[Janus]

And in moving out of the gravity well, the kinetic energy is lost to that potential energy.  It slows with a greater negative Δv than the positive Δv from the thrust, just like a roller coaster slows (exchanges kinetic energy for potential) as it moves further from Earth center.

That's an interesting idea.
The formula for Kinetic emery is: Ek = mv^2

okay

The formula for potential energy is: Ep = GMmR^2

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

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

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

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

[Dave Lev (OP)]

Thanks Janus

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

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

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

Thanks Janus

Now it is fully clear to me.

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

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

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

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

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

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

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

Et2 = Et1 + ΔEk
Et2 = Ek1 + ΔEk + Ep1

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

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

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

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

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

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

Hence

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

m(v3)^2/2- m(v1+Δv)^2/2 = GMm/(ΔR)

m(v3)^2/2- m(v1+Δv)^2/2 = GMm/(ΔR)

m(v3- v1- Δv)^2/2 = GMm/(ΔR)

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

Do you agree with that?

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

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

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

[mad aetherist]

Conclusion:
We have actually increased the Ek by ΔEk (due to tidal) and therefore got an increased velocity by Δv.
This increase Δv set an increased radius ΔR.
This change in the radius increased the Ep.
Now we want that this new Ep will help us not just to eliminate that ΔEk which we have added (due to tidal), but also more that that in order to achieve our goal that V3 must be lower than V1.
Actually, lower velocity by itself is not good enough.
Too low is also problematic.
We must get a very specific velocity to meet the new gravity force.
Is it a realistic goal?
Only if we set full calculation we can verify if it works or not.

I havent followed all of the math, but is the new gravity force due to change in R or in G or in M or in m?
If it is due to change in G then i fear that this might introduce some problems. It might depend on how one defines mass. Its complicated.

[Halc]

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

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

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

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

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

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

If you're including tidal thrust, Et is no longer constant.  That thrust is adding total energy to the orbit, taking energy (and more) away from the sun.  Momentum is conserved in that transfer, but plenty of energy is lost to heat.  But total energy of the orbit Et is increasing.

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

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

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

Et2 = Et1 + ΔEk
Et2 = Ek1 + ΔEk + Ep1

This is the new starting point (just before increasing the radius)."

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

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

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

At that moment we can claim that:

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

What do you mean by = constant?  The total stays the same, but both kinetic and potential energy are changing as the R changes, so those terms are not going to be constant.

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

Hence,
Due to the radius increase:

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

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

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

I followed all that.  The last line seems to be a copy of the line above it.

Hence

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

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

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

Do you agree with that?

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

Conclusion:
We have actually increased the Ek by ΔEk (due to tidal)

Due to a momentary increase of v actually.  Tides never do that since thrust is continuous, and Ek never goes up, not even momentarily.  The whole path just forms a spiral pattern.

and therefore got an increased velocity = Δv.
This increased Δv set an increased radius = ΔR.
This increased ΔR increased the Ep by ΔEp.
Now we expect that this ΔEp will help us not just to decrease the Ek by ΔEk (in order to get back to v1), but also more than that in order to achieve our goal that v3 must be lower than v1.
Even in a perfect system, we can't request to get back all the energy that had been invested.

Pretty much you can.  There is no friction except in the transfer of energy from sun to Earth, where much energy is lost to the friction slowing the spin of the sun, and more lost to slowing of spin of Earth.  Both apply thrust similarly.  The moon/Earth system only has significant  friction on the Earth side since the moon is tidal locked.
But in terms of orbits, the friction is negligible.  You can assume there is no loss.

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

Yea, sure.

Even if we assume that it is realistic, how do we know that we get a perfect new velocity?
Too low is also problematic.

It isn't perfect now, so don't expect it to be perfect.  Today, the Earth is moving faster than it should and its orbit (while still moving inward) is accelerating outward.  It should bottom out at max speed in 30 days and start moving out again.  The orbit is hardly circular, but for the sake of our computations, you can assume one.

We must get a very specific velocity to meet the new gravity force (at R2 = R1 +ΔR).

If you want a circular orbit, yes, but you started out circular and gave it a single jolt thrust.  It will indeed be going that perfect velocity at that perfect R2, but it won't be moving tangential at that time, so R is not going to stay at R2.  Such is the non-circular orbit you've put it into.

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

You're not actually computing ballistics here, so your equations should need to bother with the elliptical orbit part.  Yes, R2 can be computed, and the v3 out there should be perfect, and not too difficult to compute.  You don't have a figure for how much energy you intend to add to your system via tidal thrust.  Not sure if you need it.  You just have an arbitrary v2 that is bigger than v1.  That is enough to define the new orbit it seems.  You don't even need to know the mass of Earth since the same orbit change will happen to a small pebble given that same Δv.

[Dave Lev (OP)]

Thanks Halc

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

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

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

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

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

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

[Halc]

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

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

Yes.  The quote you give from the article describes tidal friction quite well.

Is there any possibility to increase the velocity of car by pressing the brakes?

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

Therefore, if we transform some of this energy into heat, than there must be less kinetic energy. Less kinetic energy means less velocity.

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

So, how could it be that a brake/resistance system (like a tidal friction) can increase the orbital velocity instead of decreasing it?

The road is moving faster than the car, so the car is speeding up as it hits the brakes.

[Dave Lev (OP)]

Thanks Halc

It is angular velocity in this case, but yes.  The Earth is always slowing its spin due to tidal friction.  That's why they keep having to add leap-seconds now and then.  The day used to be about 10 hours long. 

That is fully clear to me.
However, that phenomena is due to the impact of the gravity force on the object itself.
Based on Newton's law of universal gravitation:
https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation
Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
So, every particle at the earth attracts every other particle in the Universe due to gravity force.
That is the ultimate answer for the tidal friction on any orbital system.
However, Newton didn't specify any information about the matter in the object or how it behaves.
It can be a gas/water/metal Star/Planet/Moon or even made of rice. In the formula we only look at the mass.
It can spin in ultra velocity or stay locked
Therefore, there is no proof for the following statement by Newton gravity formula:

The spin of Earth is the road, so the friction is the brakes, speeding up the moon.

If we think that the spin of the Earth can speed up the moon, than by definition we must change the Newton's law of universal gravitation so it will also include that impact.
How can we claim that Newton's law of universal gravitation is fully correct if there is no reflection in the formula for the type of the matter or the spin of the objects/particles?
Based on the current Newton's law of universal gravitation, "every particle attracts every other particle in the universe with a force..."
So, any particle on Earth might go/down or spin faster/slower/locked due to the local impact of gravity force (tidal activity).
However, based on Newton's law of universal gravitation that local activity in one object (Earth) can't have in turn any impact on a particle in another object (Moon).
If we believe that the spin of a particle in one object can affect the velocity of a particle at a far end object than:
Why don't we change the Newton's law of universal gravitation formula in order to represent this breakthrough understanding?

[Dave Lev (OP)]

Thanks
Based on Wiki the highest point of the tidal bulge is 54 Cm:
https://en.wikipedia.org/wiki/Tide
"The theoretical amplitude of oceanic tides caused by the Moon is about 54 centimetres (21 in) at the highest point"

The near bulge is ahead of the moon, and thus pulls the moon forward.  The far bulge rotates to behind the moon and pulls it backwards.  The near one is closer, so that forward pull is greater than the backwards pull.  Net effect is gravitational thrust on the moon.

The Earth moon distance is 384,400 km. The radius of the Earth is 6,371 km.
We can try to calculate the impact of each bulge.
In order to do so, we need to estimate the mass of each bulge with reference to the Earth and verify the total impact.
However, do you agree that 6,371 X2 is neglected with regards to 384,400 km?
Do you also agree that the net mass in that 54 centimeters Bulge is also neglected with regards to the total mass of the Earth?
If so, it is quite clear that the impact of the bulges is virtually zero or close to zero.

In any case, it seems to me that the key point in our discussion is the center of mass based on Newton's second law for the description of the motion of extended objects:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"The utility of Newton's second law for the description of the motion of extended objects is the key to its general practical usefulness. The motion of any real object may involve rotations as well as linear motion, but the motion of the center of mass of the object can be described by an application of Newton's second law in the following form:"
In our case,

Tidal forces cause a bulge on the near and far sides of Earth.

If the bulges at the near side and at the far side are similar, do you agree that there is no change in the Earth center of mass location with regards to the moon.
Hence, based on this idea, those bulges shouldn't have any impact on the Moon orbital cycle as they cancel each other.
Never the less, even if we ignor the far end bulge, what might be the impact of that near side bulge to the Earth center of mass?
I would assume that less than Pico cm.
So, I wonder how this Pico cm can set the requested extra thrust on the moon which is needed to push the moon away from Earth?

However, based on your following answer, you don't like the idea of using the Earth as a point-mass.

Exactly.  And the thrust on the moon can be explained if you apply the law that way.  Earth is not a point-mass.  Point-masses cannot be susceptible to tidal friction.

Hence, let's assume that we can find that those bulges increase the total gravity force on the moon by 0.00...01.
Is it enough?
However, as the bulges are moving with the moon orbital cycle, than the moon gets this tinny extra gravity force constantly (So, there is no temporary/transient thrust).
Do you agree that it should have a similar impact as we increase the effective mass of Earth by a relative quantity - Let's say by ΔM (M=Earth mass)?
The outcome is that the moon's gravity force will be based on M+ ΔM instead of M.
This effective Earth mass (M+ΔM) should increase the velocity of the Moon.
So, instead of orbiting at velocity v, it will orbit at v +Δv.
However, as it is a constant gravity force, the orbital velocity should stay at the same amplitude.
Hence, I don't see any transient thrust which temporarily increases the velocity. 
If there is no transient increase in the velocity, do you agree that there is also no transient increase in R?
Therefore, without this transient increase in R how can we justify the whole idea of pushing the moon away from Earth?

[Halc]

Thanks
Based on Wiki the highest point of the tidal bulge is 54 Cm

Sounds like an ideal tide: If Earth was entirely covered with deep water and barely rotating, the friction would be minimized and the tides would have that amplitude.  Sounds about right.

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

The Earth moon distance is 384,400 km. The radius of the Earth is 6,371 km.
We can try to calculate the impact of each bulge.
In order to do so, we need to estimate the mass of each bulge with reference to the Earth and verify the total impact.
However, do you agree that 6,371 X2 is neglected with regards to 384,400 km?
Do you also agree that the net mass in that 54 centimeters Bulge is also neglected with regards to the total mass of the Earth?

Neglected?  By what?  No real clue what you mean by that choice of words.

If so, it is quite clear that the impact of the bulges is virtually zero or close to zero.

The impact is real, so what ever 'neglected' means, I think we should not neglect those numbers.
I would have run the computations in terms of torque, but you method works as well.

In any case, it seems to me that the key point in our discussion is the center of mass based on Newton's second law for the description of the motion of extended objects:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html

That link treats all objects as point masses.  That isn't useful in describing tidal effects.  It shows a wrench, but makes no effort to show the torque put on a thrown wrench due to tidal forces.

If the bulges at the near side and at the far side are similar, do you agree that there is no change in the Earth center of mass location with regards to the moon.

Yes, I agree with that.  Tidal forces are not about center of mass.  They're about deviations from it.

Hence, based on this idea, those bulges shouldn't have any impact on the Moon orbital cycle as they cancel each other.

Doesn't follow.

Never the less, even if we ignor the far end bulge, what might be the impact of that near side bulge to the Earth center of mass?

None.  You need to consider what each bulge does as a force acting on the moon CoM.
I put two identical masses on either side of you, one nearby to the front, and the other further away to the rear.  You accelerate towards the front one because the force from that one is greater, given the smaller separation.  That's what is pulling the moon forward.

Hence, let's assume that we can find that those bulges increase the total gravity force on the moon by 0.00...01.

They don't change the total gravity.  But they change the direction of it.  It is no longer straight towards the center of Earth because the Earth is neither a point mass nor a sphere.

Do you agree that it should have a similar impact as we increase the effective mass of Earth by a relative quantity - Let's say by ΔM (M=Earth mass)?

You want to dump more mass on Earth?  Sure, that would increase the gravitational force and pull the moon closer in.  This is assuming that new mass comes from outside the orbit of the moon.  A trillion low orbit satellites falling out of the sky will have zero effect since they were already effectively part of mass of Earth.

The outcome is that the moon's gravity force will be based on M+ ΔM instead of M.
This effective Earth mass (M+ΔM) should increase the velocity of the Moon.

Because it is pulled down to a new lower orbit, yes.  Even if it stayed at its current orbit, it would need to move faster, but dumping new mass on Earth doesn't give it the new energy it would need to accelerate like that, so it just fall to that lower orbit.

So, instead of orbiting at velocity v, it will orbit at v +Δv.
However, as it is a constant gravity force, the orbital velocity should stay at the same amplitude.

What?  It isn't constant gravity force.  You just added mass to M, which changes the force.

Hence, I don't see any transient thrust which temporarily increases the velocity.

The additional velocity in this case is gained by falling to a lower orbit, just like you personally gain velocity when falling off a ladder.  What do you mean by transient?  The higher velocity from adding mass will be permanent, or it will at least last until the mass is taken away again, or until the tides slow it down over time.

If there is no transient increase in the velocity, do you agree that there is also no transient increase in R?

Transient?  Are we still talking about the case where mass was added to Earth?  You kind of lost me with this question.

Therefore, without this transient increase in R how can we justify the whole idea of pushing the moon away from Earth?

I don't think we're talking about adding mass any more.  Not sure when that was abandoned.

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

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

If there was a prize given to the known object with the most circular gravitational orbit, I wonder which object would get the prize?

[Dave Lev (OP)]

Neglected?  By what?  No real clue what you mean by that choice of words.
The impact is real, so what ever 'neglected' means, I think we should not neglect those numbers.
I would have run the computations in terms of torque, but you method works as well.

O.K
Let's set a calculation.
The main tidal impact is on oceans.
I assume that it also doesn't work from pole to pole.
The pick is 54 cm.
Somehow it seems to me that if we take the two bulges and try to spread their total mass over the whole planet, it won't be higher than few centimeters.
However, just for the calculation let's assume 10 Cm or 1 Decimeter.
The radius of the Earth is 6,371 km =
r = 6,371,000 m or 63,710,000 decimeters.
The volume:
V (Earth) = V(r = 63,710,000) = 4/3 π r^3 = 4/3 π 63,710,000^3 = 4/3 π 2.586 10^23
V(Bulges) = V(r = 63,710,001) - V(r = 63,710,000) = 4/3 π (2.58596615 - 2.58596603) 10^23 = 4/3 π 0.00000013 10^23
Hence, the ratio is:
V(Bulges) / V (Earth) =0.00000013 / 2.586 = 5.02 10 ^(-8)
V (bulge) = 5.02 10 ^(-8) /2 * V (Earth) = 2.51 10 ^(-8) * V (Earth)
M = Earth mass
Mb = Bulge mass = V(Bulges) / V (Earth) * M = 2.51 10 ^(-8) M
Please also be aware that we mainly discuss on water in the bulges. However, the core of the Earth is made of metal which should be quite heavier than water. So, the ratio in mass should be higher.
In any case, it seems to me that the bulges mass is less than the total mass in the Mountains or even in one big chain of Mountains.
Now, we might think to add those two points of mass at the sea level of the planet, one at the front side and the other at far end side from the moon.
This is not fully correct. If we move closer to the pole, the effective radius is shorter and therefore the impact of the gravity bulges is lower.
However, for this calculation let's assume the worst case and set those two bulges points of mass at the at the maximal distance between them (2 x R).
The Gravity force of the Earth is:
F = G M m / R^2
m = Moon Mass
The Earth to the moon distance, R= 384,400 km
F = Earth Gravity force = G M m / R^2 = G M m / (147.7 10^9) = G M m 6.99 * 10^(-3) * 10^(-9) =  G M m 6.99 *  10^(-12)
Bulges distance to the moon:
The bulges don't point directly to the Moon.
Therefore, the effective distance is less than full Earth Radius.
However, I use again the full radius just as a worst case.
Hence:
The front Bulge to the moon distance = R(front) =  384,400 km - 6,371 km = 378,029 Km
The far end Bulge to the moon distance = R(front) =  384,400 km + 6,371 km = 390,771 Km.
F(front bulge) = G Mb m / R(front)^2 = G Mb m / (142.9 10^9)
F(rear bulge) = G Mb m / R(front)^2 = G Mb m / (152.7 10^9)
F(Bulge Total) = F(front) - F (rear) = G Mb m * 10^(-9) * (1/142.9 -1/152.7) =  G Mb m * 10^(-9) * 4.49 * 10^(-4)
= G 2.51 10 ^(-8) M m * 10^(-9) * 4.49 * 10^(-4) = G M m * 4.49 * 10 ^ (-21)
The ratio in the gravity force between the Bulge and the Earth is:
F(Bulge Total)/F(Earth) = G M m * 4.49 * 10 ^ (-21) / G M m 6.99 *  10^(-12) = 0.643 * 10^(-9).
Hence
F(Bulge Total) =  0.643 * 10^(-9) * F(Earth) = G M m 0.643 * 10^(-9) /R^2

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

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

[Halc]

Neglected?  By what?  No real clue what you mean by that choice of words.

O.K
Let's set a calculation.

See, you never tell me what 'neglected' meant, so I am left unable to parse the prior post.

Mb = Bulge mass = V(Bulges) / V (Earth) * M = 2.51 10 ^(-8) M
However, the core of the Earth is made of metal which should be quite heavier than water. So, the ratio in mass should be higher.

Agree, so closer to 1e-8

In any case, it seems to me that the bulges mass is less than the total mass in the Mountains or even in one big chain of Mountains.

Arguably so, since said mountains cover much less area, but are much taller than 54 cm and are made of rock.  They don't have much a gravitational difference since they're made of lighter rock and float upwards on the mantle.  The Earth mass is just denser on average between the mountain ranges.

F = Earth Gravity force = G M m / R^2 = G M m / (147.7 10^9) = G M m 6.99 * 10^(-3) * 10^(-9) =  G M m 6.99 *  10^(-12)

I got more like 6.77e-12, not 6.99, which was the inverse of the 142 distance to the nearest bulge.

Bulges distance to the moon:
The bulges don't point directly to the Moon.
Therefore, the effective distance is less than full Earth Radius.
However, I will use the full radius just as a worst case.

I would have used a nice round 6000 km, but fine.  Not like we know the actual numbers here.  I like how you're going about it.

Hence:
The front Bulge to the moon distance = R(front) =  384,400 km - 6,371 km = 378,029 Km
The far end Bulge to the moon distance = R(front) =  384,400 km + 6,371 km = 390,771 Km.

I think you mean R(rear) here.

F(front bulge) = G Mb m / R(front)^2 = G Mb m / (142.9 10^9)
F(rear bulge) = G Mb m / R(front)^2 = G Mb m / (152.7 10^9)
F(Bulge Total) = F(front) - F (rear) = G Mb m * 10^(-9) * (1/142.9 -1/152.7) =  G Mb m * 10^(-9) * 4.49 * 10^(-4)

or about 4.5e-13GMbm.  Seems a bit early to be subtracting the two force values? That makes little sense.  F total is just the sum of F(front) + F(rear), and this is only the F component perpendicular to the orbit.  We need to compute the tangential component to get the tidal thrust, and for tangential, F total is indeed the difference between the two tangential components, not the sum.  So OK, you're computing a difference here, but you've not yet done the vector trigonometry to compute the forward and resisting thrust.  OK, so you have a difference of full force here, so it is indeed valid to do the trig on just that one force.  I don't see the vector arithmetic anywhere in your post.

= G 2.51 10 ^(-8) M m * 10^(-9) * 4.49 * 10^(-4) = G M m * 4.49 * 10 ^ (-21)

OK, I think you substituted the value computed for Mb here.

The ratio in the gravity force between the Bulge and the Earth is:
F(Bulge Total)/F(Earth) = G M m * 4.49 * 10 ^ (-21) / G M m 6.99 *  10^(-12) = 0.643 * 10^(-9).
Hence

No, you subtracted the bulge forces.  The ratio of bulge force to the rest-of-earth would be computed from the sum of the bulge forces, not the difference.

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

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

Conclusions:
I have calculated the gravity force impact of the Bulges.

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

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

So, why do you call it "thrust"?

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

Yes, the mountains may mass more than the bulges, but they have no cumulative effect since they are ahead as often as behind the COM line, so the net effect is zero.

[Halc]

I will try to illustrate some of the vector work, using the guess that friction pushes the bulges 1000 km off center.

The Earth to the moon distance, R= 384,400 km
...
The front Bulge to the moon distance = R(front) =  384,400 km - 6,371 km = 378,029 Km
The far end Bulge to the moon distance = R(front) =  384,400 km + 6,371 km = 390,771 Km.
F(front bulge) = G Mb m / R(front)^2 = G Mb m / (142.9 10^9)
F(rear bulge) = G Mb m / R(front)^2 = G Mb m / (152.7 10^9)

Those forces are respectively GmMb * 7.00e-12 and  GmMb * 6.55e-12

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

Difference is 1.755 e-15 GmMb cumulative forward thrust forever adding energy/angular momentum to the orbit of the moon.

[Dave Lev (OP)]

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

Thanks for all your excellent remarks.

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

Never the less, I think that I understand the source for your statement:
Please see Figure 7.24 in the following article:
https://lifeng.lamost.org/courses/astrotoday/CHAISSON/AT307/HTML/AT30706.HTM
"Figure 7.24 The tidal bulge raised in Earth by the Moon does not point directly at the Moon. Instead, because of the effects of friction, the bulge points slightly "ahead" of the Moon, in the direction of Earth's rotation. (The magnitude of the effect is greatly exaggerated in this diagram.) Because the Moon's gravitational pull on the near-side part of the bulge is greater than the pull on the far side, the overall effect is to decrease Earth's rotation rate."
So, if I understand it correctly,  as the bulge points slightly "ahead" of the Moon, we believe that it pulls in the direction of motion (energy increase)"
Is it correct?

[Halc]

However, I still wonder why you insist to call it "Thrust".

Because it is forward, in the direction of motion.  I turn the wheel of my car and it accelerates it to the left, but that isn't thrust, and so the car turns without gaining speed.  But the engine supplies force in the forward direction, adding mechanical energy to the car as it does so, resulting perhaps in more speed, or perhaps just helping the car up a hill.  The car might be slowing, but the engine is still providing thrust/energy.  So we call that force thrust, and not the force that turns the car left.

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

The meaning of Thrust by Google is : "the propulsive force of a jet or rocket engine." "Push (something or someone) suddenly or violently in the specified direction."

Well, it might be kilotons of thrust, but it still isn't very violent.

I couldn't find any sort of "engine" in the Bulges activity. It is just increases the gravity force.

No, it doesn't.  Overall force is the same, but tides change the direction of that force.

Never the less, I think that I understand the source for your statement:
Please see Figure 7.24 in the following article:

Oh lovely.  They show the bulges dragged off to either side like I've been describing.  They'd be straight at the moon if there was no friction.

"Figure 7.24 The tidal bulge raised in Earth by the Moon does not point directly at the Moon. Instead, because of the effects of friction, the bulge points slightly "ahead" of the Moon, in the direction of Earth's rotation. (The magnitude of the effect is greatly exaggerated in this diagram.)

So much for my 1000 km estimate I bet.

Because the Moon's gravitational pull on the near-side part of the bulge is greater than the pull on the far side, the overall effect is to decrease Earth's rotation rate."
So, if I understand it correctly,  as the bulge points slightly "ahead" of the Moon, we believe that it pulls in the direction of motion (energy increase)"
Is it correct?

Yes.  I've been saying that for how many posts now?  Energy increase (thrust) to the moon, and energy decrease to Earth's rotation rate, as the comment states.  The decrease is greater than the increase.  Momentum is preserved here, but not energy (2.5 terawatts), almost all of which is lost to friction heat.  Only about 1/20th of that energy (around 120 gigawatts) is transferred to the moon.

[Dave Lev (OP)]

Thanks
So,  we believe that as the bulge points slightly "ahead" of the Moon, it pulls in the direction of motion (energy increase)".
In order to understand the real impact of that offset is, let's verify the following conditions:

1. There is no offset (0 degree) and the bulges are pointing directly to the moon
Do you agree that the effective distance between the bulges to the moon is maximal?. Therefore, can we assume the diffrence between their gravity force is maximal. However, can we assume that the thrust is Zero?
2. The offset is maximal (90 degree). The bulge are located at/almost the poles. therefore, the distance between the front bulge to the moon is actually equal to the distance of the rear bulge to the moon. Hence, can we assume that they have the same gravity force? However, what is the expected thrust? Is it zero or infinite?
3. The offset is 60 degree.
Cos (60) = 0.5
Hence, the effective radius is 0.5 * r = 0.5 * 6371 = 3185.5 K.m (Let's assume 3000 Km)
In this case, the effective distance to the moon is as follow:
R(front) =  384,400 km - 3,000 km = 381,400 Km
R(rear) =  384,400 km + 3,000 km = 387,400 Km.
So, it's easy to calculate the gravity force of each one, however, I can't understand why there will be any thrust.
I wonder if it is related to the idea that there are two bulges.
Hence, if there was just only one bulge, (let's assume only the front bulge), does it mean that the thrust will be zero?
In the same token, if we will set only the rear bulge, can we assume that the thrust will be of zero?
So, only if we have them both, we get the impact of thrust?
I have read your following answer in thread 40 and couldn't understand the source of the thrust:

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

Difference is 1.755 e-15 GmMb cumulative forward thrust forever adding energy/angular momentum to the orbit of the moon.

In order to understand the idea.
Let's eliminate the Earth.
Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?
I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?

[Halc]

So,  we believe that as the bulge points slightly "ahead" of the Moon, it pulls in the direction of motion (energy increase)".
In order to understand the real impact of that offset is, let's verify the following conditions:

1. There is no offset (0 degree) and the bulges are pointing directly to the moon
Do you agree that the effective distance between the bulges to the moon is maximal?. Therefore, can we assume the diffrence between their gravity force is maximal. However, can we assume that the thrust is Zero?

No thrust, right.  The difference between their force is irrelevant then.

2. The offset is maximal (90 degree). The bulge are located at/almost the poles.

No, 90° eastward and westward, not at the poles.  The bulges move around the equator, not from pole to pole.  Not exactly.  The orbital plane is about level with the solar system plane, not with the tilt of Earth's axis, so each bulge actually moves a ways north to south and back twice a day, regardless of offset angle.

therefore, the distance between the front bulge to the moon is actually equal to the distance of the rear bulge to the moon. Hence, can we assume that they have the same gravity force? However, what is the expected thrust? Is it zero or infinite?

Still zero since they're equal and opposite.

I think I used about 10 degrees in my example.  Possibly still too much.

3. The offset is 60 degree.
Cos (60) = 0.5
Hence, the effective radius is 0.5 * r = 0.5 * 6371 = 3185.5 K.m (Let's assume 3000 Km)
In this case, the effective distance to the moon is as follow:
R(front) =  384,400 km - 3,000 km = 381,400 Km
R(rear) =  384,400 km + 3,000 km = 387,400 Km.
So, it's easy to calculate the gravity force of each one, however, I can't understand why there will be any thrust.
I wonder if it is related to the idea that there are two bulges.

Did you read my description of the vector arithmetic in post 40?  Each bulge is off to the side by Sin(60) in this case (about 5500 km) so the force is not straight to Earth but has significant forward/backward components respectively.

Hence, if there was just only one bulge, (let's assume only the front bulge), does it mean that the thrust will be zero?

I cannot comment on that since it would change the center of gravity of Earth and hence not really be a bulge.
Thrust from each bulge is very much not zero.  They are in opposition, so the net thrust is much less.

I have read your following answer in thread 40 and couldn't understand the source of the thrust:

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

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

In order to understand the idea.
Let's eliminate the Earth.
Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?

They don't.  They're in orbit, so they'll go around each other every X many hours, not every month.  They'd not add any cumulative thrust to the moon.  The 3-body problem would probably make the whole system unstable after not much time.

I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?

It is only gravity, yes.  Work out the forces in 2 dimensions, not as a scalar.  Force is a vector, not a scalar.

If I am on a skateboard going down a slope, I speed up.  That means gravity force is providing me with thrust, increasing my speed.  Is it so unimaginable that it might do this?  If I was moving up a slope, that same gravity would slow me down, and not be thrust, but rather a braking action (or be a negative thrust if you will).  If I am on a perfectly level lot, there is no speed change, so the same gravitational force results zero thrust to my skateboard.

[Dave Lev (OP)]

Thanks
I'm quite confused with the answers about the offset/thrust.
Please let me know if the following is correct:
The Thrust is a direct outcome of the offset. (While the offset was a direct outcome of the tidal friction).
Based on the offset phase you have calculated the thrust in thread 40:

The 1.8517 figure is 700 force * 1000km / 378029 km, or essentially the F multiplied by the sin of the angle formed at the moon between the COG line and the line to the near bulge.
Difference is 1.755 e-15 GmMb cumulative forward thrust forever adding energy/angular momentum to the orbit of the moon.

However, when I have tried to focus in a general case of the offset, you have informed me that there is no thrust due to offset:

Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?
I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?

They don't.  They're in orbit, so they'll go around each other every X many hours, not every month.  They'd not add any cumulative thrust to the moon.  The 3-body problem would probably make the whole system unstable after not much time.

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

It is only gravity, yes.  Work out the forces in 2 dimensions, not as a scalar.  Force is a vector, not a scalar.
If I am on a skateboard going down a slope, I speed up.  That means gravity force is providing me with thrust, increasing my speed.  Is it so unimaginable that it might do this?  If I was moving up a slope, that same gravity would slow me down, and not be thrust, but rather a braking action (or be a negative thrust if you will).  If I am on a perfectly level lot, there is no speed change, so the same gravitational force results zero thrust to my skateboard.

I like the idea of skateboard.
This shows that we can convert gravity force into thrust.
Don't forget that Newton have used the example of falling Apple to find the whole idea of gravity.
However, we can't limit the skateboard idea just for tidal friction explanation.
If there is a possibility to convert Gravity into thrust due to the offset in a tidal system, than please set a formula for thrust for general case of offset.
This is a breakthrough concept in gravity.
Newon, kepler and Einstein didn't offer any solution for converting gravity force into thrust in orbital system.
Our scientists believe that it is feasible to get thrust due for tidal friction offset as follow:

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

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

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

[Halc]

However, when I have tried to focus in a general case of the offset, you have informed me that there is no thrust due to offset:

Let's assume that we have only two points of mass (Each one is 0.5 Earth Mass) which orbits in a fixed orbital cycle (At radius r= 6,371 Km, and at a fixed maximal distance from each other = 2r).
Try to put them at any offset as you wish (With regards to the Moon).
How can they set any sort of thrust on the moon?
I only see gravity force.
Do we base the idea of thrust on Newton gravity force? How?
If no, which law proves that there is a thrust?

They don't.  They're in orbit, so they'll go around each other every X many hours, not every month.  They'd not add any cumulative thrust to the moon.  The 3-body problem would probably make the whole system unstable after not much time.

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

There is thrust in that case, but the forces will move your objects around, so the thrust might last minutes at best, after which the pieces will rearrange and produce negative thrust.  It all cancels out.  You've removed the large middle piece which is the source of the inertia and friction.

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

I like the idea of skateboard.
This shows that we can convert gravity force into thrust.
Don't forget that Newton have used the example of falling Apple to find the whole idea of gravity.
However, we can't limit the skateboard idea just for tidal friction explanation.
If there is a possibility to convert Gravity into thrust due to the offset in a tidal system, than please set a formula for thrust for general case of offset.
This is a breakthrough concept in gravity.
Newon, kepler and Einstein didn't offer any solution for converting gravity force into thrust in orbital system.

I beg to differ.  It all follows from Newtonian mechanics.  Kepler's laws are specifically about orbits and derive from Newton's equations, and he was quite aware of tidal effects.

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

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

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

We already did that.  The linear offset is computed from the sin of the angular offset, which is closer to 10 degrees than to 60.  The distance of each bulge to the moon is the mean distance R ± cos(angular offset).
That linear offset produces a forward and retrograde force component on the moon.   Forward thrust is total force(GMmr², where M is the bulge mass) * linear offset / distance of bulge to moon.

However, If thrust works for tidal offset idea, it should also work for the example which I have offered.
If it doesn't work for this example, than it should not work also for tidal offset.

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

How can we come with idea for a special case and close it only for that case?
Do you agree that if we can't open the idea for any offset in orbital system, than we might have a problem with this idea?

Remove time from your example.  Then it is not an orbital system, but merely a bunch of masses at specific locations, with forces acting between each of them.  You can close it for that case.  With time eliminated, we need not consider anything's velocity or acceleration.  Only forces need be computed.
Ever take a statics class in college?  That's what the course is about:  Forces in the absence of time, and finding balances between them.

[Dave Lev (OP)]

Thanks Halc

The information about tidal friction is fully clear.
 

[Dave Lev (OP)]

I assume that by now we all understand how the tidal friction set the thrust which is needed to push away the Moon from the Earth.
However, the water bulges in oceans are very unique for the Earth/Moon system.
In one hand, this is the only planet in the solar system with so much water and on the other hand, the moon is relatively big enough to form the bulges.
We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?
They don't have water. They don't have relatively big moon around them. They don't form those water bulges.
So, if tidal friction can't be the answer for their drifting outwards activity, what could be the answer for that?