[Dave Lev (OP)]

Earth moves a tiny bit further away, and slows down because it is now moving up hill out of the gravitational well.
V goes down as R goes up.  Throw a ball in the air and watch V go down as the ball rises.
You seem to be neglecting the fact that things slow down when moving away from a gravitational source.  Kinetic energy is being lost to acquired gravitational potential energy.

Thanks
Based on Newton first law:
https://en.wikipedia.org/wiki/Newton%27s_laws_of_motion
First law:   In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.[2][3]"
Let's start by asking the following:
What will happen if we eliminate completely and at once the gravity force from the Earth/sun orbital system?
If I understand it correctly, the Earth will "continue to move at a constant velocity".
Therefore, do you agree that for orbital system it is incorrect to assume that "things slow down when moving away from a gravitational source"?
Hence, do you agree that the Earth will not change its inertial velocity "unless acted upon by a force"?
Based on your explanation - the Tidal is not a force. It just pushes the Earth further away from the Sun.

As we increase R we actually decrease the gravity force.
Gravity force is the power which holds the Earth in the orbital cycle around the Sun.
It must be fully synchronized with the inertial velocity of the earth in order to hold it in the orbital cycle.

It is clear that as Tidal increases R it actually decreases the gravity force.
However, based on your explanation, Tidal doesn't set any force. Therefore, by definition it doesn't set any negative force which slows down the Earth velocity as it increases R.
Therefore, once the Tidal push away the Earth from the Sun, there is less gravity to hold the Earth in a balance with it's current inertial velocity/force.
Hence, Less gravity force to hold the Earth on the orbital track means that the Earth is moving further away from the Sun.
That increasing radius, decreases the gravity force more and more.
Therefore, do you agree that from Orbital point of view, once we push away the Earth from the Sun, the Earth is starting to move down the hill?
At some point it must be totally disconnected from the sun gravity and  "continues to move at a constant velocity" as explained by Newton first law.

Unfortunately, I couldn't find in your explanation how tidal sets a negative force which slow down the Earth velocity in order to compensate less gravity force due to increasing in R.

Can you please explain this issue?

Tides just apply thrust.  Thrust adds (or removes) energy, and the new energy level finds a balance between those two things.  V goes down as R goes up.

Would you kindly show by mathematics how tidal sets thrust that adds energy which pushes the Earth away from the Sun, while somehow the Earth reduces its velocity without implying an external force?

[Halc]

Based on Newton first law:   In an inertial frame of reference, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force.[2][3]"
Let's start by asking the following:
What will happen if we eliminate completely and at once the gravity force from the Earth/sun orbital system?
If I understand it correctly, the Earth will "continue to move at a constant velocity".
Therefore, do you agree that for orbital system it is incorrect to assume that "things slow down when moving away from a gravitational source"?

There is nothing moving away from a gravitational source in your example.  You took gravity away.
You can do this by considering two small rocks at the same positions and velocities of the earth/sun system.  They’re too small to have significant gravity pull between them, so they each continue to move pretty much in a straight line.

Hence, do you agree that the Earth will not change its inertial velocity "unless acted upon by a force"?

Newton’s law says that nothing does.  It is called velocity BTW.  ‘Inertial velocity’ isn’t something different, or if it is, you need to tell me what you mean by that.
So yes, I agree with Newton’s law.  Earth will not change its velocity unless acted upon by a force.

Based on your explanation - the Tidal is not a force. It just pushes the Earth further away from the Sun.

It actually pushes the Earth tangential to the orbit, which is perpendicular to a vector away from the sun.  If it pushes or pulls, it’s a force.  That’s what force is.

As we increase R we actually decrease the gravity force.
Gravity force is the power which holds the Earth in the orbital cycle around the Sun.

Force is not power.  Gravity is the force which holds the Earth in the orbital cycle around the Sun, yes.

It must be fully synchronized with the inertial velocity of the earth in order to hold it in the orbital cycle.

It is clear that as Tidal increases R it actually decreases the gravity force.

However, based on your explanation, Tidal doesn't set any force. Therefore, by definition it doesn't set any negative force which slows down the Earth velocity as it increases R.[/quote]Tital force is a force.  I didn’t say it ‘doesn’t set any force’.  It pushes with the motion, so it actually directly acts to increase speed of Earth, but that velocity slows as the Earth pulls further out of the gravity well.  The net effect is a slower orbital speed.

It’s like a coin in one of those funnel machines, slowly spiraling faster and faster into the center.  You give the coin some thrust, and it moves further away from the center and ends up going slower than before you gave it that push.  In the absence of interference, friction is a force against its motion, and yet the coin is speeding up as it goes deeper into the gravity well.

Therefore, once the Tidal push away the Earth from the Sun, there is less gravity to hold the Earth in a balance with it's current inertial velocity/force.
Hence, Less gravity force to hold the Earth on the orbital track means that the Earth is moving further away from the Sun.

Less gravity means the Earth has already moved further away from the sun.  You seem to be confusing cause and effect here.  That gravitational force doesn’t go down until the Earth has already moved further out.

That increasing radius, decreases the gravity force more and more.
Therefore, do you agree that from Orbital point of view, once we push away the Earth from the Sun, the Earth is starting to move down the hill?

No, away from gravity is up hill.  Surely you know this.  Downhill gets you closer to the source of the gravity (the sun in this case).

At some point it must be totally disconnected from the sun gravity and  "continues to move at a constant velocity" as explained by Newton first law.

Given enough thrust, that is true, but the tidal thrust decreases with distance, so I don’t think it is possible for one object to spin away one of its satellites in isolation.  But things are not in isolation.  Given enough time and high spin, the Earth could theoretically push the moon beyond Earth’s hill radius and the moon would depart Earth and go into its own orbit about the sun.  In isolation (just sun and Earth say), there is no hill radius.

Unfortunately, I couldn't find in your explanation how tidal sets a negative force which slow down the Earth velocity in order to compensate less gravity force due to increasing in R.

It doesn’t.  It is a positive force tangential to the orbit, pushing forward, not out.  If it was a negative force, things would slow down and drop to a lower orbit.
You seem to continue to envision the tides being an outward thrust instead of a forward one.  That would not have any cumulative effect since it would not affect speed and hence no energy transfer.

Can you please explain this issue?

I have, and you don’t seem to read what I say.
Tidal forces push forward, not out.  It doesn’t directly increase radius.  The increased speed is too much for the current orbit, so it moves further out, which slows the system since that is moving against the primary gravitational force.

Consider at a comet that passes the sun at the same radius as Mercury, but much much faster.  That’s what tide forces do is make something faster.  So the comet moves to a higher orbit, and in doing so, it ends up nearly stopped as it gets so far away from the sun.  Moving away from the sun slows it down.

Tides just apply thrust.  Thrust adds (or removes) energy, and the new energy level finds a balance between those two things.  V goes down as R goes up.

Would you kindly show by mathematics how tidal sets thrust that adds energy which pushes the Earth away from the Sun, while somehow the Earth reduces its velocity without implying an external force?

Tides create a bulge on the sun that resist its spin, putting negative torque on the sun.  Torque is FR (force * radius).  That force is balanced by positive torque on the sun/Earth system, so same math but much greater R and much less F.  This conserves angular momentum.

So that F is the thrust, pushing in the direction of orbit, not outward.  That speeds up the planet, which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.  That takes Earth further away from the sun, so it gains potential energy and loses kinetic energy.  Potential energy E = force*distance, so compute the weight (not mass) of Earth (using the gravitational formula GMmR²  to get the force, and multiply that by the say millimeter it moves away.  From that you get a chunk of energy E.  Kinetic energy is mv², so the reduction in v from the orbital change is Δv = E√(1/r) where r is the change in orbital radius ΔR and E is the kinetic energy lost to potential energy.  That negative Δv is greater than the positive Δv you get from enough tidal thrust to increase Earth orbit by a millimeter, so net effect is a slower Earth.

I chose one millimeter ΔR as my example, but it takes perhaps 1000 years for the solar tides to do that.  A somewhat larger effect pushing us away is that the mass of the sun is decreasing faster than new material falls in from deep space, so gravity is slowly decreasing.

[Dave Lev (OP)]

So that F is the thrust, pushing in the direction of orbit, not outward.  That speeds up the planet, which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.  That takes Earth further away from the sun, so it gains potential energy and loses kinetic energy.

Thanks again for your great effort!
I do appreciate.

However, would you kindly explain why the Earth is losing kinetic energy due to Tidal?

It was stated that the tidal trust speeds up the orbital velocity of the Earth.
That speeds up the planet (Lets assume by Δv), which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.
That takes the Earth further away from the sun, so it gains potential energy.
So far so good!

However, with regards to the kinetic energy.
The kinetic energy of the Earth just before speeding up the planet is: mv^2.
After increasing the speed by  Δv, the  updated kinetic energy should be: m(v+Δv )^2

Therefore, by definition due to the increase in speed, we have found that the Earth is gaining higher Kinetic energy.

So, would you kindly explain why you assume that the Earth "loses kinetic energy"?

[Halc]

However, would you kindly explain why the Earth is losing kinetic energy due to Tidal?

It was stated that the tidal trust speeds up the orbital velocity of the Earth.
That speeds up the planet (Lets assume by Δv), which is now going a bit too fast for its orbit, so the path diverges outward from the original circle path.
That takes the Earth further away from the sun, so it gains potential energy.
So far so good!

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.

However, with regards to the kinetic energy.
The kinetic energy of the Earth just before speeding up the planet is: mv^2.
After increasing the speed by  Δv, the  updated kinetic energy should be: m(v+Δv )^2

If the Δv is imparted all at once without giving Earth a chance to move, yes.  There is as yet no change in R and thus no change in potential energy.  But it cannot stay at that R now since it is moving too fast.  So it moves away, which slows it.  In reality, it takes time (centuries) for that Δv to be applied, and the orbit rises steadily and the kinetic energy drops as the R increases.  All the energy and more goes into additional potential energy, not kinetic, for a net loss of kinetic energy.  If Earth got so much thrust that it ends up way out by Neptune, then it would have a small fraction of the kinetic energy it has now, but far more total energy.

[Dave Lev (OP)]

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

That's an interesting idea.
The formula for Kinetic emery is: Ek = mv^2
The formula for potential energy is: Ep = GMmR^2
I could not understand why an increasing in potential energy (Ep) should decrease Kinetic energy (Ek).
Can you please explain the rational of this idea for orbital system?
Do we have any formula which shows the energy transformation?

[Halc]

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.

You say this like it isn't obvious.  Not sure if you're making fun of me.
You've never noticed that a roller-coaster, or a wheel on a hill goes faster the closer it is to the gravity source (down), and slower the further it moves away (up)?

I cannot explain any simpler.

The formula for Kinetic emery is: Ek = mv^2
The formula for potential energy is: Ep = GMmR^2
I could not understand why an increasing in potential energy (Ep) should decrease Kinetic energy (Ek).
Can you please explain the rational of this idea for orbital system?
Do we have any formula which shows the energy transformation?

Yes, you show them just there.  There is conservation of energy.  R is going up, so you know the energy needed to do that, and the kinetic energy is the only place it comes from in this case.  The force that slows the planet can be computed with simple vector arithmetic.  All you need is the force involved (gravity formula to yield the weight of Earth) and the slope of the motion relative to the flat tangential line that would be a steady state circular orbit.
That negative force has greater magnitude than the positive force of tidal thrust from the sun, so the net is a lower speed.

[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.

[Halc]

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

Thanks Janus.  I didn't even look that close to see that one.
I had put Ep=force* distance in post 21, but even then I got the sign wrong, and I didn't bother to integrate since the force is pretty constant in the millimeter the orbit of Earth changes every 1000 years.  That simplification would never do when computing the orbit of say the comet.

[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?

[Halc]

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.

That actually doesn't mention friction at all.  When I rub my hands together, it isn't gravity that makes them warm up.

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.

Newton's formula for gravitational force also says nothing about motion or spin.  It is a static formula for force at a given moment.  So I agree, Newton's formula says nothing about my statement there.
Do you deny that the 30 rotations per month angular velocity of Earth is greater than the 1 rotation per month angular velocity of the moon?  If Earth's spin slows, that angular momentum needs to be transferred elsewhere.  Conservation of angular momentum demands that, so says Newton.

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.

It already does if you treat Earth as a set of particles instead of a single point mass, as you suggest above.

Based on the current Newton's law of universal gravitation, "every particle attracts every other particle in the universe with a force..."

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.

So, any particle on Earth might go/down or spin faster/slower/locked due to the local impact of gravity force (tidal activity).

Points don't have spin.  Gravity does not concern itself with spin, speed, or any other dynamic.  It is a static formula for force at a given moment.  Multiple particles can exert torque due to gravitational forces.  Torque is a static value.

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:

No claim of the spin of a particle has been made.

[Halc]

Short description.

Tidal forces cause a bulge on the near and far sides of Earth.  Gravity does that, and it can be derived from Newton's formula.  The rotation of Earth, coupled with friction, pushes those bulges into positions no longer directly towards and away from the moon.  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.

That's tidal thrust explained in gravitational terms.

[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.