[Dave Lev (OP)]

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

Ok
Let me offer the following:
You agree that it works perfectly in a spherically symmetric cycle.
Now, let's look again on the following diagram:
https://en.wikipedia.org/wiki/Shell_theorem#/media/File:Shell-diag-1.svg
1. If we divide the spherically symmetric cycle into two halves -One in the front and one in the back - Do you agree that the center of mass should stay at the same point?
2. If we push away the two halves from each other on the same radius line from the planet - So one will be pushed back by r Km from the center of mass, while the one in the front will be pushed forward also by r km) - do you agree that the center of mass should also stay at the same point as before?
3. If so, let's use this spited spherically symmetric cycle to represent the symmetric bulge (one in front and one in the back)
So, one spherically symmetric cycle will represent the moon minus the mass of the bulge, while the bulge will be represented by splited spherically symmetric cycle - one in the front and one in the back.
Why the two spherically symmetric cycles can't fully represents the moon + the bulge with the center of mass at the same point as before?

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.

So, if you agree that the Bulge do not change the total mass of the moon and it also do not change the center of mass of the moon, than how can you claim that there is a change in the total gravity force between the planet and the moon?

[Kryptid]

The bulges are not symmetrical: https://physics.mercer.edu/hpage/tidal%20asymmetry/asymmetry.html

[Dave Lev (OP)]

The bulges are not symmetrical: https://physics.mercer.edu/hpage/tidal%20asymmetry/asymmetry.html

This calculation might not be fully correct.
The gravity force is transformed to Taylor's series:
"Next, expand F(r) in a Taylor's series as follows:"
In order to get closer to the reality we must take this Taylor's series to the infinity.
If we don't take it to infinity, than by definition there is a room for error.
Any none perfectly assumption or set up could also cause some minor error.
They even claim:
"The gradient expression for the tidal force is a poor approximation in the Fig. 3 case,"
So, I really don't know what might be the magnitude of the error in this kind of calculation but it is there for sure.
That error could be reflected as the unrealistic change in the center of mass or in the gravity force.
Newton has used different approach.
I prefer to use Newton solution.
If the splited spherically symmetric cycle can represent the symmetric bulge, than there shouldn't be any change in the center of mass point.

[Kryptid]

In order to get closer to the reality we must take this Taylor's series to the infinity.

What is that even supposed to mean?

If we don't take it to infinity, than by definition there is a room for error.

What do you mean by "take it to infinity"? How can you take anything to infinity?

Any none perfectly assumption or set up could also cause some minor error.

Then show where the error in the math is.

Newton has used different approach.
I prefer to use Newton solution.

When did Newton show anything that was counter to what is said on this page? The method mentioned on the page used Newtonian math to arrive at the result:

The conventional treatment of the tidal field of the Moon is an approximation that uses calculus differentials, As the distance of a moon from its parent decreases progressively, the assumptions of this first-order approximation become ever more unacceptable. It is then necessary to apply Newton's universal law of gravity to virtually all components of the body.

If the splited spherically symmetric cycle can represent the symmetric bulge

That's just it, though. The bulges aren't symmetrical. Using the equations on that page, you can find that the difference between nadir and zenith (the far bulge and the near bulge, respectively) is around 5% for the Earth, about 1.4% for the Moon and 1.3% for Io.

[Dave Lev (OP)]

In order to get closer to the reality we must take this Taylor's series to the infinity.
If we don't take it to infinity, than by definition there is a room for error.

What do you mean by "take it to infinity"? How can you take anything to infinity?

If you convert something (gravity) into Taylor's series than you have to use infinity no. in this series in order to get high accuracy.

When did Newton show anything that was counter to what is said on this page? The method mentioned on the page used Newtonian math to arrive at the result:

They are using the same picture as in Newton's shell theorem:
https://en.wikipedia.org/wiki/Shell_theorem#/media/File:Shell-diag-1.svg

That's just it, though. The bulges aren't symmetrical. Using the equations on that page, you can find that the difference between nadir and zenith (the far bulge and the near bulge, respectively) is around 5% for the Earth, about 1.4% for the Moon and 1.3% for Io.

Ok
I was not aware about that difference in size between the front bulge to the rear bulge.
However - Do you agree that if they were exactly at the same size - the center of mass had to be located exactly at the center of the cycle as explained by Newton's shell theorem?
If so, let's look at the Earth.
It's front bulge is bigger by 5% from the rear one.
That for sure will set the center of mass closer to the moon (with regards to spherically symmetric bulge)
That will increase the gravity force between the Moon -Earth comparing to spherically symmetric bulge.
So, now we have stronger gravity force.
However, as there is no change in this ratio (assuming that it is constantly at 5%), there is also no change in the location of center of mass and therefore - there is no change in gravity force.
Hence, as long as the 5% stay, there is no change in the center of mass and no change is the gravity force.
Without a change in those two segments, there will be no change in the total rotation energy.
Therefore, the bulge by itself doesn't change the center of mass point, the gravity force or the total rotation energy. Only a change in the bulge ratio can change those values.
However, they can go up or down.
If the ratio be get to 4% the center of mass will be shifted backwards and therefore les gravity force and less total rotation energy.
If the ratio will get to 6%, the center of mass will be shifted inwards and therefore more gravity force and more total rotation energy.
Therefore - the bulge does not consume energy from the total rotation energy - it just change the location of center of mass and therefore, it sets the amplitude of that gravity force or the total energy based on the bulge rear/front ratio.
.

[Kryptid]

Ok
I was not aware about that difference in size between the front bulge to the rear bulge.
However - Do you agree that if they were exactly at the same size - the center of mass had to be located exactly at the center of the cycle as explained by Newton's shell theorem?
If so, let's look at the Earth.
It's front bulge is bigger by 5% from the rear one.
That for sure will set the center of mass closer to the moon (with regards to spherically symmetric bulge)
That will increase the gravity force between the Moon -Earth comparing to spherically symmetric bulge.
So, now we have stronger gravity force.
However, as there is no change in this ratio (assuming that it is constantly at 5%), there is also no change in the location of center of mass and therefore - there is no change in gravity force.
Hence, as long as the 5% stay, there is no change in the center of mass and no change is the gravity force.
Without a change in those two segments, there will be no change in the total rotation energy.
Therefore, the bulge by itself doesn't change the center of mass point, the gravity force or the total rotation energy. Only a change in the bulge ratio can change those values.
However, they can go up or down.
If the ratio be get to 4% the center of mass will be shifted backwards and therefore les gravity force and less total rotation energy.
If the ratio will get to 6%, the center of mass will be shifted inwards and therefore more gravity force and more total rotation energy.
Therefore - the bulge does not consume energy from the total rotation energy - it just change the location of center of mass and therefore, it sets the amplitude of that gravity force or the total energy based on the bulge rear/front ratio.

Actually, the 5% thing is an average value based on the semi-major axis. The difference in size of the bulges actually does change over time because of the eccentricity of the orbit. The difference will be larger at perigee and smaller at apogee. To do some example calculations for the size difference of the bulges on the Moon:

Moon distance at apogee: 405,400,000 meters

a_z = (2RGm)/(r³)(1+(3R/2r)
a_z = (2(1,737,100)(6.674 x 10⁻¹¹)(7.342 x 10²²)/((1,737,100)³)(1+(3(1,737,100)/2(405,400,000,000)
a_z = (1.702375648936 x 10¹⁹)/(5.241727755811 x 10¹⁸)(1.006427355698)
a_z = (3.24773763)(1.006427355698)
a_z = 3.2686119969964766

a_n = -(2RGm)/(r³)(1-(3R/2r)
a_n = -(2(1,737,100)(6.674 x 10⁻¹¹)(7.342 x 10²²)/((1,737,100)³)(1-(3(1,737,100)/2(405,400,000,000)
a_n = -(1.702375648936 x 10¹⁹)/(5.241727755811 x 10¹⁸)(0.9935726443)
a_n = -(3.24773763)(0.9935726443)
a_n = -3.226863265

The difference between those two values is 1.293786% at apogee.

Moon distance at perigee: 362,600,000 meters

a_z = (2RGm)/(r³)(1+(3R/2r)
a_z = (2(1,737,100)(6.674 x 10⁻¹¹)(7.342 x 10²²)/((1,737,100)³)(1+(3(1,737,100)/2(362,600,000,000)
a_z = (1.702375648936 x 10¹⁹)/(5.241727755811 x 10¹⁸)(1.007186)
a_z = (3.24773763)(1.007186)
a_z = 3.2710759299

a_n = -(2RGm)/(r³)(1-(3R/2r)
a_n = -(2(1,737,100)(6.674 x 10⁻¹¹)(7.342 x 10²²)/((1,737,100)³)(1-(3(1,737,100)/2(362,600,000,000)
a_n = -(1.702375648936 x 10¹⁹)/(5.241727755811 x 10¹⁸)(0.992813982)
a_n = -(3.24773763)(0.992813982)
a_n = -3.22439933

The difference between those two values is 1.4476% at perigee. So the difference does change over time.

Because it does.  Do the math.  It is trivial.  Compute the force on a 2m mass all at one point, and then a pair of 1m masses that are connected but separated by some distance, but have the same center of mass.  The combined force on the two 1m masses will be larger than that of the 2m mass. All you need is F=GMm/r².

As a matter of fact, I'll do exactly that. For the case of a 100 kilogram sphere orbiting 10,000,000 meters above Earth's surface:

F = (GMm)/r^²
F = ((6.674 x 10⁻¹¹(5.97237 x 10²⁴)(100))(10,000,000)²
F = (3.985959738 x 10¹⁶)/10¹⁴
F = 398.6 newtons

Now I'll consider a scenario where there are two spheres, each with a mass of 50 kilograms. One is orbiting at 11,000,000 meters and the other is orbiting at 9,000,000 meters:

F = (GMm)/r^²
F = ((6.674 x 10⁻¹¹)(5.97237 x 10²⁴)(50))(11,000,000)²
F = (1.992979869 x 10¹⁶)/(1.21 x 10¹⁴)
F = 164.7 newtons

F = (GMm)/r^²
F = ((6.674 x 10⁻¹¹)(5.97237 x 10²⁴)(50))(9,000,000)²
F = (1.992979869 x 10¹⁶)/(8.1 x 10¹³)
F = 246 newtons

Add those two forces together and you get a total force of 410.7 newtons, which is larger than the 398;6 newtons of the single sphere scenario. So asymmetrical tidal bulges aren't even necessary to get an increase in total force.

[Dave Lev (OP)]

an = -(2RGm)/(r3)(1-(3R/2r)

As I have already explained this formula is INCORRECT.
Gravity force should only be represented by the following formula:

F = (GMm)/r²

Gravity is all about mass and center of mass.
The tidal bulge only changes the location of the center of mass.
As the front bulge is bigger by 5% from the rear one - than it should shift the center of mass inwards.
Let's look at our moon.
Its face is locked with the Earth.
What is the reason for that?
I think that the answer is as follow:
The Moon's center of mass had shifted inwards (to the Earth) due to Tidal.
Therefore, the moon had lost its spherical Symmetrical shape.
It is similar to an extra weight in a cube.
That extra weight should force the cube to fall always at the same side - as the center of its mass had shifted to that side in the cube.
In the same token; the tidal bulge shifts the location of the moon's center of mass and now it can't continue to rotate.
We can also think about a Gyro or wheel.
The rotation momentum will keep the spin (assuming that there is no friction).
However, if we will set a bulge in that wheel - that is always pointed to one direction, that bulge should decrease the rotation momentum over time
Therefore, the tidal bulge acts as some sort of a friction in the rotation Momentum.
So, I agree with your idea that the tidal energy should be taken from something, but it has to be taken from the rotation momentum and not from the total orbital energy.
There is another issue with gravity.
Gravity works locally.
In other words - The impact of local gravity force is much higher than the impact of far away gravity.
The moon prefers to orbit around the Earth, even as the gravity force from the sun is stronger by more than twice than the gravity from the earth.
So we are lucky that we orbit at the first gravity layer around the moon.
As the moon's tidal energy dispassion on the Earth is relatively quite low (the mass of the Earth is much higher than the mass of the moon), the earth is still keeping its rotation momentum.
However, As the Earth's tidal energy dispassion on the moon is relatively quite high (again - the mass of the Earth is much higher than the mass of the moon), the moon has totally lost its rotation momentum.
Other planets without moons - were not so lucky as the earth. They have lost their rotation momentum as their first gravity layer is based on the Sun. As the Sun mass is much bigger than their own mass, they have totally lost all their rotation momentum.
So, Tidal energy works on the Rotation momentum/energy and not on Total gravity energy!!!

However, I have a question with regards to none perfect orbit (eccentricity less than 1).
You claim that:

The difference in size of the bulges actually does change over time because of the eccentricity of the orbit. The difference will be larger at perigee and smaller at apogee.

Let me focus on:
"The difference will be larger at perigee and smaller at apogee" - "Over time"
Do you agree that as there are changes in the gravity force between perigee to apogee, with or without tidal bulge the gravity should be go down over time?
If you agree with that - than you have solved one more enigma that I'm trying to solve.

[Kryptid]

As I have already explained this formula is INCORRECT.

Your "explanation" is what was incorrect.

Gravity force should only be represented by the following formula:

Those two formulas are telling us different things. The formula I posted tells what the force at the tidal bulges is, whereas the one you posted tells what the overall force is between two spherical objects. Each formula is giving information that the other does not provide.

But even if you don't like the formula I provided, the one you posted still allows for an increase in net gravitational force without a change in the center of gravity. I posted such an example calculation at the end of post #862. That increase in force, caused by Io's distorted shape as it moves away from Jupiter, provides a braking effect that lowers its total orbital energy.

but it has to be taken from the rotation momentum and not from the total orbital energy.

It can be taken from rotation as well, but it can also definitely be taken from orbital energy. I already explained in a prior post how a braking effect from Jupiter lowers the total orbital energy of Io.

So, Tidal energy works on the Rotation momentum/energy and not on Total gravity energy!!!

It's both, actually.

So, I agree with your idea that the tidal energy should be taken from something

Then we can end this thread here and now, since that was the point all along. You can't make an infinite energy factory using tidal forces.

Do you agree that as there are changes in the gravity force between perigee to apogee, with or without tidal bulge the gravity should be go down over time?

No, the force of gravity as felt by a satellite should actually go up over time because the loss of eccentricity will bring it closer to the planet.

[Halc]

but it has to be taken from the rotation momentum and not from the total orbital energy.

It can be taken from rotation as well, but it can also definitely be taken from orbital energy. I already explained in a prior post how a braking effect from Jupiter lowers the total orbital energy of Io.

Rotation momentum, if transferred to the satellite, increases its total orbital angular momentum, and thus energy.
This is the case with Io, and our moon: the orbital radius is increasing as the rotational momentum of Jupiter is transferred to the orbital momentum of Io, an angular acceleration effect, not a braking effect.
This is because Jupiter rotates about 4x the speed of Io's orbit. Phobos is losing altitude because its orbit has a shorter period than the rotation rate of Mars.

This was discussed heavily in the early posts of this thread, and Dave was in denial of all of it, and now I see he's pushing different (but still wrong) ideas.

[Kryptid]

Rotation momentum, if transferred to the satellite, increases its total orbital angular momentum, and thus energy.
This is the case with Io, and our moon: the orbital radius is increasing as the rotational momentum of Jupiter is transferred to the orbital momentum of Io, an angular acceleration effect, not a braking effect.

That is technically true, but both of those effects need to be taken into account. If Jupiter is transferring rotational energy to Io faster than it is losing orbital energy due to a braking effect, then it is true that its semi-major axis will increase over time. However, The orbital eccentricity should still decrease over time anyway, thus lowering the rate of tidal heating.

[Dave Lev (OP)]

As I have already explained this formula is INCORRECT.

Your "explanation" is what was incorrect.

My explanation is fully correct.
It was a severe mistake to use Taylor_series
https://en.wikipedia.org/wiki/Taylor_series
"In mathematics, a Taylor series is a representation of a function as an infinite sum of terms that are calculated from the values of the function's derivatives at a single point"
It is some sort of mathematical fiction.
We take something - gravity force, convert it to infinite something by Taylor series and than use finit something just in order to prove other something - Tidal negative impact on total rotation energy.

But even if you don't like the formula I provided,

It's not an issue that I don't like it.
It is just a totally incorrect formula.
We shouldn't use it  - Never and ever.

I posted such an example calculation at the end of post #862. That increase in force

As a matter of fact, I'll do exactly that. For the case of a 100 kilogram sphere orbiting 10,000,000 meters above Earth's surface:
F = (GMm)/r²
F = ((6.674 x 10−11(5.97237 x 1024)(100))(10,000,000)2
F = (3.985959738 x 1016)/1014
F = 398.6 newtons

Now I'll consider a scenario where there are two spheres, each with a mass of 50 kilograms. One is orbiting at 11,000,000 meters and the other is orbiting at 9,000,000 meters:

F = (GMm)/r²
F = ((6.674 x 10−11)(5.97237 x 1024)(50))(11,000,000)2
F = (1.992979869 x 1016)/(1.21 x 1014)
F = 164.7 newtons

F = (GMm)/r²
F = ((6.674 x 10−11)(5.97237 x 1024)(50))(9,000,000)2
F = (1.992979869 x 1016)/(8.1 x 1013)
F = 246 newtons

Add those two forces together and you get a total force of 410.7 newtons, which is larger than the 398;6 newtons of the single sphere scenario. So asymmetrical tidal bulges aren't even necessary to get an increase in total force.

You have a severe mistake in this calculation!
As I have stated -

Gravity is all about mass and center of mass.

So, it was a mistake to set two diffrent calculations - One for the object at 11,000,000 meters and other one for the object at 9,000,000 meters.
You had to find the center of mass of those two objects and than find the gravity force to the main body.
As an example -
Look at the Moon/Earth/Sun system.
We do not set a separate gravity force calculation for the Moon/Sun and Earth/Sun
We set a calculation for the Earth/moon center of mass and just than set the gravity calculation for that center of mass while it orbits around the Sun.

quote]but it has to be taken from the rotation momentum and not from the total orbital energy./quote]
It can be taken from rotation as well, but it can also definitely be taken from orbital energy.

No.
It can be taken ONLY from the rotation energy.
It is a severe mistake to assume that it take anything from the orbital energy!

So, I agree with your idea that the tidal energy should be taken from something

Then we can end this thread here and now, since that was the point all along. You can't make an infinite energy factory using tidal force

Yes, I was expecting for this answer from you.  I thought about it before.
However, I'm going with my discovery even if it contradicts key idea in my theory.
That shows you that the true is more important for me than to prove a key section in my theory.
In any case, I still don't think that it contradicts the theory.
The main reason for that is that we don't know how the BH or the SMBH really works from inside.
I must say that I still need to think about it.
So, let's set the activity and verify if we can overcome that milestone:

Gravity from nearby objects set tidal forces on the SMBH.
Those Tidal forces increase the internal heat energy.
However, they also decrease the rotation momentum of the SMBH.
So, how can we overcome that issue?
My idea is based on the following explanation from Halc with minor adaptation:

the rotational momentum of Jupiter is transferred to the orbital momentum of Io

The Accretion disc orbits at 0.3c. That must have an impact of the SMBH orbital rotation momentum.
Actually, at the moment of creation, the pair particles are moving at almost the speed of light.
We know that as they get to the outer side of the accretion disc, their orbital velocity is decreasing to 0.3c
So, we can claim that some of the decreasing orbital energy is transformed back to the SMBH and maintain its orbital momentum.
You might claim that somehow new energy must be added in order to maintain this activity. Otherwise, it won't work.
My answer for that will be - "Gravity force".
We have already agreed that the new pair particles are created due to the Magnetic force + the gravity force.
However, some of the energy for that creation is taking also from the gravity force.
So, the SMBH's gravity force contributes some new energy to our Universe.
In any case, new energy must be created by the SMBH – For sure
That SMBH's extra new energy is used to create the mass in the accretion disc and then the SMBH is using some of this new mass to form stars planets and moons.

[Kryptid]

It is some sort of mathematical fiction.

So where in the Wikipedia article you cited does it say that it's "fiction"?

We take something - gravity force, convert it to infinite something by Taylor series and than use finit something just in order to prove other something - Tidal negative impact on total rotation energy.

Then you must think calculus itself is wrong, since it also involves infinite sums.

It's not an issue that I don't like it.
It is just a totally incorrect formula.
We shouldn't use it  - Never and ever.

Why do I get the feeling that your understanding of mathematics doesn't go beyond simple algebra?

So, it was a mistake to set two diffrent calculations - One for the object at 11,000,000 meters and other one for the object at 9,000,000 meters.
You had to find the center of mass of those two objects and than find the gravity force to the main body.

Then let's start off by considering each sphere in isolation. Now we have two separate objects that are not connected to each other. The lower sphere is at an altitude of 9,000,000 meters and has a mass of 50 kilograms. If I did the math correctly, then Newton's equation predicts that the Earth will attract the sphere with a force of 246 newtons.

The higher sphere is at an altitude of 11,000,000 meters and has a mass of 50 kilograms. If I did the math correctly, then Newton's equation predicts that the Earth will attract the sphere with a force of 164.7 newtons. The sum total of the forces acting on both spheres is therefore 410.7 newtons.

Now we imagine taking a weightless (or sufficiently light-weight) string and connecting the two spheres. What was once two objects has now become a single object with a mass equal to the single, 100 kilogram sphere with a center of gravity located at the same altitude (10,000,000 meters). If your reasoning is correct, this now-singular object is suddenly and magically feeling 12.1 newtons of force less than it did when the objects were separate (because you claim that these two tethered spheres should feel the same force as a single sphere of identical mass and center of mass). How can the mere addition of a string cause the Earth to pull on the spheres less than it did before?

The answer is that it can't: strings don't have magical antigravity properties. We can therefore safely conclude that your reasoning is incorrect.

Tidally-distorted objects don't satisfy shell theorem. Your equation is only valid for points and spheres. The Wikipedia article on shell theorem even shows that different equations are used for rings and disks, despite the fact that they can just easily have the same center of mass and total mass as a sphere, Different shapes require different equations: https://en.wikipedia.org/wiki/Shell_theorem#Derivation_of_gravitational_field_outside_of_a_solid_sphere

Nothing on a tidally-distorted object is sphere-shaped or even hemisphere-shaped. The whole object is shaped like an ellipsoid, basically a chicken egg. So the standard gravitational equation is insufficient to accurately calculate gravitational forces acting on it.

That shows you that the true is more important for me than to prove a key section in my theory.

Then stop trying to violate the law of conservation of energy.

So, how can we overcome that issue?

You can't. The law of conservation of energy won't let you.

So, we can claim that some of the decreasing orbital energy is transformed back to the SMBH and maintain its orbital momentum.

Do you seriously not know how to do simple addition and subtraction? If an energy value of "1" was taken from the black hole to create the particles, then those particles can only give at most an energy unit of "1" back to the black hole. If it gives less than that back (which it would have to in order to continue existing), then the energy received back by the black hole must be less than it expended creating the particles. You're not going to cheat the law of conservation of energy.

My answer for that will be - "Gravity force".

Force is not energy.

However, some of the energy for that creation is taking also from the gravity force.

No, it isn't. Gravity does not create net energy.

So, the SMBH's gravity force contributes some new energy to our Universe.

No, it doesn't.

In any case, new energy must be created by the SMBH – For sure

Not if the law of conservation of energy has anything to say about it.

That SMBH's extra new energy is used to create the mass in the accretion disc and then the SMBH is using some of this new mass to form stars planets and moons.

Except that it doesn't. Again, conservation of energy...

[Dave Lev (OP)]

Then let's start off by considering each sphere in isolation. Now we have two separate objects that are not connected to each other. The lower sphere is at an altitude of 9,000,000 meters and has a mass of 50 kilograms. If I did the math correctly, then Newton's equation predicts that the Earth will attract the sphere with a force of 246 newtons.

The higher sphere is at an altitude of 11,000,000 meters and has a mass of 50 kilograms. If I did the math correctly, then Newton's equation predicts that the Earth will attract the sphere with a force of 164.7 newtons. The sum total of the forces acting on both spheres is therefore 410.7 newtons.

Now we imagine taking a weightless (or sufficiently light-weight) string and connecting the two spheres. What was once two objects has now become a single object with a mass equal to the single, 100 kilogram sphere with a center of gravity located at the same altitude (10,000,000 meters). If your reasoning is correct, this now-singular object is suddenly and magically feeling 12.1 newtons of force less than it did when the objects were separate (because you claim that these two tethered spheres should feel the same force as a single sphere of identical mass and center of mass). How can the mere addition of a string cause the Earth to pull on the spheres less than it did before?

The answer is that it can't: strings don't have magical antigravity properties. We can therefore safely conclude that your reasoning is incorrect.

Thanks for that explanation.
Your calculation is correct by 100%.
In order to prove it, please use the Moon/Earth/Sun system.
Please calculate the gravity force between the Moon/Sun and Earth/Sun systems.
Please use your following idea to set one common center of mass to the Earth moon system:
"Now we imagine taking a weightless (or sufficiently light-weight) string and connecting the two spheres."
Make the gravity forces calculations for the two scenarios. I promise you that you would be surprised to find that the results are almost identical to the systems that you have offered!!!
Therefore, your final outcome is incorrect.

We take something - gravity force, convert it to infinite something by Taylor series and then use finite something just in order to prove other something - Tidal negative impact on total rotation energy.

Then you must think calculus itself is wrong, since it also involves infinite sums.

Well, the issue is very simple.
Taylor series is an excellent tool to get close to the solution.
Theoretically, we could take a square wave and convert it to infinite no of sign waves.
However, as we can't use in reality infinite no of waves, somehow there must be a small error in our calculation.
Let's assume that this Taylor series represents the reality by 99%.
That by itself is an excellent estimation. However, there is still a delta or error of 1%.
In that formula, our scientists actually are focusing on that delta.
So, they take the delta and set a new formula for gravity which shows the connection between the tidal to the delta.
This was their severe mistake.

However, some of the energy for that creation is taking also from the gravity force.

No, it isn't. Gravity does not create net energy.

Well, I thought about it in the last few hours and I have found a solution.
 I Fully agree that the new created pair particles energy must be taken from the SMBH's energy by magnetic field
All the 100% energy!.
That is clear to all of us.
However, Gravity force still set its contribution.
The SMBH's mighty gravity force accelerates those new pair creation into the speed of light.
So, that acceleration comes for free due to the SMBH's mighty gravity force.
That acceleration sets the inwards particles rings in the accretion disc at an orbital velocity of almost the speed of light.
That ultra high orbital velocity of the particles/plasma in the accretion disc is transformed back to the SMBH's and increases its internal rotation.
That activity compensates the degradation of its rotation due to the tidal forces.
So, now we have found an excellent solution for the new energy cycle as follow:
Gravity set the Tidal force in the SMBH.
The Tidal force increases the Energy in the SMBH, and decreases its rotation velocity.
Some of that Tidal heat/energy is transformed by the magnetic field to create new pair particles.

However, each new particle get a present from the SMBH - an orbital velocity at the speed of light.
All the antiparticles fall into the SMBH, increasing its mass and its orbital velocity.
The other particles are drifted outwards into the accretion disc.
Their ultra high velocity in the accretion disc - is also transformed back into the SMBH and sets some more compensation in its rotation velocity.

In other words - Gravity only contributes rotation velocity - But that is good enough to create new energy/mass in our Universe.
 

[Kryptid]

Make the gravity forces calculations for the two scenarios. I promise you that you would be surprised to find that the results are almost identical to the systems that you have offered!!!

You realize that would make me right, not wrong, don't you?

Attraction between the Earth and Sun:

F = (GMm)/r²
F = ((6.674 x 10⁻¹¹(5.97237 x 10²⁴)(1.9885 x 10³⁰))(149,598,023,000)²)
F = (7.926080939013 x 10⁴⁴)/(2.2379568485508529 x 10²²)
F = 3.5418308070902173002872574484922 x 10²² newtons

Attraction between the Moon and Sun:

F = (GMm)/r²
F = ((6.674 x 10⁻¹¹(7.342 x 10²²)(1.9885 x 10³⁰))(149,982,422,000)²)
F = (9.7437510158 x 10⁴²)/(2.2494726908986084 x 10²²
F = 4.3315711523075275820924842308266 x 10²⁰

Sum of the two forces: 3.5851465186132925761081822908005 x 10²² newtons

Attraction between the Sun and a hypothetical Earth-Moon combination object:

F = (GMm)/r²
F = ((6.674 x 10⁻¹¹(6.04579 x 10²⁴)(1.9885 x 10³⁰))(149,602,691,137)²)
F = (8.023518449171 x 10⁴⁴)/(2.2380965195362677778850495607477 x 10²²)
F = 3.5849742757445814209639045480079 x 10²² newtons.

Those two values for force are different from each other by about 1.722 x 10¹⁸ newtons. This, of course, means that I was right and you don't get the right answer when you try to calculate the gravitational force of a non-spherical object with an equation that was made for spherical and point-like objects.

Taylor series is an excellent tool to get close to the solution.

Then what was that nonsense you were saying when you called it "mathematical fiction"?

Let's assume that this Taylor series represents the reality by 99%.
That by itself is an excellent estimation. However, there is still a delta or error of 1%.
In that formula, our scientists actually are focusing on that delta.
So, they take the delta and set a new formula for gravity which shows the connection between the tidal to the delta.
This was their severe mistake.

If the calculation for zenith was off by about 1%, then the calculation for nadir would also be off by about 1%. So a difference between the forces at nadir and zenith would still show up (especially if the calculated difference was much higher than 1%, such as is the case for the Earth, which has a tidal bulge difference of about 5%).

The SMBH's mighty gravity force accelerates those new pair creation into the speed of light.
So, that acceleration comes for free due to the SMBH's mighty gravity force.

There are two scenarios where you would get gravitational acceleration: (1) where both particles are falling into the black hole, or (2) where the particles are using a kind of gravitational slingshot effect to boost their speed. In the first scenario, obviously the particles aren't leaving the black hole. In the second scenario, the energy gained by the particles from the gravitational slingshot is subtracted from the hole's total energy. So the hole still loses energy and therefore mass.

That ultra high orbital velocity of the particles/plasma in the accretion disc is transformed back to the SMBH's and increases its internal rotation.

If the rotation of the disk is spinning up the black hole, then the rotational energy of the disk must decrease as a result. Energy is being transferred, not created.

The ultra high velocity of the new created particles in the accretion disc - is transformed back into the SMBH and sets the compensation in its rotation velocity.

No. It. Doesn't.

Do the math. The black hole can't get back more energy back than it gave out in the first place. Again, do you know what simple addition and subtraction are?

In other words - Gravity only contributes rotation velocity - But that is good enough to create new energy in our Universe.

No it doesn't. Conservation of energy won't let you. Give it up already. What part of "energy cannot be created or destroyed" do you not understand?

[Dave Lev (OP)]

In other words - Gravity only contributes rotation velocity - But that is good enough to create new energy in our Universe.

No it doesn't. Conservation of energy won't let you. Give it up already. What part of "energy cannot be created or destroyed" do you not understand?

OK
My job is to prove that the SMBH's gravity force contributes more energy than the energy that is needed to create those new particle pairs.

Let's set the calculation and verify the real contribution of the mighty SMBH's Gravity force to the total energy.

Let's start with the Virtual particle pair that orbit at almost the speed of light.
Before the creation they have no mass and therefore they have no Kinetic Energy.
At the moment of the creation, they actually get all their mass energy from the SMBH (by the magnetic field).
However, they also get kinetic energy from the SMBH's gravity force.
In other words-
The SMBH's internal energy is transformed into real particle and antiparticle, while the Kinetic energy for those particles are contributed by the SMBH's gravity force.
So, the calculation should be as follow:
M - Represents the mass of the particle or Antiparticle
Eout - The total energy that the SMBH have lost in order to generate those particle and Antiparticle air
Eout = 2M
Ek - The kinetic energy of each particle at the moment of creation. As their orbital velocity is almost the speed of light then:
Ek= 1/2 M c^2
Ep - The potential energy of a particle at the moment of creation
Er = The Total rotation energy of the particle at the moment of creation:
Er = Ek + Ep
The Ep is transformed into kinetic energy as the antiparticle is falling into the SMBH.
Therefore, if the orbital velocity of the antiparticle at the moment of creation is the speed of light, than as it drifts inwards to the SMBH than theoretically, its orbital velocity should be higher than the speed of light.
However, as it gets below the event horizon, even at that Ultra high velocity, it can't go out any more.
However, at the moment of the collision between the antiparticle to the SMBH, all the total rotation energy is transformed into new Energy in the SMBH.

The SMBH creates two particles (Particle + Antiparticle). However, only the antiparticle is falling in.
So, we need to verify if its total rotation energy is greater that the energy that was needed to create the pair.
Therefore, we need to prove that:
Er is greater than Eout
Er = Ek + Ep
For this simple calculation lets ignore the impact of the Ep
So, let's assume that:
Er = Ek = 1/2 M c^2 
Eout = 2M
Er = n * Eout
n = represents how bigger is Er from Eout
Er = 1/2 M c^2 = n 2M
n = 1/4 c^2
Conclusions:
For any energy that the SMBH contributes to create the Pair particle, it gets back at least 1/4c^2 times the "invested" energy.
This does not include:
1. The extra mass of the in falling antiparticle
2. It's Potential energy
3. The particle that is drifted outwards to the accretion disc
4. The transformation of the energy from the accretion disc back to the SMBH.
Therefore, I have proved that only the Kinetic energy of the in falling antiparticle contributes much more energy that was needed to set the whole creation activity.

You realize that would make me right, not wrong, don't you?

Attraction between the Earth and Sun:

F = (GMm)/r2
F = ((6.674 x 10−11(5.97237 x 1024)(1.9885 x 1030))(149,598,023,000)2)
F = (7.926080939013 x 1044)/(2.2379568485508529 x 1022)
F = 3.5418308070902173002872574484922 x 1022 newtons

Attraction between the Moon and Sun:

F = (GMm)/r2
F = ((6.674 x 10−11(7.342 x 1022)(1.9885 x 1030))(149,982,422,000)2)
F = (9.7437510158 x 1042)/(2.2494726908986084 x 1022
F = 4.3315711523075275820924842308266 x 1020

Sum of the two forces: 3.5851465186132925761081822908005 x 1022 newtons

Attraction between the Sun and a hypothetical Earth-Moon combination object:

F = (GMm)/r2
F = ((6.674 x 10−11(6.04579 x 1024)(1.9885 x 1030))(149,602,691,137)2)
F = (8.023518449171 x 1044)/(2.2380965195362677778850495607477 x 1022)
F = 3.5849742757445814209639045480079 x 1022 newtons.

It is totally a diffrent senario from the one that you have used.
In this case, you have set all the objects at the same radius of 149,602,691,137.
However, in  your first example you have set one object at minimal distance of 9,000,000 m

The lower sphere is at an altitude of 9,000,000 meters and has a mass of 50 kilograms.

While the other at 11,000,000 m:

The higher sphere is at an altitude of 11,000,000 meters and has a mass of 50 kilograms.

After merging the two objects, their center of mass had been set at 10,000,000 m:

Now we imagine taking a weightless (or sufficiently light-weight) string and connecting the two spheres. What was once two objects has now become a single object with a mass equal to the single, 100 kilogram sphere with a center of gravity located at the same altitude (10,000,000 meters).

I can promise you that if in your example you have used the same radius for all objects - you would probably get fully balanced forces.
Therefore, please use the minimal and maximal distance/radius from the Earth Moon to the Sun (maximal tidal) and then set your recalculation.

Taylor series is an excellent tool to get close to the solution.

Then what was that nonsense you were saying when you called it "mathematical fiction"?

Mathematical fiction is a direct outcome if we use the error/delta in the Taylor series to prove something which is totally based on that error/delta.

[Kryptid]

My job is to prove that the SMBH's gravity force contributes more energy than the energy that is needed to create those new particle pairs.

Gravity doesn't contribute any energy. All it can do is transfer or transform energy that already exists.

However, they also get kinetic energy from the SMBH's gravity force.

Every single bit of that kinetic energy will come from that black hole's rotation, reducing the black hole's rotation (and therefore its total energy) by the same amount.

Therefore, we need to prove that:
Er is greater than Eout

So you're trying to prove one of the laws of physics wrong despite the fact that you have said many times earlier that you accept the laws of physics...

For this simple calculation lets ignore the impact of the Ep

If you do that, you'll get the wrong answer. The gravitational potential energy of a particle near a super massive black hole is nowhere close to zero.

Er = Ek = 1/2 M c^2

This is wrong for two reasons. Firstly, you can't use the Newtonian kinetic energy equation when considering relativistic velocities. Secondly, particles with a non-zero rest mass like electrons and positrons can't move at the speed of light no matter how much kinetic energy you give them. If you are talking about electrons and positrons moving near the speed of light, then you have to use the relativistic kinetic energy equation: E_k = (mc²)/√(1-(v²/c²)), where,

E_k is the energy in joules
m is the mass in kilograms
v is the velocity of the particle in meters per second, and
c is the velocity of light in meters per second

So your math is going to be wrong until you fix this problem.

Eout = 2M

This will have to include the particles' potential energy, which means you can't ignore it in the calculation.

Therefore, I have proved that only the Kinetic energy of the in falling antiparticle contributes much more energy that was needed to set the whole creation activity.

All you have proven is that you don't know how to use the correct equations. Gravitational potential energy contributes to the total energy. All the infalling particle is doing is converting potential energy that it already had into kinetic energy as it falls into the hole. It isn't giving the hole any more energy than it had to begin with. The particles don't get kinetic energy out of nowhere. They are draining it from the black hole's spin. Any kinetic energy gained by the outgoing particle is lost by the black hole. Again, do you not know what "energy cannot be created or destroyed" means?

In this case, you have set all the objects at the same radius of 149,602,691,137.

Look again. I have the Earth set at a distance of 149,598,023,000 meters from the Sun, the Moon is set at a distance of 149,982,422,000 meters from the Sun and the Earth-Moon barycenter at 149,602,691,137 meters from the Sun. All three of those are different numbers.

I can promise you that if in your example you have used the same radius for all objects - you would probably get fully balanced forces.
Therefore, please use the minimal and maximal distance/radius from the Earth Moon to the Sun (maximal tidal) and then set your recalculation.

I already did put the proper distances into the equations. You probably just saw the "149" and assumed that the rest of the number was the same in all three cases, didn't you?

Mathematical fiction is a direct outcome if we use the error/delta in the Taylor series to prove something which is totally based on that error/delta.

When was the Taylor series ever proving anything by using an error? Can you find me a source that states this?

[Dave Lev (OP)]

Gravity doesn't contribute any energy. All it can do is transfer or transform energy that already exists.
Every single bit of that kinetic energy will come from that black hole's rotation, reducing the black hole's rotation (and therefore its total energy) by the same amount.

Do you mean that if the black hole's rotation is zero, than its gravity force should also be zero?
Do you really mean the gravity force is only some sort of energy transformation?
If so, why Newton didn't call it Gravity Field instead of Gravity Force?
You have to agree that once an object have lost its rotation energy, than it also should lose its gravity force.
How could it be?
It is a severe contradiction to the Newton Gravity Law.
Newton didn't set any connection between the gravity force to the rotation energy.
If there is a formula that connects the rotation energy to gravity force - Would you kindly offer it?
We all know that a function of energy is force times distance.
Therefore, Gravity force by itself can generate energy.
That energy is a direct outcome of the force. Therefore, the force is not a transformation. It generates energy based on Newton gravity force/Law.
Let's take the moon as an example.
Its face is looked with the earth.
So, it lost its ability to rotate.
Can we claim that the moon has no ability to converts its gravity force into energy just because it does not rotate any more?
If an object will come closer to the moon, its movement will not be affected by the Moon's gravity force?

[Kryptid]

Do you mean that if the black hole's rotation is zero, than its gravity force should also be zero?

No. The fact of the matter is that you've been arguing that a black hole's rotation is what is creating the particles since at least post #809 and I agreed that such a thing can work and that it was even mentioned in theory a long, long time ago. Rotating holes radiate particles. Tidal forces and gravity are not the source of energy, rotation is.

Do you really mean the gravity force is only some sort of energy transformation?

No, what I'm saying is that energy transformation is all that gravity can do. It can't create energy.

If so, why Newton didn't call it Gravity Field instead of Gravity Force?

Gravitational fields can produce a force. They are not mutually exclusive.

You have to agree that once an object have lost its rotation energy, than it also should lose its gravity force.
How could it be?

It will not lose all of its gravity. The rotation of an object contains kinetic energy. Due to mass-energy equivalence, that energy also has a mass. That resulting mass increases the strength of the gravitational field. This additional gravity will be extremely tiny compared to the gravity produced by the object's rest mass in most cases (like for planets and stars). In the case of a black hole with the maximum possible rotation rate, however, 29% of its mass-energy will be contained in that rotation (which is fairly significant). So if some of the rotation (and therefore energy) of the hole is extracted, the overall energy content (and therefore mass) of the hole is reduced.

It is a severe contradiction to the Newton Gravity Law.
Newton didn't set any connection between the gravity force to the rotation energy.
If there is a formula that connects the rotation energy to gravity force - Would you kindly offer it?

I never said that an object's entire gravitational field is generated by rotation, so this is a straw-man.

We all know that a function of energy is force times distance.
Therefore, Gravity force by itself can generate energy.

No it can't. Energy can't be created or destroyed. All "force times distance" is telling you in a gravitational field is how much of that gravitational potential energy has been transformed into kinetic energy (if the object is falling) or how much kinetic energy is being transformed into gravitational potential energy (if it is being lifted against gravity). The total energy is unchanged.

That energy is a direct outcome of the force.

Nonsense. Force cannot create energy.

Therefore, the force is not a transformation.

Force is not a transformation, but it causes transformations of energy.

It generates energy based on Newton gravity force/Law.

If that was true, then we would have recognized long ago that conservation of energy was wrong. But that's not true. Force doesn't generate energy. It only changes its form.

Can we claim that the moon has no ability to converts its gravity force into energy just because it does not rotate any more?

It never had that ability in the first place. All the Moon's gravitational field can do is transform existing energy from one form to another.

If an object will come closer to the moon, its movement will not be affected by the Moon's gravity force?

It will be affected, but the object's total energy will be unchanged. All that is happening is that some of the object's gravitational potential energy is being converted into kinetic energy.

[Dave Lev (OP)]

Tidal forces and gravity are not the source of energy, rotation is.

Sorry
I see it differently.
Gravity force and tidal creates energy.
Rotation is a direct outcome of gravity force.
Newton has set the gravity force as:
F=GMm/r^2
The Kinetic energy is directly affected by the orbital velocity.
While that orbital velocity is fully affected by the gravity force.
F = M V^2 / r
So, Newton gravity force which had been set ONLY by mass and radius (and the constant G), has a direct impact on the orbital velocity and that orbital velocity sets the Kinetic orbital energy.
Therefore, The kinetic energy is a direct outcome of the gravity force and not the other way.
Same issue with the Potential kinetic
Ep = m G h
So, even the potential energy is a direct outcome of mass and distance (or radius). As the gravity force is a direct outcome of mass and radius - the potential energy is also an outcome of gravity force.

No, what I'm saying is that energy transformation is all that gravity can do. It can't create energy.

I disagree
Gravity force can create new energy

You have to agree that once an object has lost its rotation energy, than it also should lose its gravity force.
How could it be?

It will not lose all of its gravity. The rotation of an object contains kinetic energy. Due to mass-energy equivalence, that energy also has a mass. That resulting mass increases the strength of the gravitational field. This additional gravity will be extremely tiny compared to the gravity produced by the object's rest mass in most cases (like for planets and stars). In the case of a black hole with the maximum possible rotation rate, however, 29% of its mass-energy will be contained in that rotation (which is fairly significant). So if some of the rotation (and therefore energy) of the hole is extracted, the overall energy content (and therefore mass) of the hole is reduced.

How did you get the 29% for the BH?
What should be the gravity lost in the Moon which have already lost its rotate energy?
Please show the formula which links the rotate energy to gravity force.

I never said that an object's entire gravitational field is generated by rotation,

Again - Please show the formula which links between the two.

If that was true, then we would have recognized long ago that conservation of energy was wrong. But that's not true. Force doesn't generate energy. It only changes its form.

It seems to me that you are using the conservation of energy at the wrong place.
I have already discussed about Newton formula that shows that Gravity generates energy.

 

It will be affected, but the object's total energy will be unchanged. All that is happening is that some of the object's gravitational potential energy is being converted into kinetic energy.

If I understand you correctly, you claim that the total rotation energy (Kinetic + potential) is fixed.
So, the gravity just transfers the energy from one to the other.
This is incorrect.
Gravity works locally.
So, if we take an object to the infinity - we can also claim that its potential energy is infinite.
Just based on this idea we can claim that the energy in our Universe is infinity.
In reality - if it is far enough, it will not be affected by the gravity force any more.
So, although the potential energy should be infinite, the gravity force between those two objects at a distance of infinite is virtually zero and therefore there is no meaning for that Potential energy.
On the other way, I could claim that the virtual pair are coming from the infinity.
So, at the moment of creation, their infinite potential energy had been transformed into Kinetic energy.
I assume that if you give the possibility to transform energy between kinetic and potential than you should also agree with that sort of idea.

[Kryptid]

Sorry
I see it differently.

Then you disagree with the laws of physics.

Gravity force and tidal creates energy.

Not according to the law of conservation of energy, it doesn't. Do you think you are some kind of higher authority than the laws of physics themselves?

Rotation is a direct outcome of gravity force.

Gravity isn't needed at all for rotation. Take a CD player, for example.

The Kinetic energy is directly affected by the orbital velocity.

Kinetic energy isn't merely "affected" by velocity, it is what causes the velocity to be what it is. Movement of any kind requires kinetic energy.

So, Newton gravity force which had been set ONLY by mass and radius (and the constant G), has a direct impact on the orbital velocity and that orbital velocity sets the Kinetic orbital energy.

And that kinetic energy had to come from somewhere. In the case of an asteroid captured in orbit around a planet, that kinetic energy came in part from the gravitational potential energy of the asteroid and partly from the kinetic energy that the asteroid already had.

Therefore, The kinetic energy is a direct outcome of the gravity force and not the other way.

You are right in the sense that gravity is converting gravitational potential energy into kinetic energy.

So, even the potential energy is a direct outcome of mass and distance (or radius). As the gravity force is a direct outcome of mass and radius - the potential energy is also an outcome of gravity force.

Yes, but gravity is not a source of unlimited potential energy. The amount of potential energy that an object has in a gravitational field is finite and any attempt to increase that potential energy requires an input of energy from an outside source. Raising an object against gravity, for example, increases its gravitational potential energy. But that rise in potential energy had to come from somewhere. In the case of a rocket, that energy came from the rocket engines.

I disagree
Gravity force can create new energy

Then you disagree with the law of conservation of energy. I'm still waiting for you to cite some authoritative source that agrees with you that gravity does not obey conservation of energy. The only evidence you have provided so far is  your own misunderstanding of how gravity works. That, of course, is not evidence at all.

How did you get the 29% for the BH?

Look at page 29: http://www.phys.unm.edu/~gbtaylor/astr421/lectures/24_A421_BlackHolesnew.pdf

What should be the gravity lost in the Moon which have already lost its rotate energy?

That would depend upon how fast it was rotating in the past.

Please show the formula which links the rotate energy to gravity force.

First, you calculate the mass equivalent of the rotational kinetic energy using E=mc² (or rather, the rearranged version: m = E/c²). Then you put that resulting mass into the standard gravitational attraction equation if what you have is a spherical body (as well as the mass and distance of whatever object is being affected by the field): F = G((m₁m₂)/r²). That will tell you how much gravitational force is contributed by the rotational energy.

Again - Please show the formula which links between the two.

See above.

It seems to me that you are using the conservation of energy at the wrong place.

There is no such thing as "the wrong place" for conservation of energy.

I have already discussed about Newton formula that shows that Gravity generates energy.

If you think that's correct, then give a link to an authoritative source that agrees with you that gravity is an exception to the conservation of energy. If you are correct, then that shouldn't be difficult, given how incredibly important such a discovery would be.

If I understand you correctly, you claim that the total rotation energy (Kinetic + potential) is fixed.

You don't understand correctly. Rotation can slow down (but only if that rotational energy is transferred elsewhere. It can't disappear into nothingness due to conservation of both energy and momentum).

So, if we take an object to the infinity - we can also claim that its potential energy is infinite.

No you can't. The potential energy at infinity is limited. This is because gravity becomes weaker and weaker as you get further away from its source. In calculus, this is called approaching a limit. The fact that gravitational potential energy is finite at infinity is the basis for the concept of the escape velocity. An object traveling beyond escape velocity has more kinetic energy than the same object's gravitational potential energy at infinity, which means that it can keep moving away from a source of gravity forever without ever being slowed down to zero speed.

The energy required to move an object against a gravitational field out to infinity is (GMm)/r, where G is the gravitational constant, M is the mass of the planet, m is the mass of the object and r is the radius of the planet. For the Earth, moving a 1 kilogram mass out to infinity would require ((6.674 x 10⁻¹¹)(5.97237 x 10²⁴)(1))/(6,371,000) = 62,564,114.55 joules of energy. That is very much a finite number: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/gravpe.html

Just based on this idea we can claim that the energy in our Universe is infinity.

No, we can't.

In reality - if it is far enough, it will not be affected by the gravity force any more.

This is also wrong. Gravity is a force of infinite range. It becomes increasingly weak with distance, but the force never falls all the way to zero.

So, although the potential energy should be infinite

It isn't.

the gravity force between those two objects at a distance of infinite is virtually zero

At infinity, the force would be literally zero. But that's only a mathematical consequence, since it's impossible to move any object an infinite distance away from any other object in a finite span of time.

and therefore there is no meaning for that Potential energy.

Since the potential energy actually has a finite value, it still does have meaning.

On the other way, I could claim that the virtual pair are coming from the infinity.

No you can't. That wouldn't make any sense.

So, at the moment of creation, their infinite potential energy had been transformed into Kinetic energy.

They don't have infinite potential energy.

I assume that if you give the possibility to transform energy between kinetic and potential than you should also agree with that sort of idea.

Absolutely not. Potential energy never becomes infinite, no matter what the distance.