[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?

[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).

[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.

[Dave Lev (OP)]

Thanks Halc.

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.

Why water isn't necessary?
Please remember that the tidal on Earth is based on water.
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"
If I remember correctly, at land the tidal is just few cm (2-3 Cm?).
The mass ratio between Earth/moon is significantly higher than any other Planet/moon system.
Did we try to measure the Tidal impact on other planets?
If I understand it correctly, the offset is also due to water.
Therefore, without verifying minimal bulge amplitude and offset, I really can't understand why we are so sure that there is a minimal thrust that can push or pull the moon.
With regards to gas planets -
Let's look at Jupiter as it is the biggest planet in the solar system:
https://solarsystem.nasa.gov/moons/jupiter-moons/overview/?page=0&per_page=40&order=name+asc&search=&placeholder=Enter+moon+name&condition_1=9%3Aparent_id&condition_2=moon%3Abody_type%3Ailike
"Jupiter has 53 named moons and another 26 awaiting official names. Combined, scientists now think Jupiter has 79 moons."
It Radius is: 69,911 km
It's biggest moon is Ganymede:
https://www.google.com/search?q=Biggest+Jupiter+MOON&oq=Biggest+Jupiter+MOON&aqs=chrome..69i57j0l5.6920j0j8&sourceid=chrome&ie=UTF-8
"Ganymede's mean radius is 1,635 miles (2,631.2 km). Although Ganymede is larger than Mercury it only has half its mass, classifying it as low density.
Therefore, it is clear that the ratio between Jupiter and this biggest moon is very low.
I would assume that at this ratio, the tidal impact on land in Earth might be less than one millimeter
Jupiter has 79 moons. Therefore, there is good chance that those moons cancel the tidal impact of each other.
There is one more issue.
For some moons we believe that there is a negative thrust.
This negative thrust should pull those moons inwards.
This is the theory.
However, did we measure if they are pulled inwards?
What is the chance that they will not be so cooperative with our theory?
Why we are so sure with our theory while we only have real measurements of only Earth/Moon and Sun/Earth system (Both drifts/Pushed outwards?

[Halc]

Why water isn't necessary?
Please remember that the tidal on Earth is based on water.

Water has the greatest drag on Earth's spin since it has a lot of that, but atmosphere and crust also contribute.

If I remember correctly, at land the tidal is just few cm (2-3 Cm?).

Sounds like strain to me.  It takes energy to do that, and that energy is lost to heat.

The mass ratio between Earth/moon is significantly higher than any other Planet/moon system.

Really?  The ratio is about 1.2%, far less than the 11% ratio of Charon to Pluto.  OK, Pluto isn't exactly a planet.

Did we try to measure the Tidal impact on other planets?

Apparently you don't read my posts.

If I understand it correctly, the offset is also due to water.

Most planets don't have free flowing water.  Europa does, even it there's a crust that has some inhibiting effect to its tides.  I think Europa is tide locked, so no matter.

Therefore, without verifying minimal bulge amplitude and offset, I really can't understand why we are so sure that there is a minimal thrust that can push or pull the moon.

I think those particular figures have quite been verified.  Earth tide offset isn't a fixed figure.  It varies all over the place due to geographic features and ressonance.  It's the friction that counts, and that friction is greatest in shallow areas like around England.

"Ganymede's mean radius is 1,635 miles (2,631.2 km). Although Ganymede is larger than Mercury it only has half its mass, classifying it as low density.
Therefore, it is clear that the ratio between Jupiter and this biggest moon is very low.

Big time, yes.  All the easier for Jupiter to push it along.  Easy to push small things.

I would assume that at this ratio, the tidal impact on land in Earth might be less than one millimeter

Probably, yes.  The tide raised by the ISS is waaaay less than that millimeter, but the ISS is super light, so it accelerates (negative) just as much as a large thing would at that altitude.  That acceleration rate has little to do with the mass of the orbiting thing.'

Jupiter has 79 moons. Therefore, there is good chance that those moons cancel the tidal impact of each other.

They orbit at different periods, so there is zero chance of this.  Only a moon's own tides affect that moon, not the tides of other bodies, which have random offsets and thus cancel completely in the long run.

There is one more issue.
For some moons we believe that there is a negative thrust.

Like Phobos and most of Jupiters moons, yes.

However, did we measure if they are pulled inwards?

Jupiters outer moons are so far out that the tides might not have measurable impact.  The two inner ones very much do have measurable orbit degradation.  Phobos has massive degradation, and has only some tens of millions of years left in its life.

What is the chance that they will not be so cooperative with our theory?

Pretty much nill.  It isn't exactly an untested theory.

Why we are so sure with our theory while we only have real measurements of only Earth/Moon and Sun/Earth system (Both drifts/Pushed outwards?

Making up your facts I see.  We've plenty of measurements of the others, at least the things near their primaries, which have significant forces acting on their orbits.  Most of Jupiter's outer moons have only been discovered in the last 15 years, which is not much time to measure the trivial orbital changes put on them at their very distant and very eccentric orbits.

[Dave Lev (OP)]

For some moons we believe that there is a negative thrust. This negative thrust should pull those moons inwards. This is the theory. However, did we measure if they are pulled inwards?

Like Phobos and most of Jupiter's moons, yes
Jupiters outer moons are so far out that the tides might not have measurable impact.  The two inner ones very much do have measurable orbit degradation.  Phobos has massive degradation, and has only some tens of millions of years left in its life.
Making up your facts I see.  We've plenty of measurements of the others, at least the things near their primaries, which have significant forces acting on their orbits.

Why do we ignore the distance between the moon/rings to the planet?
https://en.wikipedia.org/wiki/Phobos_(moon)
"Phobos orbits 6,000 km (3,700 mi) from the Martian surface, closer to its primary body than any other known planetary moon. "

So, Phobos orbits 6000 Km from the surface of Mars while the radius of mars is about 3400 km. so the ratio is 1 to 1.76.
As an example:
The Earth Radius is 6370 Km. If Phobos would orbit around the Earth at the same ratio, its orbital distance from the Surface should be about 11,000 Km.
That orbital radius is lower than the altitude range of navigation satellites:
https://en.wikipedia.org/wiki/Medium_Earth_orbit
"The most common use for satellites in this region is for navigation, communication, and geodetic/space environment science.[1] The most common altitude is approximately 20,200 kilometres (12,552 mi)),"

If I understand it correctly, we expect that a navigation satellite should be pulled inwards.
If so, why is it so big surprise that Phobos is also pulled inwards?

With regards to Jupiter moons/Ring:
https://en.wikipedia.org/wiki/Rings_of_Jupiter
"Main ring - The narrow and relatively thin main ring is the brightest part of Jupiter's ring system. Its outer edge is located at a radius of about 129000 km (1.806 RJ;RJ = equatorial radius of Jupiter or 71398 km) and coincides with the orbit of Jupiter's smallest inner satellite, Adrastea.[2][5] Its inner edge is not marked by any satellite and is located at about 122500 km (1.72 RJ).[2]
So, Jupiter radius is 71398 Km.
Therefore, If this ring would orbit the Earth at the same ratio (1.806), its orbital distance from the Surface of earth should be about 11,500 Km.
So, don't you see a similarity between satellite around the Earth, Phobos around Mars and Main ring around Jupiter?
Don't you agree that with tidal or without it, all of them must eventually fall down?
Therefore, how can we use an object which it's orbital cycle is so close to the host to prove a negative thrust due to tidal???
Can you please find one moon (only one) in the whole solar system that is located long enough from its planet which is pulled inwards due to negative thrust (But please - real prove for that)?

With Regards to Europa:

Most planets don't have free flowing water.  Europa does, even it there's a crust that has some inhibiting effect to its tides.  I think Europa is tide locked, so no matter.

https://www.space.com/15498-europa-sdcmp.html
Distance from Jupiter: Europa is Jupiter's sixth satellite. Its orbital distance from Jupiter is 414,000 miles (670,900 km). It takes Europa three and a half Earth-days to orbit Jupiter. Europa is tidally locked, so the same side faces Jupiter at all times.
The surface of Europa is frozen, covered with a layer of ice, but scientists think there is an ocean beneath the surface. "
So, It is located far enough from Jupiter. Ratio of about 1:10.
It is covered with a layer of ice, therefore, the chance to set any significant bulges due to tidal is quite minimal (even if it has ocean beneath the surface).
So, I don't know what kind of information we can extract from Europa:
Does it have tidal bulges? Do those bulges set the positive/negative offset? Do we know if it is pushed outwards or pulled inwards?
I assume that the answer is - No, we don't know as it is too far away to measure.
So, if all the moons are too far away from us, how do we know that all of them must obey to tidal friction idea?

[Halc]

Why do we ignore the distance between the moon/rings to the planet?

I don't ignore that.  Jupiter's two innermost moons are falling because the distance from them to the planet is less than the geosync radius of Jupiter.

As an example:
The Earth Radius is 6370 Km. If Phobos would orbit the Earth at the same ratio, its orbital distance from the Surface should be about 11,000 Km.
A commercial airplane is normally fly at about 10,000 Km.
If it is expected that airplane should come/fall down, why is it so big surprise that Phobos is also coming/falling down?

Airplanes fly at 10 km, not at 10000.  The ISS is only about 400 km up.
Airplanes fall because gravity accelerates them downward, and their orbits are much much smaller than the radius of Earth, so they'll hit the ground without effort to keep them aloft.  The ISS needs no wings because its orbital path does not include the ground in its way.

Anyway, phobos falls for the same reason that an airplane slows if it runs out of fuel: Friction with some moving slower than itself.

With regards to Jupiter moons/Ring:
https://en.wikipedia.org/wiki/Rings_of_Jupiter
"Main ring - The narrow and relatively thin main ring is the brightest part of Jupiter's ring system. Its outer edge is located at a radius of about 129000 km (1.806 RJ;RJ = equatorial radius of Jupiter or 71398 km) and coincides with the orbit of Jupiter's smallest inner satellite, Adrastea.[2][5] Its inner edge is not marked by any satellite and is located at about 122500 km (1.72 RJ).[2]
So, Jupiter radius is 71398 Km.
Therefore, If this ring would orbit the Earth at the same ratio (1.806), its orbital distance from the Surface of earth should be about 11,500 Km. (Same distance at commercial airplane on Earth).

No airplanes there.  Rings form when moons pass below the Roche limit and are torn apart by tidal forces.  That ring on Jupiter has Metis in it as well, and Adrastea is probably a chunk torn off Metis, and the ring is all the shrapnel from that destruction.

So, don't you see a similarity between airplane orbit around the Earth, to Phobos around Mars and Main ring around Jupiter?

Phobos is already starting to break up, so it will form a ring around Mars.  Plenty of similarity there.  Airplanes do no fly in outer space.  A suborbital ballistic airliner might (they don't have any right now), but not anywhere near that high up.

Don't you agree that with tidal or without it, all of them must fall down?

Without tides, neither Phobos nor Adrastea nor an airplane orbiting at 10000 km will ever fall down.  With tides, all of them will eventually.

Therefore, how can we use an object which it's orbital cycle is so close to the host to prove a negative thrust due to tidal???
Can you please find one moon (only one) in the whole solar system that is located long enough from its planet which is pulled inwards due to negative thrust (But please - real prove for that)?

You seem to want to simply dismiss any explanation as 'not proof'.  It is easy to demonstrate violations of conservation of angular momentum and make an infinite energy engine if tides produce no thrust on moons.  They already have power generators that harness tidal energy.  Where do you think it comes from then?

With Regards to Europa:

Most planets don't have free flowing water.  Europa does, even it there's a crust that has some inhibiting effect to its tides.  I think Europa is tide locked, so no matter.

Europa is tidally locked, so the same side faces Jupiter at all times.
The surface of Europa is frozen, covered with a layer of ice, but scientists think there is an ocean beneath the surface. "
So, It is located far enough from Jupiter. Ratio of about 1:10.
It is covered with a layer of ice, therefore, the chance to set any significant bulges due to tidal is quite minimal (even if it has ocean beneath the surface).

Those tides below the ice have managed to halt Europa's spin, so I'd hardly call that minimal.  The forces putting thrust on its orbit are the tides raised by Europa on Jupiter's atmosphere, not those raised on Europa.  Yes, a moon's own spin contributes to this, but most moons have lost that spin already.

Does it have tidal bulges?

A permanent deformation is not tidal, so no.  The deformation on Earth that makes sea level at the equator a larger radius than at the poles is not considered a bulge because it is permanent.  Tides are strain, some kind of back and forth motion that requires energy to maintain.  The bulge on Europa is permanent, which is why I said 'no matter'.

Do those bulges set the positive/negative offset?

The offset of Europa's bulges averages zero.  That's because it is tide locked.

Do we know if it is pushed outwards or pulled inwards?

Outward.  Geosync of Jupiter is around 170,000 km, and Europa is beyond that, and moving with the spin.  So thrust is positive.

I assume that the answer is - No, we don't know as it is too far away to measure.

The answer is yes because it would violate conservation laws for it to be otherwise.

So, if all the moons are too far away from us, how do we know that all of them must obey to tidal friction idea?

The laws of physics work everywhere, not just where humans confirm them in court.

[Dave Lev (OP)]

Airplanes fly at 10 km, not at 10000

By the time that I fixed my message, I have got your answer.
So yes, you are absolutely correct.

[Dave Lev (OP)]

The laws of physics work everywhere, not just where humans confirm them in court

Based on Newton's Shell Theorem:
https://www.math.ksu.edu/~dbski/writings/shell.pdf
The gravitational field outside a spherical shell having total mass M is the same as if the entire mass M is concentrated at its center (Center of mass).
So, if that is correct, than the gravitational field of all the mass of Earth (including the bulges), must be concentrated at the center (center of mass).
Do you agree that this law contradicts the whole idea of tidal friction, as there is no positive or negative thrust if we set the whole mass at the center?
We are using this Newton's Shell Theorem in order to explain many aspects of the Universe.
So, if the laws of physics work everywhere, why this law can't be used here?
Why I can't use it in order to prove that there is no positive or negative thrust due to offset?

[Dave Lev (OP)]

Based on Newton's Shell Theorem:
https://www.math.ksu.edu/~dbski/writings/shell.pdf
The gravitational field outside a spherical shell having total mass M is the same as if the entire mass M is concentrated at its center (Center of mass).
So, if that is correct, than the gravitational field of all the mass of Earth (including the bulges), must be concentrated at the center (center of mass).

Right, but the theorem works only for a very regular object, a sphere in particular.  All kinds of funny things can be done if the objects are irregular.  I can take two objects and put the centers of gravity very close to each other and actually get them to repel each other.  I just can't do it with spheres.

How could it be that we can't use Newton's Shell Theorem for the Earth, while the orbital velocity of stars around the galaxy is based on Newton's Shell Theorem.
http://www.astronomy.ohio-state.edu/~ryden/ast162_7/notes30.html
"In the above equation, M is the total mass in a sphere of radius a, centered on the galactic center. (The mass outside the sphere doesn't have any net effect on the star's orbit).
Since the mass M includes the mass of the suppermassive black hole at the galactic center, M is guaranteed to be much greater than M*, the mass of a single star."
So, based on that theory we took all the matter (Stars/dust/SMBH/Dark Matter...) in the orbital radius and set the calculation as all the mass is located at the very center of the galaxy.
Hence, from the galaxy point of view, we have used all that variety of matter in the Sun' orbital radius as a very regular sphere.
However, when we come look at the Earth, suddenly it is not a regular object.
Is it real?
How could it be that the none regular matter in the sun' orbital radius is more regular than the Earth itself.
Somehow, it seems to me that we are using the idea of "regular" to prove an object which is by definition none regular (as the sphere inside the orbital radius of the Sun), while for a real regular object (as the Earth) we claim that it is not regular enough.
Sorry, that isn't the way that we have to work in science.
If the Earth isn't regular enough for the Newton's Shell Theorem, than by definition the matter in the orbital radius of the Sun shouldn't be considered as a regular.
How can we use the Newton's Shell Theorem to prove the orbital velocity of the Sun, while we reject this idea just because it contradicts our theory about tidal friction?
Therefore, If the none regular matter in the orbital radius of the Sun should be considered as a regular, than by definition the Earth with or without the Bulges should be considered as a real regular.
Don't you see that severe contradiction???

[Halc]

How could it be that we can't use Newton's Shell Theorem for the Earth, while the orbital velocity of stars around the galaxy is based on Newton's Shell Theorem.
http://www.astronomy.ohio-state.edu/~ryden/ast162_7/notes30.html
"In the above equation, M is the total mass in a sphere of radius a, centered on the galactic center. (The mass outside the sphere doesn't have any net effect on the star's orbit).

That quote is not Newton’s shell theorem, which concerns spherical objects.  That law is a similar one showing how orbital speeds are directly related to the mass of the material orbited, and the orbital radius.  The moon follows that law.  The law says nothing about additional forces resulting from the non-uniformity of the local gravitational field. Stars are flung out of their normal orbits all the time due to close encounters with passing objects.  The law you quote doesn’t prevent that.

So, based on that theory we took all the matter (Stars/dust/SMBH/Dark Matter...) in the orbital radius and set the calculation as all the mass is located at the very center of the galaxy.

The calculation of the speed of the orbit if it is a reasonably circular orbit, yes.  It is not a calculation of the local forces making changes to that orbit.

Hence, from the galaxy point of view, we have used all that variety of matter in the Sun' orbital radius as a very regular sphere.
However, when we come look at the Earth, suddenly it is not a regular object.

The galaxy isn’t regular either.  Again, consider stars being diverted out of their orbits, which happens a lot, especially in the crowded places near the halo.

Is it real?
How could it be that the none regular matter in the sun' orbital radius is more regular than the Earth itself.

Earth is a very small object and far less symmetric than the galaxy, just like a rock in my yard is hardly a sphere.  Our planet has this giant moon, making it even less symmetric.  Apples to Oranges to try to compare that to the far more uniform galaxy.  Even then, it takes only one passing rouge star to disrupt our entire solar system.  We’re just luck we’re so far out that those kinds of events are really rare.

BTW, the galaxy is not a uniform sphere.  Not even close.  Newton’s law just doesn’t apply, but the 2nd lay you quote above still does.

Somehow, it seems to me that we are using the idea of "regular" to prove an object which is by definition none regular (as the sphere inside the orbital radius of the Sun)

There is no spherical object inside the orbital radius of the sun.  The sphere is a mathematical one.  All the crap inside this radius, and not the stuff outside.  That stuff is anything but a sphere.  It resembles more of a middle-heavy pancake.
The Earth on the other hand is actually a sort of sphere, but with significant deviations from being symmetric.  The galaxy below us has no such significant deviations until one pass by at least.

How can we use the Newton's Shell Theorem to prove the orbital velocity of the Sun,

We didn’t.  We used that 2nd rule about only the matter within the orbit contributing to the velocity.  You can use that rule on the Earth/moon system as well.  It works just great.  It means you need to take into account all the little stuff in low orbit around Earth to account for the moon’s speed, but you don’t need to to account for the moon when computing the orbital speed of the ISS.  That’s what the rule says.

while we reject this idea just because it contradicts our theory about tidal friction?

It doesn’t go into tidal forces at all.  Tidal forces are not a violation of the rule.  The orbital speed of the moon never deviates from what the rule predicts.  It is at all times based on the (essentially fixed) mass of the Earth and its satellites, which is the list of all the stuff inside the Moon’s orbit.
The rule has no requirement that the distribution of that matter be regularly arranged.  Newton’s law does require that uniformity.

[Dave Lev (OP)]

How can we use the Newton's Shell Theorem to prove the orbital velocity of the Sun,

We didn’t.  We used that 2nd rule about only the matter within the orbit contributing to the velocity.  You can use that rule on the Earth/moon system as well.  It works just great.  It means you need to take into account all the little stuff in low orbit around Earth to account for the moon’s speed, but you don’t need to to account for the moon when computing the orbital speed of the ISS.  That’s what the rule says.

What do you mean by: "2nd rule about only the matter"?
Is it Newton's second law?
If so, this law isn't relevant for our discussion.
It doesn't say anything about the mass in the sphere of orbital cycle of an object:

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:"
Please advice.

[Dave Lev (OP)]

What do you mean by: "2nd rule about only the matter"?
Is it Newton's second law?

The first rule you quoted was Newton's Shell Theorem concerning only uniform spherical objects.
The 2nd rule you quoted was the one from the ohio-state site concerning the mass that has net effect on an orbit and the mass that doesn't.
I don't know the name of that rule or who came up with it, but you quoted it, and then you misrepresented what it means.

The "second rule" is just a direct outcome from the "first rule" which is the Newton's Shell Theorem.
With all the respect to ohio-state (and I have a respect...), they can't just invent new rules for gravity.
So, the second statement is not a second rule, it is just a logical outcome from Newton's Shell Theorem.
They also explain how to extract M (M = mass inside star's orbit (in solar masses) ) from kepler law.
"Each star in the disk is on a very nearly circular orbit, anchored by all the mass enclosed within its orbit, whether it's luminous or not. Thus, the amount of mass within a star's orbit can be determined from Kepler's Third Law:"

Hence, by Kepler law we can calculate the total mass which is requested to meet the orbital velocity of the sun around the galaxy, while Newton’s First Theorem tell us that: "M is the total mass in a sphere of radius a, centered on the galactic center. (The mass outside the sphere doesn't have any net effect on the star's orbit)" and therefore, we also can understand that:
"Since the mass M includes the mass of the suppermassive black hole at the galactic center, M is guaranteed to be much greater than M*, the mass of a single star."
In any case, if that was not clear enough, please see the following statement:
http://users.math.cas.cz/~krizek/cosmol/pdf/B102.pdf
Newton’s First Theorem - If the density distribution of a ball of mass M is spherically symmetric, then the size of the force between the ball and a point mass m, that lies outside the interior of the ball, is given by the left-hand side of (1), where r is the distance between the point and the center of the ball."

[Dave Lev (OP)]

"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:"
Please advice.

I don't see the relevancy of that law to this portion of the discussion.  It says that the center of gravity of something like Earth follows a smooth curve even if the various parts (like your mailbox) don't.

This law is very relevant
It actually confirms that the shape of the object is none relevant for its central point of mass.
As long as all the masses in the object are fully connected the center of mass of this object is none relevant with its shape. (Spanner, dog, cat or even elephant).
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"A set of masses connected by springs will follow a path such that its center of mass moves along the same path that a point mass of the same total mass would follow under the influence of the same net force."
So, even if that spanner has an offset, it won't set any extra thrust. In our calculation we just need to focus on its center of mass.
In the same token, as the bulges are fully connected to the Earth, they are part of the whole Earth body. Therefore, in our calculation we need only to follow the center of the mass of the earth.
In other words - We won't get any extra negative or positive thrust due to mountains, oceans or any sort of bulges with or without offset while the whole masses are connected together.
That law by itself proves that the idea of thrust due to bulges is totally incorrect.
Conclusion -
Based on the following laws (each one by itself):
1. Newton's Shell Theorem
2. Newton's second law
I have proved that the shape of the object can't issue any sort of extra thrust. Newton didn't specify that idea in his laws. Therefore, the hypothetical idea of extra thrust due to the offset (or special shape of the object - with or without bulges) is incorrect.
Our scientists must find better idea why all the planets and all the far enough moons are drifting outwards.