stress on the bodies orbiting each other, and those stresses produce strain of one sort or another, and changing strain is movement that produces heat from friction.  These forces are strong enough to have tide-lock (nearly??) every moon of every planet.  Water isn't necessary.  The tides are quite significant in amplitude on planets/stars that are not rocks.  Venus has a thick atmosphere to drag around.  Pluto and Mercury are the only planets with nothing but solids to work with, and both those have become tide locked with the most significant gravity source nearby.

Tidal friction is based on Bulges.
If there is no bulges there is no offset. If there is no offset there is no thrust. If there is no thrust there is no energy to push or to pull the moon.[/quote]
None of these assert that the deformation needs to be measurably confirmed from a distance, so yea sure.
Not all deformations manifest as something that can be classified as a bulge.  I can put a foam cube out there spinning, and tidal forces will deform it, but the corners will always stick out the furthest, and hence a given deformation will not necessarily be a bulge.  Even Earth has mountains that 'bulge' out far further than does the water.

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So, do you agree that without confirmed bulges per planet, there is a problem with this hypothetical idea?
No.  The forces involved do not need confirmation in order to exist.  Their effects are measurably confirmed.  Most of the moons nearby their primaries have measurable changes to their orbits, Phobos probably being the top of the list, despite being very unlikely to ever produce a measurable deformation of Mars' atmosphere or crust until it physically hits them.

Are you just posting here as a denier of such forces?  You'd have to explain how one object can put stress on another object without producing strain.  That would be quite a piece of new physics to propose.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/12/2018 20:56:17
Thanks Halc.

Water isn't necessary.  The tides are quite significant in amplitude on planets/stars that are not rocks.  Venus has a thick atmosphere to drag around.  Pluto and Mercury are the only planets with nothing but solids to work with, and both those have become tide locked with the most significant gravity source nearby.
Why water isn't necessary?
Please remember that the tidal on Earth is based on water.
https://en.wikipedia.org/wiki/Tide
"The theoretical amplitude of oceanic tides caused by the Moon is about 54 centimetres (21 in) at the highest point"
If I remember correctly, at land the tidal is just few cm (2-3 Cm?).
The mass ratio between Earth/moon is significantly higher than any other Planet/moon system.
Did we try to measure the Tidal impact on other planets?
If I understand it correctly, the offset is also due to water.
Therefore, without verifying minimal bulge amplitude and offset, I really can't understand why we are so sure that there is a minimal thrust that can push or pull the moon.
With regards to gas planets -
Let's look at Jupiter as it is the biggest planet in the solar system:
https://solarsystem.nasa.gov/moons/jupiter-moons/overview/?page=0&per_page=40&order=name+asc&search=&placeholder=Enter+moon+name&condition_1=9%3Aparent_id&condition_2=moon%3Abody_type%3Ailike
"Jupiter has 53 named moons and another 26 awaiting official names. Combined, scientists now think Jupiter has 79 moons."
It's biggest moon is Ganymede:
"Ganymede's mean radius is 1,635 miles (2,631.2 km). Although Ganymede is larger than Mercury it only has half its mass, classifying it as low density.
Therefore, it is clear that the ratio between Jupiter and this biggest moon is very low.
I would assume that at this ratio, the tidal impact on land in Earth might be less than one millimeter
Jupiter has 79 moons. Therefore, there is good chance that those moons cancel the tidal impact of each other.
There is one more issue.
For some moons we believe that there is a negative thrust.
This negative thrust should pull those moons inwards.
This is the theory.
However, did we measure if they are pulled inwards?
What is the chance that they will not be so cooperative with our theory?
Why we are so sure with our theory while we only have real measurements of only Earth/Moon and Sun/Earth system (Both drifts/Pushed outwards?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 13/12/2018 00:58:08
Why water isn't necessary?
Please remember that the tidal on Earth is based on water.
Water has the greatest drag on Earth's spin since it has a lot of that, but atmosphere and crust also contribute.
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If I remember correctly, at land the tidal is just few cm (2-3 Cm?).
Sounds like strain to me.  It takes energy to do that, and that energy is lost to heat.

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The mass ratio between Earth/moon is significantly higher than any other Planet/moon system.
Really?  The ratio is about 1.2%, far less than the 11% ratio of Charon to Pluto.  OK, Pluto isn't exactly a planet.

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Did we try to measure the Tidal impact on other planets?
Apparently you don't read my posts.

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If I understand it correctly, the offset is also due to water.
Most planets don't have free flowing water.  Europa does, even it there's a crust that has some inhibiting effect to its tides.  I think Europa is tide locked, so no matter.

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Therefore, without verifying minimal bulge amplitude and offset, I really can't understand why we are so sure that there is a minimal thrust that can push or pull the moon.
I think those particular figures have quite been verified.  Earth tide offset isn't a fixed figure.  It varies all over the place due to geographic features and ressonance.  It's the friction that counts, and that friction is greatest in shallow areas like around England.

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"Ganymede's mean radius is 1,635 miles (2,631.2 km). Although Ganymede is larger than Mercury it only has half its mass, classifying it as low density.
Therefore, it is clear that the ratio between Jupiter and this biggest moon is very low.
Big time, yes.  All the easier for Jupiter to push it along.  Easy to push small things.

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I would assume that at this ratio, the tidal impact on land in Earth might be less than one millimeter
Probably, yes.  The tide raised by the ISS is waaaay less than that millimeter, but the ISS is super light, so it accelerates (negative) just as much as a large thing would at that altitude.  That acceleration rate has little to do with the mass of the orbiting thing.'

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Jupiter has 79 moons. Therefore, there is good chance that those moons cancel the tidal impact of each other.
They orbit at different periods, so there is zero chance of this.  Only a moon's own tides affect that moon, not the tides of other bodies, which have random offsets and thus cancel completely in the long run.

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There is one more issue.
For some moons we believe that there is a negative thrust.
Like Phobos and most of Jupiters moons, yes.

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However, did we measure if they are pulled inwards?
Jupiters outer moons are so far out that the tides might not have measurable impact.  The two inner ones very much do have measurable orbit degradation.  Phobos has massive degradation, and has only some tens of millions of years left in its life.

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What is the chance that they will not be so cooperative with our theory?
Pretty much nill.  It isn't exactly an untested theory.

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Why we are so sure with our theory while we only have real measurements of only Earth/Moon and Sun/Earth system (Both drifts/Pushed outwards?
Making up your facts I see.  We've plenty of measurements of the others, at least the things near their primaries, which have significant forces acting on their orbits.  Most of Jupiter's outer moons have only been discovered in the last 15 years, which is not much time to measure the trivial orbital changes put on them at their very distant and very eccentric orbits.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/12/2018 14:34:25
For some moons we believe that there is a negative thrust. This negative thrust should pull those moons inwards. This is the theory. However, did we measure if they are pulled inwards?
Like Phobos and most of Jupiter's moons, yes
Jupiters outer moons are so far out that the tides might not have measurable impact.  The two inner ones very much do have measurable orbit degradation.  Phobos has massive degradation, and has only some tens of millions of years left in its life.
Making up your facts I see.  We've plenty of measurements of the others, at least the things near their primaries, which have significant forces acting on their orbits.

Why do we ignore the distance between the moon/rings to the planet?
https://en.wikipedia.org/wiki/Phobos_(moon)
"Phobos orbits 6,000 km (3,700 mi) from the Martian surface, closer to its primary body than any other known planetary moon. "

So, Phobos orbits 6000 Km from the surface of Mars while the radius of mars is about 3400 km. so the ratio is 1 to 1.76.
As an example:
The Earth Radius is 6370 Km. If Phobos would orbit around the Earth at the same ratio, its orbital distance from the Surface should be about 11,000 Km.
That orbital radius is lower than the altitude range of navigation satellites:
https://en.wikipedia.org/wiki/Medium_Earth_orbit
"The most common use for satellites in this region is for navigation, communication, and geodetic/space environment science.[1] The most common altitude is approximately 20,200 kilometres (12,552 mi)),"

If I understand it correctly, we expect that a navigation satellite should be pulled inwards.
If so, why is it so big surprise that Phobos is also pulled inwards?

With regards to Jupiter moons/Ring:
https://en.wikipedia.org/wiki/Rings_of_Jupiter
"Main ring - The narrow and relatively thin main ring is the brightest part of Jupiter's ring system. Its outer edge is located at a radius of about 129000 km (1.806 RJ;RJ = equatorial radius of Jupiter or 71398 km) and coincides with the orbit of Jupiter's smallest inner satellite, Adrastea.[2][5] Its inner edge is not marked by any satellite and is located at about 122500 km (1.72 RJ).[2]
So, Jupiter radius is 71398 Km.
Therefore, If this ring would orbit the Earth at the same ratio (1.806), its orbital distance from the Surface of earth should be about 11,500 Km.
So, don't you see a similarity between satellite around the Earth, Phobos around Mars and Main ring around Jupiter?
Don't you agree that with tidal or without it, all of them must eventually fall down?
Therefore, how can we use an object which it's orbital cycle is so close to the host to prove a negative thrust due to tidal???
Can you please find one moon (only one) in the whole solar system that is located long enough from its planet which is pulled inwards due to negative thrust (But please - real prove for that)?

With Regards to Europa:
Most planets don't have free flowing water.  Europa does, even it there's a crust that has some inhibiting effect to its tides.  I think Europa is tide locked, so no matter.
https://www.space.com/15498-europa-sdcmp.html
Distance from Jupiter: Europa is Jupiter's sixth satellite. Its orbital distance from Jupiter is 414,000 miles (670,900 km). It takes Europa three and a half Earth-days to orbit Jupiter. Europa is tidally locked, so the same side faces Jupiter at all times.
The surface of Europa is frozen, covered with a layer of ice, but scientists think there is an ocean beneath the surface. "
So, It is located far enough from Jupiter. Ratio of about 1:10.
It is covered with a layer of ice, therefore, the chance to set any significant bulges due to tidal is quite minimal (even if it has ocean beneath the surface).
So, I don't know what kind of information we can extract from Europa:
Does it have tidal bulges? Do those bulges set the positive/negative offset? Do we know if it is pushed outwards or pulled inwards?
I assume that the answer is - No, we don't know as it is too far away to measure.
So, if all the moons are too far away from us, how do we know that all of them must obey to tidal friction idea?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 14/12/2018 15:38:59
Why do we ignore the distance between the moon/rings to the planet?
I don't ignore that.  Jupiter's two innermost moons are falling because the distance from them to the planet is less than the geosync radius of Jupiter.

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As an example:
The Earth Radius is 6370 Km. If Phobos would orbit the Earth at the same ratio, its orbital distance from the Surface should be about 11,000 Km.
A commercial airplane is normally fly at about 10,000 Km.
If it is expected that airplane should come/fall down, why is it so big surprise that Phobos is also coming/falling down?
Airplanes fly at 10 km, not at 10000.  The ISS is only about 400 km up.
Airplanes fall because gravity accelerates them downward, and their orbits are much much smaller than the radius of Earth, so they'll hit the ground without effort to keep them aloft.  The ISS needs no wings because its orbital path does not include the ground in its way.

Anyway, phobos falls for the same reason that an airplane slows if it runs out of fuel: Friction with some moving slower than itself.

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With regards to Jupiter moons/Ring:
https://en.wikipedia.org/wiki/Rings_of_Jupiter
"Main ring - The narrow and relatively thin main ring is the brightest part of Jupiter's ring system. Its outer edge is located at a radius of about 129000 km (1.806 RJ;RJ = equatorial radius of Jupiter or 71398 km) and coincides with the orbit of Jupiter's smallest inner satellite, Adrastea.[2][5] Its inner edge is not marked by any satellite and is located at about 122500 km (1.72 RJ).[2]
So, Jupiter radius is 71398 Km.
Therefore, If this ring would orbit the Earth at the same ratio (1.806), its orbital distance from the Surface of earth should be about 11,500 Km. (Same distance at commercial airplane on Earth).
No airplanes there.  Rings form when moons pass below the Roche limit and are torn apart by tidal forces.  That ring on Jupiter has Metis in it as well, and Adrastea is probably a chunk torn off Metis, and the ring is all the shrapnel from that destruction.

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So, don't you see a similarity between airplane orbit around the Earth, to Phobos around Mars and Main ring around Jupiter?
Phobos is already starting to break up, so it will form a ring around Mars.  Plenty of similarity there.  Airplanes do no fly in outer space.  A suborbital ballistic airliner might (they don't have any right now), but not anywhere near that high up.

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Don't you agree that with tidal or without it, all of them must fall down?
Without tides, neither Phobos nor Adrastea nor an airplane orbiting at 10000 km will ever fall down.  With tides, all of them will eventually.

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Therefore, how can we use an object which it's orbital cycle is so close to the host to prove a negative thrust due to tidal???
Can you please find one moon (only one) in the whole solar system that is located long enough from its planet which is pulled inwards due to negative thrust (But please - real prove for that)?
You seem to want to simply dismiss any explanation as 'not proof'.  It is easy to demonstrate violations of conservation of angular momentum and make an infinite energy engine if tides produce no thrust on moons.  They already have power generators that harness tidal energy.  Where do you think it comes from then?

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With Regards to Europa:
Most planets don't have free flowing water.  Europa does, even it there's a crust that has some inhibiting effect to its tides.  I think Europa is tide locked, so no matter.
Europa is tidally locked, so the same side faces Jupiter at all times.
The surface of Europa is frozen, covered with a layer of ice, but scientists think there is an ocean beneath the surface. "
So, It is located far enough from Jupiter. Ratio of about 1:10.
It is covered with a layer of ice, therefore, the chance to set any significant bulges due to tidal is quite minimal (even if it has ocean beneath the surface).
Those tides below the ice have managed to halt Europa's spin, so I'd hardly call that minimal.  The forces putting thrust on its orbit are the tides raised by Europa on Jupiter's atmosphere, not those raised on Europa.  Yes, a moon's own spin contributes to this, but most moons have lost that spin already.

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Does it have tidal bulges?
A permanent deformation is not tidal, so no.  The deformation on Earth that makes sea level at the equator a larger radius than at the poles is not considered a bulge because it is permanent.  Tides are strain, some kind of back and forth motion that requires energy to maintain.  The bulge on Europa is permanent, which is why I said 'no matter'.

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Do those bulges set the positive/negative offset?
The offset of Europa's bulges averages zero.  That's because it is tide locked.
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Do we know if it is pushed outwards or pulled inwards?
Outward.  Geosync of Jupiter is around 170,000 km, and Europa is beyond that, and moving with the spin.  So thrust is positive.

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I assume that the answer is - No, we don't know as it is too far away to measure.
The answer is yes because it would violate conservation laws for it to be otherwise.
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So, if all the moons are too far away from us, how do we know that all of them must obey to tidal friction idea?
The laws of physics work everywhere, not just where humans confirm them in court.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/12/2018 15:56:02
Airplanes fly at 10 km, not at 10000
By the time that I fixed my message, I have got your answer.
So yes, you are absolutely correct.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/12/2018 16:28:53
The laws of physics work everywhere, not just where humans confirm them in court

Based on Newton's Shell Theorem:
https://www.math.ksu.edu/~dbski/writings/shell.pdf
The gravitational field outside a spherical shell having total mass M is the same as if the entire mass M is concentrated at its center (Center of mass).
So, if that is correct, than the gravitational field of all the mass of Earth (including the bulges), must be concentrated at the center (center of mass).
Do you agree that this law contradicts the whole idea of tidal friction, as there is no positive or negative thrust if we set the whole mass at the center?
We are using this Newton's Shell Theorem in order to explain many aspects of the Universe.
So, if the laws of physics work everywhere, why this law can't be used here?
Why I can't use it in order to prove that there is no positive or negative thrust due to offset?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 14/12/2018 18:52:07
Based on Newton's Shell Theorem:
https://www.math.ksu.edu/~dbski/writings/shell.pdf
The gravitational field outside a spherical shell having total mass M is the same as if the entire mass M is concentrated at its center (Center of mass).
So, if that is correct, than the gravitational field of all the mass of Earth (including the bulges), must be concentrated at the center (center of mass).
Right, but the theorem works only for a very regular object, a sphere in particular.  All kinds of funny things can be done if the objects are irregular.  I can take two objects and put the centers of gravity very close to each other and actually get them to repel each other.  I just can't do it with spheres.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/12/2018 21:18:04
Based on Newton's Shell Theorem:
https://www.math.ksu.edu/~dbski/writings/shell.pdf
The gravitational field outside a spherical shell having total mass M is the same as if the entire mass M is concentrated at its center (Center of mass).
So, if that is correct, than the gravitational field of all the mass of Earth (including the bulges), must be concentrated at the center (center of mass).
Right, but the theorem works only for a very regular object, a sphere in particular.  All kinds of funny things can be done if the objects are irregular.  I can take two objects and put the centers of gravity very close to each other and actually get them to repel each other.  I just can't do it with spheres.

How could it be that we can't use Newton's Shell Theorem for the Earth, while the orbital velocity of stars around the galaxy is based on Newton's Shell Theorem.
http://www.astronomy.ohio-state.edu/~ryden/ast162_7/notes30.html
"In the above equation, M is the total mass in a sphere of radius a, centered on the galactic center. (The mass outside the sphere doesn't have any net effect on the star's orbit).
Since the mass M includes the mass of the suppermassive black hole at the galactic center, M is guaranteed to be much greater than M*, the mass of a single star."
So, based on that theory we took all the matter (Stars/dust/SMBH/Dark Matter...) in the orbital radius and set the calculation as all the mass is located at the very center of the galaxy.
Hence, from the galaxy point of view, we have used all that variety of matter in the Sun' orbital radius as a very regular sphere.
However, when we come look at the Earth, suddenly it is not a regular object.
Is it real?
How could it be that the none regular matter in the sun' orbital radius is more regular than the Earth itself.
Somehow, it seems to me that we are using the idea of "regular" to prove an object which is by definition none regular (as the sphere inside the orbital radius of the Sun), while for a real regular object (as the Earth) we claim that it is not regular enough.
Sorry, that isn't the way that we have to work in science.
If the Earth isn't regular enough for the Newton's Shell Theorem, than by definition the matter in the orbital radius of the Sun shouldn't be considered as a regular.
How can we use the Newton's Shell Theorem to prove the orbital velocity of the Sun, while we reject this idea just because it contradicts our theory about tidal friction?
Therefore, If the none regular matter in the orbital radius of the Sun should be considered as a regular, than by definition the Earth with or without the Bulges should be considered as a real regular.
Don't you see that severe contradiction???
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/12/2018 01:44:35
How could it be that we can't use Newton's Shell Theorem for the Earth, while the orbital velocity of stars around the galaxy is based on Newton's Shell Theorem.
http://www.astronomy.ohio-state.edu/~ryden/ast162_7/notes30.html
"In the above equation, M is the total mass in a sphere of radius a, centered on the galactic center. (The mass outside the sphere doesn't have any net effect on the star's orbit).
That quote is not Newton’s shell theorem, which concerns spherical objects.  That law is a similar one showing how orbital speeds are directly related to the mass of the material orbited, and the orbital radius.  The moon follows that law.  The law says nothing about additional forces resulting from the non-uniformity of the local gravitational field. Stars are flung out of their normal orbits all the time due to close encounters with passing objects.  The law you quote doesn’t prevent that.

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So, based on that theory we took all the matter (Stars/dust/SMBH/Dark Matter...) in the orbital radius and set the calculation as all the mass is located at the very center of the galaxy.
The calculation of the speed of the orbit if it is a reasonably circular orbit, yes.  It is not a calculation of the local forces making changes to that orbit.

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Hence, from the galaxy point of view, we have used all that variety of matter in the Sun' orbital radius as a very regular sphere.
However, when we come look at the Earth, suddenly it is not a regular object.
The galaxy isn’t regular either.  Again, consider stars being diverted out of their orbits, which happens a lot, especially in the crowded places near the halo.

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Is it real?
How could it be that the none regular matter in the sun' orbital radius is more regular than the Earth itself.
Earth is a very small object and far less symmetric than the galaxy, just like a rock in my yard is hardly a sphere.  Our planet has this giant moon, making it even less symmetric.  Apples to Oranges to try to compare that to the far more uniform galaxy.  Even then, it takes only one passing rouge star to disrupt our entire solar system.  We’re just luck we’re so far out that those kinds of events are really rare.

BTW, the galaxy is not a uniform sphere.  Not even close.  Newton’s law just doesn’t apply, but the 2nd lay you quote above still does.

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Somehow, it seems to me that we are using the idea of "regular" to prove an object which is by definition none regular (as the sphere inside the orbital radius of the Sun)
There is no spherical object inside the orbital radius of the sun.  The sphere is a mathematical one.  All the crap inside this radius, and not the stuff outside.  That stuff is anything but a sphere.  It resembles more of a middle-heavy pancake.
The Earth on the other hand is actually a sort of sphere, but with significant deviations from being symmetric.  The galaxy below us has no such significant deviations until one pass by at least.

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How can we use the Newton's Shell Theorem to prove the orbital velocity of the Sun,
We didn’t.  We used that 2nd rule about only the matter within the orbit contributing to the velocity.  You can use that rule on the Earth/moon system as well.  It works just great.  It means you need to take into account all the little stuff in low orbit around Earth to account for the moon’s speed, but you don’t need to to account for the moon when computing the orbital speed of the ISS.  That’s what the rule says.

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while we reject this idea just because it contradicts our theory about tidal friction?
It doesn’t go into tidal forces at all.  Tidal forces are not a violation of the rule.  The orbital speed of the moon never deviates from what the rule predicts.  It is at all times based on the (essentially fixed) mass of the Earth and its satellites, which is the list of all the stuff inside the Moon’s orbit.
The rule has no requirement that the distribution of that matter be regularly arranged.  Newton’s law does require that uniformity.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/12/2018 05:26:43
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How can we use the Newton's Shell Theorem to prove the orbital velocity of the Sun,
We didn’t.  We used that 2nd rule about only the matter within the orbit contributing to the velocity.  You can use that rule on the Earth/moon system as well.  It works just great.  It means you need to take into account all the little stuff in low orbit around Earth to account for the moon’s speed, but you don’t need to to account for the moon when computing the orbital speed of the ISS.  That’s what the rule says.
What do you mean by: "2nd rule about only the matter"?
Is it Newton's second law?
If so, this law isn't relevant for our discussion.
It doesn't say anything about the mass in the sphere of orbital cycle of an object:

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The utility of Newton's second law for the description of the motion of extended objects is the key to its general practical usefulness. The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:"
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/12/2018 05:47:04
What do you mean by: "2nd rule about only the matter"?
Is it Newton's second law?
The first rule you quoted was Newton's Shell Theorem concerning only uniform spherical objects.
The 2nd rule you quoted was the one from the ohio-state site concerning the mass that has net effect on an orbit and the mass that doesn't.
I don't know the name of that rule or who came up with it, but you quoted it, and then you misrepresented what it means.

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http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The utility of Newton's second law for the description of the motion of extended objects is the key to its general practical usefulness. The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:"
I don't see the relevancy of that law to this portion of the discussion.  It says that the center of gravity of something like Earth follows a smooth curve even if the various parts (like your mailbox) don't.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/12/2018 07:51:02
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What do you mean by: "2nd rule about only the matter"?
Is it Newton's second law?
The first rule you quoted was Newton's Shell Theorem concerning only uniform spherical objects.
The 2nd rule you quoted was the one from the ohio-state site concerning the mass that has net effect on an orbit and the mass that doesn't.
I don't know the name of that rule or who came up with it, but you quoted it, and then you misrepresented what it means.

The "second rule" is just a direct outcome from the "first rule" which is the Newton's Shell Theorem.
With all the respect to ohio-state (and I have a respect...), they can't just invent new rules for gravity.
So, the second statement is not a second rule, it is just a logical outcome from Newton's Shell Theorem.
They also explain how to extract M (M = mass inside star's orbit (in solar masses) ) from kepler law.
"Each star in the disk is on a very nearly circular orbit, anchored by all the mass enclosed within its orbit, whether it's luminous or not. Thus, the amount of mass within a star's orbit can be determined from Kepler's Third Law:"

Hence, by Kepler law we can calculate the total mass which is requested to meet the orbital velocity of the sun around the galaxy, while Newton’s First Theorem tell us that: "M is the total mass in a sphere of radius a, centered on the galactic center. (The mass outside the sphere doesn't have any net effect on the star's orbit)" and therefore, we also can understand that:
"Since the mass M includes the mass of the suppermassive black hole at the galactic center, M is guaranteed to be much greater than M*, the mass of a single star."
In any case, if that was not clear enough, please see the following statement:
http://users.math.cas.cz/~krizek/cosmol/pdf/B102.pdf
Newton’s First Theorem - If the density distribution of a ball of mass M is spherically symmetric, then the size of the force between the ball and a point mass m, that lies outside the interior of the ball, is given by the left-hand side of (1), where r is the distance between the point and the center of the ball."
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/12/2018 11:25:06
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"The utility of Newton's second law for the description of the motion of extended objects is the key to its general practical usefulness. The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:"
I don't see the relevancy of that law to this portion of the discussion.  It says that the center of gravity of something like Earth follows a smooth curve even if the various parts (like your mailbox) don't.
This law is very relevant
It actually confirms that the shape of the object is none relevant for its central point of mass.
As long as all the masses in the object are fully connected the center of mass of this object is none relevant with its shape. (Spanner, dog, cat or even elephant).
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"A set of masses connected by springs will follow a path such that its center of mass moves along the same path that a point mass of the same total mass would follow under the influence of the same net force."
So, even if that spanner has an offset, it won't set any extra thrust. In our calculation we just need to focus on its center of mass.
In the same token, as the bulges are fully connected to the Earth, they are part of the whole Earth body. Therefore, in our calculation we need only to follow the center of the mass of the earth.
In other words - We won't get any extra negative or positive thrust due to mountains, oceans or any sort of bulges with or without offset while the whole masses are connected together.
That law by itself proves that the idea of thrust due to bulges is totally incorrect.
Conclusion -
Based on the following laws (each one by itself):
1. Newton's Shell Theorem
2. Newton's second law
I have proved that the shape of the object can't issue any sort of extra thrust. Newton didn't specify that idea in his laws. Therefore, the hypothetical idea of extra thrust due to the offset (or special shape of the object - with or without bulges) is incorrect.
Our scientists must find better idea why all the planets and all the far enough moons are drifting outwards.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/12/2018 13:10:03
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What do you mean by: "2nd rule about only the matter"?
Is it Newton's second law?
The first rule you quoted was Newton's Shell Theorem concerning only uniform spherical objects.
The 2nd rule you quoted was the one from the ohio-state site concerning the mass that has net effect on an orbit and the mass that doesn't.
I don't know the name of that rule or who came up with it, but you quoted it, and then you misrepresented what it means.
The "second rule" is just a direct outcome from the "first rule" which is the Newton's Shell Theorem.
With all the respect to ohio-state (and I have a respect...), they can't just invent new rules for gravity.
So, the second statement is not a second rule, it is just a logical outcome from Newton's Shell Theorem.
I'm not disputing the rule.  It is a known thing, yes.  I just don't know the name of it, or how it is derived from Newton's sphere law.

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Hence, by Kepler law we can calculate the total mass which is requested to meet the orbital velocity of the sun around the galaxy, while Newton’s First Theorem tell us that: "M is the total mass in a sphere of radius a, centered on the galactic center.
You're applying the 2nd rule there, not the first theorem.  The galaxy is not a sphere nor is it spherically symmetric, so the first theorem does not immediately apply.  The nameless law does however, so we're good.

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Newton’s First Theorem - If the density distribution of a ball of mass M is spherically symmetric, then the size of the force between the ball and a point mass m, that lies outside the interior of the ball, is given by the left-hand side of (1), where r is the distance between the point and the center of the ball."
Slightly generalized version, and thus a better one, yes.  It doesn't require a sphere, but merely something spherically symmetric.  But the galaxy isn't, so this law doesn't directly apply.   The galaxy is roughly modeled as a disk, and the force vector of a point out of the plane of a disk does not point to the center of mass of the disk like Newton's theorem says it must if it were spherically symmetrical.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/12/2018 14:07:14
This law is very relevant
It actually confirms that the shape of the object is none relevant for its central point of mass.
I didn't say otherwise.  The shape of Earth for instance, with its bulges, has no effect on Earth's center of mass.  I agree with that.

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As long as all the masses in the object are fully connected the center of mass of this object is none relevant with its shape. (Spanner, dog, cat or even elephant).
All the others, yes, and even a spraying garden hose, but not the cat, which has properties outside physics.

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http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"A set of masses connected by springs will follow a path such that its center of mass moves along the same path that a point mass of the same total mass would follow under the influence of the same net force."
OK.  That one is pretty obvious too, since it derives directly from the computation of center of mass.  The springs are not even necessary.  It can be a collection of 13 random stars from nowhere in particular, and that collection of 13 unrelated stars will have a center of mass that will follow this rule.  What it doesn't mean is that an object at that center of mass will follow the same path.
What it especially doesn't mean is that any force exerted by that collection of spring-attached objects is going to be the same as if all the mass was concentrated at their center of gravity.  This is very easy to demonstrate.

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So, even if that spanner has an offset, it won't set any extra thrust.
The rules don't say that at all.  They talk about net force and those net forces acting on the center of gravity of each object in question.  So if the spanner puts a net force on an object that is anything but perpendicular to its motion, it will be exerting thrust to it.

The Earth/Sun system could be considered one such object which is not a sphere, and thus it puts massive thrust on another object (the moon), all because of its offset.  According to your assertions, that should not happen, and the moon should wander off around the sun on its own in a nice circular orbit around the center of mass of the combined Earth/Sun object.

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In our calculation we just need to focus on its center of mass.
Wrong!!  Center of mass has no angular momentum, and this tidal thrust effect is all about net forces resulting from transfer of angular momentum.  None of the laws above describe angular effects on the tumbling spanner and such.

I throw a rapidly spinning pool noodle, and it is spinning far slower before it hits the ground.  The net forces on the noodle do indeed determine the path of its center of gravity, but do not in any way describe the loss of spin.  That rules is inadequate for the situation being described.

The orbit about the galaxy is less about angular momentum and forces since there is no significant transference going on.  You can treat a lot of things as point masses on that scale, but not the galaxy as a whole since it is not spherically symmetrical.

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Conclusion -
Based on the following laws (each one by itself):
1. Newton's Shell Theorem
2. Newton's second law
You can't take laws by themselves.  Let's take these two laws.  Consider me on Earth with the sun directly overhead, and Mercury and Venus don't exist.  In a sphere of radius R where R is the distance from the sun to me, only the mass in that radius determines my orbit.  It is the Sun which is spherically symmetric.  So the first law applies.
Now let's apply the 2nd law.  I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon), and the mass outside the radius R has no effect on my orbit, therefore I should drop in towards the sun, getting sucked into the sky.  The Earth cannot hold me down because it is outside R when the sun is directly overhead, and thus does not contribute to my orbit, per the 2nd law.

This is the sort of proofs I'm seeing from you.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/12/2018 15:27:26
Thanks Halc.
I really appreciate all your efforts.
However, please try to use some solid evidences (mathematics, Newton/kepler.. laws) in order to prove your statements.

The rules don't say that at all.  They talk about net force and those net forces acting on the center of gravity of each object in question.  So if the spanner puts a net force on an object that is anything but perpendicular to its motion, it will be exerting thrust to it.

Can you please prove that a spanner can put a net force on an object just by pointing to some offset?

You can't take laws by themselves.  Let's take these two laws.  Consider me on Earth with the sun directly overhead, and Mercury and Venus don't exist.  In a sphere of radius R where R is the distance from the sun to me, only the mass in that radius determines my orbit.  It is the Sun which is spherically symmetric.  So the first law applies.
Now let's apply the 2nd law.  I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon), and the mass outside the radius R has no effect on my orbit, therefore I should drop in towards the sun, getting sucked into the sky.  The Earth cannot hold me down because it is outside R when the sun is directly overhead, and thus does not contribute to my orbit, per the 2nd law.

That example is not clear to me.
If the sun is just above me while I am waking on Earth, than based on Newton second law, I'm an integrated mass of Earth.
Therefore, the Sun has no impact at all about my location.
If there is no Earth or moon, and it is all about me and the Sun, than my orbital velocity must be a direct outcome of gravity force based on R. In this case, the Earth and the Moon are not there to have any impact on my orbital cycle around the Sun.

With regards to the following message:
"I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon)"
This is a severe mistake.
If I have to orbit around the Sun, there is no way to slow down due to that Earth spin velocity cancelation.
So, you don't have to maintain your orbital velocity. The Sun gravity force works for you. This is a key element.
Newton didn't specify even one word about the impact of spinning velocity.
This is a new idea which had not been confirmed. It seems to me as a new idea which we have invented just in order to offer a nice answer for: Why the Moon and the earth are drifting outwards?
If you believe that there is a possibility to slow down the earth orbital velocity "because Earth's spin velocity cancels some of its orbital velocity at noon", than please prove it.

Wrong!!  Center of mass has no angular momentum, and this tidal thrust effect is all about net forces resulting from transfer of angular momentum.  None of the laws above describe angular effects on the tumbling spanner and such.

I throw a rapidly spinning pool noodle, and it is spinning far slower before it hits the ground.  The net forces on the noodle do indeed determine the path of its center of gravity, but do not in any way describe the loss of spin.  That rules is inadequate for the situation being described.

The orbit about the galaxy is less about angular momentum and forces since there is no significant transference going on.  You can treat a lot of things as point masses on that scale, but not the galaxy as a whole since it is not spherically symmetrical
We do not discuss about rapidly spinning pool of noodle.
We discuss about Sun/earth/Moon orbital cycles.

If the angular momentum or the revolving speed of the Moon or the Earth can slow down the orbital velocity of the Erath or the Moon - than please prove it.
I still can't understand how any sort of offset in bulges can set any sort of thrust.
Please use the spanner as an example. you are more than welcome to force the spanner in any sort of offset as you wish. Try to prove why by doing so we shall get extra trust on the orbital object.
Please try to prove this idea by mathematics.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/12/2018 18:23:08
So far we have just discussed about the impact of the orbital cycles of Planets and Moons.
However, what about the stars? What about our star - the Sun?
Why all our scientists are positively sure that during all his life time the Sun had to keep the same orbital radius?
How could it be that all the moons and Planets are drifting outwards, (or inwards based on unproved tidal idea) while the sun is fixed at the same radius?
How could it be that in one hand our scientists claim that the SMBH increases its mass by eating stars and gas clouds, while on the other hand they don't consider an option that stars must migrate/drift inwards in order to supply the requested food for the SMBH monster?
If the SMBH has 4 x 10^6 sun mass, (while our scientists believe that this mass had been taken from the stars in the galaxy) than somehow 4 x 10^6  stars had to drift inwards.
So why our scientists are so sure that the Sun was always at the same distance from the center of the galaxy???
What makes our star so unique that it had to stay so far away from the monster at the galaxy center?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/12/2018 22:05:16
Can you please prove that a spanner can put a net force on an object just by pointing to some offset?
I didn't say it could do it by pointing to an offset.
The spanner is tumbling, and that tumblingis slowing measurably.  It couldn't do that if it was treated as a point mass as you are attempting to do.  You can't put torque on a point.

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Quote from: Halc
You can't take laws by themselves.  Let's take these two laws.  Consider me on Earth with the sun directly overhead, and Mercury and Venus don't exist.  In a sphere of radius R where R is the distance from the sun to me, only the mass in that radius determines my orbit.  It is the Sun which is spherically symmetric.  So the first law applies.
Now let's apply the 2nd law.  I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon), and the mass outside the radius R has no effect on my orbit, therefore I should drop in towards the sun, getting sucked into the sky.  The Earth cannot hold me down because it is outside R when the sun is directly overhead, and thus does not contribute to my orbit, per the 2nd law.
That example is not clear to me.
If the sun is just above me while I am waking on Earth, than based on Newton second law, I'm an integrated mass of Earth.
You're not integrated with Earth since you are completely detached from it. Seatbelts are cheating.  I'm just considering the gravity of the sun on your orbit.  According to that 2nd law, or at least according to the way you are using it, nothing outside that R sphere has a net effect on my motion, therefore, since Earth is entirely outside that sphere, its gravity doesn't affect me.

I know it's wrong, but that's how you're interpreting the rule: by oversimplifying and not considering deviations from the uniformity.  The Earth is a huge deviation.  Deviations matter.  The tidal bulges are deviations, and the change the direction of the force vector acting on the moon just as Earth changes the force vector acting on me at high noon.  But you cherry pick which deviations matter and which do not, depending on what purpose you desire.  That's very fallacious reasoning.

I haven't figured out your purpose in all this.  Clearly not here to learn.  You seem bent on finding contradiction in simple orbital mechanics, but not sure why.

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Therefore, the Sun has no impact at all about my location.
If there is no Earth or moon, and it is all about me and the Sun, than my orbital velocity must be a direct outcome of gravity force based on R. In this case, the Earth and the Moon are not there to have any impact on my orbital cycle around the Sun.
So rule #2 suddenly doesn't matter when you find it inconvenient?

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With regards to the following message:
"I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon)"
This is a severe mistake.
If I have to orbit around the Sun, there is no way to slow down due to that Earth spin velocity cancelation.
My velocity relative to the sun varies daily, fastest at midnight and slowest at noon.  Somewhere in between is proper orbital speed at this radius.  At midnight the speed is too high, and at noon it is too low, as per Kepler's 3rd law.  The speeds are easy to compute:  About 30 km/s ± about 0.5 km/s speed when the sun is directly below and above me respectively.

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So, you don't have to maintain your orbital velocity. The Sun gravity force works for you. This is a key element.
Newton didn't specify even one word about the impact of spinning velocity.
This is a new idea which had not been confirmed.
What, that your speed relative to the sun is lower at noon?  That's pretty obvious I'd think.

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If you believe that there is a possibility to slow down the earth orbital velocity
My speed, not Earth's speed.  I'm the one that suppose to fall closer to the sun because I'm moving slower than Kepler says I should.
Of course Earth speed is slowest every full moon, for the same reason.  The difference isn't as much as ±0.5 m/sec.

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We do not discuss about rapidly spinning pool of noodle.
OK, you don't know what a pool noodle is.  It's like the spanner, but more susceptible to angular forces.

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If the angular momentum or the revolving speed of the Moon or the Earth can slow down the orbital velocity of the Erath or the Moon - than please prove it.
That was demonstrated many posts ago.  I should the vector math quite a ways back.  The force vector on the moon is not perpendicular to its motion, therefore there is work being performed (energy transfer).

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I still can't understand how any sort of offset in bulges can set any sort of thrust.
Understanding it would cause your argument to fall apart, therefore nothing I can say will make you understand it.  I'm fine with that.

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Please use the spanner as an example. you are more than welcome to force the spanner in any sort of offset as you wish. Try to prove why by doing so we shall get extra trust on the orbital object.
Please try to prove this idea by mathematics.
OK, there is a spanner suspended exactly through its center of gravity on a frictionless axis.  It should be able to spin freely, and be balanced.
I give it a permanent stationary offset of 45°.  The spanner is 40 cm long, 20 on each side of the axis.  The lower side is 6372000 meters from the center of earth and has a force of GmM/40602384000000 acting on it.  The upper end is a tiny bit further away from earth and has a force of around GmM/40602384382320 resulting in a 1e-6 % difference in force. It will start to rotate to vertical because the bottom is attracted more than the top, being closer to Earth.  That is a torque force being put on an object despite its center of gravity never moving or the force on that center of gravity ever changing.
You can get rid of the axis suspending the spanner if you just put it in orbit and let its speed hold it up there.

The numbers get more significant if the spanner is a bit longer, like say over 12000 km long, even if the 12000 km spanner has an offset far less than 45°
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/12/2018 06:49:42
Thanks
Great example
I give it a permanent stationary offset of 45°.  The spanner is 40 cm long, 20 on each side of the axis.  The lower side is 6372000 meters from the center of earth and has a force of GmM/40602384000000 acting on it.  The upper end is a tiny bit further away from earth and has a force of around GmM/40602384382320 resulting in a 1e-6 % difference in force.

However, I really don't understand why there is a difference in forces.
Let's set the whole Earth at the shape of spanner.
So, the Earth will look like an extended object with all of its mass while the length of each side is 10,000 Km.
Based on Newton's Second Law for an Extended Object
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:
F = M a
In order to understand the calculation, we need to look at the Following "Newton's Second Law for a System of Particles"
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html#c2
"Newton's Second Law for a System of Particles:
The form of Newton's second law for a system of particles will be developed with the understanding that the result will apply to any extended object where the particles are in face connected to each other.
The center of mass of a system of particles can be determined from their masses and locations."
So, Our dear Newton set a complicated calculation in order to get the outcome of:
M a = F
Newton actually tells us that the shape of the object and its offset can't contribute any extra force as long as all the particles of the object are connected.
So, if you still think that there is an error in Newton calculation, please offer the updated calculation to prove why each side can contribute different force (while all the particles are fully connected).
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 16/12/2018 13:24:06
However, I really don't understand why there is a difference in forces.
I modeled the spanner as two masses separated by 30 cm (I did not put all the mass at the ends of the 40 cm spanner), and applied Newton's F=GMm/r² to get the force, and then F=ma to get the acceleration of spanner that sets it in motion.
In addition, if the Earth under it was entirely stationary and even perfectly spherical at first, it too begins to rotate by application of Newton's conservation of angular momentum law.  The moon does not do this to the Earth because unlike the spanner, it has no net offset and thus exerts no net torque directly on Earth.  All Earth's torque is due to friction, and our spanner example does not involve friction.  We can put some air around it so it stops rotating after a while.

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Let's set the whole Earth at the shape of spanner.
So, the Earth will look like an extended object with all of its mass while the length of each side is 10,000 Km.
Based on Newton's Second Law for an Extended Object
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:
F = M a
Yes, the motion of the center of mass will follow that law, just like it says.  I never said otherwise.

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In order to understand the calculation, we need to look at the Following "Newton's Second Law for a System of Particles"
We need to come up with different names for all these laws, because you're calling them all Newton's second law of something, which is confusing.  Newton's second law typically refers to the second law of motion, which is F=ma.  So lets just call the one above the 'extended object law' and this one below the 'law for a system of particles'.

Quote
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html#c2
"Newton's Second Law for a System of Particles:
The form of Newton's second law for a system of particles will be developed with the understanding that the result will apply to any extended object where the particles are in face connected to each other.
The center of mass of a system of particles can be determined from their masses and locations."
That's not the law you quoted before.  It's just a statement of the fact that the center of mass of any set of particles (connected or not) can be computed.  This lacks any description of how that computation is done (which isn't hard), and it doesn't mention net forces like the last time you mentioned it.  Maybe this is a different law.
Anyway, this wording just says there is a COM and that is it computable.  Hardly a revolutionary statement, and not something that I would call a 'law'.
The 'law' does not say that the set of particles will exert gravitational force as if they were all at the center of mass, or that the center of mass will follow the same path as the object with all its mass concentrated there.  This is very easy to demonstrate with just a two-particle object.
And yet I find you here misrepresenting these laws and attempting to do just that.

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So, Our dear Newton set a complicated calculation in order to get the outcome of:
M a = F
Newton actually tells us that the shape of the object and its offset can't contribute any extra force as long as all the particles of the object are connected.
He said no such thing. That does not follow from any of the laws you've quoted.

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So, if you still think that there is an error in Newton calculation, please offer the updated calculation to prove why each side can contribute different force (while all the particles are fully connected).
Particles are never connected.  Two things cannot touch.  None of Newtons laws about sets of objects require connectivity for this reason.  The center-of-mass of the Earth-moon system follows an ellipical path around the sun, which the center of mass of just Earth does not because the moon is yanking back and forth each month.  So treating the two as one unit gives a far smoother curve, even though the two are not physically connected.

This is not always the case.   Earth and Venus could be connected with a thin spidery thread and the center-of-mass of the single object would very much obey that law you quote (you quoted a much better version of it in a prior post), but that COM does not follow the same path as would an object that was actually all at that spot, and does not generate a gravitational field that is identical to a single mass at the COM.

Point is, the connection is not necessary.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/12/2018 15:59:18
Particles are never connected.  Two things cannot touch.  None of Newtons laws about sets of objects require connectivity for this reason.
This is a sever mistake.
All particles on earth are fully connected due to gravity.
In the following law it is stated clearly:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html#c2
"The form of Newton's second law for a system of particles will be developed with the understanding that the result will apply to any extended object where the particles are in face connected to each other."
It is also stated:
"Since nothing we have done addressed whether the particles are connected or not, this result generally applies to a system of discrete particles or to an extended object consisting of connected mass elements."
So please - why don't you agree with those simple and clear evidences???

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Let's set the whole Earth at the shape of spanner.
So, the Earth will look like an extended object with all of its mass while the length of each side is 10,000 Km.
Based on Newton's Second Law for an Extended Object
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:
F = M a
Yes, the motion of the center of mass will follow that law, just like it says.  I never said otherwise.
So, you agree that the center of mass of any kind of object at any shape (even if it is 20,000 Km spanner) must follow that law.
If so, you agree that there is only one center point of mass for any kind of object.
Therefore, there in nothing to disagree.
With only one center point of mass there is no way to get any extra thrust.
I hope that you agree with that.

Conclusions:
I didn't invent those laws.
You are more than welcome to call them at any name. However, those laws prove that there is one center of mass to any shape of object and at any offset.
Therefore, the assumption that there is an extra thrust due to offset contradicts those fundamental laws for gravity.
You didn't offer any alternative mathematic calculation in order to reject those laws and support the hypothetical idea of extra thrust due to bulges offset.
Hence, our scientists must look for better explanation why all the planets and moons are drifting outwards.
I'm specifically using the words "drifting" as I have proved that there is no extra thrust that can push them outwards.
I have no more questions about tidal friction.

With your permission, I would like to focus now on the Sun orbital cycle.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 16/12/2018 18:12:38
All particles on earth are fully connected due to gravity.
So is the moon.

In the following law it is stated clearly:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html#c2
"The form of Newton's second law for a system of particles will be developed with the understanding that the result will apply to any extended object where the particles are in face connected to each other."
It is also stated:
"Since nothing we have done addressed whether the particles are connected or not, this result generally applies to a system of discrete particles or to an extended object consisting of connected mass elements."
So please - why don't you agree with those simple and clear evidences???[/quote]
I totally agree with all that.  The former was an example (spanner) of a connected object, and the latter was an example of the same law working for unconnected objects.

I've never disagreed with any of that.  I disagree with the way you are using it.

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So, you agree that the center of mass of any kind of object at any shape (even if it is 20,000 Km spanner) must follow that law.
If so, you agree that there is only one center point of mass for any kind of object.
Yes and yes.

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Therefore, there in nothing to disagree.
With only one center point of mass there is no way to get any extra thrust.
I hope that you agree with that.
Where does any of the info on that page say that?   Why would you expect me to agree with that?

If I put forward force on my car, that's thrust.  F=ma.  The law is all about force and acceleration.  Of course it allows thrust.  If I put force on the spanner, or net force on the system of completely disconnected objects, that will move the center of gravity, which is thrust.

You make it sound like something on that page proves that nothing can change speed.

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Conclusions:
I didn't invent those laws.
You are more than welcome to call them at any name.
Turns out they're all just different applications of Newton's second law of motion: F=ma.  They're not separate laws.

[/quote]However, those laws prove that there is one center of mass to any shape of object and at any offset.
Therefore, the assumption that there is an extra thrust due to offset contradicts those fundamental laws for gravity.[/quote]
I just love how you get the 2nd line from the first one.  The first line makes zero statement about what forces might exist or not on a system of objects.  It is just a statement that an object has a center of mass.  The lower line simply does not follow from it.
You keep repeating the same assertions over and over.  We're getting nowhere.

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You didn't offer any alternative mathematic calculation in order to reject those laws and support the hypothetical idea of extra thrust due to bulges offset.
I never rejected the laws.  But I showed angular acceleration of an object where its center of mass never moved, and I showed acceleration of a set of objects without those objects ever not having a center of mass.  Tossing a handful of sand in the air illustrates it nicely.  At no point does that handful of sand not have a center of mass, and absent friction, that COM will even follow an elliptic trajectory until it hits the ground.  Plenty of thrust going on since the speed of the sand is always changing.

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Hence, our scientists must look for better explanation why all the planets and moons are drifting outwards.
Or maybe you could actually understand what the material on you linked page means.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/12/2018 19:14:23
I hope that you agree with the following:
1. The spanner has a center point of mass.
2. If we won't set an external force of the spanner there is no extra thrust.
3. In our case the spanner represents the Earth.
If you agree with all the above let's move to the following:

You claim that:
If I put force on the spanner, or net force on the system of completely disconnected objects, that will move the center of gravity, which is thrust.

Now
1. Let's use the idea that all the particles on Earth are connected.
In this case, we will assume that the tidal is so strong that it coverts totally the shape of the earth from ball shape into spanner. (Instead of just two bulges)
So, what?
The Earth in a spanner shape has exactly the same center point of mass as the Earth in ball shape (as all the particles are connected).
So, that external force (tidal) didn't change at all the location of the center point of mass (although it changed the earth shape).
Therefore, there is no way to set extra thrust on the moon.
Do you agree with that?

2. Let's assume that the particles on earth aren't connected.
So, the tidal is so strong that it split the earth into two totally separated objects.
Each object is represented by center point of mass. We can set those two center points of mass exactly where the bulges are located.
In this case I agree that the gravity force between each point to the moon is not equal. and the sum of their gravity force is different from just one central point of mass.
However, even in this case, there is no extra thrust.
(Assuming that we can hold those two points of center of mass at a constant distance and offset with regards to the moon, than the moon will orbit around those points without getting any sort of extra thrust.

3. We will split the Earth into infinite separated points of mass while we hold them at the same spot and the same offset from the moon.
Even in this case there is no thrust on the moon. However, the moon will orbit around infinite number of center of mass points instead of only one or two.

Maybe it is my limited understanding, but somehow I really don't see any way to transfer any thrust from the Earth to the moon due to tidal.
With your permission, let's move on.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 16/12/2018 23:08:15
I hope that you agree with the following:
1. The spanner has a center point of mass.
2. If we won't set an external force of the spanner there is no extra thrust.
3. In our case the spanner represents the Earth.
Agree.
#2 is true per Newton's first law.  If there is no force at all on the spanner, it will continue to move as an inertial object (stay stopped or move in a straight line).  This is true of the COM of the spanner, but not the rest of it.

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If you agree with all the above let's move to the following:

You claim that:
If I put force on the spanner, or net force on the system of completely disconnected objects, that will move the center of gravity, which is thrust.
Thrust being acceleration in that context, yes.  F=ma.  Any net force must result in net acceleration, and any acceleration is thrust in the frame of the object with the force acting upon it.  It doesn't mean any particular portion of the object will accelerate, but it means the COM of the object must accelerate.

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1. Let's use the idea that all the particles on Earth are connected.
In this case, we will assume that the tidal is so strong that it coverts totally the shape of the earth from ball shape into spanner. (Instead of just two bulges)
So, what?
The Earth in a spanner shape has exactly the same center point of mass as the Earth in ball shape (as all the particles are connected).
No.  The Earth in a spanner shape has exactly the same center point of mass as the Earth in ball shape, period.  The fact that it is connected or not is irrelevant to that.  Gluing two adjacent rocks together does not alter their mutual center of mass.
You can split Earth in two equal halves and move each chunk a lightyear in opposite directions, and the COM of Earth will remain exactly where it is now.  Any net force acting on either half will move that COM linearly as if the force was applied at that point, all per Newton's 2nd law.

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So, that external force (tidal) didn't change at all the location of the center point of mass (although it changed the earth shape).
Yes
Quote
Therefore, there is no way to set extra thrust on the moon.
I love how you just suddenly assert this when the facts above do not in any way suggest it.  No, I've repeatedly said I don't agree with this.

If you want to make it easy, use a rigid 770,000 km dumbbell shaped object holding two massive earth-halves connected by a bar.  Give that object an offset of 2°, not 10 or 45, and give it a spin of say once per 400 hours. The acceleration on the moon will be so high it will rip apart.

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2. Let's assume that the particles on earth aren't connected.
So, the tidal is so strong that it split the earth into two totally separated objects.
Each object is represented by center point of mass. We can set those two center points of mass exactly where the bulges are located.
They'll fall into each other.  The system isn't stable, so its immediate force on the moon will not be cumulative.  For the moon to split Earth in half, it would need to be far larger, enough to get Earth inside its Roche limit.  That's what is happening to Phobos right now.  It is cracking, almost ready to separate into two moons.

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In this case I agree that the gravity force between each point to the moon is not equal. and the sum of their gravity force is different from just one central point of mass.
However, even in this case, there is no extra thrust.
Unstable, but that configuration for that moment very much produces a lot of thrust on the moon.  That will lessen as the two parts fall into each other, leaving an Earth back in once piece but a day that is many hundreds of hours long.

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(Assuming that we can hold those two points of center of mass at a constant distance and offset with regards to the moon, than the moon will orbit around those points without getting any sort of extra thrust.
Now you're assuming an external application of angular momentum, which adds energy and momentum to the whole setup.  So much for conservation laws.  Newton's 3rd law says there needs to be a reaction to every action, so if there is angular momentum being added to the moon, it needs to come from somewhere.  You're supplying it with an external force holding these two Earth pieces in their unstable positions.

Anyway, if you don't see thrust from this obvious example, you need to take a class on doing vector arithmetic.  I did the math in post 40, where a positive force was computed for the forward component of forces acting on the moon.

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3. We will split the Earth into infinite separated points of mass while we hold them at the same spot and the same offset from the moon.
What spot?  The two halves were at two spots.  I can't comment on this because I don't know what spot you're talking about.

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Even in this case there is no thrust on the moon. However, the moon will orbit around infinite number of center of mass points instead of only one or two.
I think you mean different spots.  The moon will orbit the COM of the collection, just like it does now, assuming your arrangement doesn't change that COM.

I think you mean you slow the spin of Earth and have it turn once a month with the moon, holding the bulges out at their current offset.  External force will be needed to keep it like that, just like the example with two disconnected masses.

How do you expect a torque force to be exerted on the system and yet not have it gain angular energy?

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Maybe it is my limited understanding, but somehow I really don't see any way to transfer any thrust from the Earth to the moon due to tidal.
With your permission, let's move on.
You need my permission?  If you don't understand vector arithmetic, I don't think your new question is something you're likely to get right.  The Earth/sun system is pretty simple.  We can approximate it with a circle.  If you wonder why the Earth orbital radius slowly increases, we're back to the tidal example that you clearly find a need to deny.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/12/2018 04:20:28
Thanks halc

Let's agree that we don't agree on the idea that there is a constant thrust on the moon due to tidal bulges offset on the Earth.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/12/2018 15:09:07
With regards to the Sun

Why are we so sure that during all of his life time the Sun had to keep the same orbital radius?
How could it be that all the moons and Planets are drifting outwards, (or inwards based on the tidal idea) while the sun is fixed at the same radius?
How could it be that in one hand our scientists claim that the SMBH increases its mass by eating stars and gas clouds, while on the other hand they don't consider an option that stars must migrate/drift inwards in order to supply the requested food for the SMBH monster?
If the SMBH has 4 x 10^6 sun mass, (while our scientists believe that this mass had been taken from the stars in the galaxy) than somehow 4 x 10^6  stars had to drift inwards.
So why our scientists are so sure that the Sun was always at the same distance from the center of the galaxy???
What makes our star so unique that it had to stay so far away from the monster at the galaxy center?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 17/12/2018 21:18:16
With regards to the Sun

Why are we so sure that during all of his life time the Sun had to keep the same orbital radius?
Well not at the exact same radius since it is under the effect of all the local stars and such.  But it is moving around the galaxy in a pretty good imitation of a circle (having done around 20 laps so far), and if it had been seriously deflected by some large passing object, it would very unlikely be deflected onto this fairly perfect orbit.  So it seems to have been free of that sort of thing the whole time.  That's a large reason why we're here: We've had a nice stable environment that wouldn't have existed if that large mass had gone through and disrupted the solar system.

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How could it be that all the moons and Planets are drifting outwards, (or inwards based on the tidal idea) while the sun is fixed at the same radius?
We have no proof that the radius has been fixed all this time.  The clean orbit tells us that we've never had a large close encounter.  The stars that have had one typically become halo objects.
As for tides, you can't raise a tide on a black hole, and I don't think the tides raised by Sgr-A on the sun have any significant impact on its orbital radius.  Tidal forces follow a inverse-cube law, and that means Sgr-A probably has less tidal effect on the sun than I do.

Tides cannot pull our sun in.  The solar system has positive angular momentum (inclination +63°), so if any of that is affected by Sgr-A induced tides, it will push us further out.  Even if we had negative angular momentum, there isn't enough of it to drop us anywhere close to Sgr-A.

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How could it be that in one hand our scientists claim that the SMBH increases its mass by eating stars and gas clouds, while on the other hand they don't consider an option that stars must migrate/drift inwards in order to supply the requested food for the SMBH monster?
4 million stars in 15 billion years is one every 4000 years.  Hardly a hungry monster.
Spacetime is bent near the SMBH, and things cannot orbit at all near them.  You need to move at lightspeed to orbit at a radius of 1.5 the radius of the event horizon.  Anything inside that is dragged in (or would need thrust to stay out), and stuff outside that but close still spirals in.  These are relativistic effects, not Newtonian effects.

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If the SMBH has 4 x 10^6 sun mass, (while our scientists believe that this mass had been taken from the stars in the galaxy) than somehow 4 x 10^6  stars had to drift inwards.
A lot of that mass was already there before it formed into stars, but I suppose it all was outside the event horizon at some distant point, just like our solar system was all spread out before gravity pulled all that matter close enough together to make a star.

Things at the center of the galaxy are relatively congested.  Two stars pass close to each other and one goes faster and the other loses kinetic energy.  The slow star falls closer to the SMBH and the fast one shoots out and becomes a halo object.  Few things have a clean orbit that close in.

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So why our scientists are so sure that the Sun was always at the same distance from the center of the galaxy???
They're not.  There is no history we can consult.  We seem to be on the road (a clean orbit) still, so that's evidence that we've not left it, because it is hard to get back if you fall off the road, or at least it takes a lot more than 20 orbits. That's pretty strong evidence, but not proof.

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What makes our star so unique that it had to stay so far away from the monster at the galaxy center?
Most of the stars out this far are like that.  We're in the suburbs, and we're quite ordinary, not unique at all.  The big ones burn up faster, and the little ones burn much slower.  Scientists are not claiming that all the stars out here are heading for Sgr-A except us.
Look near Cassiopeia constellation for our doom.  That's the close encounter that will likely fling us out of our nice circular orbit in less time that it took life to evolve a human.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/12/2018 05:09:31
There is no history we can consult.  We seem to be on the road (a clean orbit) still, so that's evidence that we've not left it, because it is hard to get back if you fall off the road, or at least it takes a lot more than 20 orbits. That's pretty strong evidence, but not proof.
The solar system has positive angular momentum (inclination +63°),

The Sun has currently positive angular momentum (inclination +63°). In other words, it is moving upwards from the Galactic Disc lane while it orbits around the center of the galaxy.
It is expected that once it gets to the pick it should get down and cross the galactic lane downwards.
Hence, the Sun is Bobbing up and down while it orbits the center of the galaxy.
This is not unique for the Sun. Actually all the stars in the galaxy bobbing up/down, in/out or in between.
If I understand it correctly, the Sun should move up and down at least four times before it set one complete orbital cycle.
So, how can we call it "clean orbit"?
Do you think that Newton or keler would accept that bobbing orbital activity as a "clean orbit"?
Can you please find one planet or moon that is bobbing up and down while it orbits around its main host (four times per cycle)?
How can we ignore that incredible positive angular momentum???
Well not at the exact same radius since it is under the effect of all the local stars
So, you even claim that "it is under the effect of all the local stars".
Hence, why don't we accept the idea that what we see is due to the gravity impact of all the local stars?
We all know that gravity works locally.
As we stay on Earth, we are under the gravity force of the earth (Not the Sun, Not the moon and not under the gravity force of the whole galaxy ).
The moon also works under the gravity of the Earth (Although the Sun gravity on the Moon is stronger by at least twice)
The Earth (or actually - the Earth/moon center of mass) works under the gravity of the Sun (While it ignore the gravity of the whole galaxy, and so on.)
So, why the Sun doesn't orbit under the effect of all the local stars???
In other words, what is the real host of the Sun?
If the Sun goes up and down, could it be that it actually orbits around some sort of a center of mass which is the equivalent center of all the local stars? So, could it be that it doesn't directly orbit around the galaxy, but it orbits around a local center of mass while this local center of mass orbits around the galaxy.
It is the same phenomenon as the Moon orbits around the Sun (12 times per cycle).
If we shut down the light at the Earth, we might see that the moon orbits around the Sun while it's bobbing inwards and outwards. (As the Moon orbits almost horizontally to the earth/sun disc). If the Moon was orbiting vertically around the earth, we would see that it is bobbing up and down as it orbits around the Sun. Almost identical to the Sun bobbing activity.
So, why do we reject the idea that the bobbing activity shows clearly that the Sun doesn't orbit directly around the center of the galaxy, but around some local center of mass which had been set by the "effect of all the local stars"?

If you don't agree with this idea:
Would you kindly show the formula of gravity which can support that strange bobbing activity or "clean orbit" of the Sun?
I assume that you might claim that it is bobbing due to the gravity of the galactic disc lane. If so, please prove this hypothetical idea by real mathematics based only on Newton and kepler.

In my opinion, this bobbing activity is the smoking gun which we are looking for.
I really can't understand how can we ignore so important activity.
Once we agree that the Sun orbits around a local center of mass (Which is "under the effect of all the local stars" - as the Moon/sun orbit), we will get a clear answer for the Spiral galaxy enigma.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/12/2018 15:10:48
Quote from: Halc
The solar system has positive angular momentum (inclination +63°),
The Sun has currently positive angular momentum (inclination +63°). In other words, it is moving upwards from the Galactic Disc lane while it orbits around the center of the galaxy.
Inclination is the tilt angle and has nothing to do with where we are going.  0° means the axis of the solar system is the same as the axis of the galaxy.  It isn't, it is off by 63°, which means it is more on its side than not.  Any inclination over 90° would give that solar system a negative angular momentum relative to the galactic axis.

What I said has nothing to do with us moving towards the center.  As I've said, our path is on average somewhat circular: the eccentricity is low enough to put us in the disk somewhere at a fairly narrow range of radius, but is otherwise pretty meaningless.  We take a wobbly path around the galaxy and have a speed somewhere between 225 and 250 km/sec most of the time.  The forces that change that are not tidal.

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It is expected that once it gets to the pick it should get down and cross the galactic lane downwards.
I know of nobody (except you perhaps) that expects this sort of thing to ever happen.

Quote
Hence, the Sun is Bobbing up and down while it orbits the center of the galaxy.
This is not unique for the Sun. Actually all the stars in the galaxy bobbing up/down, in/out or in between.
So, how can you call it "clean orbit"?
You're right, its not very clean.  But we're in the plane and moving generally at the velocity needed to be at this current radius, so that's clean enough.

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Do you think that Newton or keler would accept that orbit as a "clean orbit"?
The orbit is described as being non-Keplarian, so no.

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If I understand it correctly, it should move up and down at least four times before it set one complete cycle.
Can you please find one planet or moon that is bobbing up and down while it orbits around its host (four times per cycle)?
Earth bobs up and down around 13 times each lap.  4 times?  No, I cannot think of one with that cycle.  I'd not heard of it for the galactic orbit.  Is there something nearby that we go around every 50 million years or so?  Is that a known thing?

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How can you ignore that incredible positive angular momentum???
Same way I ignore the moon's incredible change in positive solar angular momentum every 14 days.  It all averages out in the end.

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Well not at the exact same radius since it is under the effect of all the local stars
So, you even claim that "it is under the effect of all the local stars".
Yes, that's right.  We don't exactly orbit Sgr-A, which is tiny compared to the mass of the galaxy inside our orbital radius.  Not at all like the Earth, dominated by that one huge mass in the middle with nothing but small objects giving minor but regular deflections from that perfect elliptical path.

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The Earth works under the gravity of the Sun (While it ignore the gravity of the center of the galaxy, and so on.)
Earth accelerates due to the galaxy exactly as much as does the sun.  Gravity is never ignored.  We are inside the hill radius of the sun, so the galaxy in general isn't going to separate the two of us.  A single passing object might.

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So, why the Sun doesn't orbit under the effect of all the local stars???
The sun is within the hill radius of no 'local' dominant mass.  Perhaps there is a cluster that holds together with mutual attraction, but then the orbit of the cluster still goes around the galaxy.  Earth has also made 20 trips around the galaxy, despite the continuous change in velocity.

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In other words, what is the real host of the Sun?
Don't know.  Don't know if there is one before the galaxy as a whole.  Something must hold the galactic arms together, but that's not really an orbit.

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If the Sun goes up and down, could it be that it actually orbits around some sort of a center of mass which is the equivalent center due to all the stars/SMBH in the galaxy?
That would be orbiting the galaxy, and it wouldn't go up and down due to that, except for normal orbital eccentricity just like Earth and every other planet.  Mercury radius changes an awful lot each orbit, but that is no indication that it is orbiting some mass other than the sun.

If there is some local mass around which all the nearby stars cling, then what really counts is the orbital speed and radius of the center of mass of that local group, and not the individual velocity of any particular object that is part of that group.

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So, could it be that it is not directly orbits around the galaxy, but it orbits around a local center of mass while this local center of mass orbits around the galaxy.
Exactly.

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It is the same phenomenon as the Moon orbits around the Sun (12 times per cycle).
If we shut down the light at the Earth, we might see that the moon orbits around the Sun while it's bobbing inwards and outwards. (As the Moon orbits almost horizontally to the earth/sun disc).
Right, but that bobbing in and out doesn't mean the moon has any chance of falling into the sun.  OK, it will, but only because the sun will come out here at grab it, not because the moon will fall in.

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So, why do we reject the idea that the bobbing activity shows clearly that the Sun doesn't orbit directly around the center of the galaxy, but around some local center of mass which is under direct "effect of all the local stars"?
I didn't assert that one way or the other.  The local masses are pretty well known, and I don't know if our neighbors are always our neighbors, or if they're just nearby right now.  A good text would say.  I'm pretty sure this sort of thing is known.
Not sure if our motion relative to this local mass would be considered 'orbital' since there is no central mass to stabilize it.  It is sort of a 3-body problem with similar masses: the movements of the masses is chaotic, not at all following the rules of orbital mechanics.  Any one of the bodies might get flung away from the group, bringing the remaining members of the group closer together.  New masses might come from outside and lose enough speed to become new members of the group.

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If you don't agree:
Would you kindly show the formula of gravity which can support that strange bobbing activity or "clean orbit" of the Sun?
A=GM/r²
That describes the acceleration of any body given any number of masses spaced around it.

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Prove your hypothetical idea for that bobbing activity by real mathematics based only on Newton and kepler.
Integrate the formula above and you get the bobbing activity.  Kepler doesn't much come into play here.  It's all that simple Newton formula.

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In my opinion, this bobbing activity is the smoking gun which we are looking for.
Actually, I'm not really sure what you're looking for here.
Some single central mass (closer than SGr-A) around which we orbit?  No, we're not within any object's hill radius, so our local motion is not really best described as 'orbital'.

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Once we agree that the Sun orbits around a local center of mass (Which is "under the effect of all the local stars" - as the Moon/sun orbit), we have got the answer for the Spiral galaxy enigma.
Enigma?  The 'bobbing' is quite expected.  There are masses nearby, but none close enough for an orbit.  A local center of gravity may well come into play, loosely holding a group of sister stars together, but our motion around that is chaotic, not orbital.
I call them sister stars because most of the the local ones are probably born of the same parent supernova, which explains their somewhat similar mutual velocity as a group.  We're a 2nd generation solar system.  The 1st generation ones lack planets like our inner ones.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/12/2018 19:49:46
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So, could it be that it is not directly orbits around the galaxy, but it orbits around a local center of mass while this local center of mass orbits around the galaxy.
Exactly.
Wow

If we agree that the sun orbits around a local center of mass while this local center of mass orbits around the galaxy, than we have to agree that all the stars in the spiral galaxy has a similar orbital motion.
Based on that, now we can get better understanding about the Spiral galaxy.
For example -
When we look at the nearby stars, we might see that each one is moving at a different direction and velocity.
We might think that very soon they all will get out from the spiral arm.
This is incorrect. all the nearby stars will stay with us. Each one of them orbits around it's unique center of mass, while the center of mass is moving with the arm. and that center of mass is a direct product of all the other nearby stars...
Let's look at S2.
We think that it orbits around the SMBH. That is big mistake.
If we look carefully, we also might find that it's orbital cycle isn't clean.
In other words, it also orbit around some center of mass with this center of mass orbits around the SMBH.
This is a very important issue.
Let's take the example of the Moon/sun orbital cycle.
If we ignore the earth, than just by monitoring the orbital cycle of the moon and its mass estimated, we might think that the Sun mass is very low.
So, we can't really extract the real mass of the sun directly from the moon orbit.
We must first find the mass of the earth, and then extract the Sun mass.
In the same token, if we want to extract the real mass of the SMBH we must first find the estimated mass of the S2 center of mass point and based on that data we can extract the real value of the SMBH.
We might find that the mass of the SMBH is significantly higher than our faulty calculation.
Do you agree with that?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/12/2018 21:48:04
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So, could it be that it is not directly orbits around the galaxy, but it orbits around a local center of mass while this local center of mass orbits around the galaxy.
Exactly.
Wow

If we agree that the sun orbits around a local center of mass while this local center of mass orbits around the galaxy, than we have to agree that all the stars in the spiral galaxy has a similar orbital motion.
That's like saying that because we know about one white swan, we have to agree that all swans are white.  This is unlikely here.  Some stars are the biggest kahuna around and find nothing local around which it can orbit.
I doubt the sun's motion can be designated as an orbit.  It moves about the local center of mass, but doesn't follow any path as described by Kepler's laws, and this is IF the sun in fact is gravitationally bound to this local group.  I think there probably is such a thing, but I'm not sure.

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Based on that, now we can get better understanding about the Spiral galaxy.
For example -
When we look at the nearby stars, we might see that each one is moving at a different direction and velocity.
Small differences, sure, but nothing that takes any of them out of the general path that takes them around the galaxy at the same pace as us.
So sure, the stars around us have different velocities, but a relative speed of only a few km/sec maybe.  As a group, we're all moving thataway at 230 km/sec ± that single digit variation.  The general path is therefore pretty much all the same way like a herd of sheep.
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We might think that very soon they all will get out from the spiral arm.
I don't think anybody suggests that.  Depends on the definition of 'soon'.  Yes, it is forecast, but about 4 billion years away.  Wake me up. I want to watch.  Until then, we go around a bunch more times.

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This is incorrect. all the nearby stars will stay with us. Each one of them orbits around it's unique center of mass, while the center of mass is moving with the arm. and that center of mass is a direct product of all the other nearby stars...
Right.  We go around as a group just like the planets don't leave the solar system despite every one of them having a different velocity than the sun at all times.

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Let's look at S2.
We think that it orbits around the SMBH. That is big mistake.
If we look carefully, we also might find that it's orbital cycle isn't clean.
In other words, it also orbit around some center of mass with this center of mass orbits around the SMBH.
Pretty tall conjecture if you ask me.  They'd notice that in a moment.  They detect dark objects as S2 passes by them.  If it orbited one of them, it would have a regular deviation from its Keplerian path.  It doesn't.  So who's making the big mistake?

I find it impressive that they can watch it that well despite all the clutter between us and it that supposedly obscures the view of the galactic core.  Here also is the fastest known ballistic orbit of anything, and the gravity of the SMBH out where this star is is comparable to that of the moon.  Our black hole is a wuss...

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This is a very important issue.
Let's take the example of the Moon/sun orbital cycle.
If we ignore the earth, than just by monitoring the orbital cycle of the moon and its mass estimated, we might think that the Sun mass is very low.
So, we can't really extract the real mass of the sun directly from the moon orbit.
Yes you can. The moon is about at the same orbital radius as Earth, and goes around in the same period.  That makes the Sun's mass perfectly consistent with the orbit of the moon about it, even if we were unaware of Earth being there.
If Earth was not actually there, the moon would go around in the same orbit (speed and radius) as Earth does now, just without the monthly wiggle since it would be one object instead of two.

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We must first find the mass of the earth, and then extract the Sun mass.
In the same token, if we want to extract the real mass of the SMBH we must first find the estimated mass of the S2 center of mass point and based on that data we can extract the real value of the SMBH.
We might find that the mass of the SMBH is significantly higher than our faulty calculation.
Do you agree with that?
None of it.  It is based on the moon orbit not being consistent with the mass of the sun.  It matches perfectly, but it has a regular wiggle, meaning it orbits something, even if you can't see it.  S2 doesn't have a regular wiggle like that, and if it did, it would still go around SGr-A every 16 years like it does now.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/12/2018 23:47:46
Orbital speed is a function of 'reduced mass', which for any relationship except a binary star, is approximated by the mass of the primary.
V = √(Gμ/R) where G is the gravitationl constant, μ is the reduced mass of the pair of objects, and R is the radius.

Reduced mass μ for two objects is: μ = (m1-1 + m2-1)-1 which is pretty much the mass of the sun for Earth's orbit, and SGr-A (plus a couple hundred dark objects) for S2.

That means that the moon would orbit at about the same speed as Earth if it was by itself, and S2 would orbit at the same speed even if it was the only visible part of some larger object that it orbited.  None of these objects have enough mass to affect their respective reduced masses.

M31 (the object that is going to throw our sun on some totally new path) has a central SMBH designated as P2 with a mass that's around 10x that of Sgr-A.  'Nearby' orbits P1 which is the sort of cluster that you describe, a heavy collection of stars and such seemingly not held together with a single central mass.  It seems a strong enough gravity well to hold itself together, but it is far further from its SMBH than is S2.  If P1 were that close, tidal forces would separate the material into a ring.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/12/2018 17:05:16
Orbital speed is a function of 'reduced mass', which for any relationship except a binary star, is approximated by the mass of the primary.
V = √(Gμ/R) where G is the gravitationl constant, μ is the reduced mass of the pair of objects, and R is the radius.

Reduced mass μ for two objects is: μ = (m1-1 + m2-1)-1 which is pretty much the mass of the sun for Earth's orbit, and SGr-A (plus a couple hundred dark objects) for S2.

That means that the moon would orbit at about the same speed as Earth if it was by itself, and S2 would orbit at the same speed even if it was the only visible part of some larger object that it orbited.  None of these objects have enough mass to affect their respective reduced masses.

Yes.
You are absolutely correct.
Sorry for my mistake.
However, with regards to the SMBH:
Why do we ignore the accretion disc?
Based on our measurements, the velocity of the plasma in that disc is 0.3 c (speed of light)
If we know the radius of the accretion disc (at the verified velocity), we can easily extract the real mass of the SMBH.
So, why do we insist on S2 which doesn't have a clean orbit instead of the accretion disc which has a perfect cycle orbit?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 19/12/2018 19:52:55
However, with regards to the SMBH:
Why do we ignore the accretion disc?
Based on our measurements, the velocity of the plasma in that disc is 0.3 c (speed of light)
If we know the radius of the accretion disc (at the verified velocity), we can easily extract the real mass of the SMBH.
So, why do we insist on S2 which doesn't have a clean orbit instead of the accretion disc which has a perfect cycle orbit?
They follow S2 because it is an object that can be followed.  If they measure velocity of plasma at a specific point, that doesn't indicate the orbit, just the speed of something at one moment.
By similar thinking, one could significantly overestimate our own sun from a distance if all we measured was the occasional comet when it grows a nice visible tail, and moves near Mercury orbit but far faster than mercury.  That observation might lead one to believe that the orbital speed at that radius is much higher than it is.  The can't follow the comet through a full orbit any more than we can follow the plasma through a full orbit.  With S2 we can.  Sure, it isn't a circular orbit, but that isn't needed to do the calculation.  It just makes the calculation a little less trivial.

They also follow S2 because they can see it deviate from its orbit, giving them an idea of the density of other dark objects that it passes, which is very useful information.  There is a lot of research going into dark matter, and these sorts of things help them estimate the MACHO component of dark matter (things that don't show light like Mars).
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 19/12/2018 20:57:23
V = √(Gμ/R) where G is the gravitationl constant, μ is the reduced mass of the pair of objects, and R is the radius.
This is wrong. I substituted μ for the wrong mass.  μ is the reduced mass of the secondary, not the primary, so  V=√(GM/R), but skewed by greater acceleration (and hence speed) of the secondary. a=F/m so for orbital acceleration, a=F/μ.
For small masses (pretty much everything except the moon and Charon), μ is essentially m.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/12/2018 06:15:07
There is a lot of research going into dark matter, and these sorts of things help them estimate the MACHO component of dark matter (things that don't show light like Mars).
O.K.
As you have mentioned that issue - Dark Matter

Why Newton didn't add that brilliant idea of dark matter in his formulas?
Why he didn't say that any orbital system is based on dark matter?
Our science communities try to prove their hypothetical ideas by using Newton laws. As they have failed to explain the orbital velocity of our Sun, they came with this idea.
Now they even try to set more effort in order to show that this none realistic idea is realistic.
Let me give you an example -
Let's assume that we want to swim in a pool without water.
So, I will tell you that there is water, but the water is dark water.
You can't see it, you can't feel it, you can't smell it or drink it, but it is there because I said that it there. Would you believe me???

So, this is my personal opinion:
Dark matter is a solid proof for the failure of our scientists to show how spiral galaxy really works.
As they have failed to understand the real impact of none "clean orbit", as they have failed to understand the real impact of the Earth/moon "Drifting outwards", as they have failed to understand the real impact of the Ultra high magnetic power around the accretion disc, they couldn't explain how the gravity really works in Spiral galaxy.
I can do it.
I can explain how the whole universe works without any need for dark matter, for dark energy and for any sort of dark magic.
Based on Newton law I can easily explain:
How spiral galaxy really works?
Why there are spiral arms in spiral galaxy?
Why the stars in spiral arms have almost constant orbital velocity at any radius?
Why we see that all the far end galaxies are moving away from us at almost the speed of light, while our observable universe seems to be full with matter.
If we will come back 100,000 billion years (or even 10^Billion years) from now, we will see a similar universe with same density and almost the same numbers of galaxies per observable universe. We might not find our solar system, but there will be many similar.
Again - no need for dark matter or dark energy - just Newton law and simple common sense.
However, in order to do it I need the following basic understanding:
1. "Drifting" outwards
It's very difficult to verify a "drifting" outwards of few cm per cycle while the orbital radius is 1 Arc.  However, all the objects (assuming that they are far enough) are drifting outwards from their host. That is correct for all the objects including: Moons, planets, stars... There is no need for tidal support to explain this Phenomenon.
Therefore, any orbital cycle is by definition spiral shape cycle. So, even if it drifts 1 micro meter each cycle, it is still has spiral orbital shape.
2. Clean orbital cycle.
All the orbital cycles in the Universe must be clean. There is no room in our universe for none clean cycle (unless there is an interruption.)
So, the Sun must set a very clean orbital cycle around its center of mass. That must be correct for any star in the universe even if it is very massive star or object (unless there is no interruption).
3. The magnetic power around the accretion disc, push any matter upwards (or downwards) at a speed of 0.8 c (speed of light) that prevents from any gas cloud, star or even atom to drift inwards in order to be eaten by the SMBH monster.

That's all I need.
Based on those three elements, I can easily explain how our universe works without any sort of dark magic.
If you agree I will show how it works.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 20/12/2018 12:35:17
As you have mentioned that issue - Dark Matter

Why Newton didn't add that brilliant idea of dark matter in his formulas?
Why he didn't say that any orbital system is based on dark matter?
Newton didn't need to.  His formulas do not apply only to matter that emits light.  Jupiter will orbit the future white dwarf that the sun will become at the same radius as the black dwarf it will be even later on.  The latter is dark matter.

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Our science communities try to prove their hypothetical ideas by using Newton laws. As they have failed to explain the orbital velocity of our Sun, they came with this idea.
The idea of there being burnt out stars and rouge planets is hardly new.  Nobody suspected these things didn't exist.  They measure the density of lit-up stars and conclude they don't have enough collective mass to explain the acceleration of the sun, so they know there must be more matter that is not young stars.

The new idea was not large objects (MACHO) like star cores, but small ones (WIMP).  Those are like neutrinos, except with much more mass.

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Now they even try to set more effort in order to show that this none realistic idea is realistic.
Let me give you an example -
Let's assume that we want to swim in a pool without water.
So, I will tell you that there is water, but the water is dark water.
You can't see it, you can't feel it, you can't smell it or drink it, but it is there because I said that it there. Would you believe me???
Sure.  It explains how I can swim in a pool that is apparently lacking water.  OK, that probably means I can feel it since swimming involves applying thrust with my limbs, but in the same way the sun 'feels' the dark matter by its acceleration vector.

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So, this is my personal opinion:
Dark matter is a solid proof for the failure of our scientists to show how spiral galaxy really works.

As they have failed to understand the real impact of none "clean orbit", as they have failed to understand the real impact of the Earth/moon "Drifting outwards", as they have failed to understand the real impact of the Ultra high magnetic power around the accretion disc, they couldn't explain how the gravity really works in Spiral galaxy.
Is there some pet alternative that you think explains the observations better?  Do I see a suggestion of magnetism there?  I've never seen magnetism work detectably beyond a few meters.

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I can do it.
I can explain how the whole universe works without any need for dark matter, for dark energy and for any sort of dark magic.
OK, you do claim to have a pet explanation, all without having a basic grasp of the application of Newton's laws.

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Based on Newton law I can easily explain:
How spiral galaxy really works?
Why there are spiral arms in spiral galaxy?
Why the stars in spiral arms have almost constant orbital velocity at any radius?

Why we see that all the far end galaxies are moving away from us at almost the speed of light, while our observable universe seems to be full with matter.
If we will come back 100,000 billion years (or even 10^Billion years) from now, we will see a similar universe with same density and almost the same numbers of galaxies per observable universe. We might not find our solar system, but there will be many similar.
Again - no need for dark matter or dark energy - just Newton law and simple common sense.
I don't think Newton's law predicts that.  1st law is that those receding galaxies will continue to recede, thus reducing the density of any given volume over time.  If your theory has new matter filling the gaps, that theory doesn't follow from Newton's laws.

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However, in order to do it I need the following basic understanding:
1. "Drifting" outwards
It's very difficult to verify a "drifting" outwards of few cm per cycle while the orbital radius is 1 Arc.
What is 1 Arc?
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However, all the objects (assuming that they are far enough) are drifting outwards from their host. That is correct for all the objects including: Moons, planets, stars... There is no need for tidal support to explain this Phenomenon.
You invoked Newton's laws.  Newton says the energy and momentum needed to do that has to come from somewhere.  Not all moons do this.  Most of Jupiters moons are not drifting outward, so the assertion is false.

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Therefore, any orbital cycle is by definition spiral shape cycle. So, even if it drifts 1 micro meter each cycle, it is still has spiral orbital shape.
Agree, except for the 'by definition' part.  That spiral is a deviation from the 'by definition' elliptical model of ideal point masses.

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2. Clean orbital cycle.
All the orbital cycles in the Universe must be clean. There is no room in our universe for none clean cycle (unless there is an interruption.)
What is clean?  A perfect ellipse?  You need to define that term, because I cannot think of anything that qualifies as having a clean orbital cycle, but maybe my concept of the term is different than yours.

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So, the Sun must set a very clean orbital cycle around its center of mass.
The sun does not orbit itself.

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3. The magnetic power around the accretion disc, push any matter upwards (or downwards) at a speed of 0.8 c (speed of light) that prevents from any gas cloud, star or even atom to drift inwards in order to be eaten by the SMBH monster.
One does not push at a speed.  One pushes with a force, or applies an acceleration.  This is a direct implication of Newton's 2nd law.  Where does the 0.8c come from?

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That's all I need.
You need somebody with a complete disregard for Newton's laws.  Yes, I suppose you could make that theory plausible to such a person.

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Based on those three elements, I can easily explain how our universe works without any sort of dark magic.
If you agree I will show how it works.
I cannot agree with the points enumerated above, so don't bother.
You can't assert any of the things above.  Those should all be conclusions demonstrated by, or observations explained by application of the basic laws.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/12/2018 09:02:04
Dark Matter
Newton didn't need to.  His formulas do not apply only to matter that emits light.  Jupiter will orbit the future white dwarf that the sun will become at the same radius as the black dwarf it will be even later on.  The latter is dark matter.

Based on Wiki:
https://en.wikipedia.org/wiki/Dark_matter
"Dark matter is a hypothetical form of matter that is thought to account for approximately 85% of the matter in the universe, and about a quarter of its total energy density."
"Many experiments to directly detect and study dark matter particles are being actively undertaken, but none has yet succeeded."
This is my opinion about the dark matter:
The galaxy rotation problem - We don't know how to explain that problem based on our current hypothesis/understanding how spiral galaxy works.
Hence, instead of reconsider our hypothetical ideas about spiral galaxies and some other issues, we have found a brilliant idea of dark matter.
However, unfortunately we didn't find any evidence for that unrealistic idea: "Many experiments to directly detect and study dark matter particles are being actively undertaken, but none has yet succeeded."
Therefore, we have two options:
1. Continue to hold our none realistic ideas about spiral galaxies and continue to look for that unrealistic dark matter.
2. Open our mind to different ideas which can perfectly explain the galaxy rotation problem and all the other unsolved issues without any need to dark matter.

However, if I understand it correctly - Our science community reject the idea that they might have a problem with understanding how spiral galaxy really works. Based on their point of view, they only need to find the evidence for that dark matter.
So, how long do we have to wait until they will claim - "Sorry, there is a chance that if we don't find it - it is not there?
One year, 10 Year? Or one billion years?
Why you don't even give a chance to look for a different concept about our universe?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/12/2018 14:27:52
Therefore, we have two options:
1. Continue to hold our none realistic ideas about spiral galaxies and continue to look for that unrealistic dark matter.
2. Open our mind to different ideas which can perfectly explain the galaxy rotation problem and all the other unsolved issues without any need to dark matter.
Both options are viable, but when noting that orbits seem to indicate that there is more stuff than what we see, it seems less magical to propose that there is more stuff we don't see than to discard Newtonian mechanics altogether and rewrite physics from scratch.
So is there a model supplied by such open-minded thinking?  Trust me, such a thing would be warmly greeted if it worked.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/12/2018 16:02:05
So is there a model supplied by such open-minded thinking?  Trust me, such a thing would be warmly greeted if it worked.
Trust me.
There is a model that works by 100%.
However, I need your cooperation and good willing.
For just one moment try to forget all the wrong understanding/hypothetical ideas that we have about the galaxy (Including: Age of the Star/galaxy/Universe, dark matter/dark energy and so on).

There are three elements which are needed in this model.

1. The impact of gravity force due to local mass - The impact of gravity force due to local mass is stronger than the impact of gravity force of a very far end object. Therefore, the moon had selected to orbit around the Earth instead around the Sun, while the gravity force of the Sun/Moon is stronger by at least twice than the Earth/Moon gravity.
So, even if we shut down the light on Earth , the moon will continue to orbit around the Earth, while their mutual point of mass orbits around the Sun.
The outcome is as follow:
The moon orbits around a virtual host point (as we can't see the earth) while this virtual host point orbits around the Sun. So, the orbital cycle of the moon sets a clean orbital cycle around a virtual host point (Earth - which we can't see).
In the same token:
Every star in the galaxy must orbit around some host Point. It might be something that we see or something that we can't see. However, any star (at any size) must set a clean orbital cycle (in ellipse shape or a perfect cycle). If we can't see that host point, let's call it virtual host point. So, while the star orbits around that virtual host point, the host point orbits around the center of the galaxy. Therefore, we might think that the orbital cycle of a star around the galaxy is not clean.
With regards to our Sun
Please look at the following motion of the solar system in the galaxy:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
http://www.biocab.org/coplanarity_solar_system_and_galaxy.html

The broken gray line shows exactly the virtual host point of the Sun while it orbits around the galaxy.
So, the Sun sets a clean orbital cycle around its virtual host point. This host point is a direct outcome of the local mass gravity (Due to the nearby mass in the Orion spiral arm).
So, the total orbital motion of the Sun is 217 Km/sec however, it also moves locally upwards/downwards to the galactic disc lane at 5-7 km/sec while it moves inwards/outwards at 20 km/sec.
Each star in the spiral arm has a similar orbital motion. Therefore, when we look at the nearby stars we see the following:
http://www.basicknowledge101.com/photos/2015/perception-of-time-rotation-of-galaxy.png
We might think that they move randomly relative to each other, but in reality, each one orbits around its virtual host point, while all their host points stay together in the arm and set a nice orbital motion of about 217 Km/sec around the galaxy.

Summery -
The orbital cycle around a virtual host point is a key element in my explanation.
Please try to accept this idea as is.
If you totally can't agree with that, you are more than welcome to prove it by mathematics.
However, please don't tell a story why this idea isn't logical based on our current understanding about the Galaxy
Agree?
If you have no objection - we will set the next element.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/12/2018 17:06:29
So is there a model supplied by such open-minded thinking?  Trust me, such a thing would be warmly greeted if it worked.
Trust me.
There is a model that works by 100%.
However, I need your cooperation and good willing.
You don't need that at all if you have a good model that works, and not just a set of assertions that all things behave in a manner that they don't.  I see why you're denying the tidal forces and such.  It apparently conflicts with your assertions.

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For just one moment try to forget all the wrong understanding/hypothetical ideas that we have about the galaxy (Including: Age of the Star/galaxy/Universe, dark matter/dark energy and so on).
We must forget all of Newton's laws as well.  Shall we go back to 'impetus'?
You've supplied no laws to replace those, so all I see are requirements for this dream model of yours, but no actual model that meets those requirements.

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In the same token:
Every star in the galaxy must orbit around some host Point. It might be something that we see or something that we can't see. However, any star (at any size) must set a clean orbital cycle (in ellipse shape or a perfect cycle). If we can't see that host point, let's call it virtual host point.
This for instance violates Newton's laws, since F=GMm/r² doesn't work anymore.  Force actually goes down as distance from that host point decreases.  Sometimes it goes up.  Depends where you are.  That's what Newton's laws say anyway, but you seem to assert that these virtual host points have force of their own, not the objects themselves.  What makes some objects contribute to the force of a host point and other not?

So, while the star orbits around that virtual host point, the host point orbits around the center of the galaxy. Therefore, we might think that the orbital cycle of a star around the galaxy is not clean.

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Summery -
The orbital cycle around a virtual host point is a key element in my explanation.
Please try to accept this idea as is.
I can accept that, sure.

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If you totally can't agree with that, you are more than welcome to prove it by mathematics.
On the contrary, you need to prove, with mathematics, that it works.  So far I've seen no model, just a list of requirements.

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However, please don't tell a story why this idea isn't logical based on your our current understanding about the Galaxy
Agree?
No problem.  It seems so far not to defy its own rules, so not illogical.  Current understanding of the galaxy has been discarded, as requested.  But don't go quoting that understanding then.  That would be mathematically unsound.
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If you have no objection - we will set the next element.
Have at it!
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/12/2018 18:00:07
This for instance violates Newton's laws, since F=GMm/r² doesn't work anymore.  Force actually goes down as distance from that host point decreases.  Sometimes it goes up.  Depends where you are.  That's what Newton's laws say anyway, but you seem to assert that these virtual host points have force of their own, not the objects themselves.

The whole idea is based on Newton law!!!
I can easily prove it.
However, in order to prove it we must understand the following:

With regards to our Sun
Please look at the following motion of the solar system in the galaxy:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
http://www.biocab.org/coplanarity_solar_system_and_galaxy.html
The broken gray line shows exactly the virtual host point of the Sun while it orbits around the galaxy.
So, the Sun sets a clean orbital cycle around its virtual host point. This host point is a direct outcome of the local mass gravity (Due to the nearby mass in the Orion spiral arm).
So, the total orbital motion of the Sun is 217 Km/sec however, it also moves locally upwards/downwards to the galactic disc lane at 5-7 km/sec while it moves inwards/outwards at 20 km/sec.
Each star in the spiral arm has a similar orbital motion. Therefore, when we look at the nearby stars we see the following:
http://www.basicknowledge101.com/photos/2015/perception-of-time-rotation-of-galaxy.png
We might think that they move randomly relative to each other, but in reality, each one orbits around its virtual host point, while all their host points stay together in the arm and set a nice orbital motion of about 217 Km/sec around the galaxy

What makes some objects contribute to the force of a host point and other not?
Well, each star gravity force is affected by all the stars in the galaxy (or even in the Universe). However, as gravity works locally, the main gravity force impact is due to all the nearby stars. Therefore, the stars gather together is spiral arms. Each virtual host point sets a gravity force with all the other virtual host points of the nearby stars. They actually hold each other while they set a very high orbital motion. Therefore, a spiral arm is THE key request for their orbital velocity.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/12/2018 20:39:57
Second Element - Drifting outwards

All the stars/planets/Moons/ are drifting outwards from their host points.
In the same token each virtual host point drifts outwards from the center of the galaxy.
This element is key request for our understanding, although It is quite difficult to prove/disapprove it as we only have verified the Earth/moon and Sun/moon orbital system.
The main idea is that any orbital cycle set a spiral cycle (even if it is perfect cycle of ellipse).
This is the only issue which I think that Newton have missed in his formulas.
Newton didn't consider the impact of long time (Billion years).
He actually verified the forces at a limited time frame (years).
So, the Sun must drift outwards from the Center of the galaxy.
In the same token all the stars are drifting outwards from the center.
Even the new atoms in the accretion disc are drifting outwards.

There is no need for tidal to set a drifting outwards movement.
This is a normal mechanism of gravity.
That gives us the following important understanding: Nothing drifts inwards to the center of the galaxy.

Third Element - New Atom/Molecular creation at the center of the galaxy.
There is very powerful magnetic field around the accretion disc which prevents from any matter to drift inwards to the SMBH.
That proves that the accretion disc is actually excretion disc.
I have discussed this issue deeply in  the following thread:
What is needed to create new atoms in the Universe?
https://www.thenakedscientists.com/forum/index.php?topic=75261.40

So, new matter is created at the accretion disc and drift outwards.
As they get to the magnetic field around the accretion, they are boosted upwards at ultra high speed.
Then the new hot molecular clouds fall down to the disc lane and set the famous gas clouds that we can see from our location.
However, those hot new atoms/molecular in this gas cloud don't orbit directly any more around the SMBH.
From now on they orbit around a virtual center of mass in the gas cloud. While this virtual center of mass of the gas cloud orbits around the SMBH.
Therefore, if we look carefully at those gas clouds, we might see that the atoms/molecular orbit around the center of the gas cloud.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/12/2018 21:22:55
The whole idea is based on Newton law!!!
Nonsense.  You threw that out when you generalized a two-body problem property (nice clean orbits) to n-body problem, which has never been solved.  If some object wanders close to this host-point, it will not orbit it (at high speed since it is so close) as you seem to assert, but rather not accelerate much at all, as predicted by Newton.  Sum of the forces is a small value, so only a tiny acceleration towards that center of mass.  Only in the two body case is there guaranteed to be a mass there when you approach it, and the body will orbit at high speed as a low orbit thing does.

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I can easily prove it.
However, in order to prove it we must understand the following:

With regards to our Sun
Please look at the following motion of the solar system in the galaxy:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
Don't understand what that one is trying to depict.  It is a picture minus any description.

It shows the solar system trajectory as the straight line in the middle, with a star on it.  Then there is this nice clean blue helix that is the 'apparent motion' of the solar system.  Apparent to what if the middle line is the actual solar system?
It has some orthogonal arrows labeled with speeds, like they're velocity components of something, but not specified.

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The broken gray line shows exactly the virtual host point of the Sun while it orbits around the galaxy.
It is labeled 'solar system', not 'virtual host'.  I think it is perhaps the linear mass of the galactic arm that we go around, not a point at all.  If it didn't attract like that, the arms would soon dissolve into a more uniform cloud.
The difference is easy to tell:  If it was an orbit, the inclination would not change.  If it is motion around the arm, it will always be on an axis parallel to the arm.  I think it is the latter, and that makes it not an orbit.  There is no virtual host point, which would give us a random permanent inclination.

Anyway, the 217 km/s in that picture is too much for the apparent mass of the galaxy.  If you're not discarding Newton, then the solar system is moving way too fast at this radius.  You've made no attempt to resolve that contradiction without positing more mass than what we can see, which you seem to find offensive.  The funny host-point thing doesn't change the 217 figure one bit.

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So, the Sun sets a clean orbital cycle around its virtual host point. This host point is a direct outcome of the local mass gravity (Due to the nearby mass in the Orion spiral arm).
An arm isn't a host point, nor does it act like one.  But yes, the dynamics seem to work like that, yes.

Each star in the spiral arm has a similar orbital motion. Therefore, when we look at the nearby stars we see the following:
http://www.basicknowledge101.com/photos/2015/perception-of-time-rotation-of-galaxy.png[/quote]
Wonderful.  That shows us having the clean orbit and all the other stars moving randomly in non-clean directions, none of them having any net effect on us.

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We might think that they move randomly relative to each other, but in reality, each one orbits around its virtual host point, while all their host points stay together in the arm and set a nice orbital motion of about 217 Km/sec around the galaxy
If they all moved around the line (not a point) this way, the motion would be less random than depicted, just like the planet motions are hardly random within our solar system.

What makes some objects contribute to the force of a host point and other not?
Well, each star gravity force is affected by all the stars in the galaxy (or even in the Universe). However, as gravity works locally, the main gravity force impact is due to all the nearby stars. Therefore, the stars gather together is spiral arms. Each virtual host point sets a gravity force with all the other virtual host points of the nearby stars.[/quote]How does a nearby star's host point differ from its own position?  Each has a different host point?  Of what would that be the center of mass?
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They actually hold each other while they set a very high orbital motion. Therefore, a spiral arm is THE key request for their orbital velocity.
The spiral arm goes around a 217 km/s at this radius, which is in conflict with Newton's prediction for a galaxy with our apparent visible mass.

Going offline for a bit now and then over the holidays.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/12/2018 04:45:48
If they all moved around the line (not a point) this way, the motion would be less random than depicted, just like the planet motions are hardly random within our solar system.
Yes, that is perfectly correct.

An arm isn't a host point, nor does it act like one.  But yes, the dynamics seem to work like that, yes.
Thanks

The spiral arm goes around a 217 km/s at this radius, which is in conflict with Newton's prediction for a galaxy with our apparent visible mass.
No, there is no conflict.
We need to understand one key issue: Gravity works locally!!!

Please remember that the moon orbits around the earth, while the gravity force of the Sun/Moon is stronger more than double than the Earth/Moon gravity force. This by itself is enigma. I will explain the root for this orbital motion later on

So, the SMBH has no real impact on the Sun orbital velocity. This velocity is a direct outcome of local gravity. Hence, each section of the galaxy is affected by the local mass/gravity force.
I will explain/prove it later on.

How does a nearby star's host point differ from its own position?  Each has a different host point?  Of what would that be the center of mass?
Yes, each star has its own unique virtual host point. This issue will be clear after getting better understanding about the structure of spiral galaxy and how Newton gravity works at each section of the galaxy..
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 23/12/2018 06:21:50
We need to understand one key issue: Gravity works locally!!!

Please remember that the moon orbits around the earth, while the gravity force of the Sun/Moon is stronger more than double than the Earth/Moon gravity force. This by itself is enigma.
Is it?  The moon accelerates more due to the sun than it does due to Earth.  At no time does it accelerate away from the sun, even when between the two.  When it is there, the moon accelerates away from Earth, just like Newton says it should.

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So, the SMBH has no real impact on the Sun orbital velocity.
The sun 'orbits' the galaxy, not the SMBH.  The latter does have an impact, but nowhere near enough to give the solar system its speed.  Apparently not even the galaxy has enough.
Speed is approximated by GM/r where M is mass of the spherical thing orbited.  Plug in the mass of the SMBH into that, and you get a figure far lower.  But the galaxy is not a spherical mass, so orbital speed actually increases with radius at some ranges.

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This velocity is a direct outcome of local gravity. Hence, each section of the galaxy is affected by the local mass/gravity force.
Local gravity causes that blue helix (the 6 and 20 km/sec deviations), not the 217 km/sec speed of the arm in general.  Something needs to accelerate that line to the side, bending it into a circle around the galaxy.  Local gravity cannot do that, since there would be no reaction force.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/12/2018 17:26:03
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We need to understand one key issue: Gravity works locally!!!
Please remember that the moon orbits around the earth, while the gravity force of the Sun/Moon is stronger more than double than the Earth/Moon gravity force. This by itself is enigma.
Is it?  The moon accelerates more due to the sun than it does due to Earth.  At no time does it accelerate away from the sun, even when between the two.  When it is there, the moon accelerates away from Earth, just like Newton says it should.
Do you agree that the gravity force of the Sun/Moon is much stronger than Earth/Moon?
If so, How can we explain the idea that the moon had selected to orbit around the earth instead around the Sun while the Sun/Moon gravity is much stronger?
What Newton would say about it?

The sun 'orbits' the galaxy, not the SMBH.  The latter does have an impact, but nowhere near enough to give the solar system its speed.  Apparently not even the galaxy has enough.
Speed is approximated by GM/r where M is mass of the spherical thing orbited.  Plug in the mass of the SMBH into that, and you get a figure far lower.  But the galaxy is not a spherical mass, so orbital speed actually increases with radius at some ranges.
Yes, sure.
If we ignore how spiral galaxy really work, than yes, you are fully correct.
Hence, as our scientists ignore completely the real impact of spiral arms, and as they also ignore the great impact of local mass gravity, they have found that even the galaxy can't support the orbital velocity of the Sun. But this is a severe mistake.
How can we ignore the spiral shape of our galaxy?
How can we assume that stars are getting in and out and cross the spiral arms while they orbit around the center of the galaxy?
We are missing the whole idea of spiral shape and local mass gravity.
Our Sun is not there by itself. It is a severe mistake to verify the orbital velocity of the Sun based only of its mass.
Where is the impact of the local mass? Where is the impact of all the nearby stars???
Our sun orbits around a virtual host center. This virtual host center is there due to all the nearby masses. So it is some sort of long chain of gravity forces which holds each other while they all orbit around the galaxy.
We can see it quite clearly in the following diagram of the milky way:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
Off course, it is only schematics. We don't see the bridges and branches between the arms.
But it shows the gravity force connection in each arm.
In order to get better understanding, let's look at the last point of mass at the Persues arm. This last point of mass has only the one in front it. So, It doesn't see the whole galaxy. The galaxy can't hold this point of mass in the orbital velocity by itself. (So you are fully correct in the following statement - "Apparently not even the galaxy has enough.")
Therefore, only the gravity force between this last Point of mass to the one in front, still holds it in the arm and keeps its orbital velocity.
In each point of mass there might be few thousands of stars.
Hence, each point of mass in this chain holds itself by gravity force with the one in front, while it also holds the one in the back.
Hence - Gravity Works Locally!!!
That is the Key element of spiral galaxy.
Once you understand that, you understand how spiral galaxy works.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 24/12/2018 22:33:28
Do you agree that the gravity force of the Sun/Moon is much stronger than Earth/Moon?
If so, How can we explain the idea that the moon had selected to orbit around the earth instead around the Sun while the Sun/Moon gravity is much stronger?
It orbits both of them actually.  Whether something orbits object X or not isn't just a function of the force that X puts on it.
This is part of the 3-body problem, which has no known general solution.

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Quote from: Halc
[Orbital] speed is approximated by GM/r where M is mass of the spherical thing orbited.  Plug in the mass of the SMBH into that, and you get a figure far lower.  But the galaxy is not a spherical mass, so orbital speed actually increases with radius at some ranges.
Yes, sure.
If we ignore how spiral galaxy really work, than yes, you are fully correct.
You described it the same way, with this host point orbiting the galaxy at the same speed, unexplained by the visible mass.

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Hence, as our scientists ignore completely the real impact of spiral arms, and as they also ignore the great impact of local mass gravity, they have found that even the galaxy can't support the orbital velocity of the Sun. But this is a severe mistake.
How can we ignore the spiral shape of our galaxy?
How can we assume that stars are getting in and out and cross the spiral arms while they orbit around the center of the galaxy?
We are missing the whole idea of spiral shape and local mass gravity.
Our Sun is not there by itself. It is a severe mistake to verify the orbital velocity of the Sun based only of its mass.
Where is the impact of the local mass? Where is the impact of all the nearby stars???
Nobody ignores any of that, nor do they claim that the sun follows a clean orbit about the galaxy without that wiggle around the arms.

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Our sun orbits around a virtual host center. This virtual host center is there due to all the nearby masses. So it is some sort of long chain of gravity forces which holds each other while they all orbit around the galaxy.
... at an unexplained speed still.  Your hypothesis made no attempt to explain that.

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We can see it quite clearly in the following diagram of the milky way:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
That's a nice map.  Cool.  Hadn't seen one like that before.

In order to get better understanding, let's look at the last point of mass at the Persues arm. This last point of mass has only the one in front it. So, It doesn't see the whole galaxy. The galaxy can't hold this point of mass in the orbital velocity by itself. (So you are fully correct in the following statement - "Apparently not even the galaxy has enough.")
Therefore, only the gravity force between this last Point of mass to the one in front, still holds it in the arm and keeps its orbital velocity. [/quote]If that happened, the arms would contract into chunks instead of stretch into these pinwheel arms. If the mass in front of it played a role as you describe, it would be accelerating, not going fast.  Clockwise speed (as viewed in that diagram) is due to force towards the galaxy, not towards the stuff in front of it.  The 'one in front' contributes not at all to that.  Forces to the front would push that 'last point' outward, not forward, just as forward tidal forces make the moon orbit higher, not faster.
Anybody with familiarity of orbital mechanics knows this.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/12/2018 17:26:56
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Our sun orbits around a virtual host center. This virtual host center is there due to all the nearby masses. So it is some sort of long chain of gravity forces which holds each other while they all orbit around the galaxy.
... at an unexplained speed still.  Your hypothesis made no attempt to explain that.

The explanation is quite simple and based on a normal "drifting outwards" activity in gravity.

Please look again at the following diagram:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
Please try to focus at the Green spiral arm.
If we look directly from the Sun to the SMBH, we can see that it cross Green arm at 6KPC (let's call it star B) while if we look at the other side we can see that the green Arm starts from the edge of the Bar  (let's assume that the distance from the center is 3KPC and call it star A)
So, we have two points (or stars) on the same diameter line and on the same arm.
One point is at a distance of 3KPC and the other is at 6KPC.
We know that all the stars have almost the same velocity - V.
Hence, after time T, both will cross the same distance.
S = T * V
Both of them are drifting outwards while they orbit around the galaxy.
So, let's assume that S1 Represents the distance in the green spiral arm from Star A to Star B.
Therefore, both of them will have to cross the same distance = S1.
However, while Star A set 180 degree from the Bottom point (R=3KPC) to the Top (R=6KPC) , Star B will set only about 90 degree from Top point (R=6KPC) to the left point (R=9KPC).
That is a very simple explanation why all stars in the disc orbit at the same velocity while each one stay on the same arm and why we get the unique spiral shape of the arm.
Is it clear?
Title: Re: How gravity works in spiral galaxy?
Post by: andreasva on 25/12/2018 17:32:13
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Could it be that over time there is a change in the gravity force?

This is precisely what is happening over time, in my view.

It's quite a lengthy theory, so I'll try to be brief.

The universe is infinite, not finite.  The Big Bang never happened.  We are not expanding, or accelerating.  Dark Matter is more than likely a myth.

To understand it, you have to start from 0, and work your way in.  Physics chose the middle, and is trying to work its way out simultaneously in opposite directions.  It's too complicated and confusing.  We got a lot wrong.

You also need to consider basic fundamental equality's in math, and how that applies to the universe.

First, the whole of the universe can only be described with 3 values.

0
ι1ι

Looking at 0, it is exactly what it implies, nothing.  Should the universe ever become nothing, that's all it could ever be, because 0=0.  0 is a finite value.  It does not equal anything else but 0.  It cannot spontaneously become something else.  0≠1. Clearly we are not 0, because we exist.  I would also conclude, that the potential to be 0 is infinite, because 0, or absolute 0, is theoretically impossible.  Our existence is the empirical evidence that proves conclusively that 0 has never been, and never will be the state of the universe.  Should the universe ever reach a state of 0, it would reach a perfect equilibrium, and would remain that way forever and always.

Our universe is, > 0

The next value for the universe would be 1, but in the context of the universe its value would be absolute.  The whole of the universe would be a continuous emptiness, void of all substance.  Like 0, 1 is a finite value.  1=1.  And like 0, if the universe ever achieved this state, it would remain that way forever and always.  Also like 0, it cannot spontaneously become something else.  1=1.  Once again, our existence is the empirical evidence that proved conclusively that ι1ι has never been a state of the universe.

Our universe is, < ι1ι

These are basic fundamental equality's in math.  They don't change.  If they somehow did change, e=mc^2 would be meaningless.  Math wouldn't work.

Between 0 and 1, lies an infinite number of variables.  Or more to the point, an infinitely variable condition, which makes us an infinitely variable analog state.  We are analog.  The entire universe can be described in wave theory, which is analog.  We have always been in this state, because ∞ = ∞, and cannot be anything else.

Einstein defined half of the puzzle, but he was working from 1 to C.  Quantum mechanics is working on the other half,  from C to 0.  They are inherently different, but similar.  I'm not sure the variables directly translate though, so there is some incompatibility.  The universe is also infinite, not finite, which means the constants are also infinitely variable.  There are no physical constants, just virtual constants.  We're bound to C, so it all seems pretty stable.  We're really looking at the universe from 1 to C, in a finite manner, as Einstein defined.  However, the reality is slightly different, 0<C<1, in an infinite manner.

The big bang is pseudoscience.  The big bang takes the sum total of the entire universe and compacts it into a singularity, removing physical properties of the universe in the process.  Gravity, gone, C gone.  space-time, gone.  They've created an imaginary state of energy, a singularity, with a value of 1, and then wrapped it in nothing, or 0.  The big bang essentially claims 0=1.  0 and 1 cannot coexist.  It's wrong.  Not too surprising given the original source.  It was developed with a preconceived notion of a beginning, where there is none.

The universe is founded on two basic directions, in and out.  It is a 1-dimensional trip, in a 3-dimensional universe.  We move 3-dimensionally towards 0, but it is a destination beyond reach.  Finite values are not physically possible in our universe.  We are more or less a digital representation of an underlying analog state.  No, we aren't holograms, although I can see where it may be applicable to some extent.

What's really happening in the universe is that we're moving inward, probably at a constant rate of -C, for a lack of better terminology.  Really, there's no explanation in physics for what I'm describing.  So excuse my explanations.

As we move inwards towards a smaller state, gravity weakens between distant galaxies, and they fall back.  The gap widens over time.  It looks like expansion and acceleration, but it's mass receding inward and away.

In the early part of our existence, our region of space was dominated by matter, and was very close together.  Gravity was spread out, or diversified.  Localized effects were less pronounced.  As we (galaxies) drift apart, the effect of gravity from adjacent galaxies diminishes, while local gravity increases relative to the local mass within it.  Not sure increases is the right choice of wording.  More like, it becomes more locally concentrated or focused.  That effect is in a constant rise the further we recede from other galaxies.

In a spiral galaxy, our motion is not precisely circular.  We're literally spiraling inward towards the center, but we're also receding inwards at the same time, giving the illusion on constancy.

I suspect our most recent upcoming effort to detect Dark Matter will fail.

So, given your original statement above, absolutely.  Gravity changes over time.  It gradually becomes more locally focused as the gap between galaxies widens.

Sorry for the quick explanation.  It's pretty rough, but I hope you can get the gist of it.

And as always, I could be wrong.  I doubt it though.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 25/12/2018 20:06:26
The explanation is quite simple and based on a normal "drifting outwards" activity in gravity.
I am unaware of any such normal outward drift.  I was hoping you would describe where the energy would come from to do that.

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Please look again at the following diagram:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
Please try to focus at the Green spiral arm.
If we look directly from the Sun to the SMBH, we can see that it cross Green arm at 6KPC (let's call it star B) while if we look at the other side we can see that the green Arm starts from the edge of the Bar  (let's assume that the distance from the center is 3KPC and call it star A)
So, we have two points (or stars) on the same diameter line and on the same arm.
One point is at a distance of 3KPC and the other is at 6KPC.
OK, I know where A and B are.
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We know that all the stars have almost the same velocity - V.
Sort of....   We're pretty slow actually (~217).  Things further out have been measured at 260.  S2 has quite variable speed, but has been clocked at over 30x our speed.  So we don't 'know' that all stars have almost the same speed.  Clearly they don't.
Still, on average, this is more true than Kepler's 3rd law would predict: That things 4 times as far out will move at half the speed.  That law only works for single primary masses like our sun, and not distributed masses like the disk that is our galaxy.

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Hence, after time T, both will cross the same distance.
If they're moving at the same speed, yes.
If A is going at the same speed as B, it will go around twice for every lap made by B since it follows a path of half the length.

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Both of them are drifting outwards while they orbit around the galaxy.
So you assert, but nothing you've said supports that.  It would require energy, and you need to explain the source of that energy.

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So, let's assume that S1 Represents the distance in the green spiral arm from Star A to Star B
They are separated by about 9 KPC, linearly.  Along the curved green line of that arm, it is more like 14KPC.  So S1 is currently about 14KPS.
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Therefore, both of them will have to cross the same distance = S1.
This makes no sense.  The arms are not stationary, and the stars don't travel along the arms.  Both A and B take essentially circular paths around the galaxy.  If they move at the same speed, then A takes half the time to go around since it has a shorter path.  Nothing traverses the green path from A to B or from B to A.

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However, while Star A set 180 degree from the Bottom point (R=3KPC) to the Top (R=6KPC) , Star B will set only about 90 degree from Top point (R=6KPC) to the left point (R=9KPC).
I think you have A and B moving outward, but no force has been identified that would justify that assertion.  'A' ends up at top at 3KPC and in the same time, B ends up on the right point at 6KPC, assuming they move at the same speed.  That would have the effect of stretching the green line.  Things move clockwise in that picture remember.  You seem to be under the impression that things go the other way.

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That is a very simple explanation why all stars in the disc orbit at the same velocity while each one stay on the same arm and why we get the unique spiral shape of the arm.
Is it clear?
Clear, but quite wrong.  No energy to explain the outward motion, and you have everything moving backwards.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/12/2018 04:07:56
I am unaware of any such normal outward drift.  I was hoping you would describe where the energy would come from to do that.
There is no need for extra energy in order to set the drifting outwards activity.
This is a normal outcome of "Gravity Friction".
We know that there is a friction at almost any activity.
Somehow, we assume that there is no friction in gravity.
You call it "Tidal friction" and I call it "Gravity friction".
You think that there are moons that drifts inwards and I claim that all the moons drifts outwards due to Gravity friction.

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Both of them are drifting outwards while they orbit around the galaxy.
So you assert, but nothing you've said supports that.  It would require energy, and you need to explain the source of that energy.
As I have stated, all the stars drifts outwards due to gravity friction. So, they actually losing energy instead of gaining energy.
Let's set a simple calculation:
F = G M m / r^2
It is clear that as we increase the radius we decrease the gravity force and vice versa.
I can promise you that all the moons which you think that are drifting inwards - all of them are drifting outwards!!!
In any case, as we can't prove this issue, there is no need to argue about it.
We have agreed that we do not agree on this issue.

I think you have A and B moving outward, but no force has been identified that would justify that assertion.  'A' ends up at top at 3KPC and in the same time, B ends up on the right point at 6KPC, assuming they move at the same speed.  That would have the effect of stretching the green line.  Things move clockwise in that picture remember.  You seem to be under the impression that things go the other way.
Yes, you are absolutely correct.
I have tried to explain the basic idea of velocity adjustment while we freeze the spiral stracture.
In reality it is much more complex.
There are several elements to consider.
1. The location of the star. Is it in the Arm, in the ring, in the bulge, in the bar or outside the disc.
Currently, we only focus on the spiral arm. Starting from the ring to the last edge of the arm – (in the galactic disc). Therefore S2 isn't relevant to our current discussion. I will discuss about it later on.
2. The orbital velocity of the virtual host point of the star (at the arm) - V1
So, yes, all the stars orbit in clockwise direction (in the diagram). However, in order to keep a similar velocity at any distance (we will discuss later on about the differences in the velocities), they actually move outwards and backwards in the spiral arm. This is very critical point and I'm not sure that my explanation is fully clear.
So, the whole idea is that as a star is located outwards from the ring, we would expect it to have a faster orbital velocity (if it was a rigid disc), However, as a star drifts outwards it also set a backwards movement. This movement keeps the star in the arm while it orbits at a similar velocity as a star that is located closer to the center.
Please think about it and let me know if the point is clear to you.
3. The velocity of the star around its host point - V2
A star which is located close to the center has a shorter radius to its virtual host point. This radius increases over time due to gravity friction. Therefore, the orbital velocity of star around its virtual host point is faster when a star is located (in the arm) inwards to the galactic center and of course it is slower when a star is located further away from the center.
4. Measured orbital velocity.
The Orbital velocity of the virtual host point (V1)+ its orbital velocity around its host point (V2) sets the total orbital velocity of a star. so, if  V1 and V2 are in the same direction, the total orbital speed that we might see is V1 + v2. However, When they move in the opposite directions we see V1 - V2.
Therefore, when we measure the orbital velocity of stars we might find some variations in the orbital velocities even at the same radius although their real Virtual host point velocity is identical.
In the same token the Virtual host point of all our nearby stars might have exactly the same V1 orbital velocity. However, due to V2 we see them moving in different directions

Sort of....   We're pretty slow actually (~217).  Things further out have been measured at 260.
Yes, that is very normal.
At the edge of the arm (remember the last green point of mass at the green arm), the orbital velocity gets to its maximal pick. The Gravity force can't hold it for long time. Therefore, after some time this point of mass has to be disconnected from the arm and drifts outwards from the galactic disc. (I will discuss about that activity later on)

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 26/12/2018 17:35:13
There is no need for extra energy in order to set the drifting outwards activity.
Then you violate Newton's conservation laws.  You are asserting magic.  An object further away from a gravity well has more energy than one deeper in.  There is no normal drift to a higher energy state.  A force is required to do it, and you've not identified that force.
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This is a normal outcome of "Gravity Friction".
We know that there is a friction at almost any activity.
There is no such thing as gravitational friction.  There is physical friction, with two objects rubbing, changing kinetic energy into heat.  There is also EM friction like how they stop railroad trains, which doesn't involve physical contact, which would wear out the parts in perhaps one day.  The result is the same: heat, and loss of kinetic energy.  You've described an increase of energy that comes from nowhere by magic.  Friction creates heat.  Tidal friction is between water and the ocean floor, and it results in heating of the water.  There is no friction between the moon and Earth since the two are separated by space.

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Somehow, we assume that there is no friction in gravity.
High gravitational acceleration of massive objects radiates energy away in the form of gravitational waves.  That is a form of gravitational friction.  The Earth going around the sun loses energy at a rate of I think ~200 watts in this manner.  The result of this tiny friction causes an immeasurable loss of orbital radius, not an increase.  So yes, there is gravitational friction of a sort, but it has the effect of reducing radius, not increasing it.

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You call it "Tidal friction" and I call it "Gravity friction".
Tidal friction slows the spin of Earth.  It is between water and the ocean floor, and all that friction energy is lost to heat.  The energy/momentum transferred to the moon is not from friction.  The energy comes from the angular energy of Earth.  You've identified no such source of energy that accounts for stars gaining potential energy by moving out of the gravitational well.

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You think that there are moons that drifts inwards and I claim that all the moons drifts outwards due to Gravity friction.
Yes, you claim that, despite the violation of Newton's laws.  Physics doesn't work by making self-contradictory claims, especially ones that make assertions contrary to empirical observation.

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As I have stated, all the stars drifts outwards due to gravity friction. So, they actually losing energy instead of gaining energy.
Let's set a simple calculation:
F = G M m / r^2
It is clear that as we increase the radius we decrease the gravity force and vice versa.
You seem to be equating force with energy.  Do you know the difference?

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I can promise you that all the moons which you think that are drifting inwards - all of them are drifting outwards!!!
In any case, as we can't prove this issue, there is no need to argue about it.
The people that measure the opposite effect will appreciate your promise.  Apparently their very real measurements to the contrary don't count as proof that your assertion is quite wrong.

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Yes, you are absolutely correct.
I have tried to explain the basic idea of velocity adjustment while we freeze the spiral stracture.
In reality it is much more complex.
There are several elements to consider.
1. The location of the star. Is it in the Arm, in the ring, in the bulge, in the bar or outside the disc.
Currently, we only focus on the spiral arm. Starting from the ring to the last edge of the arm – (in the galactic disc). Therefore S2 isn't relevant to our current discussion. I will discuss about it later on.
2. The orbital velocity of the virtual host point of the star (at the arm) - V1
So, yes, all the stars orbit in clockwise direction (in the diagram). However, in order to keep a similar velocity at any distance (we will discuss later on about the differences in the velocities), they actually move outwards and backwards in the spiral arm. This is very critical point and I'm not sure that my explanation is fully clear.
That they do, as does Earth in its orbit, but the net effect is an average radius that doesn't change absent any force applying a net forward force which would account for the increase of energy.  That force would need a reaction force decreasing the energy of something else, per Newton's 3rd law.  If you don't identify that equal and opposite reaction, then your drift assertion violates Newtonian law.  This host point adds zero net force since it is sometimes in one direction, and sometimes in the opposite direction.  That adds up to a wiggle with no net change in speed or radius.

You continuously assert force by magic, without reaction.  Newton would not agree with that.

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So, the whole idea is that as a star is located outwards from the ring, we would expect it to have a faster orbital velocity (if it was a rigid disc), However, as a star drifts outwards it also set a backwards movement. This movement keeps the star in the arm while it orbits at a similar velocity as a star that is located closer to the center.
Please think about it and let me know if the point is clear to you.
The disk of the galaxy is anything but rigid.  All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them.  This is not the case of a rigid object, which would be able to spin any any number of different angular rates and still keep its components at some fixed radius.
So I don't really care how stars would move if the galaxy was a rigid disk, since it isn't one.  No, I don't really get what you're trying to convey above.  If a star stays near its arm, it is because the nearby mass of the arm attracts it.

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3. The velocity of the star around its host point - V2
A star which is located close to the center has a shorter radius to its virtual host point. This radius increases over time due to gravity friction. Therefore, the orbital velocity of star around its virtual host point is faster when a star is located (in the arm) inwards to the galactic center and of course it is slower when a star is located further away from the center.
Assuming there is such a host point, you discard Newtonian mechanics with the drifting assertion, and then assert that both gravitational potential energy (from moving away from the gravity source) and kinetic energy (from asserted greater orbit velocity out there) appear as from nowhere, totally violating energy conservation laws.  Oh yes, and even more energy from heat from 'gravitational friction'.  My, but you do pull a lot of energy from nowhere.

Let's see, you also assert that orbital velocity about the host point is a function of the distance of that star or host point from the galactic center, which makes no sense.  The galaxy affects speed around the galaxy, not speed around the host point.  The speed around the host point has no obvious formula since you've not supplied one.  Kepler's laws are of no use here since you're not describing a Kepler orbit.

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4. Measured orbital velocity.
The Orbital velocity of the virtual host point (V1)+ its orbital velocity around its host point (V2) sets the total orbital velocity of a star.
We have a second host point now?  That's fine, but you just now introduced this.  Before you said that the one host point orbited the galaxy as your picture depicted.

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so, if  V1 and V2 are in the same direction, the total orbital speed that we might see is V1 + v2. However, When they move in the opposite directions we see V1 - V2.
V usually refers to velocity, not speed, in which case it is always V1 + V2.  This simplifies the math considerably.

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Therefore, when we measure the orbital velocity of stars we might find some variations in the orbital velocities even at the same radius although their real Virtual host point velocity is identical.
Right.

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Quote from: Halc
Sort of....   We're pretty slow actually (~217).  Things further out have been measured at 260.
Yes, that is very normal.
At the edge of the arm (remember the last green point of mass at the green arm), the orbital velocity gets to its maximal pick.
The first one near the core also moves much faster, even than the 260 figure.  I don't have numbers since I don't know specific objects.  The 217 figure for our speed is just the length of a circle of 10KPS radius divided by the ~200 million years it takes for us to go around one lap.  From this speed the acceleration can be computed needed to keep our host point on that circular path, and from that acceleration the mass required to account for that acceleration can be computed.  Your host point idea doesn't change that acceleration at all, so you (in denial of additional mass) have failed to explain that acceleration.  Nothing you propose explains the force necessary for the observed acceleration.  All I see is assertions that are inconsistent with laws that have been known for many centuries.

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The Gravity force can't hold it for long time. Therefore, after some time this point of mass has to be disconnected from the arm and drifts outwards from the galactic disc. (I will discuss about that activity later on)
If the arm was moving too fast, it would not be at the radius from the galaxy it is now.  We would already be further out.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/12/2018 21:25:52
All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them.
If that is the case, than how could it be that we get the spiral shape from that random orbital activity?
How many spiral shape you can get due to this random orbital activity? Is it 0.000...0001 preset or 0.0..1 present?
Do you know that: "Spiral galaxies make up roughly 72 percent of the galaxies that scientists have observed, according to a 2010 Hubble Space Telescope survey"
https://www.space.com/22382-spiral-galaxy.html
So, do you really believe that if "All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them", they can set a spiral shape in 72 percent of the galaxies that scientists have observed???
If we see that 72% of the galaxies are spiral, don't you think that it couldn't be due to random phenomenon?
So how can we explain that incredible number of spiral galaxies?
How can we ignore the impact of the whole spiral galaxy shape?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 26/12/2018 23:08:05
Quote from: Halc
All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them.
If that is the case, than how could it be that we get the spiral shape from that random orbital activity?
How many spiral shape you can get due to this random orbital activity?
I never suggested it was random.  Everything moves as the result of net forces on them, exactly as Newton says it should.  Random movement would violate that.  Mass attracts other mass, so spinning material apparently tends to gather into arm.  Perhaps our solar system formed arms before each arm stabilized into rings and then planets.  I don't know.  I'm not an expert.  The spiral arms seem to be a fairly stable structure since many (but not all) semi-large galaxies of similar age seem to have them.

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So, do you really believe that if "All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them", they can set a spiral shape in 72 percent of the galaxies that scientists have observed???
Yes, exactly.  The spirals are perhaps not stable in the long run (since they obviously must stretch out if the inside material goes around more times than the outside stuff).  Maybe pairs of stretched arms combine into one when they get thin enough.  Our own arm seems a weak remnant of such an old arm.  Perhaps the arms form as the result of a relatively recent cannibalism of a nearby galaxy.  As I said, I'm no expert, so am just suggesting possible ways that such arms seem stable enough to be prevalent.

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If we see that 72% of the galaxies are spiral, don't you think that it couldn't be due to random phenomenon?
Clearly not.  Probably more due to gravitational effects, not random ones.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/12/2018 05:56:28
Dear Halc

In one hand you claim:
Mass attracts other mass, so spinning material apparently tends to gather into arm.
However, this is exactly what I say. Mass attracts other mass.
So, mass gravity works locally. Therefore, the mass in the Spiral arms attracts all the stars in the Arm.
Therefore, do you agree that all the stars in the Arm are directly affected by the gravity force of the Arm?

But you have stated:
All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them.
How could it be that in one hand you claim that - Mass attracts other mass in spiral arms, while on the other hand you claim - All objects are disconnected?
Do you agree that there is a contradiction?

Can you please explain how the following Milky way spiral galaxy had been formed?
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
1. Bulge:
Why it has a ball shape? Why there is no disc shape in the Bulge
2. Bar:
What is the source for that bar shape?
3. Ring
What is the resone for the Ring shape? Why it has a disc Shape instead of ball shape as we see in the Bulge?
What is the gravity impact of the Ring? Why the ring is so narrow? Why the ring doesn't continue all the way to the far end of the galaxy?
4. Spiral arm
How could it be that after the ring we see spiral arm?
Please explain why the spiral arms had been set in a flat disc and why it has this unique shape?
5. End of the Disc shape.
Why at some point, the spiral arm shape is ended? Why it doesn't last longer?
How could it be that at some point the Dark matter can't hold the stars in the disc?
Why after the far end edge of the spiral arms we see stars far above and below the galactic disc?

Please look at the following orbital velocity:
http://lempel.pagesperso-orange.fr/courbe_rot_voie-lactee.jpg
http://lempel.pagesperso-orange.fr/matiere_noire_pas_si_noire_uk.htm
How could it be that our scientists don't try to match between the spiral structure to the measured velocity???
Can you please explain why we see that measured velocity at each segment of the galaxy?

Perhaps our solar system formed arms before each arm stabilized into rings and then planets.  I don't know.  I'm not an expert.  The spiral arms seem to be a fairly stable structure since many (but not all) semi-large galaxies of similar age seem to have them.

If our scientists don't know the answers for some questions, why do you take it for granted that no one can find the real answer for Spiral galaxy enigma?
Instead of focusing on the negative aspects, why don't you help me to develop the positive ideas about the galaxy?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/12/2018 17:54:14
Mass attracts other mass.
So, mass gravity works locally. Therefore, the mass in the Spiral arms attracts all the stars in the Arm.
Therefore, do you agree that all the stars in the Arm are directly affected by the gravity force of the Arm?
Sure, but not just the mass of the arm.  All stars are directly affected by all mass.  The nearby portions of our own arm certainly have more effect than the distant parts, and the nearby parts of other arms have more effect than distant parts of our own arm.  Acceleration of the sun or a soap bubble in space can be given by a = ∑GM/r² for every object everywhere, which means the sum of acceleration due to each and every object everywhere, not just the objects in our own galactic arm, which is a pretty pathetic arm actually, not one of the main ones.

I agree to most of what you say, but gravity does not work locally.  I cannot think of any object however non-local that has mass but does not exert any gravitational influence on me.

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But you have stated:
All objects are disconnected and thus move and accelerate exactly according to the net force vectors acting on each of them.
How could it be that in one hand you claim that - Mass attracts other mass in spiral arms, while on the other hand you claim - All objects are disconnected?
Disconnected as in not attached to each other via steel cables or other means by which their trajectories might be different than what gravitational forces predict.  Two rocks attached with a string trace a different path than the same two disconnected rocks that are influenced only via their mutual gravity.

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Can you please explain how the following Milky way spiral galaxy had been formed?
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
The site would explain it better than me reading the site and deciding how much of it is known and how much is conjecture.  The link is a picture of the current configuration, and gives zero information as to the formation of that configuration.

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1. Bulge:
Why it has a ball shape? Why there is no disc shape in the Bulge
I think the bulge is the halo, which is mostly made of objects that have been displaced from their normal paths by close encounters with significant gravitational sources, thus putting them on random trajectories that take them out of the plane of the disk.
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2. Bar:
What is the source for that bar shape?
No clue.  I didn't read that far.  I think you need to go further than the wiki summary to get a good explanation of the current thinking behind that.
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3. Ring
What is the resone for the Ring shape? Why it has a disc Shape instead of ball shape as we see in the Bulge?
Again, I don't claim to have better answers than what you see on the site.  It is not a ball because it is not displaced, but normal matter that has been pulled into disk shape as does any cloud of material that rotates and condenses via mutual gravitational attraction.  A galaxy in many ways is formed similar to a solar system, but in massive slow motion, and thus far greater local gravitational effects.  Unlike a typical solar system, there is no dominant mass that cleans up the orbit of everything else.  A SMBH that is on the order of 1% of the total mass hardly seems up to the task.

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What is the gravity impact of the Ring? Why the ring is so narrow? Why the ring doesn't continue all the way to the far end of the galaxy?
The impact to us is negligible as is all mass further out and not nearby.  It has impact to any material more distant than the ring since it is part of the mass around which that material orbits.
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4. Spiral arm
How could it be that after the ring we see spiral arm?
The picture doesn't show the ring.  The site mentions it, and puts it around 50 KPCS, and the furthest arm in the picture seems to be about a third that distant, so I don't see an arm beyond the ring.
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Please explain why the spiral arms had been set in a flat disc and why it has this unique shape?
Any rotating collection of loose material under mutual attraction is going to flatten into a disk, just like any gas cloud that collects into a solar system.  Interaction friction slows much of motion outside of the plane of rotation, but does not decrease the angular momentum of the collection.

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5. End of the Disc shape.
Why at some point, the spiral arm shape is ended? Why it doesn't last longer?
The arms must be stretched as the inside points go around more often than the outside ones, so it has to get homogeneous after a while, but then again, arms can merge with larger neighbors as ours is doing, so maybe that explains how the arms persist even after all the laps we've taken.  I don't claim to know.

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How could it be that at some point the Dark matter can't hold the stars in the disc?
Who claims that?

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Why after the far end edge of the spiral arms we see stars far above and below the galactic disc?
Who knows?  New gas probably gets pulled in by the galaxy, so surely there is material out there to form stars.  A lot of the answer would be told by the trajectory of these outlier stars.  Stuff further out has had far less forces that tend to pull the material into a disk, just like our solar system's outer material like the Oort cloud is not arranged as a disk.

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Please look at the following orbital velocity:
http://lempel.pagesperso-orange.fr/courbe_rot_voie-lactee.jpg
http://lempel.pagesperso-orange.fr/matiere_noire_pas_si_noire_uk.htm
How could it be that our scientists don't try to match between the spiral structure to the measured velocity???
Don't know what you mean here.  Nothing on that chart is labeled 'spiral structure'.

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Can you please explain why we see that measured velocity at each segment of the galaxy?
Don't know what it depicts.  It shows the disk at this radius moving at about 160 km/s, quite a bit slower than the 217 figure quoted in earlier places you've linked.  It shows apparently stars moving slower than free gas, possibly because a lot of the gas is higher speed due to being ejected in explosions?  Just a guess there.

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If our scientists don't know the answers for some questions, why do you take it for granted that no one can find the real answer for Spiral galaxy enigma?
Instead of focusing on the negative aspects, why don't you help me to develop the positive ideas about the galaxy?
You've not presented any alternative explanations.   You just assert things that are in contradiction with the observed accelerations.  The scientists at least have theories that explain the observed accelerations, however distasteful you find those theories.  There has been no mathematics to back the sort of things you've been asserting, so it is not even wrong.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/12/2018 18:26:34
Please look at the following orbital velocity:
http://lempel.pagesperso-orange.fr/matiere_noire_pas_si_noire_uk.htm
That chart is poorly described.  The vertical axis is speed, not mass, yet the text says:
"This suggests the existence of an invisible heavy matter, the black matter, distributed in a halo surrounding galaxies, and representative more than 90% of the total mass of the galaxy. The graph opposite shows that."
Well the graph doesn't show mass at all, just speed.
It appears that it doesn't actually show measured speed of gas, stars, 'bulb' and dark matter, but rather speed of any due to the mass of those various things.  So actual speed is that unlabeled line at the top (217 for us), and that line is the sum of the speed effects of the various components identified in the lines below.

If you take away the dark matter, the sum of the remaining parts don't add up to the line at the top, and the observed speed is incompatible with the known mass.  This is why your assertions don't work.  You're not replacing the effect of the dark matter with something else.  These host-points of yours have no effect since speed around the galaxy of a host point is no different than the speed of any other object like a dead mouse or something.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/12/2018 15:29:11
Quote from: David
Mass attracts other mass.
So, mass gravity works locally. Therefore, the mass in the Spiral arms attracts all the stars in the Arm.
Therefore, do you agree that all the stars in the Arm are directly affected by the gravity force of the Arm?
Sure, but not just the mass of the arm.  All stars are directly affected by all mass.  The nearby portions of our own arm certainly have more effect than the distant parts, and the nearby parts of other arms have more effect than distant parts of our own arm.  Acceleration of the sun or a soap bubble in space can be given by a = ∑GM/r² for every object everywhere, which means the sum of acceleration due to each and every object everywhere, not just the objects in our own galactic arm, which is a pretty pathetic arm actually, not one of the main ones.

I agree to most of what you say, but gravity does not work locally.  I cannot think of any object however non-local that has mass but does not exert any gravitational influence on me.
Please look at the following article:
https://www.space.com/19915-milky-way-galaxy.html
"Spiral arms are like traffic jams in that the gas and stars crowd together and move more slowly in the arms. As material passes through the dense spiral arms, it is compressed and this triggers more star formation," said Camargo.

So, our scientists belive that stars don't stay at spiral Arms but they just move slower at the arms.
Therefore, all the stars actually cross the arms while they orbit around the galaxy.

Based on this idea, the Sun has to cross all the arms several times - as the orbital velocity of the Sun is faster than the orbital velocity of the Arm.
However, in the arm, the orbital velocity is lower than outside the arm.
So, based on this unrealistic idea, once we are outside the arm, we should speed up.

It is stated:
"The solar system travels at an average speed of 515,000 mph (828,000 km/h). Even at this rapid speed, the solar system would take about 230 million years to travel all the way around the Milky Way."
So, what is the expected velocity outside the arm?
How could it be that we get a spiral shape due to this explanation?
Do you think that Newton would accept the idea of several orbital velocities changing per one cycle?
How can we agree with this totally unrealistic idea about spiral arms and changing orbital velocities?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 30/12/2018 18:02:04
Please look at the following article:
https://www.space.com/19915-milky-way-galaxy.html
"Spiral arms are like traffic jams in that the gas and stars crowd together and move more slowly in the arms. As material passes through the dense spiral arms, it is compressed and this triggers more star formation," said Camargo.

So, our scientists belive that stars don't stay at spiral Arms but they just move slower at the arms.
Therefore, all the stars actually cross the arms while they orbit around the galaxy.
The article didn't say that.  It explains that the stuff moving through the arms is slowed by friction, and thus tends to more or less stay in the arms.  That seems to be why there are arms and not a homogeneous fog without structure.
Nothing in that article suggests that all stars do a certain thing.
The article says that stars and gas are slowed (by friction) to a lower speed relative to the arm, not relative to the galaxy.  The whole galaxy moves at about 217 km/sec at this radius, and things moving at a different speed tend to match the local speed due to collisions with the thick material.

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Based on this idea, the Sun has to cross all the arms several times - as the orbital velocity of the Sun is faster than the orbital velocity of the Arm.
If the sun orbital velocity around the galaxy was faster than the arm, the sun would move permanently away from the galaxy and the arm, moving perhaps closer to the next arm out.  But both are going at about 217 km/sec around the galaxy.  Yes, the sun is not centered on its arm, and thus moves in a sort of orbit around that mass, but that doesn't make the sun faster around the galaxy than the arm.

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However, in the arm, the orbital velocity is lower than outside the arm.
So, based on this unrealistic idea, once we are outside the arm, we should speed up.
You seem to be making up nonsense facts.  If you move away from mass, your speed goes down, just like the speed of a ball slows if you throw it upward.  The idea is indeed unrealistic.  The article suggests no such thing.

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It is stated:
"The solar system travels at an average speed of 515,000 mph (828,000 km/h). Even at this rapid speed, the solar system would take about 230 million years to travel all the way around the Milky Way."
So, what is the expected velocity outside the arm?
That is orbital speed around the galaxy, and you showed a graph of that vs radius.  The galactic orbital speed outside the arm is almost constant nearby, so it will be the same outside of the arm.
The first quote you give is speed relative to the arm material (for material passing through it say from the side).  That relative speed would be slowed due to the friction.

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Do you think that Newton would accept the idea of several orbital velocities changing per one cycle?
If you orbit 6 things, you have 6 orbital speeds, any of which can change due to changes in distance or friction forces or something.  The moon orbits Earth, the sun, the arm, the galaxy, and probably other things, all with different speeds and radii.  Most of those speeds are fairly constant over the period of one cycle, but not completely.

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How can we agree with this totally unrealistic idea about spiral arms and changing orbital velocities?
I didn't see any suggestion of changing orbital velocities of galactic arms.  The speeds are quite stable.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/12/2018 19:07:54
Quote from: David
Please look at the following article:
https://www.space.com/19915-milky-way-galaxy.html
"Spiral arms are like traffic jams in that the gas and stars crowd together and move more slowly in the arms. As material passes through the dense spiral arms, it is compressed and this triggers more star formation," said Camargo.

So, our scientists belive that stars don't stay at spiral Arms but they just move slower at the arms.
Therefore, all the stars actually cross the arms while they orbit around the galaxy.
The article didn't say that.  It explains that the stuff moving through the arms is slowed by friction, and thus tends to more or less stay in the arms.  That seems to be why there are arms and not a homogeneous fog without structure.
Nothing in that article suggests that all stars do a certain thing.
The article says that stars and gas are slowed (by friction) to a lower speed relative to the arm, not relative to the galaxy.  The whole galaxy moves at about 217 km/sec at this radius, and things moving at a different speed tend to match the local speed due to collisions with the thick material.
Well, it is stated: "As material passes through the dense spiral arms"
How could they stay at the arm if they passes through?
What do you understand from the idea of "passes through"?
In any case, if stars stay at the Arms (all the time from day one), than why is it?
Does it mean that they hold each other by gravity force?
If so, that exactly the whole idea which I have presented.
If they don't hold each other by gravity force, why they stay at the Arm?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 30/12/2018 19:40:08
In any case, if stars stay at the Arms (all the time from day one), than why is it?
Does it mean that they hold each other by gravity force?
If so, that exactly the whole idea which I have presented.
I never disagreed with that.  Yes, gravity (and friction) seems to be what holds them together, just like those two effects are what hold a solar system together.

You've expressed denial of dark matter, in which case the acceleration of any material in the galaxy (the arms in particular) is unexplained by your assertions.   Without dark matter, there is simply not enough gravitational force generated by the remaining matter at the observed distances to explain the acceleration needed to maintain a 217 km/sec orbital speed around the galaxy that is observed for all material in any galactic arm at this radius.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/12/2018 05:19:09
Quote from: Dave
In any case, if stars stay at the Arms (all the time from day one), than why is it?
Does it mean that they hold each other by gravity force?
If so, that exactly the whole idea which I have presented.
I never disagreed with that.  Yes, gravity (and friction) seems to be what holds them together, just like those two effects are what hold a solar system together.
So, you agree that the stars hold each other in the arm due to gravity (or friction if you like)
Therefore, do you agree that in this case each star must obey to the local gravity force in the Arm?
It is similar to the Earth/sun gravity force.
The Earth doesn't care about the gravity forces on the Sun. It is just holds/orbits the Sun by gravity force and follows it where ever it goes.
So, as we do not set the calculation why the Earth orbits around the galaxy, we also shouldn't worry why the sun orbits around the galaxy.
The Sun holds itself in the spiral arm.
Where ever the spiral arms goes - the Sun goes.
So, we have to find why/how the spiral arm orbits the galaxy and not why the Sun orbits the galaxy.
Only if the Sun is disconnected from the arm, than we must look for dark matter.
As long as is connected in the arm due to gravity, we must look for answers about the arm
We might find that Dark matter is needed for the Arm, but there is no need to find an explanation for the Sun - as long as it is connected to the arm by gravity.
In the same token - as long as the Earth is connected to the Sun by gravity force, we do not need to find an explanation why the Earth orbits around the galaxy.

As I have stated – Gravity works locally:
Another example -
An asteroid orbits around the Moon. Hence, the moon holds this asteroid by gravity.
The moon orbits around the Earth. Therefore, the Earth holds the Moon which holds the Asteroid by gravity.
So, we can continue at higher hierarchy. But at the end we might find that the asteroid orbits the Galaxy.
However, as long as the asteroid is connected by gravity to the moon, we do not try to find a direct explanation why it orbits the galaxy at supper high orbital velocity (with reference to the galaxy)
In the same token, as long as the Sun is connected to the arm by gravity force, we do not need to find a direct explanation why it orbits the galaxy.
As I have already explained - The sun orbits around a local virtual host point. Hence, we can say that the Sun holds itself to this virtual host point by gravity force (Local gravity force).
However, this virtual Sun' host point holds itself in the spiral arm by gravity force.
Therefore, at this phase, we must find the answer for – "Why the Arm orbits around the galaxy (and not why the sun orbits around the galaxy)"?
Do you agree with that?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 31/12/2018 06:42:41
Quote from: Halc
Yes, gravity (and friction) seems to be what holds them together, just like those two effects are what hold a solar system together.
So, you agree that the stars hold each other in the arm due to gravity (or friction if you like)
Not 'or friction'.  It takes both.  Without the latter, something not part of the arm may be accelerated by the gravitational pull of a nearby arm, and that thing will thus have greater than escape velocity of the arm.  It will pass through and not come back.  Friction changes that.  It bumps into stuff and slows relative to the material it hits, as described in that recent link you posted.  This keeps it from leaving, and makes it part of the arm.  Dark matter (WIMPs) in particular have no friction, and are thus not particularly bound to the arms.

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Therefore, do you agree that in this case each star must obey to the local gravity force in the Arm?
You keep asking this.  Each star's velocity and acceleration is determined by the sum of the force vectors acting upon it.  This is all things, not just local ones.
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It is similar to the Earth/sun gravity force.
Yes.  Earth is pulled by all things, not just local ones.

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The Earth doesn't care about the gravity forces on the Sun. It is just holds/orbits the Sun by gravity force and follows it where ever it goes.
Not necessarily.  It only works predictably if there are no other objects that exert gravitational force on Earth, and that is not true.  Yes, for the most part, we can reasonably assume Earth will follow the solar system.

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So, as we do not set the calculation why the Earth orbits around the galaxy, we also shouldn't worry why the sun orbits around the galaxy.
The Sun holds itself in the spiral arm.
Where ever the spiral arms goes - the Sun goes.
The arm is not an object like the sun is to Earth.  The arm is a smear of matter, all moving at different orbits about the galaxy.  Some of it goes around in much less time than parts further out.  It isn't one thing that the sun follows.

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So, we have to find why/how the spiral arm orbits the galaxy and not why the Sun orbits the galaxy.
Sure.  That's the part you cannot explain.  The arm (all parts of it) has acceleration that is incompatible with Newton's laws and your assertions.  Either your assertions are wrong, or Newton is.

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Only if the Sun is disconnected from the arm, than we must look for dark matter.
Why?  The sun has the same path as all the rest of the arm at this radius.

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As long as is connected in the arm due to gravity, we must look for answers about the arm
We might find that Dark matter is needed for the Arm, but there is no need to find an explanation for the Sun - as long as it is connected to the arm by gravity.
Nobody claims that dark matter explains the sun's attraction to the arm.  Dark matter is not concentrated in the arms, and so has little to do with the cohesive properties of the arms.

In the same token - as long as the Earth is connected to the Sun by gravity force, we do not need to find an explanation why the Earth orbits around the galaxy.

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An asteroid orbits around the Moon. Hence, the moon holds this asteroid by gravity.
The moon orbits around the Earth. Therefore, the Earth holds the Moon which holds the Asteroid by gravity.
So, we can continue at higher hierarchy.
This is fine, but if the sun was not there, you still would have to explain why the Earth goes around the nothing where the sun is supposed to be.  You're right, you don't need to discuss the asteroid going around the moon.  You can treat the whole thing as a unit.  Doing so does not explain the acceleration of the collection.

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Therefore, at this phase, we must find the answer for – "Why the Arm orbits around the galaxy (and not why the sun orbits around the galaxy)"?
Do you agree with that?
With minor edits, sure.  We must ask why the local portion of the arm orbits around the galaxy at the speed it does. Not why it goes around, but why so fast? It accelerates at about 2e-10 m/sec² and the mass of the matter we see in the galaxy is not enough to explain that acceleration.

We know why the arm orbits:  Things just do when there is a mass to orbit, such as the material that makes up the inner portion (stuff closer than us) of the galaxy.  But orbital speed is very much a function of the mass of that material, and there apparently isn't enough of it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 31/12/2018 14:05:07
You keep asking this.  Each star's velocity and acceleration is determined by the sum of the force vectors acting upon it.  This is all things, not just local ones.

Yes and No.
Yes - based on theoretically point of view.
No - based on real evidences.
The Gravity force of the Sun/Moon is stronger by more than a twice with regards to the Earth/Moon.
If the Moon had to chose its orbital path based on the sum of the force vectors acting upon it, it will had to orbit around the Sun instead around the Earth, however - this isn't the case.
So, even if the Sun/Moon gravity is stronger than Earth/moon gravity, the moon prefers to orbit around a local host (Earth).
Therefore, as I have stated, Gravity works locally.
In the same token - we can say about the Earth.
Our scientists claim that  more than 100 Billion Sun mass is needed to be at the orbital center of the Sun (in the galaxy) in order to hold the Sun in its orbital path.
If we try to calculate the Galaxy/Earth gravity force (based on that mass), we should find that it is much higher than Sun/Earth gravity force.
So, Based on a pure sum of the force vectors acting upon the earth, it had to ignore the Sun and start running around the galaxy. But again - this isn't the case.
Hence, as the Earth had selected a local mass (our Sun) instead of the galaxy, and as the Moon had selected the Earth instead of the Sun; we have to claim the following:
Each object's velocity and acceleration is mainly determined by the local mass and not by the sum of the force vectors acting upon it.
The sum of the the force vectors could set a tidal. No more than that (Assuming that there is no collision)!!!

The arm is not an object like the sun is to Earth.  The arm is a smear of matter, all moving at different orbits about the galaxy.  Some of it goes around in much less time than parts further out.  It isn't one thing that the sun follows.
We have already discussed this issue.
Each star orbits around a local virtual host point. Therefore, although we see that all the stars are moving in different directions, their host points are very stable with regards to each other.
Hence, The Arm Is an ARM. It holds all the Virtual host points at the arm due to local gravity force.
This is the key element of spiral arm.
If the arm can't hold the stars, than by definition the stars must cross the arm and get out of it.
You can't just say that the stars stay at the arm but the arm doesn't hold them by gravity.
This is none realistic statement.
In the same token - as long as the Earth is connected to the Sun by gravity force, we do not need to find an explanation why the Earth orbits around the galaxy.
Sorry - this is incorrect. We can't take it for granted. We need to find an explanation why the Earth orbit around the Sun although its gravity force is lower than the Gravity force of the galaxy.
I will give you a tip - Gravity Works locally.
This is the ultimate answer for:
Why the moon holds/orbits around the Earth instead of the Sun?
Why the Earth Holds/orbits around the Sun instead of the galaxy?
Why the Sun orbits around a virtual host point that holds itself in the arm due to Local gravity force?
If you still don't want to accept the great impact of "local gravity force" than try to explain why the moon orbits around the earth instead of the Sun.
With minor edits, sure.  We must ask why the local portion of the arm orbits around the galaxy at the speed it does. Not why it goes around, but why so fast? It accelerates at about 2e-10 m/sec² and the mass of the matter we see in the galaxy is not enough to explain that acceleration.

We know why the arm orbits:  Things just do when there is a mass to orbit, such as the material that makes up the inner portion (stuff closer than us) of the galaxy.  But orbital speed is very much a function of the mass of that material, and there apparently isn't enough of it.

The answer is very simple - Gravity works locally.
If you agree with that - you have got the answer!!!
There is no need for dark matter in the galaxy to hold the Sun in the orbital path around the center.
All we need is - Gravity works locally in the arm (and everywhere).
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 31/12/2018 17:21:05
Quote from: Halc
Each star's velocity and acceleration is determined by the sum of the force vectors acting upon it.  This is all things, not just local ones.
Yes and No.
Yes - based on theoretically point of view.
No - based on real evidences.
The theory was based on that real evidence.  If you have evidence to the contrary, that would counter 4 centuries of physics.

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The Gravity force of the Sun/Moon is stronger by more than a twice with regards to the Earth/Moon.
If the Moon had to chose its orbital path based on the sum of the force vectors acting upon it, it will had to orbit around the Sun instead around the Earth, however - this isn't the case.
Your understanding of orbital mechanics needs a lot of work.  The strongest force acting on the moon is from the sun, and that means that at all times, the moon accelerates towards the sun.  That's the implication of what I've said above.  I did not say the moon cannot orbit Earth.  The moon has no choice or preference about this.  It moves exactly as per forces as described by Newton.

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Our scientists claim that  more than 100 Billion Sun mass is needed to be at the orbital center of the Sun (in the galaxy) in order to hold the Sun in its orbital path.
If we try to calculate the Galaxy/Earth gravity force (based on that mass), we should find that it is much higher than Sun/Earth gravity force.
How do you figure that?  No scientist claims this.  It is in fact 7 orders of magnitude lower.

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So, Based on a pure sum of the force vectors acting upon the earth, it had to ignore the Sun and start running around the galaxy. But again - this isn't the case.
Wrong several ways.
First, the Earth does orbit the galaxy, having gone around nearly 20 times.
Secondly, there is no law that says a greater force will strip Earth away from the sun since the sun would also be subject to that force. This is why the moon orbits Earth despite Earth exerting less force upon it than the sun.
Thirdly, the sun contributes more to Earth's force vector than all other forces combined.

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Each star orbits around a local virtual host point.
So you claim, but you've provided no mathematics for this claim.  How much force is applied by this virtual host point?  How might that be computed?  It needs to work for more than two bodies.  The motion of a two-body system can be reduced to a one-body case and thus has a solution.
Your claim is not similarly backed and is thus no more than a hand-waving assertion.
This is typical of armchair physicists.  All magical claims but no mathematics or predictions.

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Therefore, although we see that all the stars are moving in different directions, their host points are very stable with regards to each other.
Hence, The Arm Is an ARM. It holds all the Virtual host points at the arm due to local gravity force.
This is the key element of spiral arm.
If it was the ARM, it would be the host point, and all the orbiting stuff would orbit that point.  It doesn't.  An arm is a smear with all different angular velocities about the galaxy.  This has been measured.

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If the arm can't hold the stars, than by definition the stars must cross the arm and get out of it.
By definition?  What definition would that be?
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You can't just say that the stars stay at the arm but the arm doesn't hold them by gravity.
This is none realistic statement.
But I didn't say that.

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Sorry - this is incorrect. We can't take it for granted. We need to find an explanation why the Earth orbit around the Sun although its gravity force is lower than the Gravity force of the galaxy.
Even if the galaxy did exert this greater force, Earth would still probably orbit the sun.

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I will give you a tip - Gravity Works locally.
F = GMm/r².  That is not local.  That says there is force at any distance.

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There is no need for dark matter in the galaxy to hold the Sun in the orbital path around the center.
All we need is - Gravity works locally in the arm (and everywhere).
How do you explain the acceleration of the arm?  That's the thing you keep evading, post after post.
Title: Re: How gravity works in spiral galaxy?
Post by: mad aetherist on 31/12/2018 20:59:49
The answer is very simple - Gravity works locally.  If you agree with that - you have got the answer!!!   There is no need for dark matter in the galaxy to hold the Sun in the orbital path around the center.  All we need is - Gravity works locally in the arm (and everywhere).
Gravity is the due to the acceleration of aether (ie aetheons) into mass where the aether is annihilated.
But aether does not possess any quantum mass or quantum anything -- but it is good at transmitting force to & from tween quantum mass -- the transmission being by way of reverberations of pulses that travel at over 20 billion c (hencely 1 gravity-second is equal to more than 634 light-years)(taking less than 236 sec to cross the 150,000 LY diameter of the Milky Way)(praps less than 1 sec).

Hencely the shape of the disposition of cosmic mass in the Milky Way & near the particular stars of interest is important re the action & effect & magnitude of gravity & of big G especially re orbits (moreso than it might appear by the naive use of Newton's equation) -- & the disposition of nearby galaxies has an effect.
In other words gravity is due to tension in the aether, but this is a 3D thing, & hencely the attraction tween two stars needs other stars in every direction otherwise their mutual (2D) attraction will be limited.

When Cavendish did his big G experiment he had the mass of the Earth beneath & the mass of the building next to & above.  This ruined his results.  And nowadays silly Einsteinians try to make use of big G in a galactic environment, & need to invent Dark Matter.  No, Cavendish's big G works best (sometimes) on Earth, in a lab.

The centrifuging of aether near a spinning disc or in&near a spinning spiral galaxy gives a quasi-gravity that allso helps towards bringing about a 1/R relationship rather than the naive  1/RR Newtonian.  And quasi-gravity is a major reason why galaxies (& solar systems)(& Saturn's rings) like to have a disc shape.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/01/2019 07:25:59
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The Gravity force of the Sun/Moon is stronger by more than a twice with regards to the Earth/Moon.
If the Moon had to chose its orbital path based on the sum of the force vectors acting upon it, it will had to orbit around the Sun instead around the Earth, however - this isn't the case.
Your understanding of orbital mechanics needs a lot of work.  The strongest force acting on the moon is from the sun, and that means that at all times, the moon accelerates towards the sun.  That's the implication of what I've said above.  I did not say the moon cannot orbit Earth.  The moon has no choice or preference about this.  It moves exactly as per forces as described by Newton.
O.K
"The sun attracts the moon with a force twice as large as the attraction of the earth on the moon. Why does the moon not revolve around the sun?"

Please look at the following article:
https://www.quora.com/The-sun-attracts-the-moon-with-a-force-twice-as-large-as-the-attraction-of-the-earth-on-the-moon-Why-does-the-moon-not-revolve-around-the-sun
"Here the moon and earth form a system, which is like a Binary system . If two astrologers rotate around their center of gravity together, then it is called binary system"
So, this binary system is a "local gravity force" that gives the moon/earth system the possibility to orbit around the Sun.
Hence, the noon is not their by itself, as the earth is not their by itself.
Now they both can orbit around the Sun.
But it is clear that without setting a binary system with the Earth - Just based on pure gravity force - the moon will prefer to orbit around the Sun.
It will never ever set a binary system with the earth if it was orbiting around the Sun.
https://stardate.org/astro-guide/binary-and-multi-star-systems
"A binary is a pair of stars that orbit each other. A multi-star system consists of three or more stars. The stars in a binary or a multi-star system all formed from a single cloud of gas and dust, so they are true “siblings.”"
The Earth & the Moon are real "siblings". This is a key element. I will explain later on why the current concept of how the Moon had been created is absolutely unrealistic:
https://en.wikipedia.org/wiki/Giant-impact_hypothesis
"The giant-impact hypothesis, sometimes called the Big Splash, or the Theia Impact suggests that the Moon formed out of the debris left over from a collision between Earth and an astronomical body the size of Mars, approximately 4.5 billion years ago"
I will also explain how the earth and the moon had been formed "from a single cloud of gas and dust, so they are true siblings". and why they set this binary system due to gravity force.
But in order to explain it, we must know how our galaxy really works.
The spiral arm is a key element in our discussion.
Therefore, understanding the gravity force between the objects in the arm is vital.
So, the Earth/moon is a binary system which holds them both while they orbit around the Sun.
In the same token, the arm is a multi-star-system.
Each star in the galaxy has its own virtual host point. Together, those host points set that "multi-star-system" which holds all the neaby stars in the arm.
So, the sun is not there by itself, (as the moon is not there by itself)
There is a local system that is bounded by gravity force.
Therefore, I call that Binary/multi-star-system as "local gravity force" which is the base for what we see.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/01/2019 14:46:43
"The sun attracts the moon with a force twice as large as the attraction of the earth on the moon. Why does the moon not revolve around the sun?"
As I've said, it does go around the sun, once per year, just like Earth.  So you want to know why the two (Earth and moon) don't separate due to the greater gravitational force of the sun.

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Please look at the following article:
https://www.quora.com/The-sun-attracts-the-moon-with-a-force-twice-as-large-as-the-attraction-of-the-earth-on-the-moon-Why-does-the-moon-not-revolve-around-the-sun
Ouch.  This is why I rarely go to quora for answers.  I counted well over a dozen serious errors in that article.  This gem for example:
"Earth's mass and size are big so we do not understand the rotation of the earth. But this is not the case even between Jupiter and Sun. Here the sun does not turn around like Jupiter, and the Sun also does not turn around Jupiter!"
Apparently Earth is too large to understand if it orbits the sun or not, but Jupiter is small enough that it doesn't???  What????  It doesn't help that the guy apparently lacks English skills, but really, what was he trying to say with that statement?

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"Here the moon and earth form a system, which is like a Binary system . If two astrologers rotate around their center of gravity together, then it is called binary system"
Do they have to be astrologers?  What if two physicists rotate around their common center of gravity?
If the two astrologers were anywhere near a significant mass like the moon, they would not orbit each other.  Binary systems are not always stable, and the one described there is incredibly unstable.  We need to ask why the Earth-moon system is more stable than the pair of astrologers.

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So, this binary system is a "local gravity force" that gives the moon/earth system the possibility to orbit around the Sun.
Sure.  They're the only two significant masses in that system.  The sun's gravity does not act to separate them.  The non-uniformity of the sun's gravity does, and that non-uniformity is far less between the Earth and moon that is the acceleration of the moon due to Earth.  That's why the two stay together.

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Hence, the noon is not their by itself, as the earth is not their by itself.
That article denies that the Earth is not by itself.  The orbit of Earth is depicted as a perfect circle, apparently unaffected by the pull of the moon.  Yet another mistake.  It also shows the moon curving away from the sun at times, another mistake if the sun's force is greater than the Earth's.  We can forgive these things since the diagrams are clearly not to scale.

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But it is clear that without setting a binary system with the Earth - Just based on pure gravity force - the moon will prefer to orbit around the Sun.
It will never ever set a binary system with the earth if it was orbiting around the Sun.
Not true.  It is orbiting the sun, and yet it forms a binary pair with Earth.  The sun's gravitational field is too uniform at this radius to separate the two, just like the pull of the galaxy is far too uniform to pull the arms apart.  The pull from the sun goes up as the radius drops.  Venus has more acceleration than does Earth for this reason.  Not so with the galaxy according to your graphs.  The net acceleration (proportional to net force) actually goes down as the radius decreases.  This seems to be why galaxies form arms but new solar systems don't seem to.  Maybe they do at first.  Not like we can see that.

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https://stardate.org/astro-guide/binary-and-multi-star-systems
"A binary is a pair of stars that orbit each other. A multi-star system consists of three or more stars. The stars in a binary or a multi-star system all formed from a single cloud of gas and dust, so they are true “siblings.”"
We are true siblings with many (all?) of the stars closest by, but we're not really a multi-star system.  Maybe we are.  Not sure.  The rule is not always true.  Some multi-star systems are made up of stars that didn't come from the same cloud of dust.  Maybe one system captured a passing star from somewhere else, a sort of adopted star, not a true sibling.  A single star probably cannot capture another, but any multi-star system is quite capable of it.

The stability of systems with more than two significant masses is low.  There are no 'host points', so the paths are unpredictable, and such systems are capable of ejecting stars from the family in a way that a binary system would not.

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The Earth & the Moon are real "siblings". This is a key element. I will explain later on why the current concept of how the Moon had been created is absolutely unrealistic:
Well, other moons on other planets are not real siblings in that they formed from the same material.  So their history of being siblings doesn't explain the orbit or lack of it.  Yes, we went up there and found the moon to be the same stuff as us, thus forming the idea that it is a chunk of Earth knocked off or spun off.

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"The giant-impact hypothesis, sometimes called the Big Splash, or the Theia Impact suggests that the Moon formed out of the debris left over from a collision between Earth and an astronomical body the size of Mars, approximately 4.5 billion years ago"
I will also explain how the earth and the moon had been formed "from a single cloud of gas and dust, so they are true siblings". and why they set this binary system due to gravity force.
Being formed by a whack is quite different than forming separately from a common cloud.  The latter hypothesis is unlikely.  Still, Pluto has a moon even more disproportionately large, and that one doesn't appear to be the same material.  Unclear on the history of those two.

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But in order to explain it, we must know how our galaxy really works.
The spiral arm is a key element in our discussion.
Therefore, understanding the gravity force between the objects in the arm is vital.
So, the Earth/moon is a binary system which holds them both while they orbit around the Sun.
In the same token, the arm is a multi-star-system.
Each star in the galaxy has its own virtual host point.
Together, those host points set that "multi-star-system" which holds all the neaby stars in the arm.
You've not explained this host point mechanics, and thus it is just fantasy.  It can be trivially demonstrated with a simulation of 3 or more stars, none of which will exhibit motion about a host point. You can deny the Newtonian forces upon which that simulation would run, but you've not described new laws to replaces the ones you're denying.

Sure, arms hold themselves together via gravity.  That is pretty simple since force (towards galactic center) goes up at the outside edge of the arm, and force (towards galactic center) goes down on the inside edge.  This is very different from Earth/moon where the sun's pull is greater, not less, at the inside edge.  So arm stability can be explained without this host-point nonsense.

None of how the arms hold together is relevant to the question of the acceleration of the arm itself, which is totally unexplained in your posts.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/01/2019 15:25:56
"If two astrologers rotate around their center of gravity together, then it is called binary system"
If two astrologers are in mutual orbit, how will they determine which sign is prominent at a given time?  They lack an obvious 'nighttime' to make that determination, so they only have each other to reference, meaning they'll be off 6 months from each other, despite the unspecified length of their orbital period.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/01/2019 06:11:37
Sure, arms hold themselves together via gravity.  That is pretty simple since force (towards galactic center) goes up at the outside edge of the arm, and force (towards galactic center) goes down on the inside edge.
Dear Halc
If you agree with that, than this is all is needed to explain the spiral arm structure.
As I have stated, each star is connected in the arm by Newton gravity force.
All the stars have a similar orbital speed. (More or less)
Therefore, for any time frame, they all cross the same distance.
In order to achieve it, as they drift outwards, they also drift backwards.
That activity sets by definition the spiral shape of the arms.

That is the whole idea about spiral arms.
Now, let's try to find what our scientists have to say about: "How Our Milky Way Galaxy Got Its Spiral Arms"
https://www.space.com/24642-spiral-galaxies-milky-way-shape-explained.html
Dated - February 12, 2014
"The researchers found that the universe was a very chaotic place in its infancy. The first galaxies were disks with massive, bright, star-forming clumps and little regular structure. To develop the nice spiral forms seen today, galaxies first had to settle down, or "cool," from the previous chaotic phase. This evolution took several billion years."

First contradiction - "Massive disc galaxies" - If our universe was "very chaotic place in its infancy" how could it be that we have got immidiatly disk galaxies with massive, bright, star-forming clumps and little regular structure."
Why those massive disc galaxies had been developed in the chaotic Universe? How long it should take to set the first massive disc galaxy in that unclear process? How could it be that all the billions spiral galaxies had been formed from this chaotic Universe?

"Gradually, the galaxies that were to become spirals lost most of their big clumps, and a central, bright bulge would appear; the smaller clumps throughout the galaxy would begin to form indistinct, "woolly" spiral arms.
These arms would only become very distinct arms once the universe was about 3.6 billion years old. At that age, as the galaxies had a chance to settle down, the turbulence decreased, and new stars would form in a much quieter disk. "We can see the transition from the early chaotic state to the modern, relaxed state," said Bruce Elmegreen.
These first spiral galaxies were either two-armed structures or had thick, irregular spirals with some remaining clumps. More finely structured, multi-armed galaxies like the Milky Way galaxy and its neighbor Andromeda appeared much later, when the universe was 8 billion years old."

Second contradiction - "Age" - We see spiral galaxies at the most far end of the Universe.
The estimated age of many mature spiral galaxies is more than 13.2 Billion years.
So, how could we see today very mature spiral galaxies at estimated age of only 0.6 billion years, if based on this article we need several billions (3.6 Min) and also with the assumption that we have got "Massive disc galaxies" almost immediately from the chaotic Universe?

Third contradiction – "Random process" - They don't show exactly how do we get this spiral arms from  massive disc galaxy. They just say that "Gradually, the galaxies that were to become spirals lost most of their big clumps,.."
This is a random activity by definition. They don't say why they lost the big clumps and how they really got their spiral arms based on Newton gravity.

Conclusions:
I have set a simple explanation why 400 Billion spiral galaxies have got their spiral shape from day one based on Newton gravity force.
Our scientists (even at 2014) try to find an explanation which is not correlated to the universe age time frame. This random activity can't be the base for all the billions spiral galaxies that we see in our universe.
I have set my explanations about the spiral arms structures which is much more relaibale from the last "story" from our scientists.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 02/01/2019 13:04:57
If you agree with that, than this is all is needed to explain the spiral arm structure.
It is far more complicated than either of us can explain.  You have a bowl of cake batter with a bar of dark batter across the middle.  You stir once or twice and you get a spiral structure something like the galaxy.  You stir 40 times and that structure fades to either super-thing spirals or just one color.  How are the arms still there after 40 laps with all the parts moving at different angular speeds around the galaxy?  Our own arm seems to be one of the ones that is being thinned to the point of nonexistence, possibly explaining why we've not been disturbed by the material in a thicker arm which would have disrupted life formation.

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As I have stated, each star is connected in the arm by Newton gravity force.
All the stars have a similar orbital speed. (More or less)
Too much speed, but yes, more or less the same.  Not the same angular speed, which would be needed to keep the arms intact indefinitely, but just similar linear speed.

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Therefore, for any time frame, they all cross the same distance.
Exactly, but the stars close to the center of the galaxy have far less distance to go, so they go around far more often than the stars at the far ends of the arms that might yet to have gone around even 10 times since they have so much further to travel to achieve one lap.

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In order to achieve it, as they drift outwards, they also drift backwards.
That activity sets by definition the spiral shape of the arms.
That drifting outwards and back defines the clumping of the arms.  The spinning at different angular velocities of each end defines the spiral shape, just like the cake batter.

The speed at which everything moves is unexplained by your assertions.  You never address this issue.

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Now, let's try to find what our scientists have to say about: "How Our Milky Way Galaxy Got Its Spiral Arms"
Your last link was hardly to a scientist.  This one here is at least a magazine article which presumably has some standards beyond what quora obviously has.
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https://www.space.com/24642-spiral-galaxies-milky-way-shape-explained.html
Dated - February 12, 2014
I like this one, reporting on doing the research the correct way:  Look into the past as see how the shapes evolved.  It's sort of like watching an animation of our own galaxy over billions of years, except all the frames have been dropped on the floor and need sorting.
The article is more about how the research was done than what the results of that research were.

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"The researchers found that the universe was a very chaotic place in its infancy. The first galaxies were disks with massive, bright, star-forming clumps and little regular structure. To develop the nice spiral forms seen today, galaxies first had to settle down, or "cool," from the previous chaotic phase. This evolution took several billion years."

First contradiction - "Massive disc galaxies" - If our universe was "very chaotic place in its infancy" how could it be that we have got immidiatly disk galaxies with massive, bright, star-forming clumps and little regular structure."
Why those massive disc galaxies had been developed in the chaotic Universe? How long it should take to set the first massive disc galaxy in that unclear process? How could it be that all the billions spiral galaxies formed from this chaotic Universe?
Hard to parse your comments here.  You're suggesting a contradiction, but I don't see one identified.  Higher density matter tends to group due to its mutual attraction.  That forms galaxies.  If the matter has any angular momentum, then the clump will naturally form a disk shape, just like what happens with new solar systems forming out of a gas cloud.  The disk shape is the only way to preserve angular momentum of a compressing structure.

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Second contradiction - "Age" - We see spiral galaxies at the most far end of the Universe.
The estimated age of many mature spiral galaxies is more than 13.2 Billion years.
So, how could we see today very mature spiral galaxies at estimated age of only 0.6 billion years, if based on this article we need several billions (3.6 Min) and also with the assumption that we have got "Massive disc galaxies" almost immediately from the chaotic Universe?
Nothing said that very mature spiral galaxies were of age 0.6 billion years.  That would be a young galaxy still lacking a mature spiral structure.  From where are you getting this claim of a spiral galaxy that is only 0.6 BY old?

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Third contradiction – "Random process" - They don't show exactly how do we get this spiral arms from  massive disc galaxy. They just say that "Gradually, the galaxies that were to become spirals lost most of their big clumps,.."
This is a random activity by definition. They don't say why they lost the big clumps and how they really got their spiral arms based on Newton gravity.
Missing details is not a contradiction, just not a full description.  Such details might not be clear.  They only looked closely at about 41 galaxies to get an idea of the various stages of the process.  Nobody has a video of the evolution of any one galaxy as it goes through the process.

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Conclusions:
I have set a simple explanation why 400 Billion spiral galaxies have got their spiral shape from day one based on Newton gravity force.
You have no such thing.  You just assert it, and the data from the Hubble image clearly shows that assertion to be wrong since no young galaxy has arms, and thus they're not there from 'day one'.

The article you link suggests no explanations whatsoever.  It just reports data: This is what apparently happens.  Given that, find a theory that explains it, but no such theory seems to be proposed in that article.  You have asserted having an explanation, but none is provided.  You just assert the current observed picture, and fail to explain simple observed speeds of the arms.  So your explanation falls completely short of the current views on galaxy dynamics, which are admittedly only best models at the time.  Those models are always changing, especially since none of them really provides a full explanation.

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You can accept it or reject it.
And so I have.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/01/2019 13:31:19

The speed at which everything moves is unexplained by your assertions.  You never address this issue.

Yes, I can explain it.
However, I need to see the real data.

Please look at the "Observed rotation curve" in the following article?
http://ircamera.as.arizona.edu/astr_250/Lectures/Lecture_22.htm
Observed rotation curve:
We see that at about 0.2 KPC the orbital velocity is maximal - 260 Km/sec
Than it goes down to about 190 Km/sec at 3KPC.
From this point it goes up to 210 Km/sec at 4KPC
and to 230 Km/sec at 7 KPC.
Than it starts to go down again to 200 Km/sec at 10KPC
Goes up to 235 Km/sec at 13 KPC and stay at that velocity at any further distance.
Is this correct?

If so, how could it be that we don't see the real orbital velocity near the SMBH?
We know that the orbital velocity of the plasma is 0.3 c (speed of light)
"Consider a star whose proper motion has been measured to be equivalent to 1000 Km/sec and which lies only 0.01 pc from SgrA*.
So, why we don't see the real velocity at the center?\
Why it shows that the orbital velocity at the center is almost Zero?

Why they also give this information without any connection to the galaxy shape?
I would expect to see measured velocities per radius in the following segments:
1. Accretion disc (What is the Min radius and maximal radius. What is the orbital velocity at each radius?
2. Bulge - Radius range and velocities
3. Bar - Radius range and velocities
4. Ring - Radius range and velocities
5. Disc/spiral arms - Radius range and velocities
6. From the end of the disc - Radius and velocities
Would you kindly direct me to the real measurements of orbital velocities Vs radius in the Milky Way? (From 0 - 20K Pc)
I will give full explanation once I have the real data.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 03/01/2019 14:55:16
Quote from: Halc
The speed at which everything moves is unexplained by your assertions.  You never address this issue.
Yes, I can explain it.
However, I need to see the real data.

Please look at the "Observed rotation curve" in the following article?
http://ircamera.as.arizona.edu/astr_250/Lectures/Lecture_22.htm
Observed rotation curve:
We see that at about 0.2 KPC the orbital velocity is maximal - 260 Km/sec
I get that at about 0.5 KPC.  The linear line to the left of that is consistent with the 'bar' which seems to turn as a solid object instead of a fluid collection of stars and such.  I don't know the current thinking behind what holds a bar together.  A lot of galaxies seem to have one.
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Than it goes down to about 190 Km/sec at 3KPC.
From this point it goes up to 210 Km/sec at 4KPC
and to 230 Km/sec at 7 KPC.
Than it starts to go down again to 200 Km/sec at 10KPC
Goes up to 235 Km/sec at 13 KPC and stay at that velocity at any further distance.
Is this correct?
That what it shows, yes.  The 235 is fairly constant out to where most of the measurable material stops, which is about 17 KPC.  It has to drop to a Keplerian curve eventually, else the whole universe would be orbiting our galaxy.

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If so, how could it be that we don't see the real orbital velocity near the SMBH?
The graph stops short of that radius.  We know that S2 has speeds up to 5000 km/sec, and that is not in any way reflected in the graph.  S2 is not part of the bar.

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We know that the orbital velocity of the plasma is 0.3 c (speed of light)
"Consider a star whose proper motion has been measured to be equivalent to 1000 Km/sec and which lies only 0.01 pc from SgrA*.
So, why we don't see the real velocity at the center?\
Why it shows that the orbital velocity at the center is almost Zero?
It doesn't show velocity close to the center.  The graph stops short of radius 150 PC.  Stuff close in moves fast, as you note above with this 1000 km/sec star being considered.
The graph also doesn't show anything at zero speed.  The bottom of the graph is 150 km/sec, not zero.

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Why they also give this information without any connection to the galaxy shape?
It seems there is little connection.  Objects at radius X move at the same speed regardless of being in an arm, near one side or the other, or between them.  The arm might affect regular wobble in and out, but not the tangential speed around the galaxy.

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I would expect to see measured velocities per radius in the following segments:
1. Accretion disc (What is the Min radius and maximal radius. What is the orbital velocity at each radius?
The graph actually shows exactly that.
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2. Bulge - Radius range and velocities
I suspect this is not uniform since objects have random and probably highly elliptical orbits.  Things at a given distance will have all sorts of different speeds.
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3. Bar - Radius range and velocities
The graph seems to show this as well, which is the steep part to the left of the highest point on the curve.  The ends of the bar are moving faster than anything else, just like the ends of the thrown spanner are moving faster than the rest of it.
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4. Ring - Radius range and velocities
Graph would need to be extended to the right for this, but there is not a lot of visible material via which distance and speed can be determined.
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5. Disc/spiral arms - Radius range and velocities
How is this not number 1 above?
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6. From the end of the disc - Radius and velocities
No idea what you mean by this one.
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Would you kindly direct me to the real measurements of orbital velocities Vs radius in the Milky Way? (From 0 - 20K Pc)
I will give full explanation once I have the real data.
Your link above seems to be fairly in line with similar graphs I've seen.  If you don't consider that the real data, then nothing I provide will motivate you.  How can you claim to have an explanation if it is based on data that you don't know?  Is there any part of it that is not complete fiction then?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/01/2019 17:43:44

Thanks
Your link above seems to be fairly in line with similar graphs I've seen.
So as you agree on this graph, let's set the correlation with the structure.
In order to do so, please look at the following diagram of the Milky way structure:
http://www1.ynao.ac.cn/~jinhuahe/know_base/astro_objects/galaxies/Milky-Way-Galaxy-files/logarithmic-spiral-pattern-before2001.JPG
We can see clearly that the radius of the ring is 3Kpc.
Now, please remember that the minimal orbital velocity was exactly at that radius.
Than it goes down to about 190 Km/sec at 3KPC.
So, the orbital velocity at the ring is 190 Kpc.
That shows that at the end of the bar, the orbital velocity is minimal. That evidence contradicts your following message:
The ends of the bar are moving faster than anything else, just like the ends of the thrown spanner are moving faster than the rest of it.
So, the end of the bar has the minimal orbital velocity.
This by itself must set a big red light. How could it be???
In any case, the spiral arms starts exactly at that ring (3KPC).
We don't see the end of the arms but we can assume that it ends at about 15 KPC (at about 45K light year).
Let's focus on the Bar and the bulge:
As we move inwards from the ring the orbital velocity is increasing.
It almost goes to 260 Km/s at 0.5KPC.
I get that at about 0.5 KPC.
However, as we get closer to the center from that radius, the orbital velocity gets down.
At about 0.1 KPC the orbital velocity is almost Zero!!!
This is a very important segment in the galaxy.
There is a meaning for that activity.
Our scientists focus only on the orbital velocity of the stars in the arms, while they ignore completely other important segments..
That shows that their knowledge in spiral galaxy is very poor.
I can explain each segment.
However, do you agree by now that there is high correlation between the galaxy structure and the orbital velocity?
Do you agree that your following message is incorrect?
Quote
Why they also give this information without any connection to the galaxy shape?
It seems there is little connection.  Objects at radius X move at the same speed regardless of being in an arm, near one side or the other, or between them.  The arm might affect regular wobble in and out, but not the tangential speed around the galaxy.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 04/01/2019 00:25:37
So as you agree on this graph, let's set the correlation with the structure.
In order to do so, please look at the following diagram of the Milky way structure:
http://www1.ynao.ac.cn/~jinhuahe/know_base/astro_objects/galaxies/Milky-Way-Galaxy-files/logarithmic-spiral-pattern-before2001.JPG
We can see clearly that the radius of the ring is 3Kpc.
There is no ring depicted there.  There are some dots that are part of the polar coordinates graph, but no physical ring.  The ring I think is way outside, like maybe 40 kps or something, a halo of gas not particularly dense enough to form luminous objects.

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So, the orbital velocity at the ring is 190 Kpc.
The speed of the ends of the bar.  No ring there.  Yes, the bar is larger than I posted before, and the curve would not make sense for an object that retains its shape like that.  Are the stars being measured not part of the bar?  What comprises the bar then?  I said I've not seen a description of how that supposedly works.  The site is not much help.  Hard to read the text that goes along with your picture.

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That shows that at the end of the bar, the orbital velocity is minimal. That evidence contradicts your following message:
The ends of the bar are moving faster than anything else, just like the ends of the thrown spanner are moving faster than the rest of it.
Yes, clearly the part of the curve I identified with that sort of motion is not the bar, or not all of it.

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So, the end of the bar has the minimal orbital velocity.
This by itself must set a big red light. How could it be???
Good question.  I don't know the answer to that one.  There must be some website that explains how that works, and what was being measured on that curve.

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In any case, the spiral arms starts exactly at that ring (3KPC).
We don't see the end of the arms but we can assume that it ends at about 15 KPC (at about 45K light year).
Let's focus on the Bar and the bulge:
As we move inwards from the ring the orbital velocity is increasing.
It almost goes to 260 Km/s at 0.5KPC.
However, as we get closer to the center from that radius, the orbital velocity gets down.
At about 0.1 KPC the orbital velocity is almost Zero!!!
Nothing shows that.  The min speed is 150 km/s at that radius where the graph ends.  Here's a more detailed version of the same graph, from the same site as your link above:
http://www1.ynao.ac.cn/~jinhuahe/know_base/astro_objects/galaxies/Milky-Way-Galaxy-files/rotation-curve-10kpc.PNG
That one shows the data points that make up the curve.  The data points follow the curve quite well up to about 8.5 kps, after which there are many points significantly above and below.  The curve is flat to the right because the data points are so imprecise and so few that fitting a better curve is meaningless.  On the left, I see no object slower than 200 km/sec, but the line continues the apparent trend.  All very non-Keplerian.

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This is a very important segment in the galaxy.
There is a meaning for that activity.
Our scientists focus only on the orbital velocity of the stars in the arms, while they ignore completely other important segments..
That shows that their knowledge in spiral galaxy is very poor.
No, it means that not a lot of pop-science web sites focus on that part.  I haven't tried much to actually see what the consensus is about how things move in that inner 3 kps, and how the bar keeps its shape.  Such a thing is fairly common, hardly a weird anomaly of just our own galaxy.

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I can explain each segment.
However, do you agree by now that there is high correlation between the galaxy structure and the orbital velocity?
Well I don't see it from that data.  The bar suggests one curve, and the graph shows another.  That's pretty louse correlation.  The rest of the curve (outside 3kps) doesn't have a curve that changes say in a pattern that matches the spacing of the arms, so again, little correlation.

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Do you agree that your following message is incorrect?
Quote from: Halc
It seems there is little connection.  Objects at radius X move at the same speed regardless of being in an arm, near one side or the other, or between them.  The arm might affect regular wobble in and out, but not the tangential speed around the galaxy.
I think something definitely changes around 3kps, but other than that, I stand by my comment above.  I see little correlation between structure and the shape of that curve.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 04/01/2019 04:44:18
Another nice picture from the same site shows the arm structure better:
http://www1.ynao.ac.cn/~jinhuahe/know_base/astro_objects/galaxies/Milky-Way-Galaxy-files/MW-spirals-annotated.jpg

The arms seem not to be continuous tubes of stellar material, but rather chunks arranged sort of like fallen dominoes.  It seems like the dominoes separate as the different path lengths pull them apart, and they rejoin elsewhere, thus never letting the arms thin out to nothing.  Our own particular domino has separated in this manner, more than most of them.  It is probably headed for the more dense Perseus arm to the outside.  Just my guess.

The arms form a more elliptical shape near the bar, not a nice near circular spiral like all the diagrams show.
This is not a photo of course, just an artist's rendering.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/01/2019 05:29:31
No, it means that not a lot of pop-science web sites focus on that part.  I haven't tried much to actually see what the consensus is about how things move in that inner 3 kps, and how the bar keeps its shape.  Such a thing is fairly common, hardly a weird anomaly of just our own galaxy.
Please look at the following image from NASA (which is identical to the one that you have offered)
https://solarsystem.nasa.gov/resources/285/the-milky-way-galaxy/
It shows clearly that at 3KPC there are two sides of the ring
The one which is closer to the Sun is called: Near 3KPC Arm
The one which is farther from the Sun is called: Far 3KPC Arm
Together they set the ring.
The end of the Bar is directly located just between those arms on each side.
How can you claim that:
The arms seem not to be continuous tubes of stellar material, but rather chunks arranged sort of like fallen dominoes...
Well I don't see it from that data.
The ring I think is way outside, like maybe 40 kps or something, a halo of gas not particularly dense enough to form luminous objects.
The data is clear. It is time for you to agree with the evidences.
There is a clear ring at 3KPC!!!
This ring has a great impact on the activity of the galaxy.
I will explain it later on.
But first - do you agree that :
1. There is a ring at 3KPC?
2. The orbital velocity at that ring is minimal - 190 Km Sec?
3. The Bar ends exactly at that ring?
4.The arms starts to form exactly from that Ring?
Why do you disqualify even the clear evidences which me & you represent?
Why are you so negative?
Would you kindly look again at the evidences and try to understand that spiral galaxy is not just spiral arms?
Would you kindly try to be more cooperative and help us to highlight the real activity at spiral galaxy?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 04/01/2019 13:14:20
Please look at the following image from NASA (which is identical to the one that you have offered)
https://solarsystem.nasa.gov/resources/285/the-milky-way-galaxy/
It shows clearly that at 3KPC there are two sides of the ring
That's the same image (different link) as in my prior post.
There is no ring identified or depicted.  I see the inner ends of what are the Perseus and Scutum-Centarurus (or Crux-Scutum) arms, but they're called Near and Far 3kpc arm that far in.  There is no ring there.

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The one which is closer to the Sun is called: Near 3KPC Arm
The one which is farther from the Sun is called: Far 3KPC Arm
Together they set the ring.
No, they set the inner portions of a spiral.
If you want to call it a ring, fine, but the two arms do not connect.  Each spirals out into the two major arms.  The image clearly shows this.  I notice both arms get noticeably thicker at the point halfway around when the pass relatively close by the opposite end of the bar from which they appear anchored.
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The end of the Bar is directly located just between those arms on each side.
Yes, each major arm is anchored on one end of the bar.  Your prior wiki image shows it in more graphic form:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
That image has the sun on top instead of the bottom.  It shows for instance the aqua Perseus arm just outside our position going a full lap around the galaxy where it becomes named the <near> 3kpc arm before it 'connects' to the bar.  That image fails to name the far 3kpc arm that is the bright green line.

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How can you claim that:
Quote from: Halc
The arms seem not to be continuous tubes of stellar material, but rather chunks arranged sort of like fallen dominoes...
Because that's what I see in the picture you just linked.

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The data is clear. It is time for you to agree with the evidences.
There is a clear ring at 3KPC!!!
This ring has a great impact on the activity of the galaxy.
I will explain it later on.
No site calls that a ring.  If you want to, fine. It seems to be important to you.

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But first - do you agree that :
1. There is a ring at 3KPC?
No.  It is the inner ends of the two major arms: a pair of spirals.
If you want to consider those portions not part of their arms, but a ring instead, I will entertain your idea if it actually demonstrates something.

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2. The orbital velocity at that ring is minimal - 190 Km Sec?
According to that rotation curve, it seems so, yes.  But without explanation of the bar, I'm not sure what is being measured by that rotation curve because the bar could not be stable given that curve.

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3. The Bar ends exactly at that ring?
It seems to end where each spiral starts, yes.  The opposite end of say the near 3kps arm is outside the bar and does not connect to the far 3kps arm that starts there.

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4.The arms starts to form exactly from that Ring?
They start from the bar.  The 3kps arms are the two major arms of the galaxy.  Sorry, but it seems I don't agree to much of what you see.

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Why do you disqualify even the clear evidences which me & you represent?
Why are you so negative?
You seem to need a situation to be what it isn't.  For instance you deny tidal forces because your agenda needs to have the moon drifting away naturally, without forces acting on it.  You are bending data at every step to fit a broken model instead of finding a model that fits the data.  It is a really bad way to go about learning.  It is a method of obfuscation.
You demonstrate not even minimal awareness of orbital mechanics, and yet you claim to have a theory that is better than the ones presented by people who know their stuff better than any of us.  Of course I'm going to resist what are very apparent attempts to bend observed data.
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Would you kindly look again at the evidences and try to understand that spiral galaxy is not just spiral arms?
Would you kindly try to be more cooperative and help us to highlight the real activity at spiral galaxy?
Like I said, you can present the inner portions of the arms as one ring if you like since they're not too distant from each other.  Not sure what your point is in doing that, but it isn't a major bending of what that picture shows.  I'd like to hear your point.
As you say, that curve does not fit a picture with a bar in it.  The curve seems very inconsistent with what appears to be the shape of a common galactic feature.  There is something going on that neither of us understands, so I'm not exactly claiming myself to have a better understanding of the dynamics of a galaxy.

I'd like to see an animated gif or something of a spinning galaxy.  If it had all the parts moving consistently with that rotation curve, the thing would smear out in a short time.

I found this nasa one:
https://gizmodo.com/5953694/this-stunning-nasa-simulation-shows-a-galaxys-entire-life-history
That's a really fast simulation at ~100 million years per second or so, so it is hard to see how the arms continuously merge and reform.  The bar is hard to make out near the end, but in the middle it seems to rotate the opposite way as everything else.  The roatation curve then would be the speed of all the stuff flying loose around, but not of the bar itself.  That might be one answer to that seeming inconsistency.

Most of the simulations are of long lifespans.  I wanted something a little more slow motion.  The bar seems to spin so fast that it might be the frame rate that makes it look like its going backwards.

I can find no decent simulation of a nice slow picture of say 200 million years compressed into a minute or two.  They're all much faster, making the interactions go by too fast to answer the sorts of questions we're asking here.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/01/2019 18:01:00
Quote
The one which is closer to the Sun is called: Near 3KPC Arm
The one which is farther from the Sun is called: Far 3KPC Arm
Together they set the ring.
No, they set the inner portions of a spiral.
If you want to call it a ring, fine, but the two arms do not connect.  Each spirals out into the two major arms.  The image clearly shows this.  I notice both arms get noticeably thicker at the point halfway around when the pass relatively close by the opposite end of the bar from which they appear anchored.

No - None of them spirals out. Our scientists call them 3KPC as they stay at 3KPC.
Yes - They are connected at their edges to other spiral arms, but they together set a ring or almost a ring.
I hope that the following image can give you better overview
https://crossfithartford.com/dummies_diagram_of_plant_the_milky_way.php
We can see the two 3KPC arms.
One is called near 3KPC arm and the other Far 3KPC arm.
Why do we call them both 3KPC arm?
Don't you think that the name of the 3KPC represents their radius???
It also seems that both of them are connected/almost connected.
In the following image it is quite clear:
https://solarsystem.nasa.gov/resources/285/the-milky-way-galaxy/
We see that in one side they are connected at a point which is written as galactic bar, while on the other side it is called long bar. At those points we also see the connections to the main spiral arms.
In the image it looks like elliptic ring shape.
However, don't you agree that the mane - 3KPC arm proves that this arm is located all the way at a radius of 3KPC?
They both stay there by themselves.  None of them is part of another spiral arm.
Therefore, if one arm covers the upper half/almost half of the cycle, while the other arm covers the other half/almost half of the cycle, don't you agree that in the total we get a ring/almost ring
You might call it as upper 3KPC arm + lower 3KPC arm.
However, even if you don't see a perfect ring, it is still a ring (or almost a ring...)
We know that the orbital velocity goes down to its minimal value exactly at 3KPC.
Therefore, this ring is a key element in the activity of the galaxy.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 04/01/2019 19:55:46
Please look at the following diagram about: NGC 1398: AN UTTERLY PERFECT SPIRAL GALAXY

https://www.syfy.com/syfywire/ngc-1398-an-utterly-perfect-spiral-galaxy

We clearly see a ring at the center (at the edge of the bar which isn't fully clear).
It is also stated:
"NGC 1398 is a barred spiral, with that rectangular-shaped feature running across the core and ending at the inner "ring." This is a common feature in spirals — the Milky Way has a big one — and they can affect how the stars and gas move around the galactic center (I describe this in the earlier post on NGC 1398)."
I have no idea what is the radius of that ring or its width, but it shows that there is "inner ring" in barred spiral galaxy as at the Milky Way.
It is quite clear that this "inner ring" has a significant effect on "how the stars and gas move around the galactic center" as I will explain later on.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 04/01/2019 22:10:01
Therefore, if one arm covers the upper half/almost half of the cycle, while the other arm covers the other half/almost half of the cycle, don't you agree that in the total we get a ring/almost ring
I said that for the sake of argument, I'm willing to consider that a ring. It seems important to you.

As for the rotation curve below 3kps, it seems that the speeds measured there are not tangential speeds like they are outside the bar.  The speeds are of materials moving up and down the bar I think.  That's why the bar can turn slower further in but still have higher speed stars in it.  I've seen the motion described as 'banana orbits' due to the shape of their paths within the bar, which I find to be more kidney bean shaped.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 04/01/2019 22:18:31
Please look at the following diagram about: NGC 1398: AN UTTERLY PERFECT SPIRAL GALAXY

https://www.syfy.com/syfywire/ngc-1398-an-utterly-perfect-spiral-galaxy

We clearly see a ring at the center (at the edge of the bar which isn't fully clear).
As I said, I can consider that a ring if you wish, but I see a clear spiral there.  The bar is quite clear, and there is an arm that spirals from each end of it all the way out to the edge of the galaxy, just like in ours.  Each arm makes a full lap before angling across that relatively sparse region to the thicker outer regions of the galaxy.  It looks like it has a lot of angular momentum compared to a typical galaxy.

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It is also stated:
"NGC 1398 is a barred spiral, with that rectangular-shaped feature running across the core and ending at the inner "ring." This is a common feature in spirals — the Milky Way has a big one — and they can affect how the stars and gas move around the galactic center (I describe this in the earlier post on NGC 1398)."
I have no idea what is the radius of that ring or its width, but it shows that there is "inner ring" in barred spiral galaxy as at the Milky Way.
Great!  Other people call it a ring. Let's go with that.  But that site also says this:
"I found that for a while astronomers weren't sure just what the structure of this galaxy was; some said the inner arms formed a ring, some said it's not really a ring but that the outer arms form a ring-like structure (called a pseudoring).
This image makes it clear that the inner ring is definitely just a very tightly wound set of arms
"

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It is quite clear that this "inner ring" has a significant effect on "how the stars and gas move around the galactic center" as I will explain later on.
OK
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/01/2019 00:26:03
To understand the consensus on motion of galatic features, you probably need to read up on density wave theory.  Apparently the material of the galaxy (the sun for instance) moves at ones speed (217 km/sec?) but the arm moves at a much slower speed if you watch it.  It is a wave of high density of material that doesn't move with the material, much in the same way that the wake of a boat stays with the boat despite the material (water) moving along quickly to the rear of the boat.
So a picture of the galaxy or the bar spinning does not represent the material necessarily moving at those speeds.  I've made that mistake myself in several posts.  Treating waves like moving objects is a mistake.  Realizing this helps resolve several of the seeming paradoxes that have come up in our conversation.
I'm actually learning quite a bit in the course of posting to this thread.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/01/2019 05:35:30
As I said, I can consider that a ring if you wish, but I see a clear spiral there
Great!  Other people call it a ring. Let's go with that.

Thanks!!!
So, the ring has almost a perfect cycle shape around the Bar. (Although if we look carefully inside the ring, we see clearly that it has a spiral shape.)
Once we agree on that ring, we can now start looking at the structure/velocity of barred spiral galaxy

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 05/01/2019 07:22:50

What kind of gravity force that ring could generate?
Let's assume that there is no SMBH at the center.
Let's also assume that we set Billion of stars at a perfect ring of 3KPC and hold them ALL at their current position for very long time.(Please don't tell me what will happen to the ring after one minute. This is a theoretical question about stars ring)

What kind of gravity force we might find inside the ring?
Do you agree that all the stars together will set an equivalent gravity force on any object which is located inside that ring?
Therefore:
1. If an object is located directly at the center - Do you agree that the impact of the gravity force from the stars in the ring at a distance of 3KPC in one side should cancel the impact of the stars in the other side of the ring? Therefore, the net gravity force on that object is - Zero?
2. If an object is located at 1.5KPC from the Center (Let's set it close to 12 (in clock) - The impact gravity force of the stars in 3 and 9 should cancel each other? However, the distance to 12 is now 1.5 KPC while the distance to 6 is 4.5KPC.
Therefore, do you agree that the net gravity force in the direction of 12 is quite strong?
3. If an object is located very close to 12 (let's say at 2.9 KPC). - The impact of the gravity force in 12 is almost maximal (as the distance is just 0.1KPC). However, the impact of 6 is minimal (as the distance is 5.9KPC). Remember that 3 cancel the impact of 9.
Therefore, do you agree that the impact of the ring gravity force is higher as the object is closer to the ring?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/01/2019 13:22:37
So, the ring has almost a perfect cycle shape around the Bar. (Although if we look carefully inside the ring, we see clearly that it has a spiral shape.)
Once we agree on that ring, we can now start looking at the structure/velocity of barred spiral galaxy
Ours is kind of elliptical shaped, but that 'perfect' galaxy you linked make pretty much a circle, closest I've seen.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 05/01/2019 13:57:12

What kind of gravity force that ring could generate?
Let's assume that there is no SMBH at the center.
Let's also assume that we set Billion of stars at a perfect ring of 3KPC and hold them ALL at their current position for very long time.(Please don't tell me what will happen to the ring after one minute. This is a theoretical question about stars ring)
Without reading further, a hoop like that is in equilibrium and can orbit itself, with or without a central object.  The mass within (not part of) the ring will contribute to higher speed of the ring, but a ring by itself will have some minimum speed.  It isn't totally stable, but stable enough, just like our asteroid belt, which has a central mass.  I can think of no example of a ring on its own.

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What kind of gravity force we might find inside the ring?
A hollow sphere would have a uniform field.  A body inside one would be weightless, but a ring will attract to itself.  I would fall outward down to the nearest part of the ring if I found myself stationary within it somewhere.
A stationary 'me' would have no orbit or radius, so this isn't even in violation of Newton's law that says mass outside my radius has no overall effect on my orbital speed.

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Do you agree that all the stars together will set an equivalent gravity force on any object which is located inside that ring?
I don't know what you mean by 'set an equivalent gravity force' here. Equivalent to what?

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Therefore:
1. If an object is located directly at the center - Do you agree that the impact of the gravity force from the stars in the ring at a distance of 3KPC in one side should cancel the impact of the stars in the other side of the ring? Therefore, the net gravity force on that object is - Zero?
At the center, yes.  Net of zero by symmetry.

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2. If an object is located at 1.5KPC from the Center (Let's set it close to 12 (in clock) - The impact gravity force of the stars in 3 and 9 should cancel each other?
There will be no net pull to either side, yes, again by symmetry.  The stars at 3 and 9 on the ring are both 'south' of the object that is 1.5kpc north of the center, so there is a net pull towards the center from those two stars.

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However, the distance to 12 is now 1.5 KPC while the distance to 6 is 4.5KPC.
Therefore, do you agree that the net gravity force in the direction of 12 is quite strong?
Stronger than any of the others, yes.  You need to integrate the vectors on the entire ring to get the net force on the object.  Taking these single samples will not work.

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3. If an object is located very close to 12 (let's say at 2.9 KPC). - The impact of the gravity force in 12 is almost maximal (as the distance is just 0.1KPC). However, the impact of 6 is minimal (as the distance is 5.9KPC). Remember that 3 cancel the impact of 9.
Yes on the 12max/6min, but no on the 3/9 cancelling.  Almost the entire ring pulls the 2.9kpc object downward.  This is why there is zero net force in a sphere.  To compute the net force of the ring, you need to integrate at least the vertical force vector over the entire ring.
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Therefore, do you agree that the impact of the ring gravity force is higher as the object is closer to the ring?
Not because of your faulty argument above, but yes, it is.  Your argument above would have there being a net force in a sphere, and we know that's false.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/01/2019 20:11:54
Without reading further, a hoop like that is in equilibrium and can orbit itself, with or without a central object.  The mass within (not part of) the ring will contribute to higher speed of the ring, but a ring by itself will have some minimum speed.  It isn't totally stable, but stable enough, just like our asteroid belt, which has a central mass.  I can think of no example of a ring on its own.
Thanks
The activity of the ring is clear. It is also clear that it has a great impact on any object which is located inwards (or outwards).
Let's start by looking inwards:
Let's set back  the SMBH at the center of the galaxy.
We have already found that the impact of the Ring at the center is virtually Zero.
Therefore, any object which is located very close to the center is directly affected by the SMBH. It might be a star, gas cloud or even Atom.
However, as we move further away from the center, the gravity force of the ring is more relevant.
Please remember that the SMBH mass is estimated to be in the range of few millions Sun mass. However, the total estimated mass of the ring is in the range of several billions of sun mass.
Based on that knowledge, let's look again on the orbital velocity at the range of Zero to 3KPC (about 10K LY).
We know that the orbital velocity of the accretion disc is 0.3 speed of light.
The orbital velocity of S2 which is located at a distance of 2-10 Light days from the SMBH is 5000 Km/s
If we get to a distance of 0.1 KPC (326 LY) from the center the orbital velocity gets down to 150 Km/s.
However as we get further away from this point, the orbital velocity stars to increase.
At 0.5 KPC the orbital velocity gets to its pick of almost 250 Km/sec.
So far so good.
We can assume that this increase in the orbital velocity is directly affected by the ring.
However, as we go further away from the center and get closer to the ring the orbital velocity goes down again to its minimal value at the ring - 190Km/s. This is unexpected phenomena.
Why the orbital velocity goes down again as we get closer to the ring?
In order to answer this question - Let's assume that we have only one object which orbits inside the ring.
So, we have the ring, we have the SMBH but only one star which orbits there.
We can easily calculate the estimated orbital velocity of this object at any radius from the accretion ring up to 3KPC.
However, as there are millions (or billions of stars) in the bulge + in the Bar, they also set a gravity force on each other.
Therefore, this could be the answer why we see that the velocity is going down as we get closer to the ring.
There is another factor.
The ring is connected to spiral arms.
So, it is used as a frame which holds the spiral arms and set their orbital velocity.
Therefore, it is clear that the ring is loaded by those spiral arms. This load could also have some impact on the orbital velocity of the ring and the orbital velocity of stars inside the ring.
Do you agree with that?
Please also be aware to a very critical issue.
There is no spiral shape inside the ring!!!
The spiral shape is very clear in the ring itself and in the spiral arms.
Therefore, it is clear that the ring has a significant impact of the spiral arms.
I will discuss about it later on.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 08/01/2019 00:25:10
Let's start by looking inwards:
Let's set back  the SMBH at the center of the galaxy.
We have already found that the impact of the Ring at the center is virtually Zero.
Therefore, any object which is located very close to the center is directly affected by the SMBH. It might be a star, gas cloud or even Atom.
However, as we move further away from the center, the gravity force of the ring is more relevant.
Please remember that the SMBH mass is estimated to be in the range of few millions Sun mass. However, the total estimated mass of the ring is in the range of several billions of sun mass.
Based on that knowledge, let's look again on the orbital velocity at the range of Zero to 3KPC (about 10K LY).
We know that the orbital velocity of the accretion disc is 0.3 speed of light.
I need a reference for that last one.  I don't know that.  The whole disk can't have one uniform speed, so this seems to be stated without context.
Not saying it is wrong, but I'm not sure what exactly is seen to move that that speed.

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The orbital velocity of S2 which is located at a distance of 2-10 Light days from the SMBH is 5000 Km/s
It is that speed when it is 2 light days away.  It goes much slower when it is at the other end of its orbit.  I think it goes around in about 17 years or something.

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If we get to a distance of 0.1 KPC (326 LY) from the center the orbital velocity gets down to 150 Km/s.
However as we get further away from this point, the orbital velocity stars to increase.
At 0.5 KPC the orbital velocity gets to its pick of almost 250 Km/sec.
So far so good.
We can assume that this increase in the orbital velocity is directly affected by the ring.
The ring, if pulling stuff outward, would slow orbits at higher radius within, not increase it.  That might help explain the drop from 250 to 190 as we go from 0.5 kps to 3 kps, but then it should drop more than that.  Meanwhile, it doesn't explain the increase from 150 to 250 as we go from .1 to .5 kps.  The ring doesn't do that, and a Keplerian orbit doesn't either, but then nobody is claiming a Keplerian orbit here.

I'm seeing you produce no mathematics to back whatever it is you're trying to describe.  A drop from 250 to 190 in the .5 to 3 range requires mass inside that range, not mass of the ring outside it.  The drop over that range should be a factor of about 2.5 (√6) if there is no additional mass being orbited.  The drop would be even greater than 2.5 if we take the attraction to the ring into account, which has an additional slowing effect.  So it appears that the ring might have an effect, but not the greatest effect.  There is a lot of mass between the ring and the galaxy center.

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However, as we go further away from the center and get closer to the ring the orbital velocity goes down again to its minimal value at the ring - 190Km/s. This is unexpected phenomena.
Sorry, but it is unclear.  Did you expect things there to go faster or slower?  Is the speed there unexpected, or is the fact that it seems to be at a minimum there unexpected?  What curve had been expected?

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Why the orbital velocity goes down again as we get closer to the ring?
In order to answer this question - Let's assume that we have only one object which orbits inside the ring.
So, we have the ring, we have the SMBH but only one star which orbits there.
We can easily calculate the estimated orbital velocity of this object at any radius from the accretion ring up to 3KPC.
However, as there are millions (or billions of stars) in the bulge + in the Bar, they also set a gravity force on each other.
Therefore, this could be the answer why we see that the velocity is going down as we get closer to the ring.
More stars make things go faster, not slower.  But there don't seem to be enough of them to explain any of the observed speeds.
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There is another factor.
The ring is connected to spiral arms.
So, it is used as a frame which holds the spiral arms and set their orbital velocity.
Conjecture, and not current thinking either.
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Therefore, it is clear that the ring is loaded by those spiral arms. This load could also have some impact on the orbital velocity of the ring and the orbital velocity of stars inside the ring.
Do you agree with that?
You treat the ring and arms as solid objects, talking about loads being put on them and such.  They're not.  They're collections of detached objects which don't alter paths if loads are put on other parts.  You can't alter the speed of the ring by putting a force on part of the ring.  That works only if it is an object.

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Please also be aware to a very critical issue.
There is no spiral shape inside the ring!!!
The spiral shape is very clear in the ring itself and in the spiral arms.
Therefore, it is clear that the ring has a significant impact of the spiral arms.
I will discuss about it later on.
I agree that the spiral seems to end at 3 kps, not spiraling all the way into the center.  There are some galaxies that do that, but they're in the minority.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 12/01/2019 08:16:54
You treat the ring and arms as solid objects, talking about loads being put on them and such.  They're not.  They're collections of detached objects which don't alter paths if loads are put on other parts.  You can't alter the speed of the ring by putting a force on part of the ring.  That works only if it is an object.
I disagree with you,
The ring is an object by itself as the Earth is an object by itself.
The Earth is made by many different of Atoms and molecular that are all bonded by gravity force.
In the same token, the Ring is made by many Stars and dust that are all bonded by gravity force.
In one hand you agree that spiral structure start exactly at the ring:
https://phys.org/news/2018-01-image-hubble-barred-booming-spiral.html
agree that the spiral seems to end at 3 kps, not spiraling all the way into the center.  There are some galaxies that do that, but they're in the minority.
But on the other hand you don't see the great impact of the ring.
Why is it?
Let's look carefully at the following image:
https://en.wikipedia.org/wiki/Spiral_galaxy#/media/File:PIA19341-MilkyWayGalaxy-SpiralArmsData-WISE-20150603.jpg
https://phys.org/news/2018-01-image-hubble-barred-booming-spiral.html
https://www.syfy.com/syfywire/ngc-1398-an-utterly-perfect-spiral-galaxy
We see clearly that the arms are connected exactly at the opposite sides of the ring.
In some galaxies it is quite difficult to see the ring, but we see clearly that spiral arms are connected exactly at the opposite sides of the ring and directly to the bar.
This connection between the Arm to the Bar is very important.
It shows that the spiral arm is an extension of the bar.
So, we can think about it as a long arm. Outside the ring we have the spiral shape. Inwards the ring we have the bar.

If we look carefully we in all the spiral galaxies in the Universe - we can also see that all of them have a very symmetrical view.
Spiral arms starts exactly from the opposite sides of the ring - The arm at one side is very similar to the arm in the other side. So in all the galaxies we mainly see two main arms. We might see other arms in the galaxy, but all of them are connected to those two main arms (or near by the conection point)  in a symmetrical view.
We can also ask: Is it feasible to have a spiral galaxy without a ring?
The answer is clearly - Yes, but a spiral galaxy without a ring is quite different from a one with the ring
Please look at the following image:
http://hubblesite.org/image/1636/news_release/2005-01
Do you agree that the ratio is about 1:3?
In the Milky Way the bar gets to 3 KPC while the spiral arms gets to 45 KPC.
So the ratio in the Milky way (with a ring) is 1:15.
Hence, as the ring is more developed, the ratio is higher.
Do you agree that without a ring (or with minimal ring) there is a maximal load (or maximal spiral arm size/radius) that spiral galaxy can take?

Therefore, we have to ask the following:
1. How could it be that the bar is connected to the ring while at the same spot of the connection we clearly see the starting point of the spiral arms? Our scientists only try to explain the orbital velocity in the arm, but they ignore completely the Bar. How could it be that they don't see that the bar is directly connected to the arm?
2. Why there is no spiral shape in the Bar?
3. How could it be that at the ring, the orbital velocity is at its minimal value (190Km/s)?
If we move inwards to the bar (let's assume in the Milky Way - 500 Light years inwards from the ring), the orbital velocity is higher (at about 200 Km/s), while if we move outwards into the spiral arms at the same distance, the orbital velocity is higher again (at about 200 Km/s).
4. Why spiral galaxies have a symmetrical view?
5. What is the real impact of the ring?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 12/01/2019 14:41:23
Quote from: Halc
You treat the ring and arms as solid objects, talking about loads being put on them and such.  They're not.  They're collections of detached objects which don't alter paths if loads are put on other parts.  You can't alter the speed of the ring by putting a force on part of the ring.  That works only if it is an object.
I disagree with you,
The ring is an object by itself as the Earth is an object by itself.
The Earth is made by many different of Atoms and molecular that are all bonded by gravity force.
In the same token, the Ring is made by many Stars and dust that are all bonded by gravity force.
If that were true, I would immediately drop straight through and fall to the core of the Earth because that is where gravity is trying to take me.  The fact that it doesn't happen means that the Earth is bonded by forces other than just gravity, forces that do not exist between the various objects in the ring.

Why should I continue this thread when you post something that obviously wrong?

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We see clearly that the arms are connected exactly at the opposite sides of the ring.
In some galaxies it is quite difficult to see the ring, but we see clearly that spiral arms are connected exactly at the opposite sides of the ring and directly to the bar.
This connection between the Arm to the Bar is very important.
There are no connections.  These are not solid objects.  Two smears of dust that intersect do not make physically connected dust.  If that connection is important, then your idea really falls flat, because there's no connection.  Everything in those pictures is free-falling objects interacting only by gravity, much in the same way that the atoms of Earth or a spanner are not.

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Please look at the following image:
http://hubblesite.org/image/1636/news_release/2005-01
Do you agree that the ratio is about 1:3?
About that, yes, but only if you don't include the dimmer outer parts of the arms that they cropped off the image, but you're including those parts when you say the milky way goes all the way to 45 kps:
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In the Milky Way the bar gets to 3 KPC while the spiral arms gets to 45 KPC.
So the ratio in the Milky way (with a ring) is 1:15.
If you look at the Milky Way image in your first link of the prior post, the bar ratio seems to be about 1:5 or 1:6, and a similar ratio for that 'perfect spiral' galaxy you link.
So the ratio is higher, but not a lot higher.

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Do you agree that without a ring (or with minimal ring) there is a maximal load (or maximal spiral arm size/radius) that spiral galaxy can take?
I don't know what 'load' is.  If you're treating the arms or ring as solid objects, then the question is nonsensical.
OK, you say it is a maximal galactic radius, but that is a function of how much mass and angular momentum the galaxy has.  Mass holds it together, and angular momentum flattens and spreads it out.  These numbers are different for different galaxies.  The symmetry of the galaxy has a lot to do with its recent history.  That perfect spiral galaxy has not been disturbed in quite some time, where our galaxy has been continuously eating small snack-size neighbors, and is soon to be eaten itself by a larger fish, all of which will at least temporarily destroy that nice bar/ring/ arm picture of both galaxies.

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Our scientists only try to explain the orbital velocity in the arm, but they ignore completely the Bar. How could it be that they don't see that the bar is directly connected to the arm?
Probably because it isn't.

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3. How could it be that at the ring, the orbital velocity is at its minimal value (190Km/s)?
Perhaps because there is not much additional mass inside the ring to make it go faster.
Maybe the velocity in some places is not tangential.  The graphs do not show velocity, only speed.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 13/01/2019 06:13:19
If you look at the Milky Way image in your first link of the prior post, the bar ratio seems to be about 1:5 or 1:6, and a similar ratio for that 'perfect spiral' galaxy you link.
So the ratio is higher, but not a lot higher.
Thanks
So, we agree that with a ring the ratio between the bars radius to the maximal arm radius is higher.
If so, why do we ignore the outcome of this evidence?

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Our scientists only try to explain the orbital velocity in the arm, but they ignore completely the Bar. How could it be that they don't see that the bar is directly connected to the arm?
Probably because it isn't.
Why do you claim that it isn't?
Please look again at the following image:
https://en.wikipedia.org/wiki/Spiral_galaxy#/media/File:PIA19341-MilkyWayGalaxy-SpiralArmsData-WISE-20150603.jpg
1. The starting point of the Soutum-Centaurus arm is located at the same spot where the Bar ends?
2. Exactly at the other end of the bar we see a symmetrical view. The starting point of Perseus arm is located at that spot.
3. Those two arms seems to be main arms due to their length and width.
4. The ring cross exactly at those  two edges/spots of the bar connections/locations with the those two main arms.
5. Based on this image, it is clear the bar is orbit in clock wise.
6. Near the Soutum - Centaurus arm starting point to the end bar location spot, we also see a starting point of a smaller thinner arm. However, it is located a little bit further from the clock wise orbital direction of the bar.
7. In full symmetrical view, we see another thinner arm which its starting point is located near the other Bar/main arm spot. This arm also starts a little bit further from the clock wise orbital direction of the bar.
8. So do you agree that we see clearly a symmetrical view from both end sides of the Bar?

Don't you think that there is a meaning for what we see?

As you don't like to call this spot as a "connection" point, do you estimate that the bar can move faster/slower than the starting point of the main arms and one day it should be disconnect from the arm?
If so, can you please offer an image of a Barred spiral galaxy with two main arms, while its bar (ends) had been totally disconnected from those two main arms?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 13/01/2019 14:47:23
So, we agree that with a ring the ratio between the bars radius to the maximal arm radius is higher.
If so, why do we ignore the outcome of this evidence?
Evidence of what?  A general trend?  No.  The cherry-picked dataset of 3 is not evidence of that.  A trend can be found in a survey of a lot of galaxies.
In that small dataset, I see a tightly wound set of arms, and a loosely wound set.  The one is so tight that it almost looks like a ring, just like toilet paper looks like concentric circles even though it isn't.
I suspect that the loose galaxy had less angular velocity.  Perhaps such slowly rotating barred galaxies generally have larger bar-to-galaxy ratios, but only a larger survey would be evidence of that.

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5. Based on this image, it is clear the bar is orbit in clock wise.
The bar and arms are not clearly in orbit.  The shape of them seems stable, but the image does not show motion, and the ones that do show motion do not show motion that corresponds with that rotation curve.  Thus I do not see the motion of either bar nor arms qualifying as 'orbital motion'.  Yea, they twirl around, but that's just spinning, not necessarily orbiting.

What I see in the arms is that telltale fallen-dominos signature, where material seems grouped in diagonal clumps that overlap. I can really see it in that 'perfect spiral' galaxy you linked, especially in the outer parts of the arms. It is a familiar image:
https://en.wikipedia.org/wiki/Wake#/media/File:Bodensee_at_Lindau_-_DSC06962.JPG
The duck wake appears completely stationary to a camera that is following the duck, and has that overlapping fallen-dominos signature, not just one clean wave that angles out.  The material of the waves does not follow the wave/duck however.  The water is moving in a completely different direction than is the wake.  The wake appears to be a solid object connected to the duck, but it is not connected at all.
The galaxy appear to work exactly that way.  Apparent motion of the arm waves are just wave action, not actual motion of the stars that make them up.  You cannot deduce the arm motion from the motion of the component stars and vice versa.

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8. So do you agree that we see clearly a symmetrical view from both end sides of the Bar?
Almost all galaxies look little different if rotated 180 degrees, so yes, I agree with this one.

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Don't you think that there is a meaning for what we see?
You don't identify what meaning you expect me to get from that, but probably not the sort of things you've been pushing like any of it behaving as a solid.  There are simply no forces that could account for that.  A spanner holds its shape because there are very much forces that connect the components.  Those forces are not gravitational.

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As you don't like to call this spot as a "connection" point, do you estimate that the bar can move faster/slower than the starting point of the main arms and one day it should be disconnect from the arm?
You would have images of galaxies then where that happens.  No, the [main] arms always start at the bar ends, just like the wake always starts at the duck despite not being connected .
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 13/01/2019 17:36:45
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As you don't like to call this spot as a "connection" point, do you estimate that the bar can move faster/slower than the starting point of the main arms and one day it should be disconnect from the arm?
You would have images of galaxies then where that happens.  No, the [main] arms always start at the bar ends, just like the wake always starts at the duck despite not being connected .

Thanks

So now we agree that the main arms always starts at the bar end!!!
But why is it?
What is the Bar?
Based on Wiki:
https://en.wikipedia.org/wiki/Barred_spiral_galaxy
"The creation of the bar is generally thought to be the result of a density wave radiating from the center of the galaxy whose effects reshape the orbits of the inner stars. This effect builds over time to stars orbiting further out, which creates a self-perpetuating bar structure."
As the bar had been set due to "density wave", can we claim that it is an object?
If the ring isn't an object than how could it be that the bar is an object?
If it isn't an object, how can we compare it to duck?

Now, let's look at the spiral arm? What is spiral arm?
https://earthsky.org/space/galaxies-spiral-arm
Astronomers believe that galaxies have spiral arms because galaxies rotate – or spin around a central axis – and because of something called “density waves.
So, the bar is there due to "density wave", while the spiral arms is there due to the same idea of "density wave".
Hence, how could it be that the bar has a totally different shape and structure from the spiral arm while they both had been created based on the same concept of density wave?
How could it be that the "second density wave" (spiral arm) always starts at the end of the "first density wave" (Bar)?
Why just at 3KPC (exactly at the ring) the "first density wave" (bar) had been changed its shape to "second density wave" spiral arm?
If it is all about "density wave", can we calim that the ring is also there due to density wave?
Why the galaxy needs the Bar or the ring?
Why the spiral arms can't start directly from the Bulge or even directly from the SMBH?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 13/01/2019 21:06:01
As the bar had been set due to "density wave", can we claim that it is an object?
If the ring isn't an object than how could it be that the bar is an object?
Waves are not objects.  A density wave moves through water at about 1500 m/sec despite no bit of water moving at that speed.  Waves don't have mass of their own, and thus are not subject to forces like objects are.  A wave traveling from the depths to the surface for instance is not slowed by gravity at all.  It is a mistake to treat waves as objects like the spanner where the whole spanner reacts if you exert force at only one point of it.

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If it isn't an object, how can we compare it to duck?
I was comparing everything to the wake, not the duck.

I don't know the dynamics of stellar motion within the bar.  I've seen a few pictures of paths of a single star in simulations, and the stars seem to stay within the bar, but move up and down the length of it. That wouldn't happen if it was a solid object.  But still, the star stayed in the bar (while stars in the arms seem not to stay in their arms), so in that way the bar does resemble a sort of liquid object like a blob of water floating in the space station.  The water holds itself together despite minor disturbances, but it also doesn't form a bar shape since the forces are different than the gravitational ones acting on the galactic bar.

Translation, the bar may well be an 'object' of sorts, a bit like the duck, but a liquid one at best, not a solid one. I'm willing to entertain that so long as you don't treat it like a solid spanner. The bar doesn't look like a wave, even if it formed due to density waves from the center.  The arms on the other hand look very much like density waves, like the wake of the duck.

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Astronomers believe that galaxies have spiral arms because galaxies rotate – or spin around a central axis – and because of something called “density waves.
Look up 'density wave theory' on wiki.  There are some really good illustrations that show what they're talking about.
Here's a gif showing how a wave forms moving at one speed when all the material moves at a different speed.
https://imgur.com/gallery/dtb8WrD
Follow any particular star and it doesn't stay with either arm.

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So, the bar is there due to "density wave", while the spiral arms is there due to the same idea of "density wave".
Hence, how could it be that the bar has a totally different shape and structure from the spiral arm while they both had been created based on the same concept of density wave?
I don't understand the dynamics of the bar.  It seems to not be a density wave itself like the arms are, but I could be wrong about that. They give a reference to a paper on the subject, but it is heavy reading. The site says it is formed by density waves from the center, reshaping orbits, and that effect apparently grows. I know that a lot of the 'orbits' is not stars at all, but just dense gas clouds being reshaped.  A lot of stars are born in there from that gas.
I suspect the galaxies with long bars have been undisturbed for a long time, and those with nonexistent bars are in the process of merging, which disrupts all the symmetry.  If this is true, then we should see galaxies that have recently 'eaten' and are forming new bars and reestablishing the symmetry lost during the merger.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 14/01/2019 16:09:33
I don't understand the dynamics of the bar.  It seems to not be a density wave itself like the arms are, but I could be wrong about that. They give a reference to a paper on the subject, but it is heavy reading.
I assume that even the scientist that wrote the article has no clue how the density wave creates the Bar.
If it was clear to him, he could explain it in few words. There is no need for "heavy reading" to explain something which is quite clear.
Please also be aware that he doesn't even try to explain how the ring had been formed due to density wave.

There are some really good illustrations that show what they're talking about.
Here's a gif showing how a wave forms moving at one speed when all the material moves at a different speed.
https://imgur.com/gallery/dtb8WrD
Follow any particular star and it doesn't stay with either arm.
I wonder how can we get any valid information from this illustration.
There is no ring and no bar in the illustration.
We only see stars that are moving in and out in order to set the spiral arms.
If that was correct, than by definition the 400 Billion stars that move in and out should collide with each other..
How many collisions do we see in the galaxy?
In any case, in this illustration there is no ring no bar no bulge - only spiral arms. So how can we use it as a valid illustration?
Spiral galaxy is not just spiral arms. It's time for our scientists to walk up and see what there is in spiral galaxy.
There is no way that density wave can set all of those different shapes - Bar, ring and spiral arms.

:
https://en.wikipedia.org/wiki/Wake#/media/File:Bodensee_at_Lindau_-_DSC06962.JPG
The duck wake appears completely stationary to a camera that is following the duck, and has that overlapping fallen-dominos signature, not just one clean wave that angles out.  The material of the waves does not follow the wave/duck however.  The water is moving in a completely different direction than is the wake.  The wake appears to be a solid object connected to the duck, but it is not connected at all.
We see the wake - But it has a specific shape.
For example - There is no way to get a wake in a ring shape, zig zag, or that moves in front of the duck.
So, if there is a density wave - it can't technically set all the variety shapes that we need (Bar, Ring, spiral arms, Bulge) and keep each shape at a specific radius and aria while the bar ends exactly at the starting point of the main spiral arms and this point cross the ring.
This is too challenging request from a simple idea of - "Density wave"
Therefore, I would assume that there is a severe mistake with that idea
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/01/2019 00:28:17
I assume that even the scientist that wrote the article has no clue how the density wave creates the Bar.
I don't think any of the articles you've linked have been written by scientists.  The last article claimed no author and was just a seeming effort from magazine staff or something.
I've occasionally linked to articles from actual scientists/physicists, but most of them don't write for general audiences.

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I wonder how can we get any valid information from this illustration.
There is no ring and no bar in the illustration.
It was meant to illustrate a density wave, not be a simulation of a barred galaxy.  I've seen such simulations, but the short videos made from them compress the timespan too much to observe the dynamics of waves vs stars and other material.

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We only see stars that are moving in and out in order to set the spiral arms.
If that was correct, than by definition the 400 Billion stars that move in and out should collide with each other..
By definition of what?  Most stars are pretty much moving the same way in relation to the galaxy as a whole, not in opposite directions.

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We see the wake - But it has a specific shape.
For example - There is no way to get a wake in a ring shape, zig zag, or that moves in front of the duck.
Two ducks circling each other?
The picture shows what a wake arm looks like.  Yes, galactic ones are from rotating motion, not linear motion.  There will be no V wake from that.  The shape I was illustrating was the 'fallen domino' effect, which exactly matches what I see in the arms of galaxies.

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So, if there is a density wave - it can't technically set all the variety shapes that we need (Bar, Ring, spiral arms, Bulge) and keep each shape at a specific radius and aria while the bar ends exactly at the starting point of the main spiral arms and this point cross the ring.
This is too challenging request from a simple idea of - "Density wave"
Not knowing how it works, I am in no position to declare that it cannot work.  They have simulations that reproduce it, and these simulations use only gravity acting on each thing, not physical connections.  That means it is not too challenging of a request.  They needed to posit dark matter to get it to work, but no new laws like a 5th force or something.

You've suggested no model so far, posting only observations and little actual explanation for any of them.

I think a model of a solid spinning ring and bar with solid arms attached to it would result in an asterisk-shaped galaxy with the arms sticking straight out like they would if you spun out in space a ring with attached flexible arms like that .  That is the best description I've seen so far of your model.
There would need to be a rope or some physical structure holding our sun to the arm so it isn't flung away with the spinning, but we see no structure attaching our sun to the arm.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 15/01/2019 05:47:22
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So, if there is a density wave - it can't technically set all the variety shapes that we need (Bar, Ring, spiral arms, Bulge) and keep each shape at a specific radius and aria while the bar ends exactly at the starting point of the main spiral arms and this point cross the ring.
This is too challenging request from a simple idea of - "Density wave"
Not knowing how it works, I am in no position to declare that it cannot work.  They have simulations that reproduce it, and these simulations use only gravity acting on each thing, not physical connections.  That means it is not too challenging of a request.
In all the illustration that I have found, there is no answer to the real image of Spiral galaxy as the Milky way.
I couldn't find an illustration which shows clearly the two main major arms which start directly from the end of the bar in full symmetrical view including the Bulge, ring and the Bar.

They needed to posit dark matter to get it to work, but no new laws like a 5th force or something.
I have no idea what is the meaning of the 5th force, but there is no need for it. There is also no need for dark matter.
Gravity force - That's all we need to set our galaxy.

I think a model of a solid spinning ring and bar with solid arms attached to it would result in an asterisk-shaped galaxy with the arms sticking straight out like they would if you spun out in space a ring with attached flexible arms like that .  That is the best description I've seen so far of your model.
I agree. If you try to set that kind of model without two basic elements, it won't work.

There would need to be a rope or some physical structure holding our sun to the arm so it isn't flung away with the spinning, but we see no structure attaching our sun to the arm.
There is no rope or some physical structure holding our Moon around the Earth. The same force which holds the Moon in its orbital path also holds the sun in the arm.
You've suggested no model so far, posting only observations and little actual explanation for any of them.
Agree.
Now that it is clear that our scientists have a fatal error in their illustration - it's time for me to present my model.

However, the two basic elements are:
1. There is a friction also in gravity force. This friction reduces the gravity force over time. The only way to reduce the gravity force of an object is by increasing its radius.
2. All the stars in the galaxy must orbit around some virtual host (I will explain it later on).

Once we know those two key elements – We have actually solved the Spiral galaxy enigma.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 15/01/2019 19:40:30
Quote from: Halc
They needed to posit dark matter to get it to work, but no new laws like a 5th force or something.
I have no idea what is the meaning of the 5th force, but there is no need for it. There is also no need for dark matter.
Gravity force - That's all we need to set our galaxy.
I meant no new laws.  The 5th force was just an example.  You create new laws, all of physics must be redone from scratch, which makes for a lousy theory when the existing laws suffice.

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It's time for me to present my model.

However, the two basic elements are:
1. There is a friction also in gravity force. This friction reduces the gravity force over time. The only way to reduce the gravity force of an object is by increasing its radius.
I count several violations of existing laws of physics here.  Gravitational force is described by GMm/r², and that has no term for friction or for time.  Friction requires physical contact and has an effect on momentum, and has nothing to do with gravitational force.  So you need to explain what you mean by that.
I can go down a greased slide quickly or a sand-paper slide with lots of friction, and the latter will have me going slower and bloodier at the bottom of the slide, but the gravitational force of Earth on me will be completely the same in both cases.
What else?  Oh yea:  Gravity force of an object is increased by its mass, not by its radius, and an object has no gravity force of its own.  Gravitational force is defined between two objects (the M and m above).  One object creates a field with acceleration (not force) that is a function of the object's mass (not radius of the object) and distance (r) from it.  So that's at least 3 things that seem to not agree with existing physical law in your #1.

Maybe you're just not being clear, but that's what I'm reading here.

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2. All the stars in the galaxy must orbit around some virtual host (I will explain it later on).
You've attempted explanation of this at length.  Watch out for Newton's third law, because this particular notion seems to violate it heavily.

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Once we know those two key elements – We have actually solved the Spiral galaxy enigma.
Actually describing your model would also be a plus.  A list of elements is just hand-waving if I cannot predict the Sun's motion from it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 17/01/2019 16:03:49
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It's time for me to present my model.
However, the two basic elements are:
1. There is a friction also in gravity force. This friction reduces the gravity force over time. The only way to reduce the gravity force of an object is by increasing its radius.
Friction requires physical contact and has an effect on momentum, and has nothing to do with gravitational force.  So you need to explain what you mean by that.
That is incorrect.
Based on the current idea of our scientists, there is tidal friction.
In Tidal friction there is no direct contact or friction between the Earth and the moon, however, somehow it cause the Moon to increase its orbital radius.
It shows that even our scientists believe that without direct contact there is a friction.
Therefore in orbital system there is a friction even without a direct contact.

Gravitational force is described by GMm/r², and that has no term for friction or for time.

I agree
At any given moment the formula for gravity force is as follow:
F=GMm/r²
However, Newton didn't take in his account the impact of time frame.
Somehow, our scientists believe that this force can last forever. This is incorrect.
I do believe that even gravity force must be reduced over time (While there is no change in the mass).
You can call it "Tidal friction", "Gravity friction" or "Gravity force reduced over time". The name of that phenomenon is not relevant.
The key idea is that ALL the planets and ALL the moons in the solar system are drifting outwards over time.
You would never ever find even one planet or one moon in the solar system that drifts inwards.
If we could set the measurements - we should find that ALL of them are drifting outwards over time.
Therefore, after each orbit, the orbital object must shifts/drifts outwards. It can be just few Pico millimeter per cycle or few Km per cycle - but it is there and it is real.
So, orbital cycle is spiral by definition.
This is something that our scientists had missed.
We had long discussion on that issue. So, please don't start all over again with the hypothetical idea why it isn't realistic.
I'm ready to accept your claim - If you can prove (only by real measurements) that there is even one moon or planet that drifts inwards. If you can't prove that there is a moon or planet in a solar system that drifts inwards, you can't contradict this key element phenomena in orbital system.
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2. All the stars in the galaxy must orbit around some virtual host (I will explain it later on).
You've attempted explanation of this at length.  Watch out for Newton's third law, because this particular notion seems to violate it heavily.
Please look again at the following diagram.
http://www.biocab.org/Motions_of_the_Solar_System.jpg
We see clearly that the sun orbits around some virtual host point, while this host point orbits around the galaxy (Gray Line)
We have also discussed this issue. So, I don't see a value to start all over again.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 17/01/2019 18:33:14
Quote
It's time for me to present my model.
However, the two basic elements are:
1. There is a friction also in gravity force. This friction reduces the gravity force over time. The only way to reduce the gravity force of an object is by increasing its radius.
Friction requires physical contact and has an effect on momentum, and has nothing to do with gravitational force.  So you need to explain what you mean by that.
That is incorrect.
Based on the current idea of our scientists, there is tidal friction.
...
It shows that even our scientists believe that without direct contact there is a friction.
The friction is mostly between water and the sea floor, which is direct physical contact, and is where all the heat is generated.
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In Tidal friction there is no direct contact or friction between the Earth and the moon, however, somehow it cause the Moon to increase its orbital radius.
...
Therefore in orbital system there is a friction even without a direct contact.
A fact that you have repeatedly denied in prior posts, but here you are pushing gravitational friction.  What moves the moon to a higher orbit is a positive gravitational force vector.  There is no friction in this.  The moon does not heat up, only the Earth where the friction with the water takes place.

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At any given moment the formula for gravity force is as follow:
F=GMm/r²
However, Newton didn't take in his account the impact of time frame.
Somehow, our scientists believe that this force can last forever. This is incorrect.
If you mean that they believe that the force will not change over time when none of the variables (M,m,r) change, then I agree with them.  If you're saying something else, then I need a reference for this claim.

There are theories that speed of light was different in the past, and this affects a lot of 'constants', G possibly being one of them.
If your nonexistent model involves a significant change in recent time, you are rewriting physics.
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I do believe that even gravity force must be reduced over time (While there is no change in the mass).
Yea, like that.  My point exactly.

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You can call it "Tidal friction", "Gravity friction" or "Gravity force reduced over time". The name of that phenomenon is not relevant.
It is entirely relevant because these are quite different theories and they make different predictions.
Your theory predicts all orbits should increase by the same proportion over time, but the moon orbit increases (as a ratio of its orbital radius) by 1,500,000 times more than does Earth, instead of the same ratio as you predict.
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The key idea is that ALL the planets and ALL the moons in the solar system are drifting outwards over time.
Yes, but they're not.  All of Mars' moons and over half of Jupiter's moons are losing orbit.
If you model depends on this idea, it is trivially falsified.

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You would never ever find even one planet or one moon in the solar system that drifts inwards.

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This is something that our scientists had missed.
They measured otherwise, failing to wisely realize the truth and discard such obviously biased findings.
Conservation laws were always wrong anyway.

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If you can't prove that there is a moon or planet in a solar system that drifts inwards, you can't contradict this key element phenomena in orbital system.
Orbit of Phobos is changing more than the orbital radius of any other measurable object.  It is measured as losing over 34 cm per Martian year, or 18 cm per Earth year.  You can choose to deny this of course since the scientists failed to discard this finding each and every time they measure it.  You can also choose to deny when the thing rips apart due to getting too close.

Earth is drifting outward (trivially), but due mostly to a change in M, not a change in G.
Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 17/01/2019 22:19:06
We have not been understanding gravity completely. I explain that in detail in my paper mentioned below. There-in, the GDE = Gravitation Direction of Evolution and the FDE = Fundamental Direction of Evolution.

This is an excerpt from the paper which I am also discussing in the "Is The Hubble Shift Due To Time Dilation" and "What is Space" threads in New Theories. This is from the section on Galactic Rotation Velocities:

"The masses of flattened spiral galaxy systems and spherical stellar systems have different shapes and, therefore, different shaped time dilation gradients and different effects in the time aspect.

Within a stellar system, where GR works so well, as the dilation gradient deepens more quickly as the center of the dilation "pit" is approached, all events appear to accelerate increasingly in spacetime, appearing to evolve forward faster through its apparently faster velocity “through/in” space. The dilation gradient only equalizes in an infinitesimal focal point at the center of the star, impeding the forward evolution of events in all directions, concentrating energy.

In Einstein’s 1915 paper, substituting X, Y, Z, T for his X1, X2, X3, X4, his Fundamental Metric, which can be considered the basis of the tensors describing a null gravitational field, is:

X   Y   Z   T
X   -1   0   0   0
Y   0   -1   0   0
Z   0   0   -1   0
T   0   0   0   +1

In flattened spiral galaxies, designating the Y axis as being orthogonal to the flat galactic disk, the dilation gradients along the +Y and -Y axes above and below the flat mass of the disk equalize within the disk and as the dRt → 0 along the Y axes, ∆Y→ 0.

As ∆Y = 0 at Y = 0 in the middle of the plane of the galactic disk, the Galactic Fundamental Metric within the disk of a flattened spiral galaxy is:

X   Y   Z   T
X   -1   0   0   0
Y   0   0   0   0
Z   0   0   -1   0
T   0   0   0   +1

As all the Y elements go to 0, this metric can be reduced to:

X   Z   T
X   -1   0   0
Z   0   -1   0
T   0   0   +1

As with Einstein’s Fundamental Metric, this Galactic Fundamental Metric cannot be realized in finite space as it also represents a null gravitational field without time dilation. ∆Y also never actually remains at 0 since particles oscillate above and below the plane of the galactic disk.

However, in this fundamental metric without Y elements, forward evolution can only proceed through the X and Z axes, which share a common plane, and we get circular motion around the center of the galactic mass, orthogonal to the dilation gradients. Note that the orbits in a stellar system are also orthogonal to the dilation field. As above, the GDE can only manifest orthogonal to the FDE.

There is a secondary GDE in along the edges where the Y components lose their dominance and we see an evolution inward and the formation of bars.

As the +1 in Einstein’s g44 element of the Fundamental Metric represents an invariable rate of time for all frames along all axes, the +1 of the g33 element in the Galactic Fundamental Metric represents an invariant rate of time along the X and Z axes. Within the galaxy’s dynamic metric, these time elements then change relative to the mass density along the spiral arms, and the apparent velocity relative to adjacent frames is determined by the relative rates of time along the +Y and -Y axes at Y = 0. In the absence of an accelerating gradient as in a stellar system, the relative rate of evolution within the inertial frames of the disk is primarily determined by the rate of time in the inertial frames. Thus, we see an initial rapid increase in velocities near the center of the galaxy, where the rate of time rapidly increases with distance from the massive central MECO (Magnetic Eternally Collapsing Object - the latest in black hole development).

Velocities appear slower between the arms, despite the faster rate of time, due to the shallower gradient. Within the arms, where densities are concentrated, the gradient is deeper along the Y axes and velocities appear
accelerated more as they do in a deeper gradient in a stellar system.  The dilation gradients along the Y axis decrease in slope as mass densities decrease along the arms, time goes faster at Y = 0, and we see a slightly faster evolution (apparent velocity) of the stellar systems within the continuum with distance from the galactic center.

The g33 element also varies slightly relative to the slope and depth of the gradients within the individual stellar systems. This slope effect also manifests the same as we see in a stellar system where relative acceleration increases as the gradient of the slope deepens. A test of this would be that larger masses and groups of masses should therefore appear to be evolving forward faster, and appear to have higher velocities, relative to nearby smaller masses due to their deeper, steeper, individual gradients.

Although the primary dilation gradient is along the Y axes, as the disk flattens there is also a secondary gradient looking in from the edges. The evolution in this directions forms the bars of Sb galaxies."

The full 21 page paper, General Relativity: Effects in Time As Causation, is on vixra.org at http://vixra.org/abs/1804.0109#comment-3850079405 and a final version will be published in some form by a cosmological journal in what I hope is the near future, as discussed in the other threads.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/01/2019 04:33:28
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The key idea is that ALL the planets and ALL the moons in the solar system are drifting outwards over time.
Yes, but they're not.  All of Mars' moons and over half of Jupiter's moons are losing orbit.
If you model depends on this idea, it is trivially falsified.

Can you please specify one moon (around Mars or Jupiter) that we have measured its orbital radius and verify that it is really losing orbit?
You know that we just assume that those moons are drifting outwards, but in reality we didn't confirm our hypothetical idea.
So, please don't say something which is incorrect.
You can't say for sure that those moons drift outwards without real measurements.
It was a big surprise for our scientists when they have discovered that the moon is drifting outwards from Earth. No one at that time had expected to see that increase in the orbital radius.
Therefore, it will also be a big surprise for our scientists when they will discover that ALL moons in the solar system are drifting outwards (Including All of Mars' moons and ALL Jupiter's moons).!!!

The friction is mostly between water and the sea floor, which is direct physical contact, and is where all the heat is generated.
The friction in the Earth is none relevant to the friction between the Earth to the Moon.
Just a simple example -
Let's assume that you sit at a bus while it drive at 120 Km/s
If you set a friction between your hand to the bus body, does it mean that you set a friction between the bus and the road?
In the same token, a friction between water and the sea floor on the Earth can't set a friction between the Earth and the Moon.
Orbital objects as Moon and planets are losing gravity force over time. This is a normal activity in gravity. Therefore, the Moon increases its orbital radius (around the earth) over time.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/01/2019 06:16:04
Can you please specify one moon (around Mars or Jupiter) that we have measured its orbital radius and verify that it is really losing orbit?
You know that we just assume that those moons are drifting outwards, but in reality we didn't confirm our hypothetical idea.
See my prior post, giving measured numbers for Phobos, the moon with the largest orbital change per year of anything in the solar system.  It has been repeatedly measured, not just assumed.

The friction is mostly between water and the sea floor, which is direct physical contact, and is where all the heat is generated.
The friction in the Earth is none relevant to the friction between the Earth to the Moon.[/quote]
There is no direct friction between those two.  They don't touch.  The energy added to the moon is pure gravitational.  If there was friction, it would lose kinetic energy and drop down.  Friction reduces total mechanical energy, not increases it.  I can think of no exception to this.

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Just a simple example -
Let's assume that you sit at a bus while it drive at 120 Km/s
If you set a friction between your hand to the bus body, does it mean that you set a friction between the bus and the road?

The friction between the bus and the road was already there.  My hand on the interior doesn't change that.  There is not much friction with the road since little is sliding across it.  Most of the friction would be with the air and the internal bearings of the bus and such.  These are parts that move across each other by physical contact.  Not so much the road since the tires move at the same velocity as the road where they touch, even when accelerating or braking.
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In the same token, a friction between water and the sea floor on the Earth can't set a friction between the Earth and the Moon.
I didn't claim there was friction between Earth and the moon.  They don't touch.  The dynamics between the two is simple gravitational thrust equivalent to a rock accelerating when spun around in a sling shot.

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Orbital objects as Moon and planets are losing gravity force over time. This is a normal activity in gravity. Therefore, the Moon increases its orbital radius (around the earth) over time.
I pointed out several falsification tests for this.  You ignored them, exactly as I assumed you would.  You don't do science.  You ignore any evidence against your view, and cherry pick facts out of context to support a theory that has never been described.  I see no formula for gravitational weakening over time.  You make no explanation as to why galaxies in the past seem to have the same gravity as modern ones.  This is exactly how the scientific method doesn't work.

In addition, your idea (I haven't seen a theory based on the idea) doesn't explain the sun's acceleration about the galaxy.  If gravity is weakening, all objects should be accelerating less than the scientists predict, and even they predicted an acceleration less than what is observed.  Your idea makes that problem even worse, not better.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/01/2019 10:18:42
See my prior post, giving measured numbers for Phobos, the moon with the largest orbital change per year of anything in the solar system.  It has been repeatedly measured, not just assumed.
https://en.wikipedia.org/wiki/Phobos_(moon)
Phobos has dimensions of 27 km × 22 km × 18 km,[1] and retains too little mass to be rounded under its own gravity.
So, Phobos is very small object (comparing to real moons) and it doesn't have a clear ball shape structure. It has no symmetrical shape.  It is clearly a broken object.
Therefore, why don't we call it asteroid? No more than that.
For example - Vesta — the second-largest asteroid in the solar system at 525 km (326 miles) in size and is located in the asteroid belt between Mars and Jupiter.
So, how could it be that an object at about 500 Km we call Asteroid, while a broken object about 25 Km we call moon...?

"Phobos orbits 6,000 km (3,700 mi) from the Martian surface, closer to its primary body than any other known planetary moon. It is so close that it orbits Mars much faster than Mars rotates, and completes an orbit in just 7 hours and 39 minutes."
So, this asteroid is very close to Mars.
So the ratio between mars radius to Phobos orbits from the Martian surface is 1:2.
At this ration it is clear that it is drifting inwards.
Therefore, Phobos is a wrong example.
Please try to offer a real moon (not a small broken object) with a relative longer orbital radius (similar to the Earth moon ratio).
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/01/2019 15:34:19
So, how could it be that an object at about 500 Km we call Asteroid, while a broken object about 25 Km we call moon...?
Phobos orbits a planet.  Vesta does not.  That's the difference between moon and asteroid.  Vesta isn't large enough to clean up its orbit, which is why it isn't considered a planet in itself.
The names we give to these types of objects does not have any effect on how gravity acts on them.

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"Phobos orbits 6,000 km (3,700 mi) from the Martian surface, ... "
So, this asteroid is very close to Mars.
So the ratio between mars radius to Phobos orbits from the Martian surface is 1:2.
More like 1:2.75.  Orbital radius is measured from the center, not the surface, so it is about 9400 km.  6000 km is its altitude, not its orbital radius.

There are several moons of other planets with tighter ratios than that.  Mimas (around Saturn) has a slightly lower ratio, and is quite a nice large round moon, since you seem to find that important.
Thebe (around Jupiter) is a misshapen lump about 4 times the size of Phobos and orbits at a ratio a little larger (3x) the radius.  Both these moons are increasing their orbits.

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At this ration it is clear that it is drifting inwards.
You spend pages of posts denying inward drift, and suddenly it is clear that it must, totally without explanation.  Does gravity only grow less over time if it is working on large things?  Is there a separate increasing-over-time gravity for the little stuff?
All I see is that you find you cannot continue to deny that Phobos is losing altitude.  You've denied this in several of the recent posts, but all of a sudden "it is clear" that it must.  What is happening is that your idea was falsified.  Weakening gravity would send Phobos to higher orbits if only strength of gravity determined such things.
Earth should be moving away from the sun at 1,500,000 times the value seen to keep up with the decreasing gravity that accounts for the moon movement.  Earth is a bigger round thing, so not under the just-made-up law of non-weakening of gravity for little things.

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Please try to offer a real moon (not a small broken object) with a relative longer orbital radius (similar to the Earth moon ratio).
All the big ones have positive orbital periods and lie outside geosync radius, so they all move outward per tidal forces.    Lack of a large counterexample does not explain why suddenly you find it clear that Phobos orbit must be degrading when all the prior posts asserted otherwise.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 18/01/2019 16:32:16
Lack of a large counterexample does not explain why suddenly you find it clear that Phobos orbit must be degrading when all the prior posts asserted otherwise.
Phobos is none relevant due to its size, shape and its orbital radius ratio.
However, you have stated that:
All of Mars' moons and over half of Jupiter's moons are losing orbit.
So, would you kindly offer another moon (From Mars or Jupiter) that meets the criteria and is losing orbit?
All the big ones have positive orbital periods and lie outside geosync radius, so they all move outward per tidal forces.
If I understand you correctly, there is no real evidence which can contradicts my statement that all real moons and planets are drifting outwards.
However, you claim that this process is due to "tidal forces" and I claim that it is due to normal gravity force reduction over time!!!
Please look again at the following:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
If we could monitor the orbital motion (gray line) of the Sun, we would clearly find that it drifts outwards from the center.
The assumption that the sun can orbit at the same radius for several billion years is absolutely incorrect.
The Sun' orbital motion must drifts outwards over time.
In the same token, any other star' orbital motion must drifts outwards!!!

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 18/01/2019 21:45:30
However, you claim that this process is due to "tidal forces" and I claim that it is due to normal gravity force reduction over time!!!
How much does gravity degrade over time?  Why does it only affect the moon and not Earth?
How does gravity know to degrade only for masses that Dave Lev designates as 'real moons' and not for other masses made of the exact same component materials?

Quote
Please look again at the following:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
If we could monitor the orbital motion (gray line) of the Sun, we would clearly find that it drifts outwards from the center
On the contrary.  Your theory says it should move immediately quickly from the center since it cannot hold this high acceleration needed to maintain that 8 kpc radius at a whopping 217 km/sec.
It shouldn't drift outward, but rather should never have been this far inward in the first place.  Your idea of degraded gravity makes that problem even worse.
Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 19/01/2019 05:09:44
strength of gravity
The trouble is that neither of you has defined gravity, as I do in the paper noted in my previous post. GR describes gravity in terms of the curvature of spacetime. I take it an additional step by pointing out that the curvature also represents the forward evolution of space, and the densities there-in, due to the force of the passage of time.

The passage of time is the fundamental force as it evolves space, and the events (densities) there-in, forward. When a dilation gradient is introduced, we also see an evolution down the gradient, This is the Gravitational Direction of Evolution, or GDE. The curvature of evolution we see in GR is the resultant of the GDE and Fundamental Direction of Evolutuion, or FDE, due to the simple passage of time.

The "strength of gravity" is determined by differences in the rates of time between frames, as per GR and my additional perspective.

I can't even try to read what you guys are writing as it all makes no sense as gravity, as per GR, is not defined as a force other than that due to the differences in the rates of time, Einstein's "energy components", and resultant relative lengths in space, between frames.

In spiral galaxies, this manifests as in my previous post.

Where you appear to be speaking about "friction", you should probably be looking at "frame dragging" which, again, is dependent on relative rates of time, which are also dependent on relative perspectives.....  8)

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 19/01/2019 05:43:29
How much does gravity degrade over time?
Good question.
There is no current formula for that. I assume that we have to verify the gravity degrade per many orbital objects and try to develop the requested formula for that.
Why does it only affect the moon and not Earth? How does gravity know to degrade only for masses that Dave Lev designates as 'real moons' and not for other masses made of the exact same component materials?
It effects any orbital System.(Asteroid, Moon, Planet, Star, BH, Dust, Gas cloud...)
However, the formula itself should include the orbital drifting direction and amplitude.
It should be based on the current "orbital radius ratio" (Orbital radius to Host radius ratio), Mass ratio (orbital object to host mass), "Friction" (If there is a friction. For example due to atmosphere) and some other factors (for example - Orbital structure or shape and so on).
I'm quite sure that If Newton or Einstein had the current measurements that we have today, they could find the correct formula for orbital drifting amplitude per cycle and its direction.
In any case, all orbital objects must drift in one or the other direction. Nothing can stay forever at the same radius!
Therefore, the assumption that the Sun has to stay at the same radius from day one is absolutely incorrect.
It is quite clear that all the real moons, Planets and stars must drift outwards. Even most of Asteroids drift outwards. The Oort cloud is a perfect example for billions of asteroids that had been drifted outwards from the core of the Solar system while they all orbit around the solar system. Today - all of them are still drifting outwards (even if we see that some of them are moving inwards - There is an explanation for that).
However, few of the Asteroids might drift inwards and eventually collide with their host. It happened on Earth with the big asteroid collision that had killed all the dinosaurs that have lived here (and gave us the opportunity to come). It also should be the final outcome of Phobos with Mars.
On the contrary.  Your theory says it should move immediately quickly from the center since it cannot hold this high acceleration needed to maintain that 8 kpc radius at a whopping 217 km/sec.
It shouldn't drift outward, but rather should never have been this far inward in the first place.  Your idea of degraded gravity makes that problem even worse.
Sorry. This is incorrect.
My theory doesn't say that none realistic idea.

Please, give me the chance to explain the theory.
With your permission, please let's assume that those two elements are correct and let's see how they give us the breakthrough knowledge about our Universe.
They should lead us to have excellent understanding of:
- How spiral galaxy had been created?
- Why we see the Bulge, Bar, Ring, spiral arms, and all the other mass around the galaxy.
- What is the answer for the Galaxy rotation curve problem.
- What is the real mass of the SMBH in the Milky way
- Why there is no need for dark matter to set the spiral galaxy
- Why for any star in the galaxy there is at least one outside
- Why there are hydrogen clouds strewn/bridge between Andromeda (M31) and Triangulum (M33).
- Why the far end galaxies are moving away at ultra velocity, while the density of the Universe stay at the same ratio.
- Why the universe has a blackbody radiation
- What is the age of the Universe
- What is the size of the universe
And many more questions...
All of that without any help from dark matter or dark energy.
Just pure gravity force and the two following elements:
1. All stars in the galaxy are increasing their orbital motion radius over time.
2. All stars in the galaxy must orbit around some virtual host..

Please give me the chance to explain the theory.
Let's stop the endless discussion about those two elements.
So, please from now on we will assume that those two elements are correct and try to understand what could be the outcome.
Would you kindly agree for that?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 19/01/2019 21:37:02
Please give me the chance to explain the theory.
Would you kindly agree for that?
I've agreed to this several times before.  It's not like you need permission from me.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/01/2019 06:14:39
I've agreed to this several times before.  It's not like you need permission from me.
Thanks
So, from now on we will assume that the following two elements are fully correct:
1. All stars in the galaxy must orbit around some virtual host point.
This virtual host point represents the real orbital motion point of a star around the center of the galaxy.
2. The orbital motion radius is increasing over time.
In the same token, Based on the formula (which we still have to find) any orbital object which has a high enough "orbital radius ratio" (and some other factors) is increasing its orbital radius over time. Therefore, all real Moons and planets are increasing their orbital radius over time.

I will introduce the theory step by step, based on those two elements and pure gravity force.
In each step I would mostly appreciate to get your feedback (again - based on the idea that those two elements are fully correct.)
In this theory there is no need for dark matter, dark energy, space expansion, density wave and any other none realistic hypothetical idea.
Just after the final step in the theory introduction, we can go back and verify if those two elements are correct or not.
Thanks for the cooperation in advance.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 20/01/2019 07:33:39
Let's start the theory by the focusing on the Sun' orbital motion.
http://www.biocab.org/Motions_of_the_Solar_System.jpg
We see the apparent motion of the solar system (blue dot points) the virtual host point (orange ball) and the real orbital motion of the sun around the galaxy (gray line).
Let's start with the apparent motion of the solar system. (Blue dot points).
We see that it has an elliptical shape. If I remember correctly, the ratio between the maximal length to the minimal lengths is 20 to 7.
This elliptical orbit represents the Sun orbit around its host virtual point.
Based on our understanding, it is claer that the sun must increase its orbital radius over time. Therefore, the length of this elliptical orbital cycle must increase over time.
The sun will never ever come back again to the same point in this elliptical cycle.
It will form some sort of spiral shape around this elliptical cycle.
The virtual host point (orange ball) orbits around the galaxy. This orbit (gray line) represents the real motion of the sun around the galaxy.
Please be aware that also the radius of this orbit must increase over time. Therefore we should see a spiral shape as every movement the sun is increasing its orbital motion radius.
The outcome is quite simple -
The measured sun' orbital speed represents the combination of the Sun orbit around the virtual host + the real Sun orbital motion around the galaxy.
Therefore, if we need to extract the real orbital motion of the sun around the galaxy, we must deduct from the measured sun' orbital speed the impact of the sun orbital speed around its virtual host.
Hence, I can only assume that if the 220 Km/s represents the measured orbital speed of the sun, the real orbital motion of the sun around the galaxy should be in the range of 210km/sec or less.
Please also be aware that the real orbital motion is located at the same distance from the galactic disc.
This is very important outcome from this image. We will use it later on.
Please let me know if you agree with all of that (again - based on the assumptions that the two key elements are fully correct)

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 20/01/2019 15:21:07
Let's start the theory by the focusing on the Sun' orbital motion.
http://www.biocab.org/Motions_of_the_Solar_System.jpg
You do realize that this diagram you keep referencing is a simplified model based on a series of facts, many of which are incorrect.  It comes from http://www.biocab.org/coplanarity_solar_system_and_galaxy.html
which states that the plane of the solar system is tilted at 90° WRT the galactic plane.  The actual figure is about 63°.
Here is another gem:
Quote
The second movement, which is described in most of astronomy books, is an oscillation of the Solar System from north to south and from south to north with respect to the galactic plane. The oscillation “upwards” and “downwards” is mainly established by the gravitational pull exerted by other bodies of the Solar System on the Sun, i.e. planets, asteroids, etc. The speed of this movement is 7 km/s.[\i]
They apparently seem to claim that our planets and asteroids are capable of dragging the sun back and forth at a pace of 7 km/s, and that "most of anstronomy textbooks" describe this.  Those 'textbooks' are seemingly where they are getting their facts, not from data directly from actual astronomical surveys.

Their model has the axis of rotation for this helical motion (they never call it an orbit) rotating with the position in the galaxy, always staying tangential to the curving dashed grey orbit line (that line they do call an orbit).  Orbits don't change their axis like that, so they rightly don't describe the motion as 'orbital'.  The axis of an orbit tends to be fixed over time, so our solar system has always had this 63° tilt and does not process to a different orientation when we've moved a quarter of the distance around the galaxy.

I say all this because you're seemingly using this one diagram of a poor model as some sort of measured scientific fact instead of just just a diagram of a model that it claims to be.  The site is a biology site, hardly the first place I'd look for accurate information on this subject.

The site also does not explain how the N/S motion (7 km/s) and the in/out (20 km/s) motion just happen to have the same periods.  The model is not described in detail.

Quote
Please let me know if you agree with all of that (again - based on the assumptions that the two key elements are fully correct)
You asked my in my prior post to stop with me pointing out the points with which I disagree.  I'm still doing that, but it doesn't imply that I agree with you.  If you don't want critique, don't ask me if I agree with your posts.  I'm still waiting to see where this goes.

My post here is not so much a critique of your view, but rather a critique of this diagram from which you seem to be drawing your facts.

Quote
I will introduce the theory step by step, based on those two elements and pure gravity force.
In each step I would mostly appreciate to get your feedback (again - based on the idea that those two elements are fully correct.)
I am assuming your two elements are correct.  I am staying quiet about the implications of those two elements since they've not really been spelled out.  I cannot make predictions from them yet, so I cannot falsify them.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 21/01/2019 13:27:47
I am assuming your two elements are correct.  I am staying quiet about the implications of those two elements since they've not really been spelled out.  I cannot make predictions from them yet, so I cannot falsify them.
I do appreciate your good willing to accept the two elements for this discussion.
The idea if the plane of the solar system is tilted at 90° WRT the galactic plane or 63° Isn't relevant to our discussion.
In the same token, it is also not relevant if they call the Sun orbital movement around the virtual host point (orange point) as a swinging motion.
However, the image itself shows clearly that the sun orbits around the virtual host point (orange ball) while this host orbits around the galaxy at the same amplitude (or distance) from the galactic plane.
Please look at the following image (from different article):
https://www.space.com/10532-earth-biodiversity-pattern-trace-bobbing-solar-system-path.html
It is stated that "the solar system moves around the center of the galaxy like planets around the Sun"
"The circular motion around the center is shown by the white dashed line."
This dashed line looks the same as the orbital motion of the virtual sun host point in the other article (which was represented by the gray line).
Hence, in both articles we see a similar view/outcome.

In any case, you didn't reply my questions:
Do you agree that based on the two key elements?
1. The Virtual sun host (Orange ball) is located at the same distance from the galactic disc plane while it orbits around the galaxy (gray line in one article or dashed white line in the other)?
2. The Virtual sun host orbital velocity around the galaxy is lower than "the actual motion of the sun" (Green line). Less than 210 Km/s instead of 220 Km/s?
Do you agree with that?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 21/01/2019 19:28:54
Please look at the following image (from different article):
https://www.space.com/10532-earth-biodiversity-pattern-trace-bobbing-solar-system-path.html
It shows the vertical motion, same as the other diagram, corresponding to the 7 km/sec component of the helix (the one the bioCab attributes to gravity from our own planets).  The other article didn't give a period to the wiggle, but this one says every 62 million years.  That's about 4 waves per trip around the galaxy, less than the tighter frequency depicted in their image, and far less than the dense helix depicted in the bioCab simulation.
The image does not depict an in/out motion, and so doesn't state the period of that if it exists.  The existence of that motion seems not relevant to the point this article is trying to make, so they're not denying its existence.

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It is stated that "the solar system moves around the center of the galaxy like planets around the Sun"
"The circular motion around the center is shown by the white dashed line."
This dashed line looks the same as the orbital motion of the virtual sun host point in the other article (which was represented by the gray line).
Hence, in both articles we see a similar view/outcome.
Right.  This new image has the solar system going the wrong way around, but again, that direction is irrelevant to the point of the article, so they didn't bother to check.

Quote
In any case, you didn't reply my questions:
Do you agree that based on the two key elements?
1. The Virtual sun host (Orange ball) is located at the same distance from the galactic disc plane while it orbits around the galaxy (gray line in one article or dashed white line in the other)?
2. The Virtual sun host orbital velocity around the galaxy is lower than "the actual motion of the sun" (Green line). Less than 210 Km/s instead of 220 Km/s?
Do you agree with that?
If that is the host point, then yes, that path takes a roughly circular path around the galaxy.  That the motion of the solar system in relation to that host point can be qualified as 'orbital' is something that seems to be part of your idea, but I have no data on the actual nature of that motion.  This new article gave a period to the north/south motion at least.
As for point 2, the speed 217 km/sec usually refers to motion of the host point, but slow (7-20 km/sec) motion tangential to that will alter that speed less than half a percent, so 216-218 range at best if the 217 figure is accurate.
This is trivial trigonometry. Most estimates (from a google search) put it a bit higher.
Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 22/01/2019 03:46:02
1. The Virtual sun host (Orange ball) is located at the same distance from the galactic disc plane while it orbits around the galaxy (gray line in one article or dashed white line in the other)?
2. The Virtual sun host orbital velocity around the galaxy is lower than "the actual motion of the sun" (Green line). Less than 210 Km/s instead of 220 Km/s?
Sorry, neither of these is correct so the resultant conclusion is also wrong.

Motion and velocities are only relative based upon perspective. You are ignoring that. (I would also point out that the disk is not uniform, but consists of conical spiral arms of varying densities. That complication does not need to be considered here.)

For instance, considering the perspective of the orbits of Mercury and Venus relative to the plane of the ecliptic, we assign Mercury a velocity of 47.89 km/s and Venus one of 35.03 km/s, a large difference.

But if we consider the velocity of the Sun and its forward evolution in time relative to the CMB, and the helical distances travelled by the planets we get a much different perspective:

(in the following computations the inclination of the plane of the ecliptic is ignored and:
Planetary orbital lengths and periods are as per NASA.
Orbital periods are related to 1 Earth year.
Orbital lengths are as perceived “around the Sun”.
Helical orbital lengths are computed using the following formula:
(Distance travelled by the Sun)2 + (Orbital length)2 = (Helical length)2
The distance travelled by the Sun is relative to the CMB.
Sun velocity = 368 km/s = 11.60672*109 km/yr.)

Mercury:
Orbital length: 57.909227*106 km
Orbital period = .24 yr
Orbits/yr = 4.1666
Total orbital length = 241.249839*106 km
Helical length = 11.609226961*109 km
Velocity = 368.07948 km/s vs 47.89 km/s

Venus:
Orbital length: 10.8209475*107 km
Orbital period = .62 yr
Orbits/yr = 1.6129
Total orbital length = 17.4531062*107 km
Helical length = 11.608032143*109 km
Velocity = 368.04160 km/s vs 35.03 km/s

Earth:
Orbital length: 14.9598262*107 km
Orbital period = 1 yr
Orbits/yr = 1
Total orbital length = 14.9598262*107  km
Helical length = 11.607684041*109 km
Velocity = 368.03056 km/s vs 29.79

Mars:
Orbital length: 22.7943824*107  km
Orbital period = 1.88 yr
Orbits/yr = .5319
Total orbital length = 121.2467148*106 km
Helical length = 11.607353269*109 km
Velocity = 368.02007 km/s vs 24.13

Jupiter:
Orbital length: 778.340821*106 km
Orbital period = 11.86 yr
Orbits/yr = 0.0843
Total orbital length = 65.6273879*106 km
Helical length = 11.606905535*109 km
Velocity = 368.00588 km/s vs 13.06

Saturn:
Orbital length: 142.6666422*107 km
Orbital period = 29.46 yr
Orbits/yr = 0.0339
Total orbital length = 484.27237*105 km
Helical length = 11.606821027*109 km 12576482920
Velocity = 368.00320 km/s vs 9.64

Uranus:
Orbital length: 287.0658186*107 km
Orbital period = 84.01 yr
Orbits/yr = .0199
Total orbital length = 341.70434*105 km
Helical length = 11.606770299*109 km
Velocity = 368.00159 km/s vs 6.81

Neptune
Orbital length: 449.8396441*107 km
Orbital period = 164.8 yr
Orbits/yr = 0.0060
Total orbital length = 272.96094*105 km
Helical length = 11.606752096*109 km
Velocity = 368.00101 km/s vs 5.43

From this perspective, the velocities, or rate of evolution, of Mercury and Venus are only .038 km/s different. Note also that as we increase distance from the Sun, the velocities decrease until Neptune has a velocity only .001 km/s different from the base velocity of the Sun. Relative velocities equalize with a larger perspective. If we shift out to the local group and its apparent motion relative to the CMB of 627 km/s, the difference between the Sun and Neptune’s velocity is only .00059 km/s. Viewed as a whole, the universal evolutionary rate of evolution, and apparent resultant relative velocity, is 1: all events in space evolving forward at the same rate over a steady rate of time.

This is the evolution of the quantum continuum. This is how I tie quantum and astro- physics together.

In both perspectives, the velocity and acceleration are directly related to the difference in the rate of time (dRt)/distance so are higher in steeper gradients, and this higher apparent acceleration of events in slower time frames maintains their relative positions within the overall continuum as it evolves forward as viewed from both perspectives.

This means GR is describing the forward evolution of the continuum and the events occurring within it, rather than the evolution of events through pre-existing “curved spacetime”. It is not the masses that determine relative velocities and trajectories, but the dynamics and perspectives in time.

Again, this is extracted from my paper on Relativity found here: http://vixra.org/abs/1804.0109#comment-3850079405.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/01/2019 13:40:30
1. The Virtual sun host (Orange ball) is located at the same distance from the galactic disc plane while it orbits around the galaxy (gray line in one article or dashed white line in the other)?
2. The Virtual sun host orbital velocity around the galaxy is lower than "the actual motion of the sun" (Green line). Less than 210 Km/s instead of 220 Km/s?
Sorry, neither of these is correct so the resultant conclusion is also wrong.

Motion and velocities are only relative based upon perspective. You are ignoring that.
He is not ignoring it.  There is a 'perspective' (as you put it) for each of the values specified.
Point 2 might be wrong, but not for the reasons you seem to point out.
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For instance, considering the perspective of the orbits of Mercury and Venus relative to the plane of the ecliptic, we assign Mercury a velocity of 47.89 km/s and Venus one of 35.03 km/s, a large difference.
That would be relative to the frame of the solar system, not 'relative to the plane of the ecliptic'.  A plane is not a valid reference frame.
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But if we consider the velocity of the Sun and its forward evolution in time relative to the CMB
The CMB is light, not an object relative to which one might have a velocity.  I presume you mean relative to the frame in which the CMB appears to be isotropic, but you don't say that.
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in the following computations the inclination of the plane of the ecliptic is ignored
But the inclination is not zero, so all of the number you quote are wrong, especially comments like this one:
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Note also that as we increase distance from the Sun, the velocities decrease until Neptune has a velocity only .001 km/s different from the base velocity of the Sun.
I suppose this would be the speed difference in a frame where the inclination put the axis of the solar system parallel with the direction of travel, but this is not the case in the frame you specify, so this comment is totally wrong.  You seem to base some conclusion on this fact, so the conclusion is wrong.  I could not figure out the whole point of your post.  I think it might just have been an excuse to plug the link to your paper, which I have no plans to read.
How is any of this post relevant to the discussion going on in this topic?
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Relative velocities equalize with a larger perspective.
No they don't, at least not until you get to relativistic speeds.  Even if the inclination was flat on like you suggest, the relative velocity between the sun and Neptune would be about 5.4 km/sec, not the .001 km/sec figure you suggest.
The value you quote is a difference in speed, not a difference in velocity.  I can take any two objects with arbitrarily high and random velocity differences and find frames in which they have the same speed.  So what?  The frame you selected is not one of those frames.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/01/2019 15:42:54
Thanks Halc
Now the question is - What is the estimated mass of the Sun' virtual host point?
It shows the vertical motion, same as the other diagram, corresponding to the 7 km/sec component of the helix (the one the bioCab attributes to gravity from our own planets).  The other article didn't give a period to the wiggle, but this one says every 62 million years.  That's about 4 waves per trip around the galaxy, less than the tighter frequency depicted in their image, and far less than the dense helix depicted in the bioCab simulation.
So, we know that the Sun sets one full orbit around this virtual host point in 62 Million years.
I assume that based on the radius and the orbital time we could calculate the requested mass.
Any idea?

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/01/2019 15:57:50
Sorry, neither of these is correct so the resultant conclusion is also wrong.
Motion and velocities are only relative based upon perspective. You are ignoring that. (I would also point out that the disk is not uniform, but consists of conical spiral arms of varying densities. That complication does not need to be considered here.)
It seems to me that you have missed the key point of this idea.
Please look again at the following image:
https://www.space.com/10532-earth-biodiversity-pattern-trace-bobbing-solar-system-path.html
Please try to focus on the dashed white line Vs the green line.
Do you agree that the green line is longer in its distance than the dashed White line?
It is clear that in order to set one full orbital cycle around the galaxy, an object at the green line must move faster than an object on the dashed white line.
So, do you understand by now why the virtual host velocity (dashed line) is lower than the sun actual orbital velocity (green line)?
Therefore, if the green line represents the actual orbital velocity of the sun - Let's assume that it is 220 Km/s, than the dashed white line, which represents the orbital motion of the Sun virtual host around the galaxy, must be lower than 220 Km/s.

Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 22/01/2019 16:28:19
So, do you understand by now why the virtual host velocity is lower than the sun actual orbital velocity?
Sorry. As I pointed out above, this is non-sensical. You are trying to prove the impossible, i.e., objects evolving against the direction of the time dilation gradient, or backwards gravity. Gravity only has one direction which is down the dilation gradient.

You are trying to imagine a virtual host that does not exist. The white line is meaningless. It only marks the plane of the ecliptic in the pictured galaxy. As Dave pointed out previously the scale is way off in the picture, as well. Even if your virtual host existed the difference in velocity would be meaningless.

You are also not visualizing the evolution of the continuum properly. All events are acted on by two directions of evolution, the Fundamental, in situ, evolution and the Gravitational, down gradient, evolution.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/01/2019 16:38:02
Sorry. As I pointed out above, this is non-sensical. You are trying to prove the impossible, i.e., objects evolving against the direction of the time dilation gradient, or backwards gravity. Gravity only has one direction which is down the dilation gradient.
You are trying to imagine a virtual host that does not exist. The white line is meaningless. It only marks the plane of the ecliptic in the pictured galaxy. As Dave pointed out previously the scale is way off in the picture, as well. Even if your virtual host existed the difference in velocity would be meaningless.
Sorry
In this discussion we assume that:
1. All stars in the galaxy must orbit around some virtual host point.
This virtual host point represents the real orbital motion point of a star around the center of the galaxy (Dashed white line)
2. The orbital motion radius is increasing over time.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/01/2019 17:16:01
Quote from: Halc
It shows the vertical motion, same as the other diagram, corresponding to the 7 km/sec component of the helix (the one the bioCab attributes to gravity from our own planets).  The other article didn't give a period to the wiggle, but this one says every 62 million years.  That's about 4 waves per trip around the galaxy, less than the tighter frequency depicted in their image, and far less than the dense helix depicted in the bioCab simulation.
So, we know that the Sun sets one full orbit around this virtual host point in 62 Million years.
I don't know that.  There is a motion north and south with a period of 62 million years or so.  That's what that article tells us.  It seems to have maximum acceleration at the peaks of the waves, when it is furthest from this host point, which makes the motion not really 'orbital', where max acceleration is reached when the object is closest to the point around which it orbits.

Anyway, we've not determined the period of the in/out motion (towards and away from the center of the galaxy), but if that period is also 62 million years, then at least the path is sort of elliptical and resembles an orbit in that sense, and it might make a little sense to talk about its radius.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/01/2019 17:20:32
Please look again at the following image:
https://www.space.com/10532-earth-biodiversity-pattern-trace-bobbing-solar-system-path.html
Please try to focus on the dashed white line Vs the green line.
Do you agree that the green line is longer in its distance than the dashed White line?
It is clear that in order to set one full orbital cycle around the galaxy, an object at the green line must move faster than an object on the dashed white line.
Did you calculate how much longer the green line is?  Yes, it is longer.  You have the numbers supposedly, despite my reservations on the source of the figures.  Or is established trigonometry also part of your list of theories that are to be discarded?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/01/2019 17:24:20
Quote from: captcass
You are also not visualizing the evolution of the continuum properly. All events are acted on by two directions of evolution, the Fundamental, in situ, evolution and the Gravitational, down gradient, evolution.
Dave, I suggest you just ignore these posts by captcass.  It seems to be a plug for a pet theory and pretty much off topic most of the time.  The above sentence is an off-topic word salad to me.

The white line has meaning, despite what others say.  It is not significantly shorter than the green line.
Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 22/01/2019 19:25:09
Dave, I suggest you just ignore these posts by captcass.
Sorry, I don't think that is wise. From the outset of this thread you all have been using, and trying to modify, Newtonian formulas to come up with your breakthrough. The trouble is, Newtonian formulas only approximate what General Relativity describes, which is the apparent evolution of events in the continuum due to Lorentz contractions in both time and space. People still largely use the Newtonian formulas because the Relativity formulas quickly become very complex when more than 1 body is considered. If you say you can modify the Newtonian formulas, then you are saying you can modify relativity, too. Can you demonstrate how that might work?
(I actually do this by deriving the Hubble Shift from a 2.2686"10-18 s/s acceleration in the rate of proper time that is then added to Einstein's Tensor to eliminate singularities, Big Bangs, and infinitely accelerating expansions of the universe in the "Is the Hubble Shift Due to Time Dilation" thread). .
Any discussion of gravity that does not consider the Lorentz contractions, especially the time dilation, which Einstein calls his "energy components", is off the mark. You are talking about Newtonian "forces" that do not exist. It is all about the evolution of events within the continuum, not the motion of particles "through" a pre-existing space.
That being said, I will leave you all to your debate unless someone wants to ask me a question about anything I've said here. I don't want to hijack a thread, but as I explain galactic rotation velocities in my paper, as copied above, I felt it was on topic. "On topic" doesn't always mean "on the same track" and I can see that.  :)
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/01/2019 20:38:12
If you say you can modify the Newtonian formulas, then you are saying you can modify relativity, too. Can you demonstrate how that might work?
Dave is working with 3-digit precision examples at best and I doubt the whole proposal is going to get as far a workable rewrite of Newton's work to the point of needing to worry about GR implications.
The examples are low precision without significant speed or gravity wells that we need not invoke GR in order to find the problems with the proposal.

Meanwhile, I see you invoking references to the CMB which is pretty irrelevant to a discussion of the dynamics of a local system.
Title: Re: How gravity works in spiral galaxy?
Post by: captcass on 22/01/2019 20:59:11
I see you invoking references to the CMB which is pretty irrelevant to a discussion of the dynamics of a local system.
I did that to illustrate the relative nature of apparent velocity and motion. But what I am trying to contribute does not really fit the thread so I will bow out.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/01/2019 20:12:19
It seems to have maximum acceleration at the peaks of the waves, when it is furthest from this host point, which makes the motion not really 'orbital', where max acceleration is reached when the object is closest to the point around which it orbits.
I wonder why our scientists assume that it has maximum acceleration at the peaks. Why not a sin wave?
Any idea?.
Somehow, it also seems to me that the 62 Million years per cycle it too long.
Just a brief calculation.

The nearest star is about 4 LY away from us.
Its quite clear to me that the maximal orbital radius around the virtual host must be significantly lower than that.
However, just for the calculation, let's assume that the radius is 1LY (9.461e+12 Km)
P = 2 * 3.14 * r = 6.28 * 9.461e+12 Km = 59.425 10^12 Km
In one year there is = 31536000 sec.
In the article it is stated that the orbital speed around the orbital motion (which is the virtual host) is 7 to 20 Km/s.
Just to make it easy, let's use an average speed of 10 Km/s (in a pure cycle orbit).
At a speed of 10 Km/s we can cross in one year:
L = 10 * 31536000 = 315.36 10 ^ 6 Km/Year
T = P/L = 59.425 10^12 Km / 315.36 10 ^ 6 Km/Year = 1.88 10^5 = 188,000 year
So, in order to set a full 1 LY cycle at 10Km/s, 188,000 years are needed.
Hence, how do they have got the idea for 62 Million years per cycle?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 24/01/2019 02:01:44
Quote from: Halc
It seems to have maximum acceleration at the peaks of the waves, when it is furthest from this host point, which makes the motion not really 'orbital', where max acceleration is reached when the object is closest to the point around which it orbits.
I wonder why our scientists assume that it has maximum acceleration at the peaks. Why not a sin wave?
It is something like a sin wave, which has maximum acceleration at the peaks, furthest from the source of gravity.  Regular orbital motion where the object changes distance from the gravity source (such as a comet) has max acceleration when closest to the gravity source, as you'd expect from the point-source formula of F=GMm/r².
The force is greatest when the distance r is minimum.  A sine wave does not conform to this pattern.
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The nearest star is about 4 LY away from us.
Its quite clear to me that the maximal orbital radius around the virtual host must be significantly lower than that.
You hadn't said this before.  What exactly causes the sun to orbit this point if not mass of something?  This seems to really be throwing established physics out the window if you ask me.  You seem to be describing a binary star system.
Quote
However, just for the calculation, let's assume that the radius is 1LY (9.461e+12 Km)
P = 2 * 3.14 * r = 6.28 * 9.461e+12 Km = 59.425 10^12 Km
In one year there is = 31536000 sec.
In the article it is stated that the orbital speed around the orbital motion (which is the virtual host) is 7 to 20 Km/s.
Just to make it easy, let's use an average speed of 10 Km/s (in a pure cycle orbit).
At a speed of 10 Km/s we can cross in one year:
L = 10 * 31536000 = 315.36 10 ^ 6 Km/Year
T = P/L = 59.425 10^12 Km / 315.36 10 ^ 6 Km/Year = 1.88 10^5 = 188,000 year
So, in order to set a full 1 LY cycle at 10Km/s, 188,000 years are needed.
Hence, how do they have got the idea for 62 Million years per cycle?
62/.188 = 330 light year radius circle, if your other figures are reasonably accurate, and it is a circle, etc. etc.  It obviously isn't a circle since the north/south speed is a third of the in/out speed, so more of an ellipse if the periods are the same, and just a spirograph scribble if the periods are not the same.

Anyway, the 1 light year radius is ridiculous for a unary star system.  The 330 figure sounds more reasonable.  There just isn't any significant mass a light year away to give your 'host point' the sort of pull it would need to haul the sun around at 10 km/sec at that tight radius, which would be incredible acceleration.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/01/2019 11:57:17
There just isn't any significant mass a light year away to give your 'host point' the sort of pull it would need to haul the sun around at 10 km/sec at that tight radius, which would be incredible acceleration.
It seems that you insist to criticize the key elements of my theory.
I didn't call it Host point. The correct name is "virtual host point".
So, it is a virtual point. Therefore, we don't expect to see any real mass at that Sun' "host point".
1. All stars in the galaxy must orbit around some virtual host point.
This virtual host point represents the real orbital motion point of a star around the center of the galaxy (Dashed white line)
2. The orbital motion radius is increasing over time.

Why do you keep criticize those two key elements in my theory?
Please - let me introduce the whole theory based on the two key elements and then take a decision if you accept it or reject it. Please, don't criticize it before getting the whole information about the new theory.
Let me give you an example:
Let's assume that you want to teach you kid mathematics.
For one year you tell him that the first two key elements in mathematics are:
1 + 1 = 2
1 - 1 = 0
Unfortunately, your kid doesn't agree to accept those two elements.
So, at some moment you ask him to take it as is.
However, at any step forward he tries to criticize those two basic elements.
So, how can you give him any basic understanding about mathematics, if he doesn't agree with those two key elements?
In the same token.
You have agreed to accept those two key elements (just during the introduction process of the theory)
So, please would you kindly not criticize those elements any more.
Please - try to help me to find what is the outcome due to those two key elements.

If so, there is no real mass at the virtual host point.
It just represents the equivalent mass which is needed to keep the sun in that orbital cycle around that virtual point.
It also doesn't important if the orbital cycle is elliptic or pure cycle.
Actually, in the nature most of the orbital cycles are elliptical.
There is no big difference if we need to use Newton formula or Kepler.
However, during this discussion, if possible - let's consider the orbital cycles as pure cycle ( just to make it more easy to set the calculation).
Do you agree with all of that?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 25/01/2019 13:24:36
Why do you keep criticize those two key elements in my theory?
I didn't criticize the two points.  I commented on the unlikely radius of your orbit about this VHP since you for the first time gave a number for it.  I didn't comment on the second of the two points at all.  I was hoping that by shutting up, you'd actually get to your theory.

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Let me give you an example:
Let's assume that you want to teach you kid mathematics.
For one year you tell him that the first two key elements in mathematics are:
1 + 1 = 2
1 - 1 = 0
Unfortunately, your kid doesn't agree to accept those two elements.
So, at some moment you ask him to take it as is.
However, at any step forward he tries to criticize those two basic elements.
So, how can you give him any basic understanding about mathematics, if he doesn't agree with those two key elements?
And suppose the child agrees to just take those two points as is, and yet the parent keeps harping on those two points instead of moving on.  Eventually the child has no option but to comment more on the two points since that is all that has been presented.
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If so, there is no real mass at the virtual host point.
I never claimed there has to be, but there has to be mass somewhere.  Pluto and Charon orbit a common virtual host point, and there is no mass at that point.  But each constitutes the mass that is responsible for the orbit of the other.  Without either one of those two bodies, there would be no virtual point about which the other could orbit.

I think your problem is that your two points are not premises.  They are proposed effects (not even observations), and you need to come up with viable physics that explain those effects, not come up with implications of those effects.  I'm certain that your ideas might have the implication of a galaxy behaving this or that way, but without the physics to explain those effects, the effects are not grounded in reality.  Hence my questioning them so hard.  You've presented no new physics* to explain these two proposals, and the proposals seem to completely violate existing physics.

You're telling me that 1+1=7 and asking me to take that on faith.  Fine, but eventually you need to justify how 1+1 happens to work out to 7, otherwise the implications of that assumption are grounded in nonsense.

* OK, you've presented one new piece of physics: Selective gravity that decreases over time for certain objects, but remaining constant or increasing for others.  I could build an infinite energy generator with that sort of physics.
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It just represents the equivalent mass which is needed to keep the sun in that orbital cycle around that virtual point.
It also doesn't important if the orbital cycle is elliptic or pure cycle.
Yes and yes.   But it seems there is no mass out that way, so what exactly pulls the sun into that orbital cycle if not mass?
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There is no big difference if we need to use Newton formula or Kepler.
However, during this discussion, if possible - let's consider the orbital cycles as pure cycle ( just to make it more easy to set the calculation).
Do you agree with all of that?
Actually, I do.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/01/2019 15:54:29
Actually, I do.
Many Thanks

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If so, there is no real mass at the virtual host point.
I never claimed there has to be, but there has to be mass somewhere.  Pluto and Charon orbit a common virtual host point, and there is no mass at that point.  But each constitutes the mass that is responsible for the orbit of the other.  Without either one of those two bodies, there would be no virtual point about which the other could orbit.
Yes. You are absolutely correct!
This is a very important issue. However, in my theory we actually discuss on many objects.
I will explain it.
And suppose the child agrees to just take those two points as is, and yet the parent keeps harping on those two points instead of moving on.
Yes. That is correct.
I will shift gear.
So, as our Sun orbits around a virtual host point, every star must do so.
In the same token, as that virtual host point orbits at a fixed amplitude from the galactic disc, every other virtual host point must do so.
On the other hand - please see the second element:
2. The orbital motion radius is increasing over time.
Based on that idea, all the stars, gas clouds, dust... are drifting outwards from the center. Nothing can move inwards.
So, how could it be that the center is full with mass, while all the mass is drifting in one way out?
The only logical outcome is that the galaxy must create new mass in order to compensate on all the mass that had been drifted out.
Hence, the accretion disc is actually an excretion disc.
I have already discussed deeply on the activity of new mass creation in the galaxy.
I have poved that an Atom is a cell of energy.
So, the SMBH converts its ultra high gravity force into energy. this energy is locked in new Atoms in the plasma.
It creates new Hydrogen atoms at this axcretion disc. Then due to the collisions between those new born atoms it creats all the variety of atoms and molecular.
The ultra high temp of the plasma in the accretion disc (10 ^9 c) is a clear evidence for the new born atom activity at that disc.
As all the new born molecular drift outwards from the disc - remember - all objects must drift outwards over time including those Atoms), they get to the magnetic shield around the accretion disc.
This Ultra power of magnetic disc, boost the atoms upwards at a speed of almost 0.8 c.
As they move up, they lose the boost of the magnetic force and eventually they fall back at the galactic disc plane.
They set the gas could that we see around the SMBH.
However, In the gas clouds, the molecular orbits around some virtual center of mass (as explained in element 1).
I assume that the orbital velocity in the gas cloud is quite high.
This internal orbit in the gas cloud, pulse the great impact of the nearby SMBH set the Huge New form star activity that we see.
Any new born star must drift outwards while it orbits around some virtual center of mass.
That is all for that step in my theory.
Please let me know if based on the two elements - you agree with the outcome.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 26/01/2019 00:32:37
So, as our Sun orbits around a virtual host point, every star must do so.
Does  everything orbit a virtual host point, or only stars?  If the latter, what delimits objects with a VHP vs ones that don't?  What about a small black hole, or a cold black dwarf that has less mass than some planets?
For instance, Venus doesn't seem to orbit a VHP.  It goes pretty much in a normal elliptical orbit straight around the sun with no extra wobble of its own.  All the planets actually, Earth and Pluto excepted.  Earth's host point (called its barycenter in normal terminology) has mass present at it, while the barycenter of Pluto is in empty space.  The other planets don't really have a barycenter/virtual-host-point about which they have any sort of motion that could be described as orbital.

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In the same token, as that virtual host point orbits at a fixed amplitude from the galactic disc, every other virtual host point must do so.
What do you mean by this?  They have an orbit, a wobble (sort of like the green sin wave)?  What do you mean by 'fixed'?  Do all VHP's have this same amplitude, or do you mean each is different but regular, not varying from one peak to the next?

What has this VHP thing got to do with the stuff below where the center of the galaxy creates new matter that is expelled away from there?

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On the other hand - please see the second element:
2. The orbital motion radius is increasing over time.
For some things, but not others you've said.  Do unqualified objects not increase their orbits, such as small rocks or gas clouds and such?

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Based on that idea, all the stars, gas clouds, dust... are drifting outwards from the center. Nothing can move inwards.
Not even irregular 'not real' moons?  You seem to be inconsistent with prior posts, so I'm trying to get it straight.

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So, how could it be that the center is full with mass, while all the mass is drifting in one way out?
The only logical outcome is that the galaxy must create new mass in order to compensate on all the mass that had been drifted out.
That actually sort of follow from your second point, at least when coupled with data with the mass density of the center of the galaxy.

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Hence, the accretion disc is actually an excretion disc.
I have already discussed deeply on the activity of new mass creation in the galaxy.
I have poved that an Atom is a cell of energy.
So, the SMBH converts its ultra high gravity force into energy. this energy is locked in new Atoms in the plasma.
It creates new Hydrogen atoms at this axcretion disc. Then due to the collisions between those new born atoms it creats all the variety of atoms and molecular.
The ultra high temp of the plasma in the accretion disc (10 ^9 c) is a clear evidence for the new born atom activity at that disc.
There is a formula for the temperature of the accretion disk at radius R which is something like
T(R) = √√[3GM⊕/8πσR³ (1−√(Rinner / R)) ]
Where G π and σ are constants, ⊕ is the accretion rate (which is actually an M with a dot over it, but I couldn't figure out how to write that).  Rinner is the Schwarzschild radius.  Plug your idea into that and you get a negative number for the temperature.  This is fine.  You've thrown all of known physics out the window, so there's no reason for that rule to have survived.

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As all the new born molecular drift outwards from the disc - remember - all objects must drift outwards over time including those Atoms), they get to the magnetic shield around the accretion disc.
This Ultra power of magnetic disc, boost the atoms upwards at a speed of almost 0.8 c.R
Sort of like hitting a boost pad in Mario Kart.  How does the magnetic disc know which way to send the material?  I've never heard of a magnetic field with only rotational symmetry, but I could just be ignorant of that.
At what radius does this take place?  That rotation curve has inner things moving pretty slow, and gradually moving up until maxing out at about 1KPC, after which the curve goes down again for a while.

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As they move up, they lose the boost of the magnetic force and eventually they fall back at the galactic disc plane.
They set the gas could that we see around the SMBH.
However, In the gas clouds, the molecular orbits around some virtual center of mass (as explained in element 1).
I assume that the orbital velocity in the gas cloud is quite high.
Each gas molecule has a virtual host point?

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This internal orbit in the gas cloud, pulse the great impact of the nearby SMBH set the Huge New form star activity that we see.
Any new born star must drift outwards while it orbits around some virtual center of mass.
Well, I've seen sites that describe the bar as a stellar nursery, so this seems plausible enough.
Quote
That is all for that step in my theory.
Please let me know if based on the two elements - you agree with the outcome.
The stuff-moving-out element indeed seems to require the generation of mass.  That seems to follow.
The virtual host point idea doesn't seem to contribute anything to the idea.  You haven't used that to explain anything, so I'm not sure why you have to posit this.  You describe the sun moving in a 1LY radius circle, but I see no purpose in proposing that it does that.  What gets explained by that?

Neither of the two elements seems related to this idea of magnetic fields propelling the new mass to, well, nonzero speeds.  Perhaps you need more than just the two elements.  Yes, the angular speed needs something to account for it, else you just get matter moving straight out from the center like lava from a volcano.
It seems that the speed boost from the magnetic field would account for matter moving away from the center, not something like negative gravity repulsing nearby stuff.

Do black holes in general generate new matter like that?  I ask because we don't seem to see this happening around smaller black holes.  What is special at a galaxy center that this process goes on, but we don't see it elsewhere?

Also, what happens to all this new matter after a long time?  Does it fade to nothing after a while, or does each galaxy grow indefinitely?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/01/2019 06:23:57
Thanks Halc
So, let me answer one by one.

Does  everything orbit a virtual host point, or only stars?  If the latter, what delimits objects with a VHP vs ones that don't?  What about a small black hole, or a cold black dwarf that has less mass than some planets?
For instance, Venus doesn't seem to orbit a VHP.  It goes pretty much in a normal elliptical orbit straight around the sun with no extra wobble of its own.

All the stars in the galaxy must orbit around some virtual host point. That includes all the stars in the Bulge, Bar, Spiral arms and even outside the disc. I will explain later on how the gravity impact the orbital cycle at each segment.
However, In the Solar system there is only one big star.
The Sun is the leading gravity power in the solar system.
Therefore, there is no need for a virtual host point.
However, it is quite interest to know that the moon orbits around the Earth instead around the sun although the sun gravity force on the moon is more than a twice than the Earth gravity force on the moon.
This is very important evidence.
I wonder why our scientists had neglected this observation.
In order to do so, it is clear to me that in day one of the Sun/Earth/Moon system the gravity force of the Earth/moon was much stronger than the Sun/Moon.
https://en.wikipedia.org/wiki/Hysteresis
"Hysteresis is the dependence of the state of a system on its history."
That is very clear based on my theory.
All the stars/Planets/Moon... had been formed at those molecular gas clouds at the same moment!!!
Our Sun, Earth and moon had been created at the same day (more or less) and from the same matter.
So, in day one of the solar system there were no rocky planets or moons.
All the objects were very nice round gas objects (yes even our Earth and Moon).
However, it is clear that the gas atoms (especially - Hydrogen) were dominant in all of those new formed objects.
As an example - "The sun is a big ball of gas and plasma. Most of the gas — 91 percent — is hydrogen"
If I understand it correctly, less than 2% in the gas clouds are solid molecular (Silicate, Iron... and even water)
Therefore, today the Earth & the Moon has less than 2% from the whole matter which they had in day one.
Hence, the real Earth mass in day one was bigger by 98 from its current mass value. (same issue with all the rocky planets and moons in the solar system).
Therefore, the gravity force between the Earth moon in day one was much higher than the Sun moon.
That is the History!!!
We prove it by - "Hysteresis is the dependence of the state of a system on its history."
In any case, all the rocky planets and moons were too small to hold the gas Atoms.
Due to gravity force, all the heavier molecular moved to the center of the young planets/moons while most/all the gas had been ejected out over time. Therefore, all of them have a very nice ball shape.
We would never ever get a nice ball shape if from day one all we have is a solid matter. The gas is vital for the ball shape of all the planets and moons. In other wards - there is no way that the moon had been created due to some none realistic collision with the Earth
The other gas planets were big enough to hold significant portion of the gas. Therefore, we also call them gas planets.
The Sun is the biggest one. Therefore even after all the hydrogen fusion process it has 91% of Hydrogen atoms.
Even Ceres - was a nice big ball shape planet in early time.
Unfortunately, some other big object had been collided with this planet and broke it to small pieces.
The Earth have got all its water supply from day one. (Therefore, there was no need for secial water delivery as our scientists believed.)
If I understand it correctly, there is no DNA for matter. However, if we could check it, we should find that all the matter in the solar system has the same DNA.
Actually, as all the matter in the Milky way had been created by the same SMBH, they all should carry the same DNA.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 26/01/2019 14:08:28
All the stars in the galaxy must orbit around some virtual host point. That includes all the stars in the Bulge, Bar, Spiral arms and even outside the disc. I will explain later on how the gravity impact the orbital cycle at each segment.
However, In the Solar system there is only one big star.
The Sun is the leading gravity power in the solar system.
Therefore, there is no need for a virtual host point.
You seem to contradict yourself.  You say all stars have a virtual host point, but then say now that the sun doesn't have one.  You put it a light-year away in a prior post, and now say it doesn't exist.
It is hard to figure out your VHP idea if the rules change from post to post, or in the same paragraph as has happened just above.

Yes, I agree that binary systems have stars that effectively orbit their barycenter (their mutual center of gravity).  You calling it a virtual host point just changes the terminology.  There's already one word for it.
For systems with 3 or more stars, there is very much a center of gravity which remains relatively fixed no matter the motion of the stars involved, but the motion about that point is no longer strictly orbital, depending on the dynamics of the stars.  This is what they mean by the three-body problem.  Typically pairs of stars orbit each other, and those pairs orbit other stars or other pairs, until there are only 2 bodies left.  3 or more in mutual orbit with no 2 in independent orbit is not stable.

Anyway, if you're talking only about the center of mass of multi-star solar systems, then this idea is not controversial.  Yes, that point exists, and the motions of the member stars does not move it one bit, but the point goes around the galaxy more or less as a unit.  The sun wiggles around its own point, with the point being inside the sun about half the time, and outside the other half.  The (quite predictable) path around that fixed point is anything but elliptical.  It resembles more of a scribble with no particular cycle to assign a period length.

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Gravity force on the moon is more than a twice than the Earth gravity force on the moon.
Assuming gravity works like physics says and not like you say, acceleration due to the moon itself at its surface is 0.1630 m/s, 0.0027 due to Earth, and 0.0060 due to sun. That means it is 60 times as much, which indeed is 'more than twice' the acceleration on the moon.  It is about 60x in fact.

I say acceleration because there is no force of gravity on the moon itself.  You would need an object to have a force, and no object was specified.  So if I stand on the moon, there might be a force of 135 N on me, but the 'the Earth gravity force on the moon' is about 2e20 N, which is a lot more than my 135 N.  That huge 2e20 figure is the force that Earth exerts on the moon, and is also exactly the force that the moon exerts on Earth, at least per standard physics.

I'm saying all this because I'm trying to figure out what you meant to say, as opposed to what you actually said.

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This is very important evidence.
I wonder why our scientists had neglected this observation.
I'm sorry, but I didn't get exactly which observation was neglected by scientists.  All the numbers I quoted come from those scientists, and you didn't actually quote any numbers at all, let along ones that are different than what the scientists noticed.

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In order to do so, it is clear to me that in day one of the Sun/Earth/Moon system the gravity force of the Earth/moon was much stronger than the Sun/Moon.
https://en.wikipedia.org/wiki/Hysteresis
"Hysteresis is the dependence of the state of a system on its history."
The word is used more in molecular changes than macroscopic history.  The orientation of dipoles on the seabed floor is considered hysteresis, while the nice crater in Arizona is not, despite both of them very much being a state that depends on the history of the system.
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That is very clear based on my theory.
All the stars/Planets/Moon... had been formed at those molecular gas clouds at the same moment!!!
Our Sun, Earth and moon had been created at the same day (more or less) and from the same matter.
So, in day one of the solar system there were no rocky planets or moons.
How very biblical.

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All the objects were very nice round gas objects (yes even our Earth and Moon).
However, it is clear that the gas atoms (especially - Hydrogen) were dominant in all of those new formed objects.
As an example - "The sun is a big ball of gas and plasma. Most of the gas — 91 percent — is hydrogen"
If I understand it correctly, less than 2% in the gas clouds are solid molecular (Silicate, Iron... and even water)
Therefore, today the Earth & the Moon has less than 2% from the whole matter which they had in day one.
Hence, the real Earth mass in day one was bigger by 98 from its current mass value. (same issue with all the rocky planets and moons in the solar system).
Therefore, the gravity force between the Earth moon in day one was much higher than the Sun moon.
That is the History!!!
We prove it by - "Hysteresis is the dependence of the state of a system on its history."
They actually use analysis of history in doing this sort of work, so yes, it would be interesting to see you use it yourself to demonstrate this new assertion.

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In any case, all the rocky planets and moons were too small to hold the gas Atoms.
If Earth and moon were 50 times larger on that first day, they would touch and be one thing, not two.

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Due to gravity force, all the heavier molecular moved to the center of the young planets/moons while most/all the gas had been ejected out over time. Therefore, all of them have a very nice ball shape.
We would never ever get a nice ball shape if from day one all we have is a solid matter. The gas is vital for the ball shape of all the planets and moons. In other wards - there is no way that the moon had been created due to some none realistic collision with the Earth
The other gas planets were big enough to hold significant portion of the gas. Therefore, we also call them gas planets.
The Sun is the biggest one. Therefore even after all the hydrogen fusion process it has 91% of Hydrogen atoms.

Even Ceres - was a nice big ball shape planet in early time.
Unfortunately, some other big object had been collided with this planet and broke it to small pieces.
It seems to be in one piece to me.  Or are you saying that the belt was once a single planet?  That planet is not Ceres, which is just the name of that one piece.  The hypothetical planet has sometimes been called Phaeton.

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The Earth have got all its water supply from day one. (Therefore, there was no need for secial water delivery as our scientists believed.)
If I understand it correctly, there is no DNA for matter. However, if we could check it, we should find that all the matter in the solar system has the same DNA.
Actually, as all the matter in the Milky way had been created by the same SMBH, they all should carry the same DNA.
If matter doesn't have DNA, then how can it all carry the same DNA?

My comment on all this seems to be that I don't see how any of this follows from or depends on your two primary points about VHP and stuff drifting away from gravitational sources.
You said you are addressing my questions one at a time, and this post started as the answer to my question about which objects have VHP's and which don't.  But none of this really discusses that topic except the first few lines which now says that the sun doesn't need one now, in contrast to your prior posts.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/01/2019 14:18:42
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In the same token, as that virtual host point orbits at a fixed amplitude from the galactic disc, every other virtual host point must do so.
What do you mean by this?  They have an orbit, a wobble (sort of like the green sin wave)?  What do you mean by 'fixed'?  Do all VHP's have this same amplitude, or do you mean each is different but regular, not varying from one peak to the next?
All the stars in the galaxy orbit around some virtual host point.
This host point orbits around the galaxy.
So, the stars wobble (sort of like the green sin wave), however, the virtual host point which show the orbital motion, moves at a fixed distance from the galactic disc.
If we will compare the distance between any two nearby virtual points over time (while they are at the same siral arm), we should see that the distance between the two is fixed. Therefore, the relative speed between any two nearby stars should be Zero (or almost zero).
What has this VHP thing got to do with the stuff below where the center of the galaxy creates new matter that is expelled away from there?
I do not understand the question.
Would you kindly explain the question.

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Based on that idea, all the stars, gas clouds, dust... are drifting outwards from the center. Nothing can move inwards.
Not even irregular 'not real' moons?  You seem to be inconsistent with prior posts, so I'm trying to get it straight.
As I have stated, not real moons means a broken objects.
I would assume that a broken object by definition is asteroid (even if we call it moon).
So, if the astroide is close enough to the host, it will move inwards and eventually it should collide with the host.
We must find the correct formula which shows the drifting direction over time.
In any case, as I have stated, all Stars, Planets and real moons must drift outwards.
There is a formula for the temperature of the accretion disk at radius R which is something like
T(R) = √√[3GM⊕/8πσR³ (1−√(Rinner / R)) ]
It is clear to me that in order to prove the none realistic hypothetical ideas, our scientists come with new formula. They think that by mathematics they can force the Universe to act upon their understanding.
Sorry. This is a severe mistake.
If they want to prove something, they have to use real formula by Newton, Kepler or Eisenstein.

How does the magnetic disc know which way to send the material?  I've never heard of a magnetic field with only rotational symmetry, but I could just be ignorant of that.
At what radius does this take place?
Good question.
I also think that it must be symmetrical.
So, I assume that the new born molecular are ejected upwards and downward (with regards to the galactic plane) due to the magnetic field.
I have no idea at what radius the magnetic field starts to work.
Each gas molecule has a virtual host point?
Sure.
The atom/molecular must be concentrated in order to set the gas cloud. The only possibility to do so is by orbiting around a virtual host point in the cloud.
So, as the molecular orbit around this virtual host, they start to form all the stars, planets and moons that we have in the galaxy.
However, not all the gas in the cloud is used for the new stars forming activity.
The leftover of the gas cloud is ejected from the disc plane of the galaxy.
We see it clearly in the following example:
https://www.universetoday.com/102011/hydrogen-clouds-discovered-between-andromeda-and-triangulum-galaxies/
Hydrogen Clouds Discovered Between Andromeda And Triangulum Galaxies.
Those are two spiral galaxies.
Andromeda is the mother galaxy of Triangulum.
Triangulum had started as a small black hole at the center of Andromeda.
As it drifts outwards from Andromeda center (with all the other stars) it starts to create its own new mass.
I will explain later on how this activity impacts our observable universe.
In any case, as those two galaxies move away from each other they ejects the leftover of the gas clouds.
Therefore, we see all of those gas clouds between the two galaxies.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/01/2019 14:33:17
You seem to contradict yourself.  You say all stars have a virtual host point, but then say now that the sun doesn't have one.  You put it a light-year away in a prior post, and now say it doesn't exist.
Sorry if I was not clear.
The Sun orbits around a virtual host point. But there is no need for host point in the solar system.
Therefore, all the planets and moons orbit around real objects.
I have no clue if the orbital radius around the VHP is 1 light year or 100 Light years.
The main idea was that 64 Million years for one orbital cycle of the Sun around that VHP is too long (Based on my understanding).
Yes, I agree that binary systems have stars that effectively orbit their barycenter (their mutual center of gravity).  You calling it a virtual host point just changes the terminology.  There's already one word for it.
For systems with 3 or more stars, there is very much a center of gravity which remains relatively fixed no matter the motion of the stars involved, but the motion about that point is no longer strictly orbital, depending on the dynamics of the stars.  This is what they mean by the three-body problem.  Typically pairs of stars orbit each other, and those pairs orbit other stars or other pairs, until there are only 2 bodies left.  3 or more in mutual orbit with no 2 in independent orbit is not stable.

Anyway, if you're talking only about the center of mass of multi-star solar systems, then this idea is not controversial.  Yes, that point exists, and the motions of the member stars does not move it one bit, but the point goes around the galaxy more or less as a unit.  The sun wiggles around its own point, with the point being inside the sun about half the time, and outside the other half.  The (quite predictable) path around that fixed point is anything but elliptical.  It resembles more of a scribble with no particular cycle to assign a period length.
Yes, the whole idea is based on center of mass of multi-star solar systems.
However, there are two main issues to remember:
1. Each star orbits around a virtual host center. This is a gift that each star gets when it had been formed at the gas cloud. So, the center of mass of multi-star solar systems is actually a center of the Virtual host points of all the other solar/stars systems. Therefore, when we set a calculation we actually should ignore the current location of the star itself, but focus on it's virtual host point which is represented by the white dashed line (It is also called the Star motion around the galaxy).
2. This center of mass (of the entire nearby virtual host points) is working under the impact of the galaxy' gravity. It is quite complicate, as at different segments of the galaxy it has different gravity forces. I will explain it later on.

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Gravity force on the moon is more than a twice than the Earth gravity force on the moon.
Assuming gravity works like physics says and not like you say, acceleration due to the moon itself at its surface is 0.1630 m/s, 0.0027 due to Earth, and 0.0060 due to sun. That means it is 60 times as much, which indeed is 'more than twice' the acceleration on the moon.  It is about 60x in fact.

I say acceleration because there is no force of gravity on the moon itself.  You would need an object to have a force, and no object was specified.  So if I stand on the moon, there might be a force of 135 N on me, but the 'the Earth gravity force on the moon' is about 2e20 N, which is a lot more than my 135 N.  That huge 2e20 figure is the force that Earth exerts on the moon, and is also exactly the force that the moon exerts on Earth, at least per standard physics.

I'm saying all this because I'm trying to figure out what you meant to say, as opposed to what you actually said.
You miss the whole point.
The idea is as follow:
let's assume that we break out the Moon/Earth orbital system.
Let's give the moon the possibility to chose around which one it prefers to orbit.
Under the current conditions the gravity force of the Sun/moon is twice stronger than the earth/moon.
Therefore, it is clear to me that it should chose to orbit around the Sun instead of around the moon.
However, the moon orbits around the Earth as in early time when the Moon and the earth were gas objects and the whole solar system were more concentrated, the gravity force between the Earth/moon was higher than the Moon Sun gravity force.
Is it clear?
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 26/01/2019 14:36:04
are you saying that the belt was once a single planet?  That planet is not Ceres, which is just the name of that one piece.  The hypothetical planet has sometimes been called Phaeton.
Yes. That is correct!
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/01/2019 14:58:03
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In the same token, as that virtual host point orbits at a fixed amplitude from the galactic disc, every other virtual host point must do so.
This host point orbits around the galaxy.
You didn't really answer my question.  OK, the VHP orbits around the galaxy.  Where does 'amplitude' come into that?  An orbit doesn't have an amplitude.  The star might have one, the amplitude being its distance from its host point, but how does the VHP have an amplitude?
Just trying to understand your terminology.

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In any case, as I have stated, all Stars, Planets and real moons must drift outwards.
Why is Earth not drifting measurably outward?

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It is clear to me that in order to prove the none realistic hypothetical ideas, our scientists come with new formula. They think that by mathematics they can force the Universe to act upon their understanding.
No scientist would assert that they are forcing the universe to do anything.  The universe is forcing them to find mathematics that describes it.  I've seen no mathematics from you, so they're doing much better so far.

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Sorry. This is a severe mistake.
If they want to prove something, they have to use real formula by Newton, Kepler or Eisenstein.
But you've discarded the works of all three of these people.  Their mathematics forbid the sort of thing you are describing.  Every one of the conservation laws that derive from their mathematics is violated.  If you're going to going to appeal to these people, your theory needs to conform to the physics that stems from their work.
I've been quiet about the blatant violations because I've assumed that your model would have new mathematics, but here you are suddenly appealing to established views instead of this new one you are pushing.

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Quote from: Halc
How does the magnetic disc know which way to send the material?  I've never heard of a magnetic field with only rotational symmetry, but I could just be ignorant of that.
At what radius does this take place?
Good question.
I also think that it must be symmetrical.
I said it was, but seemingly only rotational symmetry.  If it had any kind of mirror symmetry, material would be accelerated equally in both directions, which would at least conserve angular momentum.  Your idea has no angular momentum conservation.

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I have no idea at what radius the magnetic field starts to work.
So much for the good question...

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Quote from: Halc
Each gas molecule has a virtual host point?
Sure.
The atom/molecular must be concentrated in order to set the gas cloud. The only possibility to do so is by orbiting around a virtual host point in the cloud.
I can concentrate gas by squeezing it in a tube.  I can light a match this way if I squeeze hard on it.  No virtual host point needed, just compressing force.  A virtual host point I suppose would make a gas atom go in a circle, but I don't see how that helps with their being more concentrated since being a circle, the atom will just come back to its original point, or even further out since everything drifts away over time, which results in decreased concentration, not increased concentration.

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So, as the molecular orbit around this virtual host, they start to form all the stars, planets and moons that we have in the galaxy.
No they don't.  They drift away you said.  How can a star form if everything drifts away?

However, not all the gas in the cloud is used for the new stars forming activity.
The leftover of the gas cloud is ejected from the disc plane of the galaxy.
We see it clearly in the following example:
https://www.universetoday.com/102011/hydrogen-clouds-discovered-between-andromeda-and-triangulum-galaxies/
Hydrogen Clouds Discovered Between Andromeda And Triangulum Galaxies.
Those are two spiral galaxies.
Andromeda is the mother galaxy of Triangulum.
Triangulum had started as a small black hole at the center of Andromeda.
[/quote]So black holes can spawn not just new plasma material, but objects like new black holes??
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In any case, as those two galaxies move away from each other they ejects the leftover of the gas clouds.
Therefore, we see all of those gas clouds between the two galaxies.
They're actually on course for collision in 2.5 billion years, but that's according to those scientists who obviously just make stuff up.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/01/2019 15:19:44

SMBH

Do black holes in general generate new matter like that?  I ask because we don't seem to see this happening around smaller black holes.  What is special at a galaxy center that this process goes on, but we don't see it elsewhere?
There are different kinds of black holes.
For quite long time I have thought about that issue.
Let's look at the SMBH at the Milky Way.
We see clearly that there is matter in the accretion disc.
If I understand it correctly, the total mass in the accretion disc is estimated to be in the range of 3 Sun mass.
Actually, based on the orbital velocity of the mass at the accretion disc, and the estimated radius, we could easily calculate the real mass of the SMBH.
In any case, it seems to me that the SMBH must increase its mass over time.
However, I wonder how it gets new matter.
There could be two options:
1. Some of the new formed particles drift inwards.
2. In one of the article that I have found it was written that for any new particle that is created, there must be a negative particle. If that is correct, and assuming that all the positive particles are drifting outwards, than the negative must drift inwards. In this case, the SMBH is a concentration of the negative particles. (I have no clue if that is correct...)

Let's look at another kind of SMBH - Quasar:
https://www.forbes.com/sites/startswithabang/2017/07/31/universes-largest-black-hole-may-have-an-explanation-at-last/#52b55966fc55
"The brightest, most luminous objects in the entire Universe are neither stars nor galaxies, but quasars, like S5 0014+81."
So, this SMBH (we call it quasar) has no galaxy.
It creates new matter and ejects it in a stream upwards and downwards as expected.
So it works perfectly as the SMBH of any spiral galaxy should work.
New matter is created at the accretion disc. As the new molecular drifts outwards they get to the magnetic shield around the accretion disc. The ultra high magnetic power blow it upwards and downwards.
It is also clear that those molecular fall back to the disc plane of the Quasar (outwards from the magnetic shield). However, somehow, they don't form the gas cloud. Without the gas cloud there is no new born stars. Without stars there is no galaxy.
So, this is a good example for a SMBH which generates new matter without having a real spiral galaxy.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 27/01/2019 15:50:44
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In any case, as those two galaxies move away from each other they ejects the leftover of the gas clouds.
Therefore, we see all of those gas clouds between the two galaxies.
They're actually on course for collision in 2.5 billion years, but that's according to those scientists who obviously just make stuff up
No, this is incorrect.
They are quite close to each other, but they are drifting away from each other. However, both of them are moving in our direction. So the 2.5 Billion is the requested time for the collision with the Milky way.
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So, as the molecular orbit around this virtual host, they start to form all the stars, planets and moons that we have in the galaxy.
No they don't.  They drift away you said.  How can a star form if everything drifts away?
I don't see any problem with that.
Let's assume that it takes 10 Million years to set a star in a gas cloud, and in this time frame the gas cloud had increased its radius by 10%
In this case, this gas cloud will set new star and planets
Please also remember that the gas cloud itself is drifting away from the center.
So, it is clear that at some point of time the gas cloud (or the leftover from the gas cloud) will not be able to create any more stars and planets.
What is the problem with that ?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/01/2019 16:51:22
The Sun orbits around a virtual host point. But there is no need for host point in the solar system.
...
I have no clue if the orbital radius around the VHP is 1 light year or 100 Light years.
...
Yes, the whole idea is based on center of mass of multi-star solar systems.
Our sun is not part of a multi-star solar system, so it doesn't have this kind of virtual host point, and yet you talk about it having one nearby or perhaps really quite distant.  If it has one like that, then it isn't a case of the sun being part of a multi-star solar system unless you consider the entire galaxy to be a multi-star solar system.
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So, the center of mass of multi-star solar systems is actually a center of the Virtual host points of all the other solar/stars systems.
Stars in a multi-star system have each a different virtual host point, which is not the same thing as their common center of mass.  OK, so you're not giving a new name for an old thing.  You seem to be claiming that with three or more stars in one system, each will move in a circular (or at least elliptical) motion around its own particular VHP.  Have you tried simulating such a thing with say 3-5 objects?  If you apply Newton's laws, you will notice that this just doesn't happen.  I've been willing to accept this different physics, but you've not given the new mathematics for it, so I cannot create a simulation that makes predictions.

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Therefore, when we set a calculation
I don't see any setting of calculations going on.

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I'm saying all this because I'm trying to figure out what you meant to say, as opposed to what you actually said.
You miss the whole point.
I know.  I said I did, and was trying to figure out your point.

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The idea is as follow:
let's assume that we break out the Moon/Earth orbital system.
Let's give the moon the possibility to chose around which one it prefers to orbit.
Under the current conditions the gravity force of the Sun/moon is twice stronger than the earth/moon.
Therefore, it is clear to me that it should chose to orbit around the Sun instead of around the moon.
That doesn't follow at all.  You're assuming it is a function of the greatest force, and it isn't.  And the moon does orbit the sun, and the galaxy, and the sun's virtual host point if that exists.

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However, the moon orbits around the Earth as in early time when the Moon and the earth were gas objects and the whole solar system were more concentrated, the gravity force between the Earth/moon was higher than the Moon Sun gravity force.
Is it clear?
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/01/2019 17:46:53
There are different kinds of black holes.
For quite long time I have thought about that issue.
Let's look at the SMBH at the Milky Way.
We see clearly that there is matter in the accretion disc.
If I understand it correctly, the total mass in the accretion disc is estimated to be in the range of 3 Sun mass.
Actually, based on the orbital velocity of the mass at the accretion disc, and the estimated radius, we could easily calculate the real mass of the SMBH.
Yes, that's how it's done in normal physics, but you've denied that when the scientists measure the mass of the galaxy using that exact method and got a mass much larger than all the material they've measured.  The scientists say there must be more mass they can't see, while you say they are using the wrong physics to calculate the mass.

Therefore, you don't know the mass of the SMBH using the speed of the material orbiting it, since you deny the validity of that method.  You need a new way of measuring the mass of an object.

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It's not like it stops abruptly.  Technically, the whole galaxy is an accretion disk.  It just changes properties the closer to the gravity source it gets.

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In any case, it seems to me that the SMBH must increase its mass over time.
You have it spewing matter out into the galaxy.  I'd think it that something that does that might be losing mass. I don't see how stuff falls in if the stuff is supposed to tend to drift away, not towards it.

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However, I wonder how it gets new matter.
There could be two options:
1. Some of the new formed particles drift inwards.
2. In one of the article that I have found it was written that for any new particle that is created, there must be a negative particle.

If that is correct, and assuming that all the positive particles are drifting outwards, than the negative must drift inwards.
The negative particles are antimatter actually, yes.  If the antimatter falls in, the black hole actually gets smaller.  It's how Hawking radiation eventually evaporates any black hole given enough time, but I'm not sure of your idea has this sort of process in it.
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In this case, the SMBH is a concentration of the negative particles. (I have no clue if that is correct...)
That's actually a way to solve it, if the black hole is antimatter, such radiation would increase the mass, not decrease it, assuming it makes sense for the anti-halves to tend to go in and not a random distribution.  Anyway, such radiation (change in mass) is greatest for the smallest black holes and slowest for the big ones.  Sgr-A might lose a few atoms per year via Hawking radiation. Maybe not even that fast.

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Let's look at another kind of SMBH - Quasar:
https://www.forbes.com/sites/startswithabang/2017/07/31/universes-largest-black-hole-may-have-an-explanation-at-last/#52b55966fc55
"The brightest, most luminous objects in the entire Universe are neither stars nor galaxies, but quasars, like S5 0014+81."
They're just young galaxies.  There are not any nearby because nothing nearby is that young.  Only looking waaay into the past do you see the initial formation processes of a galaxy.  Our own galaxy might have been a quasar at some point, but it was probably too small in its youth to be the sort of object they're looking at.  Those are massive objects that have formed what are the proper big galaxies we now see closer by.
This is not an expert statement.  It is just me spouting off at the mouth.  I know they've found black holes that are well of 100x the size of Sgr-A, and the formation of those would have generated the sort of light that you see in a quasar.
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Without the gas cloud there is no new born stars. Without stars there is no galaxy.
You don't know how many stars it has.  Trying to spot one is like spotting a firefly sitting between a car's headlights.  These quasars are also way to far away to see individual stars, except when they go supernova.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 27/01/2019 18:01:10
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They're actually on course for collision in 2.5 billion years, but that's according to those scientists who obviously just make stuff up
No, this is incorrect.
They are quite close to each other, but they are drifting away from each other.
But accelerating inward, on schedule for a 2nd collision pass in 2.5 billion years.

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However, both of them are moving in our direction. So the 2.5 Billion is the requested time for the collision with the Milky way.
Andromeda will takes its first pass in about 4 billion years, well after triangulum galaxy is well within Andromeda, and the 2nd about 2 billion years after that.  After that, you don't really count passes anymore, but it takes even longer for the process to finally finish and our SMBH is swallowed by a much larger one.
Not sure if you support that idea since these things are suppose to drift away, not be pulled in and combined.

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I don't see any problem with that.
Let's assume that it takes 10 Million years to set a star in a gas cloud, and in this time frame the gas cloud had increased its radius by 10%
In this case, this gas cloud will set new star and planets
Your gas cloud just got 33% less dense (33% increase of volume from a 10% increase in radius).  It would seem to require gas to be condensed to form stars, else they would already have been stars when the gas cloud was tighter.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 29/01/2019 15:11:42
Your gas cloud just got 33% less dense (33% increase of volume from a 10% increase in radius).  It would seem to require gas to be condensed to form stars, else they would already have been stars when the gas cloud was tighter.

Please see the following article about: Star Formation in gas cloud Around Suppermassive Black Holes:

https://arxiv.org/ftp/arxiv/papers/0810/0810.2723.pdf
"The presence of young massive stars orbiting on eccentric rings within a few tenths of a parsec of the suppermassive black hole in the Galactic centre is challenging for theories of star formation"

"The presence of young massive stars in the vicinity of the Galactic centre is difficult to reconcile with current models of star formation where turbulent molecular clouds produce a mostly clustered population of stars with a remarkably constant distribution of stellar masses, covering stars with masses from less than a tenth to greater than 100 times the mass of the sun (6)."

We have to understand the real meaning of: "turbulent molecular clouds".
In the article there is no real explanation. (Why they didn't trace that turbulent activity?)
However, I think that there is an order in that "turbulent".
I assume that it represents several/many internal orbital cycles.
If we could look carefully, we might find that around each orbital cycle there is an activity for new star forming.
So, deep in the gas cloud each star stars to orbit around a virtual center of mass.
All the virtual centers of new born stars are orbiting according to a multi star system under the impact of the nearby SMBH.
Therefore, deep in the gas cloud, each new born star gets a gift of orbiting around a virtual Host point.
That is correct to any star in the galaxy.
Actually, if we will trace S2, we might find that it set a bobbling activity as it orbits around the SMBH. That bobbling proves that even this star orbit around a virtual center of mass.
So, the gas cloud can set new stars as it drifts outwards. It just an issue of verifying the rate of outwards drifting VS the rate of new star forming.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 29/01/2019 23:08:37
[Concerning the] article about: Star Formation in gas cloud Around Suppermassive Black Holes:

"The presence of young massive stars orbiting on eccentric rings within a few tenths of a parsec of the suppermassive black hole in the Galactic centre is challenging for theories of star formation"

"The presence of young massive stars in the vicinity of the Galactic centre is difficult to reconcile with current models of star formation where turbulent molecular clouds produce a mostly clustered population of stars with a remarkably constant distribution of stellar masses, covering stars with masses from less than a tenth to greater than 100 times the mass of the sun (6)."

We have to understand the real meaning of: "turbulent molecular clouds".
In the article there is no real explanation. (Why they didn't trace that turbulent activity?)
The article is proposing an alternate description from the typical model that dominates regions further from the black hole.  Our own sun may well have been born of a turbulent molecular cloud.  Your idea might propose otherwise.

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Actually, if we will trace S2, we might find that it set a bobbling activity as it orbits around the SMBH.
They've tracked it quite closely through more than a full orbit now.  It doesn't have any kind of regular bobbing activity.  S2 does seem to be the sort of object referred to in the article, being a large young star in a highly eccentric orbit about Sgr-A.  That one gets particularly close to it.

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That bobbling proves that even this star orbit around a virtual center of mass.
Or it would if it actually was observed to do that.

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So, the gas cloud can set new stars as it drifts outwards. It just an issue of verifying the rate of outwards drifting VS the rate of new star forming.
The article you linked described the gas cloud drifting inwards, where the compression from the SMBH heats it up and aids star formation.  Drifting outward would have a cooling effect just like how decompressing gasses is how air-conditioners create low temperatures from warm materials.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 30/01/2019 14:47:39
They've tracked it quite closely through more than a full orbit now.  It doesn't have any kind of regular bobbing activity.  S2 does seem to be the sort of object referred to in the article, being a large young star in a highly eccentric orbit about Sgr-A.  That one gets particularly close to it.
Yes it does.
S2 has a clear bobbling orbiting path.
Please look at the following diagram:
http://www.phy6.org/stargaze/Sblkhole.htm
We see clearly that the measured locations of S2 are not directly at the elliptical orbital cycle.
I do recall that in one of the articles it was stated that in 2002, the light of S2 was exactly at the same location as SMBH.
Our scientists were puzzled from this evidence.
They couldn't explain that verification. Therefore they had an idea of error in their instruments.
This was a big mistake.
There is no error in the data.
S2 orbits around a Virtual host point, while this VHP orbits around the SMBH.
Therefore, we see some Zig Zag in the location of S2 with reference to the expected orbital path.
Hence, S2 can even cross the location of the SMBH, while its VHP is moving exactly at the expected elliptical cycle.
I really don't understand why we have only few measured points on that image.
Technically we should know evry day where it is.
Why this information is not valid for all of us?
Why they hide this important information???

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 30/01/2019 17:30:24
S2 has a clear bobbling orbiting path.
Please look at the following diagram:
http://www.phy6.org/stargaze/Sblkhole.htm
We see clearly that the measured locations of S2 are not directly at the elliptical orbital cycle.
I don't see it.  I see an ellipse (viewed almost edge-on) and every measurement falling on it within the margin of error.  Nothing in the article suggests an irregularity to the orbit, or especially a regular one.  Have you run a fourier transform on the measurement data?  It would show such a host-point cycle if it had one.

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I do recall that in one of the articles it was stated that in 2002, the light of S2 was exactly at the same location as SMBH.
Same location in the sky, not the same location in space, which would be impossible.  It could not emerge back out if it was there.

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I really don't understand why we have only few measured points on that image.
There are more I think since other groups have been tracking S2 quite closely in an effort to detect dark matter.  Your host point would very much appear to be dark matter.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 01/02/2019 15:16:09
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S2 has a clear bobbling orbiting path.
Please look at the following diagram:
http://www.phy6.org/stargaze/Sblkhole.htm
We see clearly that the measured locations of S2 are not directly at the elliptical orbital cycle.
I don't see it.  I see an ellipse (viewed almost edge-on) and every measurement falling on it within the margin of error.  Nothing in the article suggests an irregularity to the orbit, or especially a regular one.
Sorry
This is the biggest mistake of our scientists!
In order to understand this important issue, let's look at the example of the Moon/Sun orbital cycle/
Let's assume that the Earth is invisible from outside. Therefore, it is actually a VHP for the Moon while this one orbits around the Sun.
So, if we try to monitor the Moon/Sun orbit, while we have no clue if the Earth is there, what shall we see?
R1 (Moon/Earth) = 384,400 Km.
R2 (Earth/Sun) = 149,600,000 Km.
The ration is:
R1/R2 = 384,400 / 149,600,000 = 0.0025
In this example let's assume that the Moon/Earth orbital cycle is located at the same Earth/Sun orbital cycle.
Therefore, if we try to monitor the Moon/Sun orbital cycle, we might see that the moon is moving in almost a perfect cycle, with very minor error.
If we randomly monitor the location of the Moon, it is clear that the error fit is in the range of Zero to 0.0025.
Just if we monitor the moon while it is at its maximal/minimal distance from the sun (R2+R1 or R2-R1) we can get that maximal error fit of 0.0025.
From statistical point of view the chance to get randomly to the maximal 0.0025 Error is very low.
The error should be mainly greater than Zero and lower than 0.0025.
Now, please, let me remind you that in this example we don't know if the Earth (which is the VHP of the Moon) is there.
So, if we give those results to our scientists, what would be the answer?
Based on your answer, our scientists should reply:  "Every measurement falling on it within the margin of error"
So, based on this error margin, the Earth doesn't exist.
Now, let's go back to the S2 orbital cycle.
Please look again at the measured. What is the estimated error margin?
http://www.phy6.org/stargaze/Sblkhole.htm
We see clearly that many measured points are located outside the Fit orbital cycle.
If we look carefully, we clearly see that the maximal measured error point took place on 1995.36
It seems to me as an error which is significantly bigger than 0.0025.
If you also look carefully, we see that the SMBH is not located at the symmetrical point in that orbital cycle.
It is located too close to the left bottom side.
That is one more evidence that S2 doesn't orbit directly around the SMBH.
What about the orbital velocity? Did we try to verify if there are small variations in the measured velocities?
So, how our scientists could afford themselves to ignore those clear evidences of none fit?
I think that it is a fatal mistake to claim:  "Every measurement falling on it within the margin of error"!!!
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 01/02/2019 17:23:03
This is the biggest mistake of our scientists!
The scientists know how to do vector arithmetic, and you don't, so the odds of them being the one making the mistakes are pretty low.
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In order to understand this important issue, let's look at the example of the Moon/Sun orbital cycle/
Let's assume that the Earth is invisible from outside. Therefore, it is actually a VHP for the Moon while this one orbits around the Sun.
You just described dark matter, something you deny.

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So, if we try to monitor the Moon/Sun orbit, while we have no clue if the Earth is there, what shall we see?
R1 (Moon/Earth) = 384,400 Km.
R2 (Earth/Sun) = 149,600,000 Km.
The ration is:
R1/R2 = 384,400 / 149,600,000 = 0.0025
Yes, and if you run a Fourier transform on the data for the position of the moon over time, there would be two huge spikes for the periods of a month and a year.  The virtual host point would be glaringly obvious from that data.

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In this example let's assume that the Moon/Earth orbital cycle is located at the same Earth/Sun orbital cycle.
I don't know what you mean by this.  A cycle doesn't have a location, and I don't think you are proposing that the Earth is where the sun is.  You want the moon to go around the Earth in one sidereal year?  That would put it at one of the Lagrange points, making it a trojan satellite, not really orbiting the Earth.

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Therefore, if we try to monitor the Moon/Sun orbital cycle, we might see that the moon is moving in almost a perfect cycle, with very minor error.
No error at all in fact.

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If we randomly monitor the location of the Moon, it is clear that the error fit is in the range of Zero to 0.0025.
Just if we monitor the moon while it is at its maximal/minimal distance from the sun (R2+R1 or R2-R1) we can get that maximal error fit of 0.0025.
We'd know the moon's position to far greater accuracy than that from the distance you indicate.
If you are appealing to margin of error, then it doesn't support your idea, it just says the data isn't accurate enough to make a determination.  No scientists are making mistakes then.  A regular perpetration would show up in a Fourier transform with enough data points, even with a margin of error far larger than the perpetration.

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From statistical point of view the chance to get randomly to the maximal 0.0025 Error is very low.
OK, your knowledge of statistics is on par with your vector arithmetic prowess I see.  Statistics is exactly how you get an arbitrarily small margin of error from data points each of which has a large margin of error.

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The error should be mainly greater than Zero and lower than 0.0025.
Now, please, let me remind you that in this example we don't know if the Earth (which is the VHP of the Moon) is there.
So, if we give those results to our scientists, what would be the answer?
If it goes around the Earth in the same period that it goes around the sun, then the dark Earth will be undetectable.  I'm not sure why you proposed that period, or even if that is what you meant.

Based on your answer, our scientists should reply:  "Every measurement falling on it within the margin of error"
So, based on this error margin, the Earth doesn't exist.[/quote]
If the moon has a period around Earth that is a month (something less than a year), and the measurements are all a worse margin of error than this 0.0025, then a sufficiently large sample of measurements would suffice to detect Earth, or the lack of it.

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If you also look carefully, we see that the SMBH is not located at the symmetrical point in that orbital cycle.
It is located too close to the left bottom side.
That's because you're looking at an ellipse nearly edge-on which distorts where the focus points are.  Our vantage point is not perpendicular to the plane of S2's orbit, but you seem to assume so.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/02/2019 06:28:50
In this example let's assume that the Moon/Earth orbital cycle is located at the same Earth/Sun orbital cycle.
Sorry if I was not clear.
The idea is:  The Moon/Earth orbital cycle disc plane is located at the same Earth/Sun orbital cycle disc plane.
Again - In this example we want to verify that just by monitoring the locations of the Sun & moon in one cycle (one year) we should find that the Earth is missing (assuming that we don't see it in our verification).
If the moon has a period around Earth that is a month (something less than a year), and the measurements are all a worse margin of error than this 0.0025, then a sufficiently large sample of measurements would suffice to detect Earth, or the lack of it.
O.K.
So, you understand the idea of this example. You also agree that if we see that our measurements of the moon has a small variations from the expected perfect orbital cycle (even if it is less than 0.0025) we should know that the moon must orbit around some other object. We call it Earth, but we can also call it VHP (as we don't see it in our example).
Therefore, I still don't understand why do you still support our scientists while they have clearly neglecting the variations of S2 locations from the expected perfect orbital cycle? They call those variations - Errors. But those variations are very important observations.
Let me use the example at a different perspective:
Let's assume that the  Moon/Earth orbital cycle disc plane is located vertically to the Earth/Sun orbital cycle disc plane and it orbits directly in the Earth/sun orbital cycle.
Let's also assume that our vantage point is perpendicular to the Earth/Sun orbital plane.
Therefore, it is clear that in all the verifications of the Moon location points we should see that it is perfectly located directly at the expected orbital cycle around the Sun.
So, without any variations (or errors) from the expected orbital cycle, we might think there is no need for the earth to explain a perfect orbital fit of the Moon around the Sun.
However, there is another key verification - Velocity.
"KEPLER’S SECOND LAW DESCRIBES THE WAY AN OBJECT’S SPEED VARIES ALONG ITS ORBIT
A planet’s orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun’s gravitational pull on it, and the faster the planet moves. The farther it is from the Sun, the weaker the Sun’s gravitational pull, and the slower it moves in its orbit."
So, in a real elliptical orbit we should find that the velocity of the planet is decreasing smoothly as the farther it is from the Sun.
However, in this example, Due to the Moon/Earth orbital path, sometimes the moon orbits in the orbital Earth/sun direction, and sometimes on the opposite direction (I hope that you understand me correctly).
Therefore, we should find the moon is not smoothly increasing or decreasing its orbital velocity as it orbits around the sun.
So, we have two ways to verify if the Earth (or the VHP) is there.
One - By verify if there is variations (error) in the verified moon location with reference to the expected orbital elliptical cycle.
Two - By verify if there is none smoothly variations in the orbital velocity.
the location of the moon.
With regards to S2 - Even with the assumption that our vantage point is not perpendicular to the plane of S2's orbit, we clearly see that S2 Does not fulfill the two requirements.
Therefore, it shows that S2 must orbit around a VHP while this one might orbit the SMBH.
There is one more key evidence for that:
Please look at the following diagram:
https://en.wikipedia.org/wiki/S2_(star)#/media/File:Galactic_centre_orbits.svg
We see several S stars that orbit around the SMBH.
I assume that our vantage point is not perpendicular to any of those S planes of orbit.
There is good chance that each orbital cycle plane might be different from all the other.
However, if we try to estimate the requested location of the SMBH for each cycle (based on kepler law), we should find that each one needs the SMBH at a different location in space.
So, that also shows that those stars do not orbit directly around the same SMBH.
Each one of them orbits around a VHP while all of those VHP orbits around the SMBH.
Why can't you agree with something which is so clear to us?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 02/02/2019 14:04:07
The idea is:  The Moon/Earth orbital cycle disc plane is located at the same Earth/Sun orbital cycle disc plane.
Oh, OK.  It already is you know.  There is virtually no tilt to it, something not true in general.
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Again - In this example we want to verify that just by monitoring the locations of the Sun & moon in one cycle (one year) we should find that the Earth is missing (assuming that we don't see it in our verification).
Earth would be detectable in this manner no matter what the tilt of the moon's orbit.
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So, you understand the idea of this example. You also agree that if we see that our measurements of the moon has a small variations from the expected perfect orbital cycle (even if it is less than 0.0025) we should know that the moon must orbit around some other object.
It could have a very large variation (called eccentricity) and we'd still be able to detect the orbit, yes.

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We call it Earth, but we can also call it VHP (as we don't see it in our example).
Yes, we presumably don't know it's there ahead of time, so we certainly don't know its name.

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Therefore, I still don't understand why do you still support our scientists while they have clearly neglecting the variations of S2 locations from the expected perfect orbital cycle? They call those variations - Errors. But those variations are very important observations.
They've probably run a Fourier transform on the data and found no regular orbit to it.  That's why asked if you had done that.  You seem to be getting your data from pictures and conclusions written by somebody else, rather than from the actual measurements taken.  If you claim a pattern to the data, take the data and demonstrate it.

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Let me use the example at a different perspective:
Let's assume that the  Moon/Earth orbital cycle disc plane is located vertically to the Earth/Sun orbital cycle disc plane and it orbits directly in the Earth/sun orbital cycle.
You mean perpendicular I presume.  A plane cannot be vertical to another one.
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Let's also assume that our vantage point is perpendicular to the Earth/Sun orbital plane.
Therefore, it is clear that in all the verifications of the Moon location points we should see that it is perfectly located directly at the expected orbital cycle around the Sun.
No, it would move back and forth in one dimension but not in the other.  So assuming the axis runs right/left from that vantage, the motion you'd see from that vantage is a continuous up/down movement, which is towards and away from the sun when Earth is on the upper or lower side of the orbit, and forwards and backwards in its orbit when Earth is on either side.
If your measurements were good, you could also detect velocity towards and away from your vantage point, which would clue you in to motion in that dimension.  Good measurements are 3D and thus better than 2D pictures.

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So, without any variations (or errors) from the expected orbital cycle, we might think there is no need for the earth to explain a perfect orbital fit of the Moon around the Sun.
I can think of no orientation of axes or vantage point that would make Earth undetectable in this way.

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However, there is another key verification - Velocity.
"KEPLER’S SECOND LAW DESCRIBES THE WAY AN OBJECT’S SPEED VARIES ALONG ITS ORBIT
A planet’s orbital speed changes, depending on how far it is from the Sun. The closer a planet is to the Sun, the stronger the Sun’s gravitational pull on it, and the faster the planet moves. The farther it is from the Sun, the weaker the Sun’s gravitational pull, and the slower it moves in its orbit."
So, in a real elliptical orbit we should find that the velocity of the planet is decreasing smoothly as the farther it is from the Sun.
However, in this example, Due to the Moon/Earth orbital path, sometimes the moon orbits in the orbital Earth/sun direction, and sometimes on the opposite direction (I hope that you understand me correctly).
Therefore, we should find the moon is not smoothly increasing or decreasing its orbital velocity as it orbits around the sun.
So, we have two ways to verify if the Earth (or the VHP) is there.
They're actually the same way.  You variations are detectable despite any orientation of axes or vantage points.  So yes, there would be no way to hide the VHP if there was one.  Earth is not a VHP in that scenario.  It is a dark host mass.  It is a real mass, not a virtual one, and is dark, simply meaning we cannot see it.  For the moon or any other object to orbit an actual VHP would mean acceleration without any of the known forces to account for it.
So if scientists were to detect such regular motion, they'd know.  They detect this sort of motion for many stars, even when the motion is very subtle (less than the radius of the star), and use the motion to detect dark masses like gas giant planets.  Our own star exhibits this sort of motion about a host point, but its motion around it is not elliptical and thus really an orbit.  Sometimes the sun is incredibly close to its host point, and sometimes further away.  It's acceleration is not a function of its distance from that host point, and in the case of Keplerian orbit, it would be a function of that distance, per Kepler's 2nd law you quoted above which only applies to simple 2-body orbits.

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One - By verify if there is variations (error) in the verified moon location with reference to the expected orbital elliptical cycle.
Two - By verify if there is none smoothly variations in the orbital velocity.
Same thing, since any variation in location or velocity would result in a variation of the other.

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With regards to S2 - Even with the assumption that our vantage point is not perpendicular to the plane of S2's orbit, we clearly see that S2 Does not fulfill the two requirements.
You clearly see what you want to.  I don't see it at all, mostly because all I have is a lousy diagram and not actual data, but the perfect orbit path depicted hits every single one of the data points within the margin of error, meaning there many be deviation from that path, but said deviation is anything but clear.

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Therefore, it shows that S2 must orbit around a VHP while this one might orbit the SMBH.
I told you how to clearly show that from the data set, or from a larger one.  You've not done that, so nothing has been shown.

There is one more key evidence for that:
Please look at the following diagram:
https://en.wikipedia.org/wiki/S2_(star)#/media/File:Galactic_centre_orbits.svg
We see several S stars that orbit around the SMBH.
I assume that our vantage point is not perpendicular to any of those S planes of orbit.
There is good chance that each orbital cycle plane might be different from all the other.
However, if we try to estimate the requested location of the SMBH for each cycle (based on kepler law), we should find that each one needs the SMBH at a different location in space.[/quote]
How so?  For it to appear be at one of the focus points of the elliptic paths, you need a perpendicular vantage point, and you admit that our vantage is anything but perpendicular.  The graph does not plot position, but rather ascension and declination, but you're interpreting the plot as position when presuming where the SMBH should be.

The graph is deceptive because they compare the appearance plot of these stars to a position plot of the orbit of several solar-system planets, none of which are presented as they appear to us, which would be nearly straight lines massively outside the narrow range of the diagram.

Anyway, have you ever watched elliptical motion from a non-perpendicular vantage point?  I have, at length even.  I had an entertaining screen saver that did just that, with a dozen objects in random orbits about a central mass, and each object left a trail behind it so the shape of the orbit was easy to see, and it looked exactly like that picture, except that your picture doesn't show where the motion is fast or slow, so you don't get a good feel for the actual point in the path where the approach to the center is closest.  S14 for instance is clearly closest on the far lower right of its path, not at the place where it goes behind the SMBH and appears to be in the same place.  It is a 2D plot of 3D motion, but you're interpreting it as 2D space when making these invalid claims.

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Each one of them orbits around a VHP while all of those VHP orbits around the SMBH.
Why can't you agree with something which is so clear to us?
The moon's path around the sun is a wobbly ellipse.  None of those lines wobble.  Maybe they're simplified, but then they've hidden the evidence you seek.  I see nothing but clean ellipses depicted, so no VHPs.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 02/02/2019 14:22:28
Keep in mind that S2 is expected to deviate from a perfect orbit. Clean orbits are a feature of 2-body systems and there are a lot more than 2 bodies within a couple light years of the galactic center. Space is not empty down there, and the path of S2 is permanently deviated every time it passes by some object by chance.  That's why they're tracking it so closely.  They project its path, and when a deviation occurs, they know it passed something, and they can estimate the mass of the object passed.  In this way they can measure the density of MACHO dark matter in the region where S2 orbits.  If there was a VHP, the deviation would be temporary and regular, with S2 always being some fixed distance from the mean orbit (the plotted line you see).  They'd notice that immediately since they do that sort of analysis on almost every star of interest to see if it is part of an unseen binary system or something.  In this way, plenty of stars are know to orbit an old burnt out half of a binary system.  These are orbits about a dark mass, not about anything virtual.  Your VHP idea violates the simplest of Newtonian physics by proposing a new force that accelerates stars and other objects into orbits around these host points.  Gravity force is computed by GMm/r² and M is zero for a VHP, so the gravitational force exerted by a VHP is zero, and thus a VHP cannot pull a star into an orbital path by gravity.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 03/02/2019 09:41:28
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Therefore, I still don't understand why do you still support our scientists while they have clearly neglecting the variations of S2 locations from the expected perfect orbital cycle? They call those variations - Errors. But those variations are very important observations.
They've probably run a Fourier transform on the data and found no regular orbit to it.  That's why asked if you had done that.  You seem to be getting your data from pictures and conclusions written by somebody else, rather than from the actual measurements taken.  If you claim a pattern to the data, take the data and demonstrate it.
Our scientists can do it better.
Please look at fig 1.4 in pg. 20
Center of Mass at Offset Position
They claim clearly, that just based on the Red line they could set the Keplerian fit. Based on the Green line there is no fit.
However, they were also forced to shift the major axis in order to get the fit.
Please see pg 78 at Figure 7.6.
They specifically indicate that the shift in major axis was very critical to get the fit.
In the following examples the couldn't get the fit:
"Exemplary three non-fitting orbits. Example of 3 orbits with an error ≥ 5 σ corresponding to: (Left) the case of Western position, a fit with 3.3×106M point mass + 0.8×106M extended component (Middle) the case of Eastern position, a fit with 3.3×106M point mass + 0.4×106M extended component, and (Right) the case of Northern position, a fit with 2.7×106M point mass + 0.3×106M extended component."
Therefore, it is clear that without changing the position of the Major axis and specifically assume that it moves on the red line, there is no fit.
I don't have to prove it. It is written by our scientists.
So, how can we take an elliptical cycle and shift its major axis???
We know that our vantage point is not perpendicular to the plane of S2's orbit. But even if we try to place that elliptical cycle in space, there is no way to shift the major axis while we keep the shape of the elliptical cycle as is.
I have tried to do it without success.
This by itself proves that based on kepler there is no fit between S2 and the SMBH. Just after all those manipulations over manipulations they claim for a fit.
Sorry this is incorrect.
Why they so deeply insist for Fit???
Those none fits are more important than a fit.
The none fit gives us deep understanding about our galaxy.
It proves that S2 doesn't directly orbit around the SMBH.
That proves that there must be something between S2 and the SMBH.
In other words, S2 must orbit around some invisible object while that invisible object might orbit around the SMBH.
I call that invisible object a Virtual Host point - VHP.
Gravity force is computed by GMm/r² and M is zero for a VHP, so the gravitational force exerted by a VHP is zero, and thus a VHP cannot pull a star into an orbital path by gravity.
I have never claimed that the VHP has Zero mass.
It all based on Multi star system, as you have explained:
Quote from: Halc
Anyway, if you're talking only about the center of mass of multi-star solar systems, then this idea is not controversial.  Yes, that point exists, and the motions of the member stars does not move it one bit, but the point goes around the galaxy more or less as a unit.  The sun wiggles around its own point, with the point being inside the sun about half the time, and outside the other half.  The (quite predictable) path around that fixed point is anything but elliptical.  It resembles more of a scribble with no particular cycle to assign a period length.
I will explain it later on how it really works in the galaxy, and how that system can set all the unique features of the spiral galaxy without any need for dark matter.
Anyhow, it seems to me that you support our scientists by all means.
So, what ever I might offer and show, you have already took a discussion to support the current incorrect ideas for good.
Is it correct?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 03/02/2019 14:45:09
Our scientists can do it better.
Please look at fig 1.4 in pg. 20
Center of Mass at Offset Position
They claim clearly, that just based on the Red line they could set the Keplerian fit. Based on the Green line there is no fit.
Careful.  You make it sound like they're getting a Keplerian fit from looking at the red and green lines, but they're getting those fits from the data, and drawing the lines according to the fit.  You are doing it the opposite way.

Anyway, yes, the green line is an apparent better fit, and it plots a non-Keplerian orbit, which is to be expected since the center of the galaxy has far more than two bodies.  Keplerian orbits only occur with two body systems.
The title of the paper indicates that this is a study of mass distribution near the center of the galaxy, so of course it is going to be focusing on the deviations from that Keplerian orbit caused by the distribution of additional mass.

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However, they were also forced to shift the major axis in order to get the fit.
Please see pg 78 at Figure 7.6.
They specifically indicate that the shift in major axis was very critical to get the fit.
In the following examples the couldn't get the fit:
"Exemplary three non-fitting orbits. Example of 3 orbits with an error ≥ 5 σ corresponding to: (Left) the case of Western position, a fit with 3.3×106M point mass + 0.8×106M extended component (Middle) the case of Eastern position, a fit with 3.3×106M point mass + 0.4×106M extended component, and (Right) the case of Northern position, a fit with 2.7×106M point mass + 0.3×106M extended component."
Therefore, it is clear that without changing the position of the Major axis and specifically assume that it moves on the red line, there is no fit.
There is no fit anyway.  I see three different attempts at a fit to the same data, each set apparently fitting a different subset of the measurements.  I don't see any assumption about assuming that a red line moves.

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I don't have to prove it. It is written by our scientists.
So, how can we take an elliptical cycle and shift its major axis???
It isn't elliptical.  It wasn't expected to be, since it isn't a two body system.  The orbit of Uranus isn't elliptical, and measuring the deviations from a pure Keplerian orbit is exactly how they knew where to look for Neptune and eventually Pluto.  That's what they're doing in this paper is learning about the nature and density of other objects using the same techniques used to find additional planets.

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We know that our vantage point is not perpendicular to the plane of S2's orbit. But even if we try to place that elliptical cycle in space, there is no way to shift the major axis while we keep the shape of the elliptical cycle as is.
I have tried to do it without success.
From just 2D point plots, it is actually quite effortless.  You can change the axis to almost anywhere you like.  From a 3D plot (one with temporal data), there is only one fit to the major axis.

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This by itself proves that based on kepler there is no fit between S2 and the SMBH.
There is always a fit.  It is the line that most closely matches the data.  The fit need not match the actual path since that's not what a fit is.
That S2 does not perfectly follow a Keplerian orbit about the SMBH is expected.  It isn't a 2 body system.

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Just after all those manipulations over manipulations they claim for a fit.
Sorry this is incorrect.
Why they so deeply insist for Fit???
It's them doing statistical analysis on the data.  It's what you do.  These guys know their statistics.

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It proves that S2 doesn't directly orbit around the SMBH.
Depends on your definition of directly.  It doesn't orbit only the mass of the SMBH, yes.  There is other mass around which it orbits.  It is a crowded place down there.

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That proves that there must be something between S2 and the SMBH.
Of course.  Why wouldn't there be??  This paper is trying to learn more about how much more something is between S2 and the SMBH, and the distribution of it, and not just prove that something is there.  It would be a far larger claim to say there was not something additional inside that orbit.

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In other words, S2 must orbit around some invisible object while that invisible object might orbit around the SMBH.
That doesn't follow at all.  Just because Venus is inside our orbit doesn't mean we orbit it.  But Venus does make the orbit of the Earth/moon system non-Keplerian, just as we deviate the orbit of Venus despite not being inside its orbit.

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I call that invisible object a Virtual Host point - VHP.
Oh, it's an object now. The story changes. How is it virtual if it's an object?  An invisible object is dark matter.

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I have never claimed that the VHP has Zero mass.
It all based on Multi star system, as you have explained:
Our sun is not a multi-star system.  Such systems are quite common, but we're not one of them.
In a binary system, it is the gravity of the other object that pulls each object into a Keplerian orbit about the other.  The VHP (center of mass of the two) does not exert any force since it is massless.

If you story is now that the VHP has mass, then it is just a dark object and not at all the center of mass of any system.

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I will explain it later on how it really works in the galaxy, and how that system can set all the unique features of the spiral galaxy without any need for dark matter.
You just posited dark matter being the unseen mass around which S2 orbits.  You just don't call it dark matter, but that's what it is: matter you can't see, and if S2 had such a secondary orbit, it would count as having been seen, just like all the other binary systems where only one of the two is luminous.

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Anyhow, it seems to me that you support our scientists by all means.
They know how to do mathematics that is completely beyond my skill set.  You've displayed a need to review high-school mathematics, and sometimes even lower.  Your ideas work in your head because you don't have the mathematics skills to see the trivial violations with your assertions.
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So, what ever I might offer and show, you have already took a discussion to support the current incorrect ideas for good.
Is it correct?
Not at all.  I'm just pointing out places where your idea doesn't hold water.  A new theory that works and makes falsifiable predictions would be warmly received.  But I've seen nothing but bad science from your postings: cherry picking data, drawing conclusions from diagrams that are not to scale instead of actual data, and complete denial of any evidence that counters your ideas.
If you want to get any attention, compute a curve of S2's path based on a VHP that matches the data points better than any the scientists have.  You'd need to find the period of the secondary orbit.  But none of the green non-Keplerian paths match a model with a secondary orbit.  You need to find that line.  Do the work, else you have zero evidence of the VHP thing.

I still have no idea why the VHP idea was necessary for the whole idea of matter being created at the SMBH and moving outward.  It seems that if it worked that way, it would work with or without everything needing a VHP.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 07/02/2019 16:08:47
Please look at the following article:
http://en.es-static.us/upl/2018/03/S2-black-hole-orbit.png
We see clearly the orbital shape of S2 and some other S stars.
https://earthsky.org/space/star-s2-s0-2-single-milky-way-monster-black-hole
It is stated:
"Until now, it was thought that S0-2 might be a double star. Two stars orbiting each other would have complicated the upcoming gravity test
But a team of astronomers led by Devin Chu of Hilo, Hawaii – an astronomy grad student at UCLA – has found that S2 doesn’t have a companion"
The question is: Why our scientists consider that there are two orbiting stars?
Could it be that there is a difficulty to explain the orbital path of S2 just by one star?
Could it be that they also have thought on a possibility that based on the real verifications, S2 should orbit around some other object as I was expecting?
So, could it be that even our scientists understand that there is a problem with S2 orbital Path?
I still have no idea why the VHP idea was necessary for the whole idea of matter being created at the SMBH and moving outward.
You don't know, but it is clear to me that our scientists know why a companion star is needed!!!
As I was expecting:
There is no need for any real companion star (or object). The VHP by itself can give a perfect explanation for what we see.

It is also stated:
"[S2] orbits Sgr A* on an ellipse that takes about 15 years to complete. The diameter of its orbit is about 300 billion km [200 billion miles], which may sound like a lot, but we’re talking about a suppermassive black hole here! That’s close!
And it gets closer. Because the orbit is an ellipse, the star drops down to a mere 18 billion km [11 billion miles] from the black hole, a positively terrifying close approach. That’s only four times farther from the black hole than Neptune is from our sun."
I wonder if that information represents the S2 elliptical orbit cycle that we see.
It was stated that our vantage point is not perpendicular to the plane of S2's orbit.
So, what we see doesn't represent the real S2 orbital cycle.
Therefore, I would like to understand what we really see and what is the real orbital cycle (based on our understanding).
It was also stated that the major axis had been shifted to the left.
Unfortunately till now I couldn't understand how the Major axis could shift to the left while we see that S2 cycle set a nice Symmetrical elliptical shape.
So, our scientists must give us full information about the real S2 orbital cycle and how it is positioned in space in order to set the expected major axis shift in the nice elliptical cycle that we see.
This is very important information.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 07/02/2019 19:05:51
It is stated:
"Until now, it was thought that S0-2 might be a double star. Two stars orbiting each other would have complicated the upcoming gravity test
But a team of astronomers led by Devin Chu of Hilo, Hawaii – an astronomy grad student at UCLA – has found that S2 doesn’t have a companion"
The question is: Why our scientists consider that there are two orbiting stars?
That's kind of the default state.  Most stars come in pairs or larger groups.  The lone star like our own is in the minority.  So until verified one way or the other, a given star is likely to be in orbit about one or more companions.

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Could it be that they also have thought on a possibility that based on the real verifications, S2 should orbit around some other object as I was expecting?
No.  The verification was finally done, and it turned up no other object.  S2 is by itself, or so says the article you quote.  Such a companion would have interfered with the smooth velocity change being measured, demonstrating relativistic red-shift.

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As I was expecting:
There is no need for any real companion star (or object). The VHP by itself can give a perfect explanation for what we see.
1) We don't see this motion you claim.  2) A VHP by itself has no mass and cannot affect the path of a star.  It is the mass of one or more companion objects that makes a star's path have a semi-regular deviation from a clean path.  S2 does not have a clean path since it passes by many other objects during its 15 year circuit, but none of those objects orbit S2, and S2 does not orbit them.

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It is also stated:
"[S2] orbits Sgr A* on an ellipse that takes about 15 years to complete. The diameter of its orbit is about 300 billion km [200 billion miles], which may sound like a lot, but we’re talking about a suppermassive black hole here! That’s close!
And it gets closer. Because the orbit is an ellipse, the star drops down to a mere 18 billion km [11 billion miles] from the black hole, a positively terrifying close approach. That’s only four times farther from the black hole than Neptune is from our sun."
I wonder if that information represents the S2 elliptical orbit cycle that we see.
If you're asking if that information describes the elliptical path we see from our vantage, then yes.

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It was stated that our vantage point is not perpendicular to the plane of S2's orbit.
I think I said that, yes.  The article didn't bother to mention that.  We're not perpendicular to any of the orbits plotted in that picture, but we seem to be nearly edge-on to at least one of them (S14).

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So, what we see doesn't represent the real S2 orbital cycle.
That we don't have a perpendicular vantage doesn't mean it isn't real.  Venus's orbit (or that of any of the planets for that matter) hardly looks like an ellipse from our vantage, but it is in fact nearly circular.  We very much can see its real orbital cycle without need for a perpendicular vantage.

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It was also stated that the major axis had been shifted to the left.
OK.  To be expected.  It is precessing, even more so due to the relativistic component of its orbit.  Even Earth does this.

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Unfortunately till now I couldn't understand how the Major axis could shift to the left while we see that S2 cycle set a nice Symmetrical elliptical shape.
It isn't a nice symmetrical ellipse.  You saw them trying to fit different curves to the data.  None of them was exact.  It is passing objects that deflect its path, as expected.  But the axis shift is probably more from precession than it is from random deflections from non-orbiting masses.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/02/2019 12:02:45
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So, what we see doesn't represent the real S2 orbital cycle.
That we don't have a perpendicular vantage doesn't mean it isn't real.  Venus's orbit (or that of any of the planets for that matter) hardly looks like an ellipse from our vantage, but it is in fact nearly circular.  We very much can see its real orbital cycle without need for a perpendicular vantage.
You do not answer the question!
What is the real Orbital cycle shape of S2???
What do you mean by: "We very much can see its real orbital cycle without need for a perpendicular vantage."
So, do you mean that what we see is almost the real orbital cycle of S2?
How can you set any sort of calculation on something that is almost correct?
OK.  To be expected.  It is precessing, even more so due to the relativistic component of its orbit.  Even Earth does this.
It isn't a nice symmetrical ellipse.  You saw them trying to fit different curves to the data.  None of them was exact.  It is passing objects that deflect its path, as expected.  But the axis shift is probably more from precession than it is from random deflections from non-orbiting masses.
The major axis shift is very dramatically.
How can you compare it to Earth?
Our scientists have to find the center of mass based on the orbital cycle of S2 (by ignoring the SMBH) and then find how far is it from the SMBH.
Than they have to prove by calculation, that without any need for a perpendicular vantage, we can move the calculated center to the location of the SMBH.
Sorry - you try to offer a solution to a problem without any real calculation.
I would expect from our scientists to offer real solution.

2) A VHP by itself has no mass and cannot affect the path of a star.  It is the mass of one or more companion objects that makes a star's path have a semi-regular deviation from a clean path.  S2 does not have a clean path since it passes by many other objects during its 15 year circuit, but none of those objects orbit S2, and S2 does not orbit them.
After all of our discussion, it seems that you have no clue what is the real meaning of the VHP.
So, let me explain it again for you:
The VHP is the center of mass for a star in a multi mega star system
Each star has a unique VHP.
"Anyway, if you're talking only about the center of mass of multi-star solar systems, then this idea is not controversial.  Yes, that point exists, and the motions of the member stars does not move it one bit, but the point goes around the galaxy more or less as a unit.  The sun wiggles around its own point, with the point being inside the sun about half the time, and outside the other half.  The (quite predictable) path around that fixed point is anything but elliptical.  It resembles more of a scribble with no particular cycle to assign a period length."
So, the VHP is a virtual point which represents the center of mass of multi star system.
There is no real object there.
But it is a virtual point in space which which is used as the center of mass for a specific star.
Therefore, each star in the galaxy (which is under a Multi star system) must have a unique VHP for itself.
Back to S2:
S2 orbital cycle is directly affected by the nearby multi star system.
Therefore, S2 must orbit around its unique VHP, while this VHP might orbit around the center of the galaxy or the SMBH.
Why is it so difficult for you to understand my simple message?
You can accept it or reject it.
But please, try to understand my explanation.
If you understand the explanation, let's see if it is feasible:
You claim that :
"S2 does not have a clean path since it passes by many other objects during its 15 year circuit, but none of those objects orbit S2, and S2 does not orbit them."
How can we verify if S2 orbits around other objects?
When we normally discuss about a multi star system - we mainly think about two, three, four...or maximal 16 stars.
If that was the case for S2, than we could easily find the other stars in the Multi star system.
However, in the center there are much more than few stars or even few thousands of stars.
Therefore, each S stars that we see must orbit around a unique VHP, while this VHP orbits according to the Multi Mega star system.
So, that multi mega star system is affected by all the nearby stars, by the SMBH and even by the 3KPC RING!!!
Yes, I do believe that this ring also has an important impact on the orbital cycles of the stars in the center.
Therefore, if we take in account all the gravities impacts we should find that S2 orbits perfectly around the impact of all the gravities and that there is no need for any dark matter.
Each S star orbits perfectly based only on real mass.
It is clear to me that you disagree with my explanation.
The key point for me is to deliver the message about the VHP.
I hope that by now you understand what I mean by that word - VHP.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 08/02/2019 14:34:45
Halo is a perfect example for the Multi mega star system:
"These halo stars are grouped together in giant structures that orbit the center of our galaxy, above and below the flat disk of the Milky Way."
So, what kind of force keeps them together? How many stars there are in those giant structures?
It is clear to me that they all group together due to internal gravity force.
So, based on multi mega star system (as halo stars), each star must orbit around all the other stars, but it is impossible to verify a single orbit around a specific star or group of stars.
Therefore, if we will try to trace one star there, we will not be able to see that it orbits any other star (or group).
In the same token, S2 orbits due to the impact of all the mass in the center - (Multi mega stars system or giant structures), while we will not be able to verify that it actually orbits around any specific stars.

We call those giant star structures as halo stars because they have ejected from the disc of the spiral arm.
Actually, as long as they are in the disc, they must be connected to one of the spiral arms.
Let's look again at the Milky way diagram:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
We see clearly that at the end of each spiral arm there are many points. It looks as a broken chain.
I assume that in each point of the chain there are Millions of stars.
Those millions of stars are grouped together due to gravity.
Each point in this chain set a local giant gravity.
Please look at the last chain in each arm.
This last chain holds itself to the one in front by gravity force.
As long as it is connected to the arm, it is part of the arm, It stays at the galactic disc plane and it follows the orbital path of the spiral arm.
However, it will not have the power to hold itself for long time.
The orbital velocity at that point is too high for the balanced gravity force in that last point of giant star structure.
Sooner or later, this last point in the chain will have to be disconnected from the arm.
However, once it is ejected from the arm, it is also ejected from the disc plane.
So, S2 which had been created at the core of the molecular cloud at the center of the galaxy must drift outwards over time.
One day S2 will arrive to the same radius from the center as the solar system.
At that time, our sun might get to the last point in the arm orbital chain.
Then, we will be ejected from the arm and be part of the Halo stars....

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 08/02/2019 23:58:32
You do not answer the question!
What is the real Orbital cycle shape of S2???
Something approximated by an ellipse, about 970 AU long and around 450 wide.  That's the actual shape, not what it looks like from here.  Like all orbits, it precesses, so it doesn't trace the exact same path each time around, even without other objects perturbing it.

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What do you mean by: "We very much can see its real orbital cycle without need for a perpendicular vantage."
I was talking about the orbit of Venus, which looks, from our vantage, like an almost flat ellipse with the sun nowhere near either foci.  That orbit is quite real, and yes, we see it, so if you mean something else when you ask if we see the real orbit, I don't get your question.
The image of S2 is delayed by over 25000 years, so in that sense it isn't real.  Maybe the thing has since been eaten and isn't real at all anymore.  Betelgeuse has the same problem.  Never sure if it is really there or we're just seeing an afterimage of a nonexistent star.

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How can you set any sort of calculation on something that is almost correct?
Tell that to the weather prediction guys.  To be completely correct, you need a full description of its state.

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The major axis shift is very dramatically.
How can you compare it to Earth?
Earth orbit is nearly circular.  OK, S2 is not as eccentric as say a comet, but it also has a relativistic orbit, and that might effect major axis shift.  I don't know how to compute that.  I know that relativity very much plays a role because they never were able to predict Mercury's orbit until all GR effects were accounted for.
S2 is also passing random objects, any of which can send it onto a new trajectory, a different ellipse.  That seems to happen multiple times per orbit, as shown by some of the pages you've been linking.

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Our scientists have to find the center of mass based on the orbital cycle of S2 (by ignoring the SMBH) and then find how far is it from the SMBH.
Sorry, but I cannot figure out this statement.  What center of mass are they seeking?  S2?  Sgr-A?  Something else?  S2's CoM is where S2 is.  Ditto for Sgr-A.  Neither seems to be a binary object.

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Sorry - you try to offer a solution to a problem without any real calculation.
Not sure what I was asked to calculate.  I certainly don't see any calculations from you, the guy who expects a different idea to be taken seriously.

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After all of our discussion, it seems that you have no clue what is the real meaning of the VHP.
So, let me explain it again for you:
The VHP is the center of mass for a star in a multi mega star system
Each star has a unique VHP.
Why wouldn't all the stars in said mega-star system have the same VHP?  How does any particular star know which stars to include in the VHP calculation and which to leave off?  Our sun for instance has stars visible in every direction.  So how might I begin to guess which of those contribute to the VHP and which do not?  Most importantly, why would you expect motion about this VHP to be anything resembling something close to a familiar elliptic orbit?

Anyway, I don't deny that our star is potentially part of a group, and that the group sort of travels as a clump in a cleaner path than any individual mass.  I just don't think our motion within that group can be characterized as being in orbit around the group's center of mass or any other virtual point.  That only works for 2 body systems, not larger ones.
For even a simple 3-body system, barring symmetrical arrangements, none of the three biodies will accelerate towards the common center of mass.  In the most simple case of 3 equal masses with unequal spacing in a line, the middle mass will accelerate away from the common center of mass.  You seem to realize none of this.

The only VHP I see is the center of the galaxy.  We also are influenced by the disk, but forces from a disk are not characterized by an orbit.  The force of the disk actually grows with distance (to a point at least), unlike the force from a given point mass.  I think the disk force contributes to that 62 million year cycle thing, but I'm no expert.
The in/out motion you showed in one of your pictures might be due to the solar system having an eccentric orbit about the galactic center. It would have a ~200 million year period if it was that. Few objects seem to have really low eccentricity trajectories.

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So, the VHP is a virtual point which represents the center of mass of multi star system.
I've already accepted this general idea, but in general, all stars in that group have essentially the same VHP then.

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There is no real object there.
But it is a virtual point in space which which is used as the center of mass for a specific star.
It is the center of mass for all the stars in the set.  The center of mass for a specific star is the star itself.
You really need to choose whether you're talking about a virtual center of mass of a set of multiple objects, or the center of mass of a particular object, which tends to be the middle of said object.

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Therefore, each star in the galaxy (which is under a Multi star system) must have a unique VHP for itself.
Yes, called its location.  That point isn't virtual, since the star is there.
Venus's center of mass is at its center, but Venus loosely orbits the center of mass of the solar system, and more accurately orbits the sun itself, but still not a perfect orbit around it.  No object's path can be accurately described by a clean orbit about a VHP which in turn orbits either the solar system or the sun.

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Back to S2:
S2 orbital cycle is directly affected by the nearby multi star system.
What multi-star systems would that be?  There are other stars in that area, but none seem to form a system with S2, and nobody knows how many of these nearby objects are part of multi-object systems (stay with each other in their primary orbit).

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Therefore, S2 must orbit around its unique VHP, while this VHP might orbit around the center of the galaxy or the SMBH.
They've actually eliminated this option as a possiblility.

Your logic is horrible.  S2 is affected by other masses, therefore it must orbit something other than its primary (the SMBH).  That simply doesn't follow.  Venus, for example, is affected by other masses (I cannot think of a mass that doesn't affect it), but it does not orbit any VHP.

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Why is it so difficult for you to understand my simple message?
You can accept it or reject it.
I understand the message, but since it is based on nonsense, I cannot accept it.

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How can we verify if S2 orbits around other objects?
I already told you this.  It would have a period to that orbit.  Orbits have periods.  Do the math that detects regular periods in the data. Making baseless assertions is not the way to go about it.

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When we normally discuss about a multi star system - we mainly think about two, three, four...or maximal 16 stars.
I can think of much larger systems than that.

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If that was the case for S2, than we could easily find the other stars in the Multi star system.
However, in the center there are much more than few stars or even few thousands of stars.
Therefore, each S stars that we see must orbit around a unique VHP, while this VHP orbits according to the Multi Mega star system.
So, that multi mega star system is affected by all the nearby stars, by the SMBH and even by the 3KPC RING!!!
Yes, all of this stuff has an effect, but none of it is a system that orbits as a unit, else all the other objects in the system would have the same 16 year period around Sgr-A, but we don't see that.  We see not one other object travelling with S2, else S2 would have a regular perturbation to its orbit.  You keep asserting a pattern that just isn't being seen.  Not saying it isn't possible, just that it isn't the case with this particular star.

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Yes, I do believe that this ring also has an important impact on the orbital cycles of the stars in the center.
What impact would that be?  I mean, I agree there is one, but I suspect you haven't thought it through.  What impact would a ring have on Earth's orbit for instance?  Suppose it is out at the asteroid belt, but having the same mass in relation to the sun as the galactic ring does to our SMBH.  Our actual asteroid belt has almost no mass compared to the sun, so we'd need to imagine a more massive one.

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Therefore, if we take in account all the gravities impacts we should find that S2 orbits perfectly around the impact of all the gravities and that there is no need for any dark matter.
But you haven't taken anything into account. You just imagine that if you did, this S2 orbit (and that of our own solar system for that matter) would fit. Imagining it isn't evidence at all.  You need to actually do the work and show how your idea makes the data fit better than the misguided ideas of all the professional astronomers who have to posit objects that cannot be seen in order to get the numbers to work.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 09/02/2019 00:32:00
Halo is a perfect example for the Multi mega star system:
"These halo stars are grouped together in giant structures that orbit the center of our galaxy, above and below the flat disk of the Milky Way."
So, what kind of force keeps them together? How many stars there are in those giant structures?
The article doesn't say if they're kept together or still spreading apart or whatever.  It doesn't give a count either, but I'm sure the count depends heavily on what sort of object counts as a star and what doesn't.  There's not exactly a clean line defining that.

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It is clear to me that they all group together due to internal gravity force.
If they stay together, then yes, gravity seems a lot more plausible than rubber bands or something.

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So, based on multi mega star system (as halo stars), each star must orbit around all the other stars
Yet again, that doesn't follow.  Each star may tend to stay with the group, but it may also be ejected by chance.  The motion within the group is not necessarily anything that resembles the elliptical path of an orbit.

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In the same token, S2 orbits due to the impact of all the mass in the center - (Multi mega stars system or giant structures), while we will not be able to verify that it actually orbits around any specific stars.
They've verified that it doesn't orbit any other specific stars.

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We call those giant star structures as halo stars because they have ejected from the disc of the spiral arm.
Actually, as long as they are in the disc, they must be connected to one of the spiral arms.
Structures between the arms are still in the disk, so this is false.

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Let's look again at the Milky way diagram:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
We see clearly that at the end of each spiral arm there are many points. It looks as a broken chain.
Those broken lines are extrapolated arms, places that we cannot see at all.  They're not broken, but they're guesses as to what's there.  It is reasonable to assume the arms continue through the regions unseen, just like we can see in other galaxies.

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Please look at the last chain in each arm.
This last chain holds itself to the one in front by gravity force.
I'm sorry, but you're asking me to consider gravity of a 'can't see it' designation on a diagram.  This makes no sense.  Those are not real clumps of stars that have been observed like that.  Read the text accompanying the diagram.

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As long as it is connected to the arm, it is part of the arm, It stays at the galactic disc plane and it follows the orbital path of the spiral arm.
You're treating arms as objects, not as what they probably are, which is waves.  The arms do not move with the stars, else they'd have been smeared out a long time ago.  Our sun (and any star) is connected to no arm, and moves from one to the next as it orbits the galaxy at an entirely different rate than the apparent motion of the density waves that form the arms.

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So, S2 which had been created at the core of the molecular cloud at the center of the galaxy must drift outwards over time.
One day S2 will arrive to the same radius from the center as the solar system.
It is a big star and even assuming it will drift like that, it will burn out long before it gets anywhere.  The big stars nearby (e.g. Betelgeuse) similarly could not have come from the center since they're far too young to have made the trip.  Either that or the big stars must drift outward at an incredible pace, making one wonder why S2 isn't getting any further away with each orbit.

Not claiming to know the answer myself.  The bar is a known stellar nursery, but I don't understand the dynamics of the bar.  S2 seems not to be part of that because it doesn't fall in that rotation curve of the galaxy that we were looking at.  I don't know anyone that can explain that curve and how it fits with named objects like S2.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/02/2019 07:19:58
The bar is a known stellar nursery, but I don't understand the dynamics of the bar.
You don't know, but I know it for sure.
The center of the galaxy is the stellar nursery for most (or even all) the stars in the galaxy/Universe!!!
A star can be created ONLY in a Molecular gas cloud which orbiting around/near a massive object as a SMBH.
So, 99.9..9% of the stars in the Milky way galaxy had been created in that stellar nursery.
Each star have got all its Planets and moons in it's first day.
Therefore, the age of the Earth & Moon is identical to the age of our Sun.
All the objects in the Solar system had been created from the same matter at the same day.
The Earth and the Moon had been also created as a compact molecular gas object.
Their mass in the first day was heavier by more than 98 times than it is today.
The planets and moons in the solar system have got all the Water supply in their first day from the molecular gas cloud.
Over time, due to their compact size, they have lost most of the gas as Hydrogen and became rocky objects.
There was a time when water flows at the surface of Mars.
At that time, Mars was orbitting the Sun at the same radius as we are today and the Earth was much closer to the Sun while it's temp was quite higher.
The big stars nearby (e.g. Betelgeuse) similarly could not have come from the center since they're far too young to have made the trip.
As I have stated, all the stars (Yes - including Betelgeuse) had been formed at the center of the Milky Way galaxy.
Betelgeuse isn't a young star. It is as old as our sun.
You have a severe misunderstanding about the age of the stars in the galaxy.
We could set the calculation how long it might take to a star to drift from the center of the galaxy all the way to our current location.
Therefore, all the stars in the center are young stars.
As we move outwards, the stars get older.
So, there is a severe error in the way that our scientists estimate the age of the stars in the galaxy.
If they stay together, then yes, gravity seems a lot more plausible than rubber bands or something.
Thanks
Gravity keeps all the stars in the Halo stellar. In the same token, gravity keeps all the stars in spiral arms.
You're treating arms as objects, not as what they probably are, which is waves.
Yes, spiral arms are objects. All the stars in the arm are connected to each other by gravity force. The same gravity force which keeps the stars at the halo stellar also keeps the stars at the spiral arms.
However, as I have stated before, there are bridges and branches between the Arms.
Therefore, stars can migrate from arm to arm by those bridges and branches.
We can think about an arm as a highway while there are bridges between the arms.
Each star may tend to stay with the group, but it may also be ejected by chance.
That is correct!
As long as the star stay at the arm or cross to the next arm by bridge, it will save its location in the galactic disc.
However, if it will dare to move away from the arm or the bridge, it will be ejected as a rocket from the galactic disc for good.
Nothing can come back to the arm or the galactic disc.
why S2 isn't getting any further away with each orbit.
Yes, it is!
However, it is very difficult to monitor that drifting outwards.
S2 orbits around its unique VHP. (It is a host point - sorry if I used before the idea of center of mass).
So, S2 set a very clean orbit around its VHP while it increases its radius by only few cm or few m. per cycle.
However, this represents the S2 orbit around its orbital motion.
Let's look again at the following diagram of the Sun Motion:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
In the same token, S2 apparent motion (blue dot points) orbits around its orbital motion (gray dashed line).
Please, you don't have to agree with that, just try to understand my message.
So, S2 orbits around a first level of VHP (let's call it VHP1) and set that nice apparent motion (Blue dot point)
Therefore, when we trace S2, we see that it doesn't move exactly on its orbital cycle:
http://www.phy6.org/stargaze/Kep3laws.htm
However, for this big orbital cycle (15 years), there is second level of VHP (Let's call it VHP2).
Therefore, if we assume that In each cycle, VHP1 drifts from VHP2 by few Km per cycle, than VHP2 drifts away from the SMBH by much more than that per cycle.
So far we have mainly discuss on this VHP (VHP2) but we have ignored the first VHP level (VHP1).
So, any star in the galaxy must orbit around a VHP1 which follows with him anywhere from its first day one in the Universe.
Therefore, when we look at the nearby stars, we shouldn't trace the physical location of the stars, but we must monitor the virtual locations of their VHP1.
If we do so, we should see that the distances between all the nearby VHP1' stars in the Orion arm are absolutely fixed!!!
So, all the VHP1 of the stars in the Sun' nearby area are moving exactly at the same velocity!!!
Therefore, no one is moving away from the arm!

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/02/2019 09:13:37
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What is the real Orbital cycle shape of S2???
Something approximated by an ellipse, about 970 AU long and around 450 wide.  That's the actual shape, not what it looks like from here.
Thanks!
So, the ratio between the major axis to the shorter axis is about 1:2
I have just found a fantastic article:
https://www.slideshare.net/NUCLIO-PT/development-and-implementation-of-student-activity-to-find-mass-of-black-hole-by-naoki-matsumoto
Please look at at slide pg. 18
It is stated that the orbital inclination angle as viewed from the Earth of S2 is 45 degree. While after the correction to 90 degree we get the final results at the attached image.
However, it is quite clear to any first year student that the SgrA* (SMBH) can't be used as the host for this orbital cycle. No way!!!
Now, please look at slide in pg 20.
The calculated Periastron date is year 2002.25 + 0.05 = 2002.3
The calculated Apastoron date is year 1994.62
The full one orbital cycle Is:
P = 15.4 Year.
Based on that data, they have found the mass of the host should be (in pg 22)
3.4 * 10^6 Sun Mass.
So far so good.
I fully agree with this calculation!
However, they also claim that after the adjustment (which they do not explain), The estimated value of the SMBH mass is
SMBH = 4.1 * 10 ^6 Sun mass.
That is a severe mistake!!!
As I have stated, S2 does not orbit around the SMBH.
S2 orbits around its VPH1 which orbits around VHP2 which orbits around the SMBH (assuming that there are no more stages in between).
Therefore:
VHP2 (of S2) = 3.4 * 10^6 Sun Mass.
Now, we have to find the exact relationship between VHP2 and SMBH.
First, based on Kepler, we have to find the perfect location of VHP2 in that S2 orbital cycle of 15.4 years.
It is clear to me that it should be located high above the current location of the SMBH.
Than we have to find how long it takes to VHP2 to set one full cycle around the SMBH and the cycle shape (VHP2' Periastron and  Apastoron)
Just after getting this information we can extract the real value of SMBH.

There is a solid prove for my explanation.
Please look at the following article:
https://alchetron.com/S2-(star)#demo
https://alchetron.com/S2-(star)
If we look carefully, we should see that S2 doesn't set a full nice cycle. At the top of this image we clearly see that there is a shift in the orbital cycle (as an open loop).
This shift is a direct outcome of the orbital motion of VHP2 around the SMBH.
Therefore, The SMBH can stay in its place while VHP2 (of S2) orbits around it and set the same famous orbital motion as the Sun does.
Please look again at the following image:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
Hence, S2 will never ever close one full cycle. It moves forwards due to the orbital motion of VHP2 around the SMBH and set some sort of open loop.
Based on this small shift we should find the orbital velocity of VHP2 around the SMBH.
This might help us to estimate the size of this orbital cycle and extract the real value of the SMBH.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 09/02/2019 22:28:30
Please look at at slide pg. 18
It is stated that the orbital inclination angle as viewed from the Earth of S2 is 45 degree. While after the correction to 90 degree we get the final results at the attached image.
However, it is quite clear to any first year student that the SgrA* (SMBH) can't be used as the host for this orbital cycle. No way!!!
I agree.  The rotation was not done on the correct axis, since if it was done correctly, SgrA would have been at the focus.

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Now, please look at slide in pg 20.
The calculated Periastron date is year 2002.25 + 0.05 = 2002.3
The calculated Apastoron date is year 1994.62
The full one orbital cycle Is:
P = 15.4 Year.
Based on that data, they have found the mass of the host should be (in pg 22)
3.4 * 10^6 Sun Mass.
Mass of host cannot be determined from just orbital period.  Jupiter has a sort of similar orbital period and yet orbits something of far less mass.  Anyway, yes, they measured the time for half an orbit and figured that the other half takes similar time.  Don't need to be a genius to figure that one out.

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I fully agree with this calculation!
They didn't show their work.  There is a reference to a 'equation 4'.  The only math done on that slide is a conversion of units.

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As I have stated, S2 does not orbit around the SMBH.
S2 orbits around its VPH1 which orbits around VHP2 which orbits around the SMBH (assuming that there are no more stages in between).
Therefore:
VHP2 (of S2) = 3.4 * 10^6 Sun Mass.
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Therefore, if we take in account all the gravities impacts we should find that S2 orbits perfectly around the impact of all the gravities and that there is no need for any dark matter.
You just posited a VHP2 of mass similar to SgrA, which is a serious boat load of dark matter.  You are contradicting yourself.
I think they would notice if S2 was orbiting a second object that massive, but you seem to have other ideas.  How far away is VHP1 from this VHP2?  What period?

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Now, we have to find the exact relationship between VHP2 and SMBH.
They're nearly the same mass and should be orbiting each other.  Other objects should be orbiting the combined mass of the two.

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First, based on Kepler
You can't really quote Kepler.  You've denied his laws completely in the prior post, and Kepler's laws only apply to two-body systems and that isn't the case here.

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we have to find the perfect location of VHP2 in that S2 orbital cycle of 15.4 years.
It is clear to me that it should be located high above the current location of the SMBH.
Which direction is 'high above'?  There's not exactly a clear direction that is 'above' out there.
You're just talking about a distance?  How much is 'high above' then?  How is SgrA not high above this VHP2, since it is so massive?  How far is VHP1 from VHP2?

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There is a solid prove for my explanation.
Do you have even the slightest idea what constitutes evidence, let alone proof?

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Please look at the following article:
https://alchetron.com/S2-(star)#demo
https://alchetron.com/S2-(star)

If we look carefully, we should see that S2 doesn't set a full nice cycle. At the top of this image we clearly see that there is a shift in the orbital cycle (as an open loop).[/quote]
Yes.  Nobody claimed its motion is Keplerian, and this orbit also has a significant relativistic component.

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This shift is a direct outcome of the orbital motion of VHP2 around the SMBH.
So you assert, but where's the proof of that?  You claimed there was a proof somewhere in it.

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Please look again at the following image:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
This image in no way reflects reality.  Nothing is to scale and the periods depicted do not reflect measured values.
Best I can tell from research, the 5-7 km/sec component has that 62 million year cycle (up and down about 3 times per trip around the galaxy, due to disk effects), and the 20 km/sec motion is one cycle per orbit, due to our eccentricity.  The blue dotted line is almost parallel with the grey line if it was drawn more correctly.

If we actually followed the blue dotted line as depicted (which is about 9x the length of the grey dashed line) and the 217 km/sec motion of the VHP was accurate, the sun would be moving at about 2000 km/sec, well above the escape velocity of the galaxy.  We'd just permanently shoot away.

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Hence, S2 will never ever close one full cycle. It moves forwards due to the orbital motion of VHP2 around the SMBH and set some sort of open loop.
I cannot figure out what you mean to convey with this description.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 16/02/2019 06:11:24
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As I have stated, S2 does not orbit around the SMBH.
S2 orbits around its VPH1 which orbits around VHP2 which orbits around the SMBH (assuming that there are no more stages in between).
Therefore:
VHP2 (of S2) = 3.4 * 10^6 Sun Mass.
Therefore, if we take in account all the gravities impacts we should find that S2 orbits perfectly around the impact of all the gravities and that there is no need for any dark matter.
You just posited a VHP2 of mass similar to SgrA, which is a serious boat load of dark matter.  You are contradicting yourself.
I think they would notice if S2 was orbiting a second object that massive, but you seem to have other ideas.  How far away is VHP1 from this VHP2?  What period?
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Now, we have to find the exact relationship between VHP2 and SMBH.
They're nearly the same mass and should be orbiting each other.  Other objects should be orbiting the combined mass of the two.
Dear Halc
After all our discussion, it seems to me that you still don't understand the real meaning of VHP.
Please, you don't have to agree with my explanation - but you have to understand my message.
So, let me explain it one more time for you:
1. VHP = Virtual Host Point
2. VHP is not a real object. It is a virtual object.
3. VHP1 - As I have already stated - ALL stars in the galaxy had been formed in a gas cloud near the SMBH. Each new born star must orbit around a virtual host point (VHP1) in the gas cloud in order to crystallize from molecular gas into real Star.
Therefore, the first level of virtual point (VHP1) is some sort of a gift which had been given to any star in the galaxy from the gas cloud in his first day. Hence, once the Star is out from the gas cloud its physical location is not relevant. We have to focus only on the Virtual point of his host (VPH1). Each star in the galaxy has a unique VHP1. This VHP1 goes with it anywhere it goes.
4. VHP can work over VHP. For example - The Moon orbits around the Earth, while The Earth orbits around the Sun. (Technically, the center of the Earth/Moon orbits around the Sun, but for this discussion, we can assume that the Earth orbits around the Sun. So, we see two levels or orbital cycles. In the same token - S2 orbits around VHP1, while this VHP1 orbits around another virtual host which is called VHP2. This VHP2 can technically orbits around other level of virtual host point - VHP3... and so on...
Therefore, if we focus on those Virtual points, we won't find any real object or any dark matter. All of them are virtual points. However, each Virtual Host Point represents mass. We can calculate the estimated mass of each VHP directly from the star mass and its orbital movement.
5. Dark matter - There is no dark matter and there is no need for dark matter. It is a pure fiction due to the simple outcome that our scientists have failed to understand how gravity really works at the galaxy.
6. Planets and Moons - They are all "by products" due to the process of  "new star creation" in the gas cloud. I can explain it later on if you wish. In any case, by definition - Each star in the galaxy gets also planets and moons as another gift from the gas cloud.

Final conclusion:
When we try to look at any nearby star, we shouldn't focus only on its real location and its mass. We have to focus on the location of its VHP1. Hence, by monitoring the movement of each star (physical location) in the galaxy, we can extract the location of its VHP1 and calculate its host mass (based on orbital cycle of the star and its mass).
From now on, this virtual host point and the calculated mass represents the first level of the orbital cycle of any real star in the galaxy. This is the most critical issue in the galaxy!!!
The shape of spiral galaxy is a direct product of VHP1 which is based on real object.

Is that all clear to you?
Please, let me know if you understand the message. Again, you don't have to agree with that. Just let me know if finely you understand the real meaning of VHP.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 17/02/2019 14:24:28
So, let me explain it one more time for you:
1. VHP = Virtual Host Point
2. VHP is not a real object. It is a virtual object.
3. VHP1 - As I have already stated - ALL stars in the galaxy had been formed in a gas cloud near the SMBH. Each new born star must orbit around a virtual host point (VHP1) in the gas cloud in order to crystallize from molecular gas into real Star.
Therefore, the first level of virtual point (VHP1) is some sort of a gift which had been given to any star in the galaxy from the gas cloud in his first day. Hence, once the Star is out from the gas cloud its physical location is not relevant. We have to focus only on the Virtual point of his host (VPH1). Each star in the galaxy has a unique VHP1. This VHP1 goes with it anywhere it goes.
This is a fairy talk unless you explain how to find the VHP.  Given mass distributed here and there, where are the respective VHP's?  Without that, you've describing fantasy.

Take 3 equal masses distributed in a triangle of sides 3, 5, and 7.  From that, you can find the center of mass of the three which is the same for all three of them.  We can call this VHPc.  In addition, each object X Y and Z has a sort of VHP that is halfway between the other two objects.

The acceleration of none of the objects is towards any of these four points, and in fact, in some arrangements, the acceleration of the object at the obtuse angle is actually away from VHPc.  The motion of none of the objects (I have not given their velocities) cannot be modeled by any number of tiered VHP's, at least not according to Newtonian physics.  Therefore you need to supply the new physics which replaces Newtonian physics (and a whole lot of other things like those governing nuclear reactions).  You have not done any of this, so until I have this, I am forced to comment only on the unbacked fairy talk.

Where is the VHP of each of the objects in question?  Give them a velocity if that matters.  Under Newtonian physics, the acceleration of each does not require knowledge of the velocities, but the trajectories do need that.
Under your rule, the mathematics should be fairly simple.  Find the VHP of each object, and then the object moves at some (to be specified) rate around that VHP.  Next find the motion of each of those VHPs around perhaps a central one this time.  Now you have a model which you can pit against real observations.
Without that, you're talking into the air.  You have no theory, just an ignorant idea.

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6. Planets and Moons - They are all "by products" due to the process of  "new star creation" in the gas cloud. I can explain it later on if you wish. In any case, by definition - Each star in the galaxy gets also planets and moons as another gift from the gas cloud.
Your theory is going to have to apparently account for exactly how mass is created from some objects and not others.  What is the rule?  Not a verbal description, but a specified law.  Without that, there is not theory.  It needs to describe the source for forces that give (apparently without reaction) angular motion to these created objects.

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Is that all clear to you?
Your idea is clear, but since I can make zero predictions from it, it is totally unclear.  You need a theory to go with your idea.  Still waiting for it.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/02/2019 12:04:18
Take 3 equal masses distributed in a triangle of sides 3, 5, and 7.  From that, you can find the center of mass of the three which is the same for all three of them.  We can call this VHPc.  In addition, each object X Y and Z has a sort of VHP that is halfway between the other two objects

That is good example.
So, it seems that you understand the meaning of VHP.
However, instead of three object, we must think on thousand, Millions or even billions objects.
Not just stars, but also gas clouds, SMBH and any other real object.
There is no dark matter in the galaxy.
This is a pure science fiction.
Our scientists had failed to understand how spiral galaxy really works, therefore, they came with this none realistic idea.
Based on the VHP idea we can get the spiral galaxy shape without any need for dark matter.
The motion of none of the objects (I have not given their velocities) cannot be modeled by any number of tiered VHP's, at least not according to Newtonian physics.
I disagree with that statement.
VHP over VHP is the key element in spiral galaxy which set the structure of spiral arm.
Please look again at the following image:
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
Please focus on the last chain point in the red spiral Arm (for example)
In this last chain point there might be thousands or millions stars.
As long as it is connected to the chain point in front, it will follow the orbital momentum of the spiral arm.
However, once it disconnected from the point in front, it will soon drift away from the disc plane.
Our scientists, ignore completely the meaning of the disc plane.
They think that this last chain point (with all the millions objects) orbits around the whole mass inwards the orbital cycle of that point. They have found that the real mass in the galaxy can't support its orbital velocity.  Therefore, they have got into conclusion that dark matter is needed.
As explained in the following:
https://www.quora.com/How-is-the-motion-of-stars-similar-to-the-motion-of-the-sun
"The stars don't reduce in orbital velocity as much as they should be doing... unless there's more mass than we can see.  Behold: meet the mysterious dark matter that must be responsible for this.  More mass due to dark matter can explain that."
That of course is a fatal mistake.
They actually ignore completely the disc plane.
Why suddenly at the end of the spiral arm the stars do not stay any more at the galactic disc?
in that article they show the different orbital cycles:
Disc star orbital - Yellow
Halo star orbits - Green
Bulge star orbits - Red.
So, only the Yellow represents the orbital star at the disc plane.
The dark matter can't give a clear explanation for that phenomenon.
Why suddenly at the end of spiral arms, the stars starts to orbits high above and below the disc (higher than the estimated disc width of about 1KLY.)
So, let me explain how it really works.
Let's assume that in this point there are 10,000 stars.
We see clearly that those stars are connected to each other.
Gravity force is the only force that can keep them all together.
In this activity, they all set a VHP which represents the center of mass of all this last chain point.
Let's call it VHP - m1 (m1 represents the total mass in the last chain point).
Now, for each point in this chain there must be also a represented total mass.
So, the one in front will be called - m2, the other one in front will be called m3 and so no.
Hence, we can see a long chain of points that are connected to each other.
m1 is connected to m2.
m2 is connected to m3.
m3 is connected to m4
mn is connected to mn+1.

However, when I say connected I mean gravity force.
In order to set a constant gravity force we must set an orbital cycle.
So, we actually have the following:
VHP m1 orbits around VHP m2.
VHP m2 orbits around VHP m3
VHP mn orbits around VHP mn+1
However, how could it be that we don't see any orbital cycle?
Because, the orbital velocity of VHP m1 around  VHP m2 is (almost) identical to the orbital velocity of VHP m2 around the galaxy.
Therefore, the orbital velocity of VHP mn around  VHP mn+1 is (almost) identical to the orbital velocity of VHP mn+1 around the galaxy.
I specifically say almost identical, as there is a room for this point to drift outwards. (As I have already explained - in real orbital cycles - orbital objects drift outwards from their host)
This is the basic idea of spiral arm and how VHP works over VHP.
However, at some point, the orbital velocity of VHPm1 can't trace any more the orbital velocity of VHP m2 around the galaxy.
Sooner or later, this last chain point should be disconnected from the spiral arm.
So, I give a clear explanation why the last chain point at the end of the spiral arm should be disconnected from the spiral arm and move away from the galactic disc in order to set the halo star.
The dark matter can't explain this phenomenon.
Do you agree that if the dark matter idea was correct, than the disc plane - or the spiral arm had to be continued further more (theoretically - to the infinity.)
I don't see any explanation why there is an end for the spiral arm based on this dark matter fiction.
So, I hope that by now you understand the idea of VHPmn over VHPmn+1 in the spiral arm which is needed to hold the stars in the arm (and on the disc plane).
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 22/02/2019 17:59:28
Rotation curve problem
Now we need to understand how could it be that the stars are connected to each other while we get the following rotation curve:
https://en.wikipedia.org/wiki/File:M33_rotation_curve_HI.gif
This is the rotation curve of M33 galaxy, but any spiral galaxy should have a similar rotation curve (more or less).
So what do we see.
Please focus only the observation graph.
The expected graph is none relevant as they assume that each star orbits around the galaxy by its own.
However, I claim that the stars are connected to each other.
So, if we think about a rigid disc, it is clear to us that the orbital velocity of a star at 20KLY from the center should be much higher than a star at a distance of 10KLY.
Actually as the radius in 20K is twice longer than in 10K the orbital ring is longer by:
2^2 = 4
Therefore, in a rigid disc, in order to keep itself in the arm, the orbital velocity at 20K should be four time faster than at 10K.
So, how can we explain this problem?
Spiral galaxy isn't a rigid disc.
The stars in the arm are connected to each other, but they also drift outwards.
Let's monitor few points:
At 5KLY - the orbital velocity is 70 Km/sec
At 10K - 90 Km/sec
At 20K - 110 Km/sec
So, let's assume that at t=0 we trace three stars at those located radius from the center.
If we will try to monitor the location of the stars after T
t=T
We should see that all of them had been drifted outwards.
Let's assume that T represents the time that is needed for a star to drift from 5KLY to 10KLY.
By this time there is high possibility that a star that was at 10KLY had been drifted to 20KLY.
due to the shape of the spiral arm, we should find that the total distance that the first star had to move from 5KLY to 10KLY might be a little bit shorter that the other star which had been drifted from 10KLY to 20KLY.
Therefore, there is no significant change in the orbital velocity as we trace stars that are located further away from the center (although - they are connected by spiral arms).
As I have already explained, new stars are formed in the center.
Those new stars replace the other stars which had been drifted out from in the spiral arms.
Therefore, at any given moment we see that all the arms are full with stars.
Therefore it is clear to me that our Sun is only temporary located at the current location.
As it drifts outwards, sooner or later it will get to the end of the spiral arm and then it will be ejected from the arm and the galactic plane.
So, the spiral galaxy acts as the biggest star sprinkler in the Universe.
Therefore, we see that for any star in the galaxy there is at least one outside.
Hence - all the stars that we see around our galaxy - in the Halo, in the dwarf galaxies... - all of them had been ejected from the galaxy.
One day we will be there too.
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 22/02/2019 20:33:50
Take 3 equal masses distributed in a triangle of sides 3, 5, and 7.  From that, you can find the center of mass of the three which is the same for all three of them.  We can call this VHPc.  In addition, each object X Y and Z has a sort of VHP that is halfway between the other two objects
That is good example.
So, it seems that you understand the meaning of VHP.
However, instead of three object, we must think on thousand, Millions or even billions objects.
No, let's start simple, and if we can get that to work, then we can attempt to extend the findings to larger collections of unequal size objects.  I gave no coordinates for the above example, so maybe I can provide one that also has a coordinate system.  The triangle is chosen so all objects are at integral locations with integral distances between them:  3 objects of mass 1 each, at X,Y,Z coordinates of 9,8,0  18,8,0  and  -27,-16,0
That puts all the objects in the Z plane since there is always a plane defined by 3 objects.  With 4 or more, we'd need nonzero Z values.  The center of mass is at the origin in this case.  Where is the VHP of each object?  If you can tell me that, we can go to the next step and define the motion of each object and the motion of the VHPs.  If you can't tell me where each VHP is, then you're just blowing smoke and your idea is baseless.
For simplicity again, we can say the initial velocity of each object is zero and see where gravity takes them.

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The motion of none of the objects (I have not given their velocities) cannot be modeled by any number of tiered VHP's, at least not according to Newtonian physics.
I disagree with that statement.
Then show me with the above example.  You disagree because you haven't tried it and verified that your idea predicts different motion than does Newton's laws.

Without that demonstration of a trivial case, the rest has no foundation and is just blather.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/02/2019 05:41:52
Then show me with the above example.  You disagree because you haven't tried it and verified that your idea predicts different motion than does Newton's laws.
With Pleasure:
Please look at the following article:
https://www.livescience.com/63698-milky-way-spiral-wobble.html
It is stated:
"While still rotating around the Milky Way's galactic center, these rogue stars also orbit around one another in a wobbly, spiral pattern that has only become more tangled over the past eon."
So, this short description fully supports my theory. (Please focus only on what they see and ignore their none realistic explanation for that key evidence!)
It is stated: " these rogue stars also orbit around one another" - So as they orbit around one another they set a center of mass which I call it - VHP.
Hence, the example had been given by a solid verification of our scientists - as you have requested.
However, it is also stated: "these rogue stars also orbit around one another in a wobbly, spiral pattern".
What does it mean "Wobbly"?
Why the stars are wobbling?
This is a solid evidence to my explanation that any star in the galaxy orbits around its unique VHP1.
The wobbling activity is a direct outcome of VHP1!!!
This VHP1 is the first element for Newton activity in the galaxy.
The Sun is wobbling. I have already proved it by an article.
All the nearby stars are wobbling. S2 star in the center of the galaxy and many others are wobbling as I have proved by several articles.

However, it is quite clear to me that you are going to reject also this evidence.
Do you really want to understand my message or do you just want to prove that whatever I say is incorrect?
Why don't you even try to understand the deep explanation which I have offered?
Why do you constantly reject any explanation, example, evidence and article which I have offered so far?
Why don't you answer my questions about the problems in using the dark matter concept?
I have found the explanation for every bit in our spiral galaxy/observable universe/real universe without any need for dark matter or dark energy, as our universe is working according to very simple roles.
So far I have introduced just the first steps from the whole theory.
Unfortunately, you force me to explain the same issue again and again and therefore we can't move on.

Hence -
Do you have real willing to open your mind and heart to this breakthrough understanding?
Is there any chance for you to focus only on real evidences and verify how my explanation perfectly fits with those verifications/evidences?
Without it, we won't be able to move forward.

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 23/02/2019 14:01:10
Then show me with the above example.  You disagree because you haven't tried it and verified that your idea predicts different motion than does Newton's laws.
With Pleasure:
Please look at the following article:
A diversion right out of the gate.  Nothing in any article is going to tell me how to determine the location of the VHPs in the simple example I provided, unless the article discusses your idea, which this one doesn't.
The origin is the common center of mass of the three.  I say this because I was doubtful that you are capable of working that out for yourself.
Is that the VHP of all 3 objects?  I ask because you say each object has its own, like they're in different places.
Given the location of the VHP, I can predict motion if I know how to compute the attraction between an object and its VHP.  You've not provided that either.  No website is going to give this answer because your idea is accepted on any website.

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However, it is quite clear to me that you are going to reject also this evidence.
I don't deny that binary systems have stars that orbit each other.  But the site just doesn't answer the question I asked.  You have to answer it, because your physics is not discussed anywhere outside this very long topic which unfortunately has no actual theory described so far.  I am trying to get to the actual theory instead of just a list of anomalies that you claim will be solved.

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I want to know how your theory works.  I gave the most trivial example I can think of to make it easy for you.  But apparently you can't do it.  You have no actual theory.  Every question is met immediately with a diversion.

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Why don't you even try to understand the deep explanation which I have offered?
You've never given a deep explanation.  If you had, I'd have been able to answer my question.
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Why do you constantly reject any explanation, example, evidence and article which I have offered so far?
Those sites are not pushing your idea, so none of them help clarify your idea.

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Why don't you answer my questions about the problems in using the dark matter concept?
You claim a solution to that problem, but without knowing your theory details, I cannot comment on how well it does that.

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So far I have introduced just the first steps from the whole theory.
You could not answer my question.  That seems to count as having provided a preface to the theory, but not even the first page of the actual theory.

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Unfortunately, you force me to explain the same issue again and again and therefore we can't move on.
No!  Don't move on.  Answer the question!  Where is the VHP of each object in the example, and how does an object's motion relate to that of its VHP?  Newton provided answers like this, where there is no VHP and objects act on each other via force computed by GMm/r².  From that info alone I can compute the Newtonian motion of those 3 objects.  I want to do so with your rules, but I don't know them.  If the paths are different, then your theory is non-Newtonian.  This is OK.  Actual theories are out there (such as MoND, or Modified Newtonian Dynamics) which reject the above formula at certain scales in order to not have to posit dark matter.  The MoND thing looks nothing like what you are describing.

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Hence -
Do you have real willing to open your mind and heart to this breakthrough understanding?
I'm trying real hard, but you cannot answer the most trivial question, and as you say:
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Without it, we won't be able to move forward.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 23/02/2019 18:20:57
Where is the VHP of each object in the example, and how does an object's motion relate to that of its VHP?
Let's focus on VHP1.
Based on Newton - it is not expected to see any wobbling activity during full orbital cycle (it could move in elliptical cycle - but not up and down from the orbital disc).
We must distinguish between "orbit around one another" to "wobbling".
In the example that I have used it was stated:
"While still rotating around the Milky Way's galactic center, these rogue stars also orbit around one another in a wobbly, spiral pattern.."
So, when we focus on the message: "orbit around one another" this represents a simple Newton solution. In this case we can say that the stars orbit around some VHP. In the same token, in your example - the stars also orbit around each other - so we can also say that the stars orbit around some sort of VHP.
In your example there is no wobbling activity. Newton didn't expect to see any wobbling activity. Newton only focus on planets and moons and there is no wobbling activity in those systems.
However, in the example which I have offered it is specifically stated that the stars are wobbling!!!
This wobbling activity can only be found in Stars!!!
Our scientists don't understand the real impact of this wobbling activity.
I claim that this wobbling activity is the key element activity for our understanding. it highlights how gravity really works in our spiral galaxy.
The wobbling activity is a direct outcome of VHP1.
So, only if we see a wobbling activity, than we know that there is a VHP1.
You won't find it in any Sun/Planet/moon activity. Therefore, you won't expect to find it based on simple Newton law.
However - if you look at any star in the galaxy, you should find that it is wobbling.
Our Sun is wobbling, the nearby stars are wobbling, S2 is wobbling all S stars are wobbling, the stars in that Halo stars are wobbling ... and so on
How can we ignore that important evidence???.
As I have already explained, this VHP1 is a direct outcome from the creation of any new star in the galaxy.
Each star had been created in the core of gas cloud.
The creation of new star from the gas cloud is based on the huge gravity impact of the nearby SMBH.
I'm quite sure that if we set that gas cloud far away from the SMBH (or even far away from the galaxy) - this gas cloud will not be able to set any sort of new star forming activity.
During the crystallize process from dust/gas/molecular into star, I assume that each star must orbit around some virtual center (I call it VHP1). Therefore, this VHP1 is a direct outcome from the nearby SMBH gravity impact during the new star forming activity.
Please be aware, that each star also comes with Planets and moons. However, those objects orbits directly around the star. Therefore, none of them has VHP1.
So, as the star emerges from the gas cloud it also comes with an integrated unique VHP1.
I'm not sure that I have the tools to deeply explain how it really works and how the SMBH gravity force set this process.
However, the wobbling activity is a solid prove for the existence of VHP1.
In your example, there is no wobbling activity. Therefore this example is none relevant to our case.
Please remember - VHP1 only exists in wobbling activity..
Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 23/02/2019 22:34:36
Where is the VHP of each object in the example, and how does an object's motion relate to that of its VHP?
Let's focus on VHP1.
Based on Newton - it is not expected to see any wobbling activity during full orbital cycle (it could move in elliptical cycle - but not up and down from the orbital disc).
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In your example there is no wobbling activity.
How do you know this?  You've not said where the VHP is for any object.

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Newton didn't expect to see any wobbling activity.
He does if gravity is exerted by any other mass than the primary.

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Newton only focus on planets and moons and there is no wobbling activity in those systems.
This is utterly false.  There is no mention in his formulas for special treatment of planets and moons.  Mass is mass.

If it helps you, you can take the 3 objects in my example and designate them as moons or possibly planets.

I still see nothing in your post that helps me identify where the VHP is for any specific object, or how the motion of that object relates to the position of that VHP.  You continue to evade the simplest question.

You do not have a theory at all.  How is your theory different than the ancient description that God is responsible for the motion of all seen objects?  Whatever you measure, God put it there.  There is no predictions, only declarations that whatever is seen is explained by the 'theory', which is that God is responsible for it.

The only difference between your theory and the God one is that the God one was quite accurate in making predictions.  The heliocentric models (things go around the sun, not the Earth) were resisted at first because their mathematical predictions were not as accurate as those make by a church with centuries of astronomers with tables and fudge factors.

Anyway, I see you taking this path.  "My theory, if it were ever to actually be spelled out, would account for everything we see, as long as it is in hindsight."  It cannot predict new things because without the actual meat of the theory, we can only guess how God wants the new thing to move.  The church for example would not be able to predict how the 3 objects in my example would move since it doesn't know God's intentions for those three objects.  Neither do you because you don't actually have a theory.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 24/02/2019 07:16:18
I still see nothing in your post that helps me identify where the VHP is for any specific object, or how the motion of that object relates to the position of that VHP.  You continue to evade the simplest question.
As you insist, let's look at a simple orbital motion of three objects by wiki:
https://en.wikipedia.org/wiki/Three-body_problem#/media/File:Three_body_problem_figure-8_orbit_animation.gif
So, we see a very nice motion. But there is no wobbling activity.
In order to understand the wobbling activity lets go back to our Sun.
Please look at the green line of
https://www.quora.com/How-is-the-motion-of-stars-similar-to-the-motion-of-the-sun
It is stated:
"But the stars and sun don't just orbit in a fixed plane like the planets orbit the sun.  They have a slight up and down wavy motion while they move along their trajectory, like so"
We see clearly that the sun is moving up and down in a sine wave while it orbits the center of the galaxy.
So, how can we explain this orbital motion?
Let's look at the other nearby stars.
it is stated:
"Stars in the local solar neighborhood move randomly relative to one another"
We see clearly that some goes up and other goes down.
It they will continue in that momentum, those stars should go upwards or downwards from the disc plane.
There are billions stars in the disc plane, how could it be that none of them continue with its momentum and move away from the disc plane?
Do you agree that this proves that all the nearby stars (and actally, all the stars in the disc plane) also "have a slight up and down wavy motion while they move along their trajectory"?
Please also look at the following diagram:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
We see clearly that the sun orbits around some sort of center (VHP1) while this center orbits around the galaxy.
Why don't you agree with that?

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Newton didn't expect to see wobbling activity.
He does if gravity is exerted by any other mass than the primary.
You claim that Newton should support this wobbling motion.
So would you kindly explain what is the scource for this wobbling motion?
Why do they have a slight up and down wavy motion while they move along their trajectory?
How could it be that this motion is based Newton?
Can you please show me one example for that motion which had been certified by Newton?
Where is the other mass which is needed for that wobbling motion (without VHP1).
Why the stars do not continue with their momentum and move upwards or downwards from the disc plane?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 24/02/2019 19:33:52
As you insist,
I was insisting you consider my simple example, and not some special case example you found on a website that doesn't support your view.  So you still have given me no clue as to how to figure out where a VHP is.
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let's look at a simple orbital motion of three objects by wiki:
https://en.wikipedia.org/wiki/Three-body_problem#/media/File:Three_body_problem_figure-8_orbit_animation.gif
So, we see a very nice motion. But there is no wobbling activity.
What do you mean by that??  Of course there is 'wobbling' activity.  If one of those objects was luminous, we'd see it wobbling vertically at twice the frequency and about a third the amplitude of the horizontal wobble, which would be a strange sight.  Such an object would not trace a clean ellipse around the galaxy.

That is a weird special case which is barely stable, and my example was not.  But it is a nice example of a wobbling activity without a VHP.  There just isn't one.  There is a center of gravity of course, but none of the objects orbit it, and when one is located at it, acceleration is at zero, not the maximum value as you would expect when an object gets closest to the virtual point about which it orbits.

It seems funny that you would link to an example that refutes what you've been asserting.

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In order to understand the wobbling activity lets go back to our Sun.
Please look at the green line of
https://www.quora.com/How-is-the-motion-of-stars-similar-to-the-motion-of-the-sun
It is stated:
"But the stars and sun don't just orbit in a fixed plane like the planets orbit the sun.  They have a slight up and down wavy motion while they move along their trajectory, like so"
We see clearly that the sun is moving up and down in a sine wave while it orbits the center of the galaxy.
Yes, that near-sine-wave motion is a easily derived from a mass effected by a planar arrangement of mass.  A solar system has almost all its mass at the center leaving little mass to have a wobble like that which has a period of anything on the order of the period of orbit.  Pluto is out of the plane and has such a wobble which manifests itself as precession since the wobble period is far long than its year.

Anyway, we do not 'see clearly' anything, because the green line in that picture is massively out of scale (so that you can distinguish it from a clean elliptical orbit).  It shows a velocity perpendicular to the plane that is the same order as the orbital velocity instead of about 2-3% of it as it should, and it shows a period that is also 4 times the actual measured period.  Your biocab link was even worse in both respects, and was actually factually wrong, not just out of scale.

So yes, this sine-motion is expected from a planar arrangement of mass, which (locally) describes the galatic disk.  The mass certainly isn't at the center like it is with our solar system.  The SMBH masses less than 0.05% of the mass of the galaxy, the vast majority taking a planar arrangement.
There is a VHP (your term) of sorts which is where the sun crosses the plane twice each wave.  It doesn't orbit that point, but just moves back and forth straight across it as would any object pulled by a planar arrangement of mass.  And yes, it is the VHP that moves at perhaps 230 km/sec (per that site) and the solar system itself at maybe 0.04% faster than that as it follows that longer green line.
They show the solar system following a perfectly circular orbit, but our orbit about the galaxy does have some eccentricity, probably around 0.05.  The path around the galaxy certainly isn't circular or even elliptic since it gets massive perturbations from moving through denser concentrations of matter in its travels, so the eccentricity value is fairly meaningless.  I'm just saying it isn't a nice orbit with constant speed like the simplified pictures show.

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Let's look at the other nearby stars.
it is stated:
"Stars in the local solar neighborhood move randomly relative to one another"
We see clearly that some goes up and other goes down.
No velocities are given, so this isn't clear at all.  The arrows are relative to the sun, not relative to the galactic disk, and they're all the same size, meaning it isn't depicting vectors.  It is just a crude picture trying to illustrate those words.  Maybe they're all moving up, just some less than us, and some more.

They might be trivial differences while we (the other stars) move as a group around the galaxy, or they might be significant differences in which case the nearby stars just happen to be nearby now, but we'll never see them close again.

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It they will continue in that momentum, those stars should go upwards or downwards from the disc plane.
Presumably, yes.  Some of them might be stationary relative to the disk, and hence not have that sort of motion.  The site hardly gives any data on this.

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There are billions stars in the disc plane, how could it be that none of them continue with its momentum and move away from the disc plane?
They almost all move relative to the plane in some way, and they (almost) all come back just like we do.  The plane is the average, not a clump of different stuff that doesn't have this motion.  Sure, some stars experience a close encounter with something big and get flung out of the disk, becoming a halo object of sorts.

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Do you agree that this proves that all the nearby stars (and actally, all the stars in the disc plane) also "have a slight up and down wavy motion while they move along their trajectory"?
Statistics says it is highly improbable that a random number is exactly zero, so this would suggest that a given moment, no star has zero motion perpendicular to the plane.  Even if it did, one moment later, gravity of some nearby object would alter this value to something nonzero.  None of the text above proves this, no.  I reached for statistics to agree with the statement.  Point is, some have more than others.  Even the sun has zero motion perpendicular to the plane about 5-6 times per lap around the galaxy, but that zero value lasts but an instant.

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Please also look at the following diagram:
http://www.biocab.org/Motions_of_the_Solar_System.jpg
We see clearly that the sun orbits around some sort of center (VHP1) while this center orbits around the galaxy.
If that picture was even remotely accurate, then I suppose it would.  These guys are not astronomers or physicists.  They're biologists.  I don't go to astronomers to ask my biology questions either.
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Why don't you agree with that?
The solar system does not move like that.  For one, we'd have to move at about 1500 km/sec to follow that blue line and still keep up with the 217 km/sec grey line.  The blue line is around 7 times longer than the grey line, but labelled with much lower speeds and no mention of periods.

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You claim that Newton should support this wobbling motion.
Of course.  His laws demand it if there is mass acting on the thing that wobbles.  The moon's path wobbles around the sun because Earth is there yanking back and forth once a month, just as Newton describes.

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So would you kindly explain what is the scource for this wobbling motion?
Mass of something other than itself or its primary.  Any third body or cloud or disk or whatever.

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Why do they have a slight up and down wavy motion while they move along their trajectory?
Any object will have this motion relative to a planar arrangement of mass.

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How could it be that this motion is based Newton?
Because his laws describe it perfectly, and he did it first, so his name got attributed to those laws.  A plane of stuff is going to attract.  You can't go around it, so you get pulled straight towards it and carry right through to the other side after which the plane pulls you the other way.  It's pretty easy to work out.

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Can you please show me one example for that motion which had been certified by Newton?
An apple falling to the ground.

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Where is the other mass which is needed for that wobbling motion (without VHP1).
The disk is the other mass.
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Why the stars do not continue with their momentum and move upwards or downwards from the disc plane?
Because the disk pulls them back.  If we're above the disk, then most of the stars are below us, pulling us downward.  Those stars (including us) make up the disk.  We don't hit the disk on the way through since the stars have space between them.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 25/02/2019 16:32:38
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Why the stars do not continue with their momentum and move upwards or downwards from the disc plane?
Because the disk pulls them back.  If we're above the disk, then most of the stars are below us, pulling us downward.  Those stars (including us) make up the disk.  We don't hit the disk on the way through since the stars have space between them.

You claim that: "Those stars (including us) make up the disk.
So, let me start by asking: why all of those stars stay at the disc from the first stage?
What kind of force keeps them all in the disc?
Why just from the starting point of the Bar (at about 1KPC) to the edge of the spiral arms (at about 10KPC) there is a disc?
How could it be that in the bulge (less than 1KPC) stars do not orbit in a disc?
How could it be that all S stars which are located very close to the SMBH do not share the same orbital disc plane?
Why each S star sets its unique orbital disc plane although they are directly affected by the mighty gravity force of the SMBH?
Our scientists claim that the sun orbits around the center of the galaxy due to the dark matter:
http://cdms.berkeley.edu/Education/DMpages/FAQ/question36.html
"The radius of the sun's orbit is about 2.5x1017 km, so the total mass of dark matter within that orbit is 6x1040 kg. This is the mass of 3x1010 (30 billion) stars like the sun! The entire galaxy only contains ~100 billion stars, so the dark matter does have a significant effect on the sun's orbit through the galaxy. For objects farther out near the edge of the galaxy, the dark matter is actually the main thing keeping them in their orbits. This is more or less how dark matter was discovered by astronomer Vera Rubin and others: the orbital speeds of galactic stars and gas clouds don't match our expectations from the visible matter."
Just think about the huge gravity force which is needed to keep the sun in his track around the center of the galaxy. It is also clear that most of the gravity force comes from the dark matter.
Based on our scientists, the dark matter is not located at the disc itself.
So, how could it be that the stars around us with almost neglected gravity force (comparing to the ultra high gravity force due to the dark matter) can hold each other at the same orbital disc plane?
What kind of force hold all the stars in the disc?
The Sun is located at the Orion arm.
The Orion Arm diameter is about 1,000KLY
In one hand our scientists claim that this arm is only a density wave. So, there are no gravity force connections between the stars.
However, now you claim that those stars pull back any star that is moving too high or too low from the disc plane.
How could it be? don't you see a contradiction?
What kind of power/force do they use in order to pull each other back?
Is it gravity force or some other force?
If it is gravity - Can you please explain how the gravity force pulls back any star which goes too high above the disc or too low below the disc??
You claim that it pull back - but in reality it works like a spring. Up and down Up and down in a constant movment.
How could it be?
If I understand it correctly, Newton gave us two options:
1. Direct contact. For example:
An apple falling to the ground.
2. Orbital motion.

However, I have never heard about Newton gravity force which could force a star to move up and down continually without orbital motion.
Based on this theory, how could it be that the local gravity force due to local mass of nearby stars can have any sort of effect comparing to a gravity force due to 30 billion solar mass?
It seems to me as an elephant which is swinging nearby a mosquito, while the mosquito is sure that it is due to is mass. Is it real?

As for example -.
Let's assume that somehow we can force all the stars in 1,000KLY segment in our spiral arm (by glue) at a fixed location in the segment.
Now, let's take a free star and push it to the most upwards side of the arm and release it.
If I understand it correctly, the gravity force of the other stars will pull it back.
But it will have two options:
One - To collide with one of the nearby stars
Two - start orbit around some sort of center of mass or around VHP which is a direct outcome of the stars in that segment.
Can you please explain how it could move up and down in a constant movement?
How could it be that gravity acts like a spring?
Now, let's assume that somehow the gravity force keeps it from moving to high from the most upwards side of the arm or too low from the most downwards side of the arm.
Why the same force can't hold it from moving away from the arm (too left or too right)?
Why only Up/down and not right/left?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 26/02/2019 19:44:34
You claim that: "Those stars (including us) make up the disk.
Stars, gas, planets, screwdrivers, etc.  Yes.  The disk is made of stuff.

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So, let me start by asking: why all of those stars stay at the disc from the first stage?
Not all do.  Some get thrown out, but that takes effort, some outside force.

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What kind of force keeps them all in the disc?
Gravity.  The list of forces isn't long, even if you are making up new ones.
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Why just from the starting point of the Bar (at about 1KPC) to the edge of the spiral arms (at about 10KPC) there is a disc?
The disk goes further out than that, but sure.  One might consider stuff closer than 1kpc as well.  It doesn't have an abrupt start or end.

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How could it be that in the bulge (less than 1KPC) stars do not orbit in a disc?
Anything that orbits does so in more or less a plane.  The orbit of no single object can be described as a disk.  A disk is a collection of material flattened by mutual angular momentum.

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How could it be that all S stars which are located very close to the SMBH do not share the same orbital disc plane?
I'm no expert, but I imagine that the close objects have a point mass around which they might orbit.  The solar system does not.  That explains an orbital path as opposed to a more characteristic disk path, but not so much the orbital plane.  I suspect that planar objects are in the plane because they've not been knocked out of it.  Those objects or gas clouds that have acquire random trajectories that take some of them by chance very close to the center of the galaxy where they take up a new orbit in the new plane into which they were deflected.

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Why each S star sets its unique orbital disc plane although they are directly affected by the mighty gravity force of the SMBH?
The SMBH is not capable of altering the orientation of the orbital axis of a moving object.

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Our scientists claim that the sun orbits around the center of the galaxy due to the dark matter:
I find that wording misleading in this context, but the context of the page is different.  Keep that in mind. I would say the sun loosely orbits the galaxy, not the center of the galaxy since, unlike any planet, the vast majority of the forces acting on us come from places other than the center.  About 99.9% of the solar system mass is at the center, but less than 1% in the case of the galaxy.  You just can't compare orbital mechanics between the two.

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Just think about the huge gravity force which is needed to keep the sun in his track around the center of the galaxy. It is also clear that most of the gravity force comes from the dark matter.
Based on our scientists, the dark matter is not located at the disc itself.
So, how could it be that the stars around us with almost neglected gravity force (comparing to the ultra high gravity force due to the dark matter) can hold each other at the same orbital disc plane?
Dark matter has no more gravity force than regular matter.  There are not different kinds of gravity.  A KG of dark matter exerts the same gravitational force as a KG rock or a KG black hole.
Dark matter (at least WIMP dark matter, not so much the MACHO parts) is not subject to friction like gas clouds are, so it doesn't tend to flatten into a disk just because it has angular momentum.  So it makes sense that it isn't concentrated in a disk like the interactive matter tends to do.

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The Sun is located at the Orion arm.
The Orion Arm diameter is about 1,000KLY
In one hand our scientists claim that this arm is only a density wave. So, there are no gravity force connections between the stars.
The stars very much attract each other by gravity.  Newton's laws demand this.  If the arm is a density wave, then it attracts stuff more than the less dense places between the arms, but as the density wave moves on, it doesn't carry the solar system or anything else with it.  The density waves pass through us, meaning the sky is more cluttered when we're in a proper arm than when we're in the low-density regions between the arms, just like a leaf in the water rising up and down with the waves without moving along with them.

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However, now you claim that those stars pull back any star that is moving too high or too low from the disc plane.
How could it be? don't you see a contradiction?
If it is gravity - Can you please explain how the gravity force pulls back any star which goes too high above the disc or too low below the disc??
I see no relevance to the arms.  Suddenly you're talking about the disk.  If the disk is below you, wouldn't you expect all that mass to pull you toward it?  You seem to suggest otherwise, which would contradict the laws of gravity.  I have no idea why you find this baffling.

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You claim that it pull back - but in reality it works like a spring. Up and down Up and down in a constant movment.
It is like a spring.  If I extend the spring, it tries to pull me back.  If that shoots me to the other side, the spring pulls me back again.  Hence the green sine wave you see in that one picture.
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How could it be?
How can it not be?  It would seem to violate Newton's laws if all that mass on one side didn't attract us back each time.
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If I understand it correctly, Newton gave us two options:
1. Direct contact. For example:
Quote from: Halc
An apple falling to the ground.
That isn't direct contact.  The apple has all the mass of the Earth on one side of it, so is attracted by gravity.  That's enough to break free of its bond on the tree and enter into orbital free fall around Earth.  I can't help it that the ground gets in the way and causes the apple to stop via forces other than gravity.  The disk isn't like that.  It is unlikely to actually strike anything large enough to prevent us from passing right through.  But we hit small stuff which slows us each time, so eventually the sine wave fades until the next time something big passes close by again to get a new sine wave going.  I think the friction with the disk is pretty minimal, so Earth will burn up before that wave fades noticeably.
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2. Orbital motion.
The apple goes into orbit when it detaches from the tree, yes.  Not a pure orbit since it is not in a vacuum.  There is a lot of friction.  It would never come back to its start point even if the ground didn't get in the way.  The apple is in orbit around a single mass.  The solar system is not in orbit about the disk in this way.

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However, I have never heard about Newton gravity force which could force a star to move up and down continually without orbital motion.
Try dropping an apple from some distance from a dense disk with a hole in it to allow the apple to pass through.  It will go back and forth forever through the hole, or so says Newton's laws.
You might consider that hole to by your VHP since the apple is always drawn to it, but the motion relative to that VHP would not be an orbit.  The closer the apple gets to that point, the smaller the force is acting on it.  With an orbit, the closer you get to the VHP, the greater the force.

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Based on this theory, how could it be that the local gravity force due to local mass of nearby stars can have any sort of effect comparing to a gravity force due to 30 billion solar mass?
No idea what you're talking about here.  What 30 billion masses?  If they're far enough away, the nearby stars have greater effect since forces in inversely proportional to r².  Still, the 30 billion solar masses, while further away, will pull on each of those local stars, accelerating them as a group if the group moves away from the the disk.  The local stars are on all sides, so their pull tends to cancel out.  The disk, when all on one side, does not have its effect cancelled out.

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It seems to me as an elephant which is swinging nearby a mosquito, while the mosquito is sure that it is due to is mass. Is it real?
I don't understand the analogy.  An elephant on a swing moves due to the gravity of Earth which dwarfs the mass of both the mosquito and the elephant.

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As for example -.
Let's assume that somehow we can force all the stars in 1,000KLY segment in our spiral arm (by glue) at a fixed location in the segment.
Glue them to what?  What is a segment?  Are you trying to glue it to the arm?  What if the arm moves at some insane speed?  What if it moves faster than light?  Waves can do that.  No, I don''t suggest arms move that fast, but they probably move much faster than any of the material that lights them up.

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Now, let's take a free star and push it to the most upwards side of the arm and release it.
If I understand it correctly, the gravity force of the other stars will pull it back.
It will probably be moving too fast to do this, and will be flung away to some larger radius from the galaxy.

OK, so let's say we release the star at the mean velocity of all the other material in the arm, but not the speed of the arm itself.  I think that's what you have in mind.  I think the density of the arm will move on and the star will stay at the radius where you put it.  The arms's don't have much pull for very long since they pull one way as they approach and pull the other way after they pass.  The effects cancel.  It is moving out of the disk (in a direction parallel to the galactic axis) that would result in an attraction back to the disk.

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But it will have two options:
One - To collide with one of the nearby stars
That can always happen, but they're small targets.  We've passed through the disk some ~110 times probably and haven't got close to one yet.

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Two - start orbit around some sort of center of mass or around VHP which is a direct outcome of the stars in that segment.
Now you've jumped to your view.  You have refused to define where a VHP is given a certain setup, so this cannot be answered intelligently.
So how about option 3: It will continue its way around the galaxy, except without the green sine wave since you put it at equilibrium at the middle of the disk so it need not move up and down at all until defected by something.

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Can you please explain how it could move up and down in a constant movement?
I think it will not, since you arrested that motion when you glued it in place.

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Now, let's assume that somehow the gravity force keeps it from moving to high from the most upwards side of the arm or too low from the most downwards side of the arm.
That makes no sense.  The arm is not an object.  Nothing can orbit it.  No matter is particularly part of it.  They're just waves that pass through the matter.  Your theory might propose otherwise, but then you have to answer your own questions, like what speed do the arms move, and why they're not completely wound tight by now.

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Why the same force can't hold it from moving away from the arm (too left or too right)?
Why only Up/down and not right/left?
The disk is an object.  The arms are not.  The arms pull in both directions as they pass with a net effect of more or less zero.  The disk never passes away.  It's always there.
Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 02/03/2019 08:30:18
Please look again at the following article:
https://www.livescience.com/63698-milky-way-spiral-wobble.html
It is stated clearly:
"While still rotating around the Milky Way's galactic center, these rogue stars also orbit around one another in a wobbly, spiral pattern that has only become more tangled over the past eon."
Why can't we trust this verification of our scientists?
It is stated clearly: "these rogue stars also orbit around one another in a wobbly, spiral pattern.."
So, do you agree that our scientists see that while the stars in these rogue stars halo orbit around one another, they also move in a wobbly pattern?
What is the meaning of that wobbly pattern?
Do you agree that it means that stars do not orbit exactly at the expected orbital cycle, but they move in a wobbly way around this expected orbital cycle?
We know that the Sun and all the other stars in the disc are wobbling while they orbit around the galaxy.
We caim that it is due to the galactic disc.
But stars also wobble high or below the disc.
I have offered also other evidences for that wobbling outside the disc -
S2:
https://www.researchgate.net/figure/Fit-to-the-orbit-of-the-S2-star-fitted-data-and-relative-errors-are-in-blue-the-red_fig1_272845577
It is stated:
Fit to the orbit of the S2 star: fitted data and relative errors are in blue, the red line is the orbit.
So we see clearly that the measured points are not located directly on the orbital path.
Our scientists call it "relative errors in blue". But those "errors" proves that our scientists see that S2 is wobbling!
Therefore, it should be clear to all of us that our scientists see wobbling path activity of S2.
Please remember that S2 is also not located at the disc.
Hence, it proves that the hypothetical idea about the Up/Down movement or wobbling due to the disc is incorrect as we see the wobbling activity also out the disc.

The disk is an object.  The arms are not.  The arms pull in both directions as they pass with a net effect of more or less zero.  The disk never passes away.  It's always there.
How could it be?
There are stars above and below the disc while there are also stars in between the arms.
Actually, the width of the disc is dictated by the width of the arm.
In all the images of the milky way, we clearly see that most/all of the stars are located in the arms.
As an example:
http://www.astronoo.com/en/articles/galactic-arms.html
What is the estimated ratio between the total stars in the arms comparing to the total stars in the disc outside the arm?
Do you agree that the arm is mainly made out of stars while the disc is mainly made out of arms?
If we take out all stars from all arms, do you think that there will be left something that can still be called disc?
Can you please explain what do you mean by: "The arms pull in both directions as they pass with a net effect of more or less zero."
If I understand it correctly, you claim that stars can cross the arm as they move vertically to the disc, but they can't cross the arm as they move horizontally to the disc.
Hence, a star can't move horizontally and cross the arm as the stars in the Arm pull it back due to gravity force from the stars in the arm.
So, if the stars in the arm prevent from a star to cross the arm vertically (outwards or downwards) due to gravity force, than why the same gravity force (Due to the stars in the arm) can't prevent from a star to cross the arm horizontally?
How could it be that the stars in the arm sets a gravity force which works horizontally, but doesn't work vertically?

With regards to the spiral galaxy structure.
In the bulge - there is no disc. Each star orbits at a different orbital disc plane - (therefore we call it bulge...). This contradicts the solar orbital system, as all the planets orbit at the same disc plane - with the exception of Ploto.
Why is it?
If we move further away from the bulge in the direction of the bar we see that stars start to orbit in a disc. (very wide disc close to the bulge, but as we go further away to the edge af the bar (which is also the starting point of the Arms - at 3KPC) we clearly see the disc shape.
Therefore, it is clear that the Arms + Bar set the disc and not vice versa.
Hence, if we call the disc an object than by definition the Arm and the Bar must be considered as an object.
I still don't understand why our scientists just focus on the disc and the arms?
Why they don't explain the Bulge and the Bar?
Why they don't show in their modeling how we could get the full spiral galaxy shape including: Bulge, Bar, Spiral Arms, Disc, Halo stars and so on.
Why they only focus in one segment of the galaxy (spiral arms + disc).
With regards to modeling:
Let's look at the following article:
https://kof.zcu.cz/st/dis/schwarzmeier/galaxy_models.html
A. "5.8  - Spherical galaxies
Our first objective is a simulation of elliptical galaxies with an eccentricity of zero (spherical galaxies). We will perform the simulation of two spherical systems – Plummer’s and Hernquist’s models."
However, the outcome was -
"Figure 5–13: A time sequence of Plummer’s model in the xy plane. Practically no evolution can be seen."
So, if they start with a spherical galaxy - they don't get a disc and they don't get the spiral arms!!!
B. However, if they start the modeling as a disc, they get some spiral arms which have gone after some time:
"5.9        Initial conditions for flattened systems
A spiral galaxy is flat and can be represented by a flattened potential. The disk of the spiral galaxy is generally very flat and we will describe them as infinitely thin. We will suppose that disks are axisymmetrical and therefore, initial conditions will be set up in cylindrical coordinates."
"Figure 5–16: A time sequence for Kepler’s disk model in the xy plane. The disk is ideally flat in the yz plane."
C. Just when they start with a very thin disc, with other parameters including the radial velocity dispersion of the solar neighborhood, they have finelly got the spiral arms.
"We will create a disk with kinematically hot (supported by random motions, rather than the rotational motion) stars. The thickness of the disk is set such that Toomre’s stability criterion is satisfied. The radial velocity dispersion  corresponds to the value of the Solar neighborhood, i.e."
But this is absolutely none realistic.
They have actually proved that when they took a realistic starting point as spherical galaxies, they didn't get by the modeling the expected shape of spiral arms and disc.
D. After the Big bang, we didn't get any disc (especially not a thin disc of stars!!!).
So, they can't just start from the point where the disc is already there with all the requested velocities. This is an error by definition.
They have to show how long it takes from the Big bang moment to set all the requested stars in a spherical galaxies and from that point to the starting point of a very thin disc.
Than they have to add it to the requested time to form a spiral disc.
"Figure 5–20: A time sequence for the hot exponential disk with  in the xy plane evolved with the Barnes-Hut N-body algorithm"
If in the following modeling 1 billion years is needed to set a spiral galaxy from thin disc, they have to add the requested time from the big bang.
So, this modeling is not realistic as they didn't add the requested time from the Big bang to that starting point.
I'm quite sure that they might find that even after 10 billion years no real spiral arm would be formed.
E. As stated, they have only showed spiral arms and disc. Not even one single word about the other sections of spiral galaxy as Bulge & Bar. This is another sever mistake.
F. Actually, this modeling proves that there is no way to get spiral arms galaxy directly from spherical galaxies.
This evidence proves that our scientists have a severe mistake in their understanding how the spiral galaxy had been evolved from time zero - Big bang.
"Figure 5–21: A time sequence for the hot exponential disk with  in the yz plane evolved with the Barnes-Hut N-body algorithm. The thickening caused by the random motion of stars can be seen"
We see that the starting point is a very thin disc.
After 1.02 Billion years the disc became quite thick.
Why they stopped after this moment?
Why they didn't continue with the modeling?
In one of the modeling they have showed clearly that after getting the spiral shape there is a possibility to lose it again.
Therefore, I'm quite sure that if they will continue with the modeling (for few more billion years) that disc will be so thick that we shouldn't see any spiral structure - maybe only spherical galaxy.

This could prove that spiral galaxy might be converted over time into spherical galaxy but not vice versa.

So, do you agree that based on those key ideas, there must be a severe problem with our current understanding about spiral galaxy and the evolvement activity of this galaxy from the Big Bang starting point?

Title: Re: How gravity works in spiral galaxy?
Post by: Halc on 02/03/2019 18:07:58
It is stated clearly:
"While still rotating around the Milky Way's galactic center, these rogue stars also orbit around one another in a wobbly, spiral pattern that has only become more tangled over the past eon."
Why can't we trust this verification of our scientists?
It is stated clearly: "these rogue stars also orbit around one another in a wobbly, spiral pattern.."
So, do you agree that our scientists see that while the stars in these rogue stars halo orbit around one another, they also move in a wobbly pattern?
What is the meaning of that wobbly pattern?
It says in the article.  They mapped the position and velocity of a huge number of objects and found this one region that had motion that was anomalous to the typical paths of the rest of them.  It apparently means somebody threw a rock into a calm pond and long after the rock had passed, the disturbance of it is still visible, centered on the passage point via 'wobbling' waves.  They even extrapolated back and found the object (the rock) that is responsible.

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Do you agree that it means that stars do not orbit exactly at the expected orbital cycle, but they move in a wobbly way around this expected orbital cycle?
A 'wobbly' way is the expected cycle.  We're not in empty space so nobody expects our path to be a nice clean circle or something.

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I have offered also other evidences for that wobbling outside the disc -
S2:
https://www.researchgate.net/figure/Fit-to-the-orbit-of-the-S2-star-fitted-data-and-relative-errors-are-in-blue-the-red_fig1_272845577
It is stated:
Fit to the orbit of the S2 star: fitted data and relative errors are in blue, the red line is the orbit.
So we see clearly that the measured points are not located directly on the orbital path.
Interesting, huh?  I also notice that the error bars abruptly get tiny on the right.  Somebody bought a new more accurate telescope during those 18 years.

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Our scientists call it "relative errors in blue". But those "errors" proves that our scientists see that S2 is wobbling!
It is indeed wobbling. It's not in empty space after all. It just doesn't have a regular one like it would if it orbited some companion.  It has no VHP.

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Hence, it proves that the hypothetical idea about the Up/Down movement or wobbling due to the disc is incorrect as we see the wobbling activity also out the disc.
What happens somewhere not in the disk is no evidence whatsoever about the dynamics of the disk.

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There are stars above and below the disc while there are also stars in between the arms.
Actually, the width of the disc is dictated by the width of the arm.
Disks have width?  You mean thickness?
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In all the images of the milky way, we clearly see that most/all of the stars are located in the arms.
Yes, arms light up because more stars are there, but that doesn't make arms objects unless the motion of the stars is the same as the motion of the arms, and it isn't.

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Do you agree that the arm is mainly made out of stars while the disc is mainly made out of arms?
Disks are made of stars and gas and other material.  That mass moves differently than the arms, so no, I would not say that the disk is made of the arms any more than I say a lake is made out of waves.  It is made out of water, and yes, there is more water where the waves are high than where the waves are low, but the wave moves on and takes none of the water with it.

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Can you please explain what do you mean by: "The arms pull in both directions as they pass with a net effect of more or less zero."
If I understand it correctly, you claim that stars can cross the arm as they move vertically to the disc, but they can't cross the arm as they move horizontally to the disc.
I didn't say that.  Stars cross the disk as they move vertically back and forth across it, and due to the gravity of it.  The arm wave seemingly passes through us as it goes by, but like a leaf on the water, it makes the leaf wobble but doesn't actually change its position in the end.

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Hence, a star can't move horizontally and cross the arm as the stars in the Arm pull it back due to gravity force from the stars in the arm.
I don't know what you mean by this.  I don't know which direction you consider horizontal.  The arm passes through us, so that's pretty much the same as us crossing it.  Yes, I imagine the higher density of the stars causes some alteration to the path we'd otherwise take.

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So, if the stars in the arm prevent from a star to cross the arm vertically (outwards or downwards) due to gravity force, than why the same gravity force (Due to the stars in the arm) can't prevent from a star to cross the arm horizontally?
Nothing is prevented from moving where it is heading.  There is material on all sides, so for the most part our progress in unhampered by that.  Think again of the leaf on the water which is not prevented from crossing the wave, and isn't even really moved by it in the long run.

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With regards to the spiral galaxy structure.
In the bulge - there is no disc. Each star orbits at a different orbital disc plane - (therefore we call it bulge...). This contradicts the solar orbital system, as all the planets orbit at the same disc plane - with the exception of Ploto.
Pluto isn't a planet, and there are plenty of objects that orbit out of the plane.  None of them happen to be a planet, which are planar because they formed out of the same rotating disk of gas as the sun and have had nothing that can divert them out of that original plane.
The solar system is a lousy model of a little galaxy since it is held together by a primary mass and the galaxy isn't.  The dynamics are very different.

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I still don't understand why our scientists just focus on the disc and the arms?
They probably focus more on stars and clouds and their interactions.  No idea what you mean by your statement there.

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Why they don't explain the Bulge and the Bar?
There are models that do.  Your model doesn't because you have no model.  You don't post any simulations of your model.

With regards to modeling:
...
So, if they start with a spherical galaxy - they don't get a disc and they don't get the spiral arms!!!
B. However, if they start the modeling as a disc, they get some spiral arms which have gone after some time:
"5.9        Initial conditions for flattened systems[/quote]
They get arms (at least briefly) with a different model.  It isn't strictly the use of different initial conditions, but the use of a different model.  The arms are very short lived.  The model isn't very predictive.

All the simulations run for only a short time, like a billion years.  Real galaxies run undisturbed for perhaps this long, but in the longer run, they're constantly bombarded by collisions which disrupt the structure.  A good simulation running for the full 13 billion years needs to include such evolution.

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C. Just when they start with a disc it with other parameters including the radial velocity dispersion of the solar neighborhood, they have finelly got the spiral arms.
Briefly, yes, and using a different model.

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They have actually proved that when they took a realistic starting point as spherical galaxies, they didn't get by the modeling the expected shape of spiral arms and disc.
That was a different model, and I don't know what angular momentum they gave it.

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D. After the Big bang, we didn't get any disc.
So, they can't just start from the point where the disc is already there with all the requested velocities. This is an error by definition.
Indeed, so how did it get that way?  Here, the accretion model of solar system formation is somewhat useful since the early stages of a solar system is distributed material with angular momentum but no central mass yet.  Disks do form naturally, at least by standard models.  Your assertions seem to suggest that fully formed solar systems and their defined planes just spring into existence out of black holes, so maybe there is something that similarly spews forth galaxies with their disks already in place.

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They have to show how long it takes from the Big bang moment to set all the requested stars in a spherical galaxies and from that point to the starting point of a very thin disc.
A model of a spherical galaxy doesn't ever become a disk.  There are galaxies that stay spherical.  There are many classes of galaxy types.

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"Figure 5–20: A time sequence for the hot exponential disk with  in the xy plane evolved with the Barnes-Hut N-body algorithm"
That's a third model now, and it still doesn't look like a real galaxy, but at least the arms seem a bit more stable.

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If in the following modeling 1 billion years is needed to set a spiral galaxy from thin disc, they have to add the requested time from the big bang.
The arms are quite distinct after 0.4 billion years.  There were sort of there in the Kepler model as well, but after a billion years, they were gone again.  They were not stable.  Kepler's model I think is one where the arms are objects instead of waves.

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I'm quite sure that they might find that even after 10 billion years no real spiral arm would be formed.
Actually, this modeling proves that there is no way to get spiral arms galaxy directly from spherical galaxies.
A model that doesn't exhibit the desired behavior proves that the model does not correspond entirely to the modelled thing.  It doesn't in any way prove that another model would not fare better.  Your grasp on logic is on the same level as your grasp of physics.

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This is evidence which proves that our scientists have a severe mistake in their understanding how the spiral galaxy had been evolved.
What mistake would that be?  Are they asserting that one of these models is the correct one?  I doubt that seriously.  They're quite aware that they don't have a full understanding of the dynamics yet.

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"Figure 5–21: A time sequence for the hot exponential disk with  in the yz plane evolved with the Barnes-Hut N-body algorithm. The thickening caused by the random motion of stars can be seen"
We see that the starting point is a very thin disc.
After 1.02 Billion years the disc became quite thick.
The initial conditions are a very thin disk, just to see what thickness would form naturally.  It isn't suggested that the disk was ever that thin.
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Why they stopped after this moment?
Why they didn't continue with the modeling?
Maybe it didn't evolve much after that.  To go longer, it would need to simulate the collisions with objects coming from the outside.  Those happen frequently at first, and still occur more than once every billion years.  I've seen animated simulations that run the whole lifespan of the galaxy and it shows all those outside objects from which most of the mass of the galaxy comes.  We were likely quite small in the beginning.

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I'm quite sure that after 13 billion years that disc will be so thick that we would see just a normal spherical galaxy.
I've not seen any simulation that did that.
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This could prove that spiral galaxy might be converted over time into spherical galaxy but not vice versa.
If there was a model, sure, but you don't have one, so there is no simulation of it.

Title: Re: How gravity works in spiral galaxy?
Post by: Dave Lev on 09/03/2019 14:38:57
Dear Halc
You have chosen to reject all the evidences which I have offered.
It seems to me that even if Newton was standing next to me and fully support my message - you would still reject any idea which contradicts the current mean stream which you fully support.
So, it is quite clear to me that whatever I will say you would reject.
Therefore, let me summarize between those none relevant ideas of the mean stream and my real theory.

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In all the images of the milky way, we clearly see that most/all of the stars are located in the arms.
Yes, arms light up because more stars are there, but that doesn't make arms objects unless the motion of the stars is the same as the motion of the arms, and it isn't.
Sorry
Our scientists have no clue how spiral arms really works.
The assumption that - The motion of the stars isn't the same as the motion of the arms, is absolutely incorrect!!!
I have deeply explained how it really works.
Based on my explanation, the arms are made of stars. Any star which will dare to move out of the arm will be ejected from the galaxy (unless it is crossing a bridges or branches between the arms).
So, the arms are real object!
I have introduced again and again the following image, but you have chosen to ignore that real image/evidence.
https://en.wikipedia.org/wiki/Milky_Way#/media/File:Milky_Way_Arms.svg
It shows that based on our verifications all/most of the stars in the milky way disc are located at the arms.
However, in the same article it is stated:
https://en.wikipedia.org/wiki/Milky_Way
"Outside the gravitational influence of the Galactic bars, the structure of the interstellar medium and stars in the disk of the Milky Way is organized into four spiral arms.[113] Spiral arms typically contain a higher density of interstellar gas and dust than the Galactic average as well as a greater concentration of star formation, as traced by H II regions[114][115] and molecular clouds.[116]"
What does it mean "higher density of interstellar gas and dust" in the following statement"?
Where are all of those nearby interstellar gas and dust?
If we look nearby, we only see G-type stars.
Why is it?
Why our scientists discuss about interstellar gas and dust while we mainly see G-type stars?
If I understand correctly - there is only one nubila in the whole Orion arm.
Don't forget that in the simulation our scientists are using stars!
So, please don't speak about interstellar gas and dust just to confuse us - speak about G-type stars (as you are using them in the simulation).
Now, let's try to verify the density of those stars.
In the following article it is stated:
http://www.solstation.com/stars3/100-gs.htm
"As many as 512 or more stars of spectral type "G" (not including white dwarf stellar remnants) are currently believed to be located within 100 light-years or (or 30.7 parsecs) of Sol -- including Sol itself. Only around 64 are located within 50 light-years (ly), while some 448 are estimated to lie between 50 and 100 light-years -- a volume of space that is seven times as large as the inner sphere within 50 ly of Sol. A comparison of the density of G-type stars between the two volumes of space indicates that the outer spherical shell has around 100 percent of the spatial density of known G-type stars as the inner spherical volume, which suggests that astronomers have identified the great majority of the G-type stars that are actually located within 100 ly of Sol, assuming the same spatial distribution in the Solar neighborhood".
So we know the density of G-type stars in a 50 LY volume: "Only around 64 are located within 50 light-years".
That fully meets by 100% the G-type stars density expectation at 50 and 100 light-years volume: "while some 448 are estimated to lie between 50 and 100 light-years"
That shows the G-type stars density at the Orion arm (There are 448 + 64 = 512  G -type stars in a 100LY around the Sun)
However, what is the G-type stars density at the same volume (100LY) between the orion arm and the next nearby arm?
Is it only 400, 200, 100 or just Zero?
I assume that they might find an area between the arms with high density of G -type stars (As it is a bridge or branch between the arms)
However, is there any possibility to find in the same volume of 100KL between the arms ZERO G -type stars?
I'm sure by 100% that there are big areas between the arms without even a single star.
That is a direct outcome from my theory.
So, if that is correct, my theory is also correct.
Wobbling
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Our scientists call it "relative errors in blue". But those "errors" proves that our scientists see that S2 is wobbling!
It is indeed wobbling. It's not in empty space after all. It just doesn't have a regular one like it would if it orbited some companion.  It has no VHP.
Thanks
So you agree that at least S2 is wobbling.
However, why are you so sure that: "It's not in empty space after all. It just doesn't have a regular one like it would if it orbited some companion?
Once you agree that stars are wobbling - you actually say that you can't explain it by normal orbital cycle.
If we look at our sun, it is not just wobbles up and down. It doesn't move horizontally to the disc. it actually move with some phase to the disc.
If that is correct, than the idea that it moves up and down due to the disc is incorrect by definition.
What is the amplitude of that wobbling activity of the sun?
I'm quite sure that it is less than 100Ly or even 50LY.
However, the diameter of the spiral arm is 1000LY.
There must be stars which are located at the most upwards side or downwards side of the arm. What is their up down wobbling amplitude. If they have the same wobbling amplitude as the sun, than by definition they don't cross the center of the disc.
Hence, if there is only one star that doesn't cross the center of the galaxy, than it is another solid proof that the stars do not wobble due to the disc.
If we look at the solar system, and monitor the moon cycle, while the earth is dark, we should see the it is wobbling as it orbits around the sun.
So, the only real explanation for that wobbling is a Virtual host point - VHP.
Even if you disagree with that - this is real!
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Do you agree that it means that stars do not orbit exactly at the expected orbital cycle, but they move in a wobbly way around this expected orbital cycle?
A 'wobbly' way is the expected cycle.&