[Dave Lev (OP)]

I'm putting a hold on all of the other matters for the moment and focusing on conservation of energy because that is the crux of the problem right now. I'm not moving on until that issue is solved first.

Yes, fully agree.

If new energy could be created from a gravitational field, that would violate conservation of energy because the energy is no longer constant.

There is no violation!
Tidal force creates new energy out of gravitational field as it has no negative impact on the gravity force.

Tidal forces transform existing orbital or rotational energy into heat energy.

This is your biggest mistake.
The formula of gravity is:
F=G m1 x m2 / r^2
This formula is correct for each and every particle in an orbital object.
The formula is fixed under any tidal force.
Hence, Tidal does not change the gravity forces and therefore tidal energy is coming for free.
If you wish to believe that tidal energy changes the gravity forces, than please offer your updated Newton formula for gravity under Tidal forces.

[Kryptid]

The formula is fixed under any tidal forces.

The formula you provided measures force for a fixed distance. Almost all natural orbits are eccentric. They change their distances over time. This change in force is part of what causes the generation of heat. If two orbiting objects are at a constant distance from each other and tidally-locked to each other, there is no change in force over time and as such there is no generation of heat.

tidal energy is coming for free.

Did you even read this?

conservation of energy Physics. a fundamental law of physics and chemistry stating that the total energy of an isolated system is constant despite internal changes. It is most commonly expressed as “energy can neither be created nor destroyed”, and is the basis of the first law of thermodynamics.

[Dave Lev (OP)]

The formula you provided measures force for a fixed distance. Almost all natural orbits are eccentric. They change their distances over time. This change in force is part of what causes the generation of heat. If two orbiting objects are at a constant distance from each other and tidally-locked to each other, there is no change in force over time and as such there is no generation of heat.

You only offer words over words.
Where is the formula that proves your understanding that tidal decreases the gravity force???

Let me offer an exaple why you are totally wrong:
https://en.wikipedia.org/wiki/Two-body_problem#/media/File:Orbit5.gif
"Two bodies with similar mass orbiting a common barycenter external to both bodies, with elliptic orbits—typical of binary stars"
So we see clearly their common barycenter (the sign + in red)
Now, from any direction that we will look at this system, the center of mass will always be the common barycenter.
The same issue works for tidal.
As long as the common barycenter of an object under Tidal force is fixed, than there is no negative impact on the gravity force.
So, please, if you think that tidal can change the gravity force or the common barycenter - than please show the formula for that!!!

[Kryptid]

Where is the formula that proves your understanding that tidal decreases the gravity force???

I never said that it did. What decreases is the rotational or orbital energy possessed by the planet-satellite system. Gravity is constant with distance (as indicated by the "r" in the equation you had in your earlier post). Changing distance (a change in "r") changes the gravitational force. That is what happens in an eccentric orbit: the distance changes over time. That change in gravitational force over time results in a change in tidal force over time. That is what generates the heat.

Force is "free", but energy isn't. Force is not energy. Force cannot be turned into energy. Force only allows one form of energy to be turned into another.

As long as the common barycenter of an object under Tidal force is fixed, than there is no negative impact on the gravity force.

Correct, but under those circumstances, we are no longer talking about an eccentric orbit. So orbital eccentricity is not generating any heating effects here. If the planet is rotating, then there is a change in distance (and therefore a change in force) between different parts of the planet and the satellite. As a beach on Earth moves closer to the Moon due to the Earth's spin, the gravitational force becomes stronger and the sea level rises. If the Earth and Moon were tidally-locked (no longer rotating relative to each other) and it their orbits had an eccentricity of zero (the distance between them is constant), then there would no longer be any form of tidal heating at all.

So, please, if you think that tidal can change the gravity force or the common barycenter - than please show the formula for that!!!

You have it backwards. It is a changing gravitational field strength or direction over time that creates tides. The distance of the Moon from some beach on Earth, for example, is changing over time. Since gravitational field strength is directly tied to distance, that causes the sea level at the beach to change in accordance with the changing distance (tides).

Tidal force creates new energy

ScienceDirect disagrees: https://www.sciencedirect.com/topics/engineering/conservation-of-energy

energy can neither be created nor destroyed

[Dave Lev (OP)]

If the planet is rotating, then there is a change in distance (and therefore a change in force) between different parts of the planet and the satellite

Dear Kryptid
Let's agree on the following:
A rotating planet by itself, has no effect on Newton gravity formula!
However, you claim that if the planet is rotating, then there is a change in distance.
A change in distance means a change in the center of mass.
So, would you kindly show the formula for that?
Is it one more wishful thinking?
If a planet is rotating, how it could affect its center of mass?
Did Newton mention that a rotating object changes the location of its center of mass or distance?
Actually this statement contradicts Newton Shell theorem:
https://en.wikipedia.org/wiki/Shell_theorem
"Isaac Newton proved the shell theorem and stated that:
1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre."
Therefore, it can rotate as fast as you want.
As long as all of it is a spherically symmetric body there is no change in its center of mass and therefore there is no change in gravity force.
It is also stated:
"2. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell."
Therefore, any object inside the shell can move at any direction - As Long as the whole body is still spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside,
Hence, again, there is no change in the center of mass for a rotating object and no net gravitational force is exerted by the shell on any object inside!
If you still believe in that unrealistic Idea of rotating body that can change the center of mass, than this time you must show the formula which links between the rotating activity to distance?

There is very much an impact on the gravity force as the radius changes periodically, and the tidal forces acting on the moon thus vary over time, and heat it, despite it being tide-locked.

Now, let's assume that instead of a spherically symmetric shell (hollow ball), we have a spherically symmetric shell (in American football shape, or tidal shape) that always point to the other body.
Do you see any change in the center of mass?
If there is no change in the center of mass, there is no change in the distance and no change in gravity force.

So, would you kindly once and for all offer the formula that supports your idea (and in the same time contradicts Newton law).

[Dave Lev (OP)]

A football shape is not spherically symmetric.

That is correct.
However, the extra symetrical matter infront and at the back don't really change the center of mass point.
Therefore, even if the tidal shape is not a spherically symmetric, as long as it face directly the other object - there is no change in the center of mass.

You haven't described its motion

If you mean the motion inside the body - than Newton didn't describe any motion.
He specifically claims that "no net gravitational force is exerted by the shell on any object inside".
Therefore, any object inside the shell can set any sort of motion without getting any net gravity force by the shell.
Therefore, its motion is none relevant to the gravity force.

If you mean the motions between the Earth/Moon:

If it represents a hollow moon that is more or less in a similar orbit about our Earth in an isolated 2-body system, they yes, it's center of mass is continuously changing since Earth exerts force on it, and force on a mass results in its acceleration, which moves its center of mass.  The moon's center of mass orbits Earth, and does not stay put.

Than this is correct with or without tidal.
Therefore, tidal has no effect on the gravity force between the two or on their orbital cycle.
So far you couldn't offer any formula that links the tidal impact on gravity force.
You couldn't show the formula that decreases the gravity force due to tidal.

[Kryptid]

It isn't correct.  1, there is no barycenter of an object.  But there is one for a system of two objects like say Earth/moon.

That was a mis-reading on my part. I thought he was talking about a pair of objects the whole time.

2 That barycenter is fixed (a non-accelerating center of gravity of the two body system) despite the fact that the moon's orbit is eccentric. 

I suppose that depends on how you define "fixed". As the Moon moves further away from the Earth in its orbit, the distance to the barycenter from both the Earth's center and the Moon's center must increase very slightly.

Dear Kryptid
Let's agree on the following:
A rotating planet by itself, has no effect on Newton gravity formula!
However, you claim that if the planet is rotating, then there is a change in distance.
A change in distance means a change in the center of mass.
So, would you kindly show the formula for that?
Is it one more wishful thinking?
If a planet is rotating, how it could affect its center of mass?
Did Newton mention that a rotating object changes the location of its center of mass or distance?
Actually this statement contradicts Newton Shell theorem:
https://en.wikipedia.org/wiki/Shell_theorem
"Isaac Newton proved the shell theorem and stated that:
1. A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre."
Therefore, it can rotate as fast as you want.
As long as all of it is a spherically symmetric body there is no change in its center of mass and therefore there is no change in gravity force.
It is also stated:
"2. If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell."
Therefore, any object inside the shell can move at any direction - As Long as the whole body is still spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside,
Hence, again, there is no change in the center of mass for a rotating object and no net gravitational force is exerted by the shell on any object inside!
If you still believe in that unrealistic Idea of rotating body that can change the center of mass, than this time you must show the formula which links between the rotating activity to distance?

You are completely misunderstanding what I am saying. I'm not saying that rotation causes a change in distance between a planet and its satellite as a whole. I said that it causes a change in distance of different parts of the planet. The distance between Orlando, Florida and the Moon's center is changing over time because the Earth is rotating. When Orlando, Florida is directly under the Moon, the Moon is closer to it than when it is on the opposite side of the Earth from the Moon. Thus, the tidal forces acting on Orlando, Florida due to the Moon is changing over time.

But this is all beside the point. The law of conservation of energy won't allow you to create energy. Period: https://www.sciencedirect.com/topics/engineering/conservation-of-energy

conservation of energy Physics. a fundamental law of physics and chemistry stating that the total energy of an isolated system is constant despite internal changes. It is most commonly expressed as “energy can neither be created nor destroyed”, and is the basis of the first law of thermodynamics.

I see that you have conveniently avoided this quote in all of your posts so far.

[Halc]

2 That barycenter is fixed (a non-accelerating center of gravity of the two body system) despite the fact that the moon's orbit is eccentric.

I suppose that depends on how you define "fixed". As the Moon moves further away from the Earth in its orbit, the distance to the barycenter from both the Earth's center and the Moon's center must increase very slightly.

Yea, but that's the Earth and Moon both moving away from the barycenter, not any motion of the barycenter itself.  Acceleration is absolute, and the barycenter does not accelerate due to the varying distance between a pair of objects with an eccentric orbit, or even if they're connected with springs and hinged counterweights and whatnot.  It can't move while it's a contained system.  That's just simple conservation of momentum.

[Kryptid]

Yea, but that's the Earth and Moon both moving away from the barycenter, not any motion of the barycenter itself.  Acceleration is absolute, and the barycenter does not accelerate due to the varying distance between a pair of objects with an eccentric orbit, or even if they're connected with springs and hinged counterweights and whatnot.  It can't move while it's a contained system.  That's just simple conservation of momentum.

Duly noted.

[Dave Lev (OP)]

That was a mis-reading on my part. I thought he was talking about a pair of objects the whole time.

You have understood me correctly and perfectly.
Halc didn't understand my intention.
Those two orbiting objects represent one Body.
In that body there could be much more than just two objects.
For example - Globular cluster:
https://en.wikipedia.org/wiki/Globular_cluster
"A globular cluster is a spherical collection of stars that orbits a galactic core. Globular clusters are very tightly bound by gravity, which gives them their spherical shapes, and relatively high stellar densities toward their centers."
Their galactic core represents their center mass. with regards to all the orbiting objects inside this cluster, this galactic core represents a "fixed point".

I suppose that depends on how you define "fixed". As the Moon moves further away from the Earth in its orbit, the distance to the barycenter from both the Earth's center and the Moon's center must increase very slightly.

So, the "fixed point" is the barycenter or galactic center to the objects in one body.
The Earth for example has finite no. of atoms.
All those atoms set the Earth center of mass. That center of mass is fixed/same with tidal or without tidal
Therfore – Tidal does not change the location of the Eath center of mass.
As gravity is all about center of mass – Tidal has no impact on the gravity forces or orbital energy

But this is all beside the point. The law of conservation of energy won't allow you to create energy. Period: https://www.sciencedirect.com/topics/engineering/conservation-of-energy
"conservation of energy Physics. a fundamental law of physics and chemistry stating that the total energy of an isolated system is constant despite internal changes. It is most commonly expressed as “energy can neither be created nor destroyed”, and is the basis of the first law of thermodynamics."

OK
I understand by now that you can't offer any formula that show that Tidal energy can reduce the gravity force.
However, You hope that the "conservation of energy Physics" will support this kind of wrong understanding.
This is a BIGGEST mistake
Gravity is force.
However, it can be represented by Orbital Energy:
https://en.wikipedia.org/wiki/Specific_orbital_energy
"In the gravitational two-body problem, the specific orbital energy {\displaystyle \epsilon }\epsilon  (or vis-viva energy) of two orbiting bodies is the constant sum of their mutual potential energy ({\displaystyle \epsilon _{p}}\epsilon_p) and their total kinetic energy ({\displaystyle \epsilon _{k}}\epsilon _{k}), divided by the reduced mass. "
So, the total orbital energy is the sum of Kinetic Energy + potential energy.
That Orbital energy is fully under conservation of energy Physics.
However, this total orbital energy isn't affected by tidal energy that represents extra heat in any orbital body.
The extra heat due to tidal is a side effect of the "fixed" orbital energy.
Hence, Tidal energy has no impact on that total orbital energy.
Again - If you still think differently - than please would you kindly offer the matematics how the gravity force (or the total orbital energy) is reduced/increased by the Tidal extra energy/heat.

[Kryptid]

I understand by now that you can't offer any formula that show that Tidal energy can reduce the gravity force.

Because I never said that. Tidal forces don't reduce the force of gravity.

That Orbital energy is fully under conservation of energy Physics.

So is tidal energy. All energy is.

However, this total orbital energy isn't affected by tidal energy that represents extra heat in any orbital body.
The extra heat due to tidal is a side effect of the "fixed" orbital energy.
Hence, Tidal energy has no impact on that total orbital energy.

Heat is a form of energy. As such, it cannot be created. Heat can only come into being if other forms of energy are transformed into heat.

Again - If you still think differently - than please would you kindly offer the matematics how the gravity force (or the total orbital energy) is reduced/increased by the Tidal extra energy/heat.

Why would I offer mathematics for a position that I don't even support? You are building a straw-man argument.

Gravity alone doesn't cause tidal heating. Tidal forces alone don't cause tidal heating either. Your house right now is under tidal forces because the floor is closer to the Earth's core than its roof is. This means that the Earth is pulling very slightly more strongly on the floor than on the roof. That is the definition of a tidal force. Force doesn't generate heat, though. If you want to generate heat using tidal forces, you have to change those tidal forces over time. One way to accomplish that is to change the distance of an object from a source of gravity over time (which is equivalent to an object taking on an eccentric orbit).

To show this, consider the equation to calculate tidal heating: https://en.wikipedia.org/wiki/Tidal_heating

E_tidal = -Im(k₂)(21/2)((R⁵n⁵e²)/G), where

"E_tidal" is the rate of tidal heating in watts
"-Im(k₂)" is the efficiency of body dissipation (a dimensionless parameter)
"R" is the radius of the body in meters
"n" is the body's mean orbital motion in radians per second
"e" is the orbital eccentricity, and
"G" is the gravitational constant

If I calculate this heating rate for something like Io:

E_tidal = -Im(k₂)(21/2)((R⁵n⁵e²)/G)
E_tidal = -(0.005)(10.5)(((1,822,000)⁵)((4.1 x 10^-5)⁵)((0.0041)²))/(6.674 x 10^-11)
E_tidal = (-10.5)((2 x 10³¹)(1.159 x 10^-22)(1.681 x 10^-5))/6.674 x 10^-11)
E_tidal = -6.13 x 10¹⁵ watts

But what happens if we modify the scenario where the tidal forces are constant? That is, what if we take away the orbital eccentricity?

E_tidal = -Im([k₂)(21/2)((R⁵n⁵e²)/G)
E_tidal = -(0.005)(10.5)(((1,822,000)⁵)((4.1 x 10^-5)⁵)((0.0)²))/(6.674 x 10^-11)
E_tidal = (-10.5)((2 x 10³¹)(1.159 x 10^-22)(0))/6.674 x 10^-11)
E_tidal = 0 watts

The power is zero watts. No heat is generated at all.

[Dave Lev (OP)]

To show this, consider the equation to calculate tidal heating: https://en.wikipedia.org/wiki/Tidal_heating

Thanks for the explanation about the Tidal heat calculation.
However, you agree that the Tidal doesn't reduce the force of gravity.

Tidal forces don't reduce the force of gravity.

Without changing the gravity force, there is no change in the total orbital energy.
Therefore, with or without Tidal - the gravity force or the total rotation energy stay the same.
Hence, tidal is only a side effect of the gravity force.
The extra tidal heat has no impact on the gravity force or the total rotation energy.
In order to get better understanding for that let me go back to the - Globular cluster:

https://en.wikipedia.org/wiki/Globular_cluster
"A globular cluster is a spherical collection of stars that orbits a galactic core. Globular clusters are very tightly bound by gravity, which gives them their spherical shapes, and relatively high stellar densities toward their centers."

Let assume that in the whole universe there are only one Globular Cluster that orbits around one main and massive body.
In this example - instead of stars we will set several millions or billions rigid balls. they collide with each other, there is also a friction between the balls, but there is no mass lost in the total globular cluster.
Therefore - the internal movement of the rigid balls and the internal friction/collision increases the internal heat in that globular cluster.
However, as there is no change in the gravity force between the globular cluster and the main body, that extra heat don't affect the total orbital energy.
Therefore, the internal heat in the globular cluster is for free!!!
By doing that new extra heat is created without decreasing other energy source or without any transformation of energy from other party..

So is tidal energy. All energy is.

Tidal energy is totally different from all the other energies.

Heat is a form of energy. As such, it cannot be created. Heat can only come into being if other forms of energy are transformed into heat.

That is correct for any kind of energy except - Tidal heat/energy
I agree, any Heat/energy can only come into being if other forms of energy are transformed into heat.
For example - Electromagnetic energy represents a transformation of energy from one source to other.
However, Tidal heat/energy it totally different as there is no transformation of energy in that activity.
Tidal is a side effect of gravity activity and not part of the total orbital energy.
Tidal don't have any impact on Newton gravity force formula or the total orbital energy.
So, there is no transformation of energy from the total orbital energy into tidal.
Hence, tidal heat/energy is only a side effect of the orbital gravity forces.
Tidal has no impact on the gravity force or the total rotation energy
Tidal heat heat/energy doesn't use any energy from the orbital system.
Therefore, as Tidal doesn't consume energy from the orbital system, and there is no need for energy transformation - Tidal heat/energy comes for free.

[Kryptid]

Without changing the gravity force, there is no change in the total orbital energy.
Therefore, with or without Tidal - the gravity force or the total rotation energy stay the same.

The strength of the gravitational field produced by each body remains the same strength, but the force felt by each body changes over time due to the eccentricity of the orbit. It becomes higher as the satellite approaches the planet and then becomes lower when it moves away in its orbit. This change in force stretches the satellite, then relaxes the stretching, then stretches it again, then relaxes it again. It is that stretching that produces tidal heating. A constant force doesn't do that. I already demonstrated that with the equation.

A good explanation for why tidal forces slow down a satellite (and therefore reduce its total orbital energy) can be found here: https://www.physicsforums.com/threads/ultimate-source-of-energy-on-io.227907/#post-1689833

To sum it up, internal friction due to the stretching of the satellite causes Jupiter to pull on it slightly harder when it tries to move away from the planet. This extra pull has a braking effect, slowing it down and lowering its orbital energy. Thus, less kinetic energy is now available to raise the satellite out and away from the planet on its way to apogee. This loss of kinetic energy means that the satellite cannot travel as far away from the planet as it could before, which means that its apogee has become slightly smaller. Since the difference between apogee and perigee has become smaller, the orbit has become less eccentric. Every time the satellite completes an orbit, the difference between apogee and perigee thus becomes smaller and smaller. This is known as tidal circularization: https://en.wikipedia.org/wiki/Tidal_circularization

As the orbit becomes less and less eccentric, the tidal heating becomes weaker and weaker. Once the orbit becomes completely circular, tidal heating ceases (as per my calculation). So tidal heating is not "for free". It comes at the cost of orbital energy, just like I said before.

This video explains that the energy for tidal heating does come from orbital energy. It states, "orbital energy is dissipated as heat":

Wikipedia says the same thing: https://en.wikipedia.org/wiki/Tidal_heating_of_Io

Orbital and rotational energy are dissipated as heat in the crust of the moon.

This website agrees: https://astrobites.org/2012/08/30/astrophysical-classics-predicting-tidal-heating-of-io/

This is because as tides distort a moon, rock is stretched and compressed. As you can imagine, there’s friction when you stretch rock, and friction releases some of the satellite’s orbital energy as heat.

Therefore - the internal movement of the rigid balls and the internal friction/collision increases the internal heat in that globular cluster.
However, as there is no change in the gravity force between the globular cluster and the main body, that extra heat don't affect the total orbital energy.

The friction between the balls that generates the heat will cause them to slow down. You should be quite aware that friction does exactly that: it slows things down. It is a conversion of the ball's energy of motion into heat energy. Since the balls are slowing down, they have less total orbital energy.

Tidal energy is totally different from all the other energies.

Please back that statement up with an authoritative source. If this was true, it would be extremely noteworthy. Yet I have never seen a single source state that "energy cannot be created or destroyed unless it's tidal energy." Every source I have ever seen simply says "energy cannot be created or destroyed". It doesn't make exceptions for any kind of energy. Energy is energy. No energy is free.

That is correct for any kind of energy except - Tidal heat/energy

So then you finally admit that you think energy can be created and therefore you disagree with the law of conservation of energy. So stop claiming that you agree with it.

[Dave Lev (OP)]

The friction between the balls that generates the heat will cause them to slow down. You should be quite aware that friction does exactly that: it slows things down. It is a conversion of the ball's energy of motion into heat energy. Since the balls are slowing down, they have less total orbital energy.

Thanks
So, your message is very clear:
There is no change in gravity force, however, the extra tidal heating/energy must be deducted from the total orbiting energy.
Let's see if it is freezable by using the example of that globular clusters balls and convert words into real formulas.
In that Globular clusters balls there are billions of balls.
All of them orbit around the center of mass point while they create tidal heating due to friction with each other.

Hence
Their total mass is estimated as m2.
The main body mass is m1.
The radius between the two centers of mass is r

Kinetic Energy = Ek = 1/2 m2 v^2
Potential Energy = Ep = m2 G h = m2 G r
Tidal Energy = E(tidal)
The total rotation energy (without tidal) = Er = Ek +Ep =1/2 m2 v^2 + m2 G r
However if we wish to deduct the E(tidal) from that Er we must set a change in one of the following argument
G, m2, v, or r
G = no option to change it
m2 = it is fixed, so there is no way to change.
Based on gravity formula
v^2 = G (m1 + m2) / r
So, if we wish to change v we must change r
Therefore, in order to decrease Er by the value of Etidal we must change r
However, by changing r we actually change the gravity force, as the formula for gravity is:
Gravity force = F = G m1 m2 /r^2

Therefore, the statement that the Tidal energy decreases the total rotation energy without changing the gravity force is technically not realistic.
Any comment?

[Kryptid]

Tidal Energy = E(tidal)

If you want to calculate the rate of tidal heating, you need to use the formula I posted earlier: E_tidal = -Im(k₂)(21/2)((R⁵n⁵e²)/G)

Therefore, the statement that the Tidal energy decreases the total rotation energy without changing the gravity force is technically not realistic.
Any comment?

I see this as a possible source of confusion. The strength of the overall gravitational field of a planet is unchanging. That is what I mean by the force of gravity being unchanging. However, I need to distinguish this from the force felt by the satellite. In the case of an elliptical orbit, the force experienced by the satellite does change over time because the distance changes over time.

Gravity force = F = G m1 m2 /r^2

The critical part of this formula is the "r". In an elliptical orbit, the value of "r" changes over time. This makes the gravitational force experienced by the satellite change over time as well. If you want specific numbers, I will calculate them for our Moon. When the Moon is at apogee, it is at a maximum distance from the Earth (405,400 kilometers). When it is at perigee, it is at a minimum distance (363,600 kilometers).

For apogee:

F = G((m₁m₂)/r²)
F = (6.674 x 10⁻¹¹)((5.97237 x 10²⁴)(7.342 x 10²²)/((405,400,000)²)
F = 2.92649 x 10³⁷/1.64349 x 10¹⁷
F = 1.78 x 10²⁰ newtons

For perigee:

F = G((m₁m₂)/r²)
F = (6.674 x 10⁻¹¹)((5.97237 x 10²⁴)(7.342 x 10²²)/((363,600,000)²)
F = 2.92649 x 10³⁷/1.322 x 10¹⁷
F = 2.21 x 10²⁰ newtons

The gravitational force experienced by the Moon is 24% stronger when it is at perigee than when it is at apogee. This is because the orbit is eccentric. If the orbit was not eccentric, perigee and apogee would be the same and there would be no change of force over time. No change of force means no tidal heating. Tidal heating causes an orbit to become less eccentric over time because the energy drained from the orbit does not allow the satellite to move quite as far away from the planet as it did on each prior orbit. This slowly decreases the eccentricity and as such slowly decreases the amount of tidal heating over time.

[Dave Lev (OP)]

If the orbit was not eccentric, perigee and apogee would be the same and there would be no change of force over time. No change of force means no tidal heating.

Thanks
So,  in case of a pure cycle orbit (If the orbit not eccentric, perigee and apogee would be the same) there is no way to get tidal forces.
Do you agree with that?
However, I assume that in the nature all the orbit cycles must have some eccentric.
 

The critical part of this formula is the "r". In an elliptical orbit, the value of "r" changes over time. This makes the gravitational force experienced by the satellite change over time as well. If you want specific numbers, I will calculate them for our Moon. When the Moon is at apogee, it is at a maximum distance from the Earth (405,400 kilometers). When it is at perigee, it is at a minimum distance (363,600 kilometers).

For apogee:

F = G((m1m2)/r2)
F = (6.674 x 10−11)((5.97237 x 1024)(7.342 x 1022)/((405,400,000)2)
F = 2.92649 x 1037/1.64349 x 1017
F = 1.78 x 1020 newtons

For perigee:

F = G((m1m2)/r2)
F = (6.674 x 10−11)((5.97237 x 1024)(7.342 x 1022)/((363,600,000)2)
F = 2.92649 x 1037/1.322 x 1017
F = 2.21 x 1020 newtons

The gravitational force experienced by the Moon is 24% stronger when it is at perigee than when it is at apogee. This is because the orbit is eccentric.

This explanation is fully clear to me.
We see the difference between the perigee to apogee
That activity represents only one full cycle.
However, do you agree that in the next cycle - we should get exactly the same output?
So, what can we learn from that?
Can we assume that there is no change in the gravity force between one full orbital cycle to the next full cycle?
I really don't understand how could it be that the total Rotation energy per cycle should be decreased due to tidal energy dissipation, while the gravity force between one full orbital cycle to the next one must stay unchanged.
It is clear to me that if we reduce the total rotation energy per cycle, r must be changed from one full cycle to the next cycle, otherwise, the total rotation energy per cycle is unchanged.
So, would you kindly set the formula for the total rotation energy per one full cycle, and show why this total rotation energy must go down due to tidal energy in the next full orbital cycle, while there is no change in gravity force between one full orbital cycle to the next full cycle.

[Kryptid]

However, do you agree that in the next cycle - we should get exactly the same output?

Nope. The satellite's apogee gets very slightly smaller each time an orbit is completed. This makes the difference between apogee and perigee smaller. That smaller difference results in a smaller difference between the gravitational forces, which consequently makes the tidal heat generated also less. It becomes smaller and smaller with each completed orbit.

I really don't understand how could it be that the total Rotation energy per cycle should be decreased due to tidal energy dissipation, while the gravity force between one full orbital cycle to the next one must stay unchanged.

That's the issue: the gravitational force felt by the satellite isn't the same on each orbit because the distances involved are becoming smaller.

So, would you kindly set the formula for the total rotation energy per one full cycle,

The link I provided earlier gives the formula to calculate the energy loss per orbit and even gives an example calculation: https://www.physicsforums.com/threads/ultimate-source-of-energy-on-io.227907/#post-1689833

and show why this total rotation energy must go down due to tidal energy in the next full orbital cycle

I posted a link to the explanation earlier. Start off with the assumption that Io is composed of a frictionless fluid. It obviously isn't, but this is to demonstrate a point. As Io nears perigee, the increase in tidal forces felt by it causes it to stretch in the direction of Jupiter. Since there is no friction, no heat is generated by the stretching. At the same time, the lack of friction means there is no lag between the time that Io feels Jupiter's increased gravity and its response to that gravity. So as Io nears Jupiter, its tidal bulge becomes larger and larger, reaching a maximum height exactly where Jupiter's gravity is the strongest (perigee). Then, at an equal rate, the tidal bulge becomes smaller and smaller as it moves away from Jupiter.

This perfect symmetry preserves the total orbital energy of Io. That is, Jupiter is pulling equally hard on Io 1 minute before it reaches perigee and 1 minute after it passes perigee. So the acceleration that Io feels as it nears Jupiter is exactly equal to the deceleration it feels as it moves away from Jupiter.

This symmetry is broken once friction is added. Now there is a lag time between when Io feels Jupiter's gravity and when Io responds to that gravity. So once Io reaches perigee, its tidal bulge has not quite reached its peak height because the stretching is being slowed down by internal friction. So Io's tidal bulge actually reaches its peak height after it has already passed perigee. This makes Io slightly more stretched 1 minute after it has passed perigee than it was 1 minute before perigee. This causes it to experience and unequal pull from Jupiter's gravity. Since the bulge is larger as Io leaves than as it approached, Jupiter decelerates Io faster than it accelerated it. This is a braking effect that robs Io of orbital energy and prevents it from traveling all the way to its previous apogee. The internal friction that caused this is simultaneously responsible for generating the tidal heat.

So the mechanism that links orbital energy loss to tidal heating is friction. Get rid of the friction and you get rid of the loss of orbital energy (while also getting rid of tidal heating).

while there is no change in gravity force between one full orbital cycle to the next full cycle.

The gravitational force does change, and that is due to the changing distances.

[Dave Lev (OP)]

The link I provided earlier gives the formula to calculate the energy loss per orbit and even gives an example calculation: https://www.physicsforums.com/threads/ultimate-source-of-energy-on-io.227907/#post-1689833

In this article it is stated:
"When a moon stretches, because of the tidal effect, it's gravitational attraction to the planet will become bigger by
M(moon) * M(planet) * G * d * r / R^4
Where d is the height of the bulge, r is the radius of the moon and R the distance to the planet. (this is actually a simplification, assuming that the mass of the moon is divided in two halves, at a distance R+r+d and R-r-d from the planet. The real effect will be smaller, but proportional to this).
Source https://www.physicsforums.com/threads/ultimate-source-of-energy-on-io.227907/#post-1689833
So, the extra gravitation due to tidal bulge = M(moon) * M(planet) * G * d * r / R^4"

This idea totally contradicts the shell theorem by Newton
https://en.wikipedia.org/wiki/Shell_theorem
"Isaac Newton proved the shell theorem[1] and stated that:
A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre."
Let's look at the following image:
https://en.wikipedia.org/wiki/Shell_theorem#/media/File:Shell-diag-1-anim.gif
m represents the planet, while the shell represents the moon.
In this example we see a pure cycle.
However, we can divide it to two halves and claim that the half at the front has more impact than the one at the back.
Newton have proved (after long calculation) that as long as the half at the front in symmetrical to the half at the back the total mass of that cycle or moon is:
"which suggests that the gravity of a solid spherical ball to an exterior object can be simplified as that of a point mass in the centre of the ball with the same mass."
Now, if we add a symmetrical bulge at the front and at the back of this cycle, don't you agree that as the half at the front is fully symmetrical with the half at the back, than the outcome should be:
That the gravity of a solid spherical ball (with symmetrical bulge) to an exterior object can be simplified as that of a point mass in the centre of the ball with the same mass.?

Actually, we could split this pure cycle two halves and set the total mass of each cycle in one new cycle.
So, we could have two balls connecting one to each while they are directly in line with the planet.
Don't you agree that even in this case, their gravity impact to an exterior object can be simplified as point of mass in the meeting point between those two balls while the total mass is the sum of the two balls?

Therefore, the assumption that a symmetrical bulge (in front and in the back) can change the center of mass of the moon is a fatal mistake.

If you still think differently, would you kindly explain why Newton proved that although the first half is closer to the planet, the total two halves set the center of the mass exactly at the center of the ball?
What is the difference between this scenario to symmetrical bulge that one is added in the front and one in the back?

Gravity is all about mass and center of mass.
The bulge doesn't change the total mass of the moon.
However, if you can prove that the Symmetrical bulge can change the Moon's center of mass than there is a change in the gravity force between the planet and the moon.

[Halc]

The link I provided earlier gives the formula to calculate the energy loss per orbit and even gives an example calculation: https://www.physicsforums.com/threads/ultimate-source-of-energy-on-io.227907/#post-1689833

In this article it is stated:
"When a moon stretches, because of the tidal effect, it's gravitational attraction to the planet will become bigger by
M(moon) * M(planet) * G * d * r / R^4
Where d is the height of the bulge, r is the radius of the moon and R the distance to the planet. (this is actually a simplification, assuming that the mass of the moon is divided in two halves, at a distance R+r+d and R-r-d from the planet. The real effect will be smaller, but proportional to this)."

That's correct.  The F=GMm/r² only works exactly for spherically symmetric bodies.

This idea totally contradicts the shell theorem by Newton

No it doesn't.

"Isaac Newton proved the shell theorem[1] and stated that:
A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre."
Let's look at the following image:
https://en.wikipedia.org/wiki/Shell_theorem#/media/File:Shell-diag-1-anim.gif
m represents the planet, while the shell represents the moon.

It does not represent a tidally distorted moon.  Only a spherical approximation.

However, we can divide it to two halves and claim that the half at the front has more impact than the one at the back.
Newton have proved (after long calculation) that as long as the half at the front in symmetrical to the half at the back the total mass of that cycle or moon is:
"which suggests that the gravity of a solid spherical ball to an exterior object can be simplified as that of a point mass in the centre of the ball with the same mass."
Now, if we add a symmetrical bulge at the front and at the back of this cycle, don't you agree that as the half at the front is fully symmetrical with the half at the back

Per the article you first linked, the two do not cancel out.  The two masses are not spherically distributed.  The part I bolded above is something you made up.  Newton could trivially demonstrate that this isn't the case.

Therefore, the assumption that a symmetrical bulge (in front and in the back) can change the center of mass of the moon is a fatal mistake.

A pair of bulges like that indeed has no effect on the center of mass of the moon.  Nobody claimed otherwise.

Gravity is all about mass and center of mass.

Obviously not, since those bulges did not change the total mass or the center of mass, yet per your first link, the gravitational force is different.  So gravity is not just about center of mass.
Given the right distribution of mass, I can have gravity actually repel a pair of objects (as measured from their respective centers of gravity).

[Kryptid]

Halc has it right. Shell theorem only applies to spherically-symmetrical objects. A tidally-distorted satellite is not spherically symmetrical.