[Dave Lev (OP)]

Further our discussion about new matter creation at the excretion disc of the Milky way:
Please see: https://www.thenakedscientists.com/forum/index.php?topic=75261.40

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?

[Ophiolite]

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.

[Dave Lev (OP)]

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.

I don't see any contradiction between this formula to your explanations.
Do you agree with that?

[Halc]

I don't see any contradiction between this formula to your explanations.
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.

[mad aetherist]

Further our discussion about new matter creation at the excretion disc of the Milky way:
Please see: https://www.thenakedscientists.com/forum/index.php?topic=75261.40

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.

[mad aetherist]

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.

This article shows the basic concepts of the argument: http://www.flight-light-and-spin.com/proof/instant-gravity.htm

Yes u must get an outwards spiral not inwards.
That link is interesting.
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.

[Dave Lev (OP)]

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?

[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.

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.

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.

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

More like a few meters.

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.

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.

 

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.

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.

[Dave Lev (OP)]

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?

[Dave Lev (OP)]

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?

[mad aetherist]

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).

[guest39538]

Electric Universe ,  that  simple ,  N is  attracted to N , I am right and science lies .

[Dave Lev (OP)]

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?

[Halc]

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.

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.

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.

(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.

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. 

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.

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.

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.

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.

[Dave Lev (OP)]

Thanks Halc

1. Centrifugal force

Quote

Quote
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).

Quote

Quote
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?
1. Increases radius - R
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?

[Halc]

Thanks Halc

1. Centrifugal force

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

[Dave Lev (OP)]

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?

[Halc]

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.

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.

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.

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.

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.

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.

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).

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.

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.

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.

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.

[Dave Lev (OP)]

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"?

[Dave Lev (OP)]

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?