[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.

[Halc]

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.

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.

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.

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.

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

Do you agree with that?

No

[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.

[Halc]

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

[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?

[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.

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.

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.

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.

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.

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.

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.

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.

[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 .

[Halc]

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.

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.

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.

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

Fine...

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.

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.

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.

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.

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.

[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

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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?
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.