[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?

[Halc]

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.

The water makes a nice difference, yes.  The size of the moon is irrelevant.  If it was just an automobile up there, it might generate a trillionth of the gravity, a trillionth of the tidal effect on the ocean, and hence a trillionth of the force, which would result in the exact same acceleration.  The automobile would move to higher orbits at the exact same pace as does the moon, but the Earth would have a trillionth of the friction, so the spin would not degrade at a measurable rate like we get from our big honkin moon.

We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?

Because they go around slower than the spin of the primary, and in the same direction.  If they orbited lower (at or below geo-sync), or if they went around the opposite way, they'd get pushed down.  Such objects tend to fall into their primaries before too long.

The rings of Saturn is a nice example of a moon that did just that.  Too close, and tidal forces pulled it within the Roche limit where it breaks up.
Another example might be Venus, whose moon, if it had ever had one, would probably have gone around the normal way, causing it to drop its orbit that is retrograde to the spin of Venus itself.

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?

It works with the planet itself.  Even the relatively solid moon has strain in the crust from tidal forces from Earth, and those forces are really small since the moon is tide-locked, but it still rocks back and forth with its elliptical orbit.

Anyway, the sun is hardly solid and each of the planets form a tide on it.  The big planets are all very liquid and susceptible to significant tides.  The amazing story is Pluto and Charon, both rocks pretty much free of stuff that would exhibit tidal strain, and yet the two of them have managed to become mutually tidal locked already.

[Dave Lev (OP)]

We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?

Because they go around slower than the spin of the primary, and in the same direction.  If they orbited lower (at or below geo-sync), or if they went around the opposite way, they'd get pushed down.  Such objects tend to fall into their primaries before too long.

Thanks
So, the tidal friction idea between Earth/Moon is actually none relevant to all the other orbital systems in the solar system.
Now there is new idea.
All the planets and moons are pushed away as they all "go around slower than the spin of the primary, and in the same direction".

1. Would you kindly explain this idea?
2. Can we prove it by mathematical calculations/formula?
3. Why our moon isn't pushed outwards due to this idea?

[Halc]

So, the tidal friction idea between Earth/Moon is actually none relevant to all the other orbital systems in the solar system.

I think I said it was relevant to all of them.  They're all getting pushed out, except for the two exceptions I mentioned, both of which resulted in no moon.

Now there is new idea.
All the planets and moons are pushed away as they all "go around slower than the spin of the primary, and in the same direction".

Not a new idea.  Been around for at least 300 years.  The ones that don't do that don't survive, so the only ones left in our solar system are the ones that go slow, and in the same direction.

1. Would you kindly explain this idea?

If they orbit faster than the primary, the tidal bulges will lag behind the orbit, and the resulting forces will put a braking action on the object.  Similarly if the orbit is retrograde (faster or slower), the force is braking, not thrust.  Energy is lost, so the moon falls down into the planet.
The ISS is slowly dropping its orbit due to this effect since it goes around every 90 minutes, 16x faster than the rotation of Earth.  Left alone long enough, and even absent friction from being too close to the atmosphere, it will eventually fall.

2. Can we prove it by mathematical calculations/formula?

Yea, the same calculation used to compute the thrust force on the moon.  That calculation yields a negative number for a low or retrograde orbit.

3. Why our moon isn't pushed outwards due to this idea?

Thrust is forward and positive, so the orbit of the moon increases.  So it is 'pushed outward' at a pace of a few cm per year.

[Halc]

We know that all the planets and Moons in the solar system are pushed/drift away. Why is it?

Because they go around slower than the spin of the primary, and in the same direction.  If they orbited lower (at or below geo-sync), or if they went around the opposite way, they'd get pushed down.  Such objects tend to fall into their primaries before too long.

Both our statements are wrong here, so I need to correct this.
I looked at Jupiter's moons as an example, and most of them are in fact not being pushed out, but have diminishing orbits.
Metis and Adrastea have orbital periods shorter than the primary, and hence have negative thrust.  They will eventually fall in turn, giving Jupiter some nice rings.
The next 13 moons including all the popular ones have growing orbits.  That's just 13 of the 63 known moons.
The outer 48 moons all have, for some reason, retrograde orbits.  I suspect the reason is that they're all captured objects, and a free object will get a speed boost when passing a gravitational source in the direction of its orbit (not in the direction of its spin).  All the Voyager spacecraft have achieved the majority of their energy from such gravitational boosts.  The effect is to be flung away all the harder, not to be captured.  But if the object passes in front of the primary and outbound, it slows, possibly enough to be captured.  I think this is how each of the moons was captured.  Each needs to come in at greater than escape velocity and lose enough of it to drop below escape velocity.
I find that odd that Jupiter has so many retrograde moons, but none for any of the other planets.

All the moons have been there long enough to become tidal locked.  All those retrograde moons are so far out that their tidal effects result in negligible braking, so their orbits will probably take many billions to trillions of years before they fall into Jupiter.

Mars has 2 moons, and Phobos is dropping (seemingly faster than tidal effects can explain), and is expected to crash in a mere 40 million years.
Saturn has destroyed its last low-orbit moon, forming the current rings.  The lowest one is outside Saturn's geosync orbit and has positive tidal thrust.  All of Saturn's 60 moons have growing orbits.
Uranus has 27 moons, 11 of which are in low orbit and 'falling'.
Neptune has 14 moons, 5 of which are in low orbit and 'falling'.

None of these have retrograde moons.

[Dave Lev (OP)]

Thanks for the explanation about the moons.

However, I have no clue why if the Moon orbits in one direction/periods it will be pushed outwards, while if it orbits in the other direction/periods, it should be pulled inwards.
In tidal friction, the offset of the bulges sets the Thrust?
In this case, there is no offset.
So, how can we get a negative or positive thrust without any offset?

Metis and Adrastea have orbital periods shorter than the primary, and hence have negative thrust.  They will eventually fall in turn, giving Jupiter some nice rings.

Can you please explain why the orbital direction/periods can set a thrust?

[Halc]

Thanks for the explanation about the moons.

However, I have no clue why if the Moon orbits in one direction/periods it will be pushed outwards, while if it orbits in the other direction/periods, it should be pulled inwards.
In tidal friction, the offset of the bulges sets the Thrust?

Yes, the offsets set the thrust.  If the offset is positive, so it the thrust.

In this case, there is no offset.

In what case?  The only case with no offset is the tides from objects in geoSync orbit.  Those have no offset, so those orbits are relatively stable.

Metis and Adrastea have orbital periods shorter than the primary, and hence have negative thrust.  They will eventually fall in turn, giving Jupiter some nice rings.

Can you please explain why the orbital direction/periods can set a thrust?

The tidal offsets are negative for those moons, as are the offsets for all the retrograde moons.

[Dave Lev (OP)]

Yes, the offsets set the thrust.  If the offset is positive, so it the thrust.

Can you please explain how the orbital direction sets the offset?
Why orbital periods that are shorter than the primary can set a negative offset?
Do we have any drawing for that idea (as we have for the water bulges on Earth)?
So, how do we get any sort of offsets due to orbital direction or orbital periods?

[Halc]

Yes, the offsets set the thrust.  If the offset is positive, so it the thrust.

Can you please explain how the orbital direction sets the offset?

It is friction.  If my car is on a road, the brakes act as friction.  If the road is moving forward faster than the car, hitting the brakes will accelerate (positive thrust) to a higher speed.  If the road is slower than the car, or moving at any speed in the opposite direction, the brakes will slow the car down (negative thrust).

Why orbital periods that are shorter than the primary can set a negative offset?

Because the friction of the slower primary drags the bulge to the rear, a negative offset.

Do we have any drawing for that idea (as we have for the water bulges on Earth)?

They all look the same.  Bulges.  Offsets.  A negative offset puts the bulges on the opposite side.  A zero offset (which you only get with geosync orbits) are directly towards and away from the orbiting thing, and result in zero friction and zero thrust.

So, how do we get any sort of offsets due to orbital direction or orbital periods?

Gravity tries to pull the offsets straight at the orbiting thing, but the spin of the primary in this case pushes the offset to one side or the other, depending which way the ground under the orbiting object appears to be moving.

If I look at Earth from a forward pointing ship in low orbit (400 km), things tend to appear in front of the ship and disappear behind it, just like the view from an airplane.   So friction with the ground below is going reduce my velocity, so steps are taken to avoid this friction as much as possible.  The airplane has optimal aerodynamics to minimize this drag.  The ISS is high enough to minimize atmospheric drag, but there's little it can do about tidal drag.

Same ship at geosync (36000 km):  The Earth below is now stationary.  I see the exact same spot below whenever I look.  Friction with that would produce zero thrust.

Same ship further out, (100000 km).  Now the Earth features appear from behind me and rotate away to the front.  Earth is spinning faster than my orbit.  Friction with that would push me forward (positive thrust).  The moon's orbit is in this last category (beyond 36000 km).

[Halc]

Similarly, I am suicide Halc, deciding to do the job right and jump out of the tallest building I can find.  There is a likely tall space elevator right on the equator that goes up 100000 km.
They actually are trying to plan such a structure since it would make it so much easier to put things in space.

I could jump out at 400 km up, and fall quickly to my death.  I could leap out at 36000 km, and just hover there forever.  If I jump out at 50000 km, I'd fall up, but at least stay in orbit.  If at 55000 km, I'd fall up and never come back.

I could not find an answer to the question of the maximum height I could jump out of a window and still hit the ground.  Jumping out at any lower point puts you in orbit.  At some height, the perihelion of that orbit is above the altitude of the atmosphere, and a jump from that altitude or above will not ever hit the ground.

[Dave Lev (OP)]

Because the friction of the slower primary drags the bulge to the rear, a negative offset.

There are nine planets in the solar system.
We all know that there are bulges in our planet.
However, do we see any sort of bulge at any other planet in the solar system?
If we don't see any bulge at any other planet, how can we think that a bulge which doesn't exist can set an offset that we wish for?
Please advice if you agree with the following:
Tidal friction is based on Bulges.
If there is no bulges there is no offset. If there is no offset there is no thrust. If there is no thrust there is no energy to push or to pull the moon.
So, do you agree that without confirmed bulges per planet, there is a problem with this hypothetical idea?

[Halc]

There are nine planets in the solar system.
We all know that there are bulges in our planet.
However, do we see any sort of bulge at any other planet in the solar system?
If we don't see any bulge at any other planet, how can we think that a bulge which doesn't exist can set an offset that we wish for?

I look at a picture of Earth and see no bulge.  They're not exactly pronounced.  Point is, the tidal forces produce stress on the bodies orbiting each other, and those stresses produce strain of one sort or another, and changing strain is movement that produces heat from friction.  These forces are strong enough to have tide-lock (nearly??) every moon of every planet.  Water isn't necessary.  The tides are quite significant in amplitude on planets/stars that are not rocks.  Venus has a thick atmosphere to drag around.  Pluto and Mercury are the only planets with nothing but solids to work with, and both those have become tide locked with the most significant gravity source nearby.

Please advice if you agree with the following:
Tidal friction is based on Bulges.
If there is no bulges there is no offset. If there is no offset there is no thrust. If there is no thrust there is no energy to push or to pull the moon.[/quote]
None of these assert that the deformation needs to be measurably confirmed from a distance, so yea sure.
Not all deformations manifest as something that can be classified as a bulge.  I can put a foam cube out there spinning, and tidal forces will deform it, but the corners will always stick out the furthest, and hence a given deformation will not necessarily be a bulge.  Even Earth has mountains that 'bulge' out far further than does the water.

So, do you agree that without confirmed bulges per planet, there is a problem with this hypothetical idea?

No.  The forces involved do not need confirmation in order to exist.  Their effects are measurably confirmed.  Most of the moons nearby their primaries have measurable changes to their orbits, Phobos probably being the top of the list, despite being very unlikely to ever produce a measurable deformation of Mars' atmosphere or crust until it physically hits them.

Are you just posting here as a denier of such forces?  You'd have to explain how one object can put stress on another object without producing strain.  That would be quite a piece of new physics to propose.