[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?

[Halc]

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.

[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.

[Halc]

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.

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.

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.

[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.

[Halc]

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.

I'm not disputing the rule.  It is a known thing, yes.  I just don't know the name of it, or how it is derived from Newton's sphere 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.

You're applying the 2nd rule there, not the first theorem.  The galaxy is not a sphere nor is it spherically symmetric, so the first theorem does not immediately apply.  The nameless law does however, so we're good.

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."

Slightly generalized version, and thus a better one, yes.  It doesn't require a sphere, but merely something spherically symmetric.  But the galaxy isn't, so this law doesn't directly apply.   The galaxy is roughly modeled as a disk, and the force vector of a point out of the plane of a disk does not point to the center of mass of the disk like Newton's theorem says it must if it were spherically symmetrical.

[Halc]

This law is very relevant
It actually confirms that the shape of the object is none relevant for its central point of mass.

I didn't say otherwise.  The shape of Earth for instance, with its bulges, has no effect on Earth's center of mass.  I agree with that.

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).

All the others, yes, and even a spraying garden hose, but not the cat, which has properties outside physics.

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."

OK.  That one is pretty obvious too, since it derives directly from the computation of center of mass.  The springs are not even necessary.  It can be a collection of 13 random stars from nowhere in particular, and that collection of 13 unrelated stars will have a center of mass that will follow this rule.  What it doesn't mean is that an object at that center of mass will follow the same path.
What it especially doesn't mean is that any force exerted by that collection of spring-attached objects is going to be the same as if all the mass was concentrated at their center of gravity.  This is very easy to demonstrate.

So, even if that spanner has an offset, it won't set any extra thrust.

The rules don't say that at all.  They talk about net force and those net forces acting on the center of gravity of each object in question.  So if the spanner puts a net force on an object that is anything but perpendicular to its motion, it will be exerting thrust to it.

The Earth/Sun system could be considered one such object which is not a sphere, and thus it puts massive thrust on another object (the moon), all because of its offset.  According to your assertions, that should not happen, and the moon should wander off around the sun on its own in a nice circular orbit around the center of mass of the combined Earth/Sun object.

In our calculation we just need to focus on its center of mass.

Wrong!!  Center of mass has no angular momentum, and this tidal thrust effect is all about net forces resulting from transfer of angular momentum.  None of the laws above describe angular effects on the tumbling spanner and such.

I throw a rapidly spinning pool noodle, and it is spinning far slower before it hits the ground.  The net forces on the noodle do indeed determine the path of its center of gravity, but do not in any way describe the loss of spin.  That rules is inadequate for the situation being described.

The orbit about the galaxy is less about angular momentum and forces since there is no significant transference going on.  You can treat a lot of things as point masses on that scale, but not the galaxy as a whole since it is not spherically symmetrical.

Conclusion -
Based on the following laws (each one by itself):
1. Newton's Shell Theorem
2. Newton's second law

You can't take laws by themselves.  Let's take these two laws.  Consider me on Earth with the sun directly overhead, and Mercury and Venus don't exist.  In a sphere of radius R where R is the distance from the sun to me, only the mass in that radius determines my orbit.  It is the Sun which is spherically symmetric.  So the first law applies.
Now let's apply the 2nd law.  I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon), and the mass outside the radius R has no effect on my orbit, therefore I should drop in towards the sun, getting sucked into the sky.  The Earth cannot hold me down because it is outside R when the sun is directly overhead, and thus does not contribute to my orbit, per the 2nd law.

This is the sort of proofs I'm seeing from you.

[Dave Lev (OP)]

Thanks Halc.
I really appreciate all your efforts.
However, please try to use some solid evidences (mathematics, Newton/kepler.. laws) in order to prove your statements.

The rules don't say that at all.  They talk about net force and those net forces acting on the center of gravity of each object in question.  So if the spanner puts a net force on an object that is anything but perpendicular to its motion, it will be exerting thrust to it.

Can you please prove that a spanner can put a net force on an object just by pointing to some offset?

You can't take laws by themselves.  Let's take these two laws.  Consider me on Earth with the sun directly overhead, and Mercury and Venus don't exist.  In a sphere of radius R where R is the distance from the sun to me, only the mass in that radius determines my orbit.  It is the Sun which is spherically symmetric.  So the first law applies.
Now let's apply the 2nd law.  I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon), and the mass outside the radius R has no effect on my orbit, therefore I should drop in towards the sun, getting sucked into the sky.  The Earth cannot hold me down because it is outside R when the sun is directly overhead, and thus does not contribute to my orbit, per the 2nd law.

That example is not clear to me.
If the sun is just above me while I am waking on Earth, than based on Newton second law, I'm an integrated mass of Earth.
Therefore, the Sun has no impact at all about my location.
If there is no Earth or moon, and it is all about me and the Sun, than my orbital velocity must be a direct outcome of gravity force based on R. In this case, the Earth and the Moon are not there to have any impact on my orbital cycle around the Sun.

With regards to the following message:
"I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon)"
This is a severe mistake.
If I have to orbit around the Sun, there is no way to slow down due to that Earth spin velocity cancelation.
So, you don't have to maintain your orbital velocity. The Sun gravity force works for you. This is a key element.
Newton didn't specify even one word about the impact of spinning velocity.
This is a new idea which had not been confirmed. It seems to me as a new idea which we have invented just in order to offer a nice answer for: Why the Moon and the earth are drifting outwards?
If you believe that there is a possibility to slow down the earth orbital velocity "because Earth's spin velocity cancels some of its orbital velocity at noon", than please prove it.

Wrong!!  Center of mass has no angular momentum, and this tidal thrust effect is all about net forces resulting from transfer of angular momentum.  None of the laws above describe angular effects on the tumbling spanner and such.

I throw a rapidly spinning pool noodle, and it is spinning far slower before it hits the ground.  The net forces on the noodle do indeed determine the path of its center of gravity, but do not in any way describe the loss of spin.  That rules is inadequate for the situation being described.

The orbit about the galaxy is less about angular momentum and forces since there is no significant transference going on.  You can treat a lot of things as point masses on that scale, but not the galaxy as a whole since it is not spherically symmetrical

We do not discuss about rapidly spinning pool of noodle.
We discuss about Sun/earth/Moon orbital cycles.

If the angular momentum or the revolving speed of the Moon or the Earth can slow down the orbital velocity of the Erath or the Moon - than please prove it.
I still can't understand how any sort of offset in bulges can set any sort of thrust.
Please use the spanner as an example. you are more than welcome to force the spanner in any sort of offset as you wish. Try to prove why by doing so we shall get extra trust on the orbital object.
Please try to prove this idea by mathematics.

[Dave Lev (OP)]

So far we have just discussed about the impact of the orbital cycles of Planets and Moons.
However, what about the stars? What about our star - the Sun?
Why all our scientists are positively sure that during all his life time the Sun had to keep the same orbital radius?
How could it be that all the moons and Planets are drifting outwards, (or inwards based on unproved tidal idea) while the sun is fixed at the same radius?
How could it be that in one hand our scientists claim that the SMBH increases its mass by eating stars and gas clouds, while on the other hand they don't consider an option that stars must migrate/drift inwards in order to supply the requested food for the SMBH monster?
If the SMBH has 4 x 10^6 sun mass, (while our scientists believe that this mass had been taken from the stars in the galaxy) than somehow 4 x 10^6  stars had to drift inwards.
So why our scientists are so sure that the Sun was always at the same distance from the center of the galaxy???
What makes our star so unique that it had to stay so far away from the monster at the galaxy center?

[Halc]

Can you please prove that a spanner can put a net force on an object just by pointing to some offset?

I didn't say it could do it by pointing to an offset.
The spanner is tumbling, and that tumblingis slowing measurably.  It couldn't do that if it was treated as a point mass as you are attempting to do.  You can't put torque on a point.

You can't take laws by themselves.  Let's take these two laws.  Consider me on Earth with the sun directly overhead, and Mercury and Venus don't exist.  In a sphere of radius R where R is the distance from the sun to me, only the mass in that radius determines my orbit.  It is the Sun which is spherically symmetric.  So the first law applies.
Now let's apply the 2nd law.  I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon), and the mass outside the radius R has no effect on my orbit, therefore I should drop in towards the sun, getting sucked into the sky.  The Earth cannot hold me down because it is outside R when the sun is directly overhead, and thus does not contribute to my orbit, per the 2nd law.

That example is not clear to me.
If the sun is just above me while I am waking on Earth, than based on Newton second law, I'm an integrated mass of Earth.

You're not integrated with Earth since you are completely detached from it. Seatbelts are cheating.  I'm just considering the gravity of the sun on your orbit.  According to that 2nd law, or at least according to the way you are using it, nothing outside that R sphere has a net effect on my motion, therefore, since Earth is entirely outside that sphere, its gravity doesn't affect me.

I know it's wrong, but that's how you're interpreting the rule: by oversimplifying and not considering deviations from the uniformity.  The Earth is a huge deviation.  Deviations matter.  The tidal bulges are deviations, and the change the direction of the force vector acting on the moon just as Earth changes the force vector acting on me at high noon.  But you cherry pick which deviations matter and which do not, depending on what purpose you desire.  That's very fallacious reasoning.

I haven't figured out your purpose in all this.  Clearly not here to learn.  You seem bent on finding contradiction in simple orbital mechanics, but not sure why.

Therefore, the Sun has no impact at all about my location.
If there is no Earth or moon, and it is all about me and the Sun, than my orbital velocity must be a direct outcome of gravity force based on R. In this case, the Earth and the Moon are not there to have any impact on my orbital cycle around the Sun.

So rule #2 suddenly doesn't matter when you find it inconvenient?

With regards to the following message:
"I am going too slow to maintain my orbit at that radius (because Earth's spin velocity cancels some of its orbital velocity at noon)"
This is a severe mistake.
If I have to orbit around the Sun, there is no way to slow down due to that Earth spin velocity cancelation.

My velocity relative to the sun varies daily, fastest at midnight and slowest at noon.  Somewhere in between is proper orbital speed at this radius.  At midnight the speed is too high, and at noon it is too low, as per Kepler's 3rd law.  The speeds are easy to compute:  About 30 km/s ± about 0.5 km/s speed when the sun is directly below and above me respectively.

So, you don't have to maintain your orbital velocity. The Sun gravity force works for you. This is a key element.
Newton didn't specify even one word about the impact of spinning velocity.
This is a new idea which had not been confirmed.

What, that your speed relative to the sun is lower at noon?  That's pretty obvious I'd think.

If you believe that there is a possibility to slow down the earth orbital velocity

My speed, not Earth's speed.  I'm the one that suppose to fall closer to the sun because I'm moving slower than Kepler says I should.
Of course Earth speed is slowest every full moon, for the same reason.  The difference isn't as much as ±0.5 m/sec.

We do not discuss about rapidly spinning pool of noodle.

OK, you don't know what a pool noodle is.  It's like the spanner, but more susceptible to angular forces.

If the angular momentum or the revolving speed of the Moon or the Earth can slow down the orbital velocity of the Erath or the Moon - than please prove it.

That was demonstrated many posts ago.  I should the vector math quite a ways back.  The force vector on the moon is not perpendicular to its motion, therefore there is work being performed (energy transfer).

I still can't understand how any sort of offset in bulges can set any sort of thrust.

Understanding it would cause your argument to fall apart, therefore nothing I can say will make you understand it.  I'm fine with that.

Please use the spanner as an example. you are more than welcome to force the spanner in any sort of offset as you wish. Try to prove why by doing so we shall get extra trust on the orbital object.
Please try to prove this idea by mathematics.

OK, there is a spanner suspended exactly through its center of gravity on a frictionless axis.  It should be able to spin freely, and be balanced.
I give it a permanent stationary offset of 45°.  The spanner is 40 cm long, 20 on each side of the axis.  The lower side is 6372000 meters from the center of earth and has a force of GmM/40602384000000 acting on it.  The upper end is a tiny bit further away from earth and has a force of around GmM/40602384382320 resulting in a 1e-6 % difference in force. It will start to rotate to vertical because the bottom is attracted more than the top, being closer to Earth.  That is a torque force being put on an object despite its center of gravity never moving or the force on that center of gravity ever changing.
You can get rid of the axis suspending the spanner if you just put it in orbit and let its speed hold it up there.

The numbers get more significant if the spanner is a bit longer, like say over 12000 km long, even if the 12000 km spanner has an offset far less than 45°

[Dave Lev (OP)]

Thanks
Great example

I give it a permanent stationary offset of 45°.  The spanner is 40 cm long, 20 on each side of the axis.  The lower side is 6372000 meters from the center of earth and has a force of GmM/40602384000000 acting on it.  The upper end is a tiny bit further away from earth and has a force of around GmM/40602384382320 resulting in a 1e-6 % difference in force.

However, I really don't understand why there is a difference in forces.
You discuss about 20 cm.
Let's set the whole Earth at the shape of spanner.
So, the Earth will look like an extended object with all of its mass while the length of each side is 10,000 Km. 
Based on Newton's Second Law for an Extended Object
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"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:
F = M a
In order to understand the calculation, we need to look at the Following "Newton's Second Law for a System of Particles"
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html#c2
"Newton's Second Law for a System of Particles:
The form of Newton's second law for a system of particles will be developed with the understanding that the result will apply to any extended object where the particles are in face connected to each other.
The center of mass of a system of particles can be determined from their masses and locations."
So, Our dear Newton set a complicated calculation in order to get the outcome of:
M a = F
Newton actually tells us that the shape of the object and its offset can't contribute any extra force as long as all the particles of the object are connected.
So, if you still think that there is an error in Newton calculation, please offer the updated calculation to prove why each side can contribute different force (while all the particles are fully connected).

[Halc]

However, I really don't understand why there is a difference in forces.

I modeled the spanner as two masses separated by 30 cm (I did not put all the mass at the ends of the 40 cm spanner), and applied Newton's F=GMm/rČ to get the force, and then F=ma to get the acceleration of spanner that sets it in motion.
In addition, if the Earth under it was entirely stationary and even perfectly spherical at first, it too begins to rotate by application of Newton's conservation of angular momentum law.  The moon does not do this to the Earth because unlike the spanner, it has no net offset and thus exerts no net torque directly on Earth.  All Earth's torque is due to friction, and our spanner example does not involve friction.  We can put some air around it so it stops rotating after a while.

Let's set the whole Earth at the shape of spanner.
So, the Earth will look like an extended object with all of its mass while the length of each side is 10,000 Km. 
Based on Newton's Second Law for an Extended Object
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html
"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:
F = M a

Yes, the motion of the center of mass will follow that law, just like it says.  I never said otherwise.

In order to understand the calculation, we need to look at the Following "Newton's Second Law for a System of Particles"

We need to come up with different names for all these laws, because you're calling them all Newton's second law of something, which is confusing.  Newton's second law typically refers to the second law of motion, which is F=ma.  So lets just call the one above the 'extended object law' and this one below the 'law for a system of particles'.

http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/n2ext.html#c2
"Newton's Second Law for a System of Particles:
The form of Newton's second law for a system of particles will be developed with the understanding that the result will apply to any extended object where the particles are in face connected to each other.
The center of mass of a system of particles can be determined from their masses and locations."

That's not the law you quoted before.  It's just a statement of the fact that the center of mass of any set of particles (connected or not) can be computed.  This lacks any description of how that computation is done (which isn't hard), and it doesn't mention net forces like the last time you mentioned it.  Maybe this is a different law.
Anyway, this wording just says there is a COM and that is it computable.  Hardly a revolutionary statement, and not something that I would call a 'law'.
The 'law' does not say that the set of particles will exert gravitational force as if they were all at the center of mass, or that the center of mass will follow the same path as the object with all its mass concentrated there.  This is very easy to demonstrate with just a two-particle object.
And yet I find you here misrepresenting these laws and attempting to do just that.

So, Our dear Newton set a complicated calculation in order to get the outcome of:
M a = F
Newton actually tells us that the shape of the object and its offset can't contribute any extra force as long as all the particles of the object are connected.

He said no such thing. That does not follow from any of the laws you've quoted.

So, if you still think that there is an error in Newton calculation, please offer the updated calculation to prove why each side can contribute different force (while all the particles are fully connected).

Particles are never connected.  Two things cannot touch.  None of Newtons laws about sets of objects require connectivity for this reason.  The center-of-mass of the Earth-moon system follows an ellipical path around the sun, which the center of mass of just Earth does not because the moon is yanking back and forth each month.  So treating the two as one unit gives a far smoother curve, even though the two are not physically connected.

This is not always the case.   Earth and Venus could be connected with a thin spidery thread and the center-of-mass of the single object would very much obey that law you quote (you quoted a much better version of it in a prior post), but that COM does not follow the same path as would an object that was actually all at that spot, and does not generate a gravitational field that is identical to a single mass at the COM.

Point is, the connection is not necessary.