APOD: Phobos: Doomed Moon of Mars (2022 Jul 03)

[APOD Robot]

Phobos: Doomed Moon of Mars

Explanation: This moon is doomed. Mars, the red planet named for the Roman god of war, has two tiny moons, Phobos and Deimos, whose names are derived from the Greek for Fear and Panic. These martian moons may well be captured asteroids originating in the main asteroid belt between Mars and Jupiter or perhaps from even more distant reaches of our Solar System. The larger moon, Phobos, is indeed seen to be a cratered, asteroid-like object in this stunning color image from the robotic Mars Reconnaissance Orbiter, with objects as small as 10 meters visible. But Phobos orbits so close to Mars - about 5,800 kilometers above the surface compared to 400,000 kilometers for our Moon - that gravitational tidal forces are dragging it down. In perhaps 50 million years, Phobos is expected to disintegrate into a ring of debris.

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[Steve Hoyle]

The disintergration of Phobos raises the question "What will be the consequences for earth when this happens?" A rather academic question given that it is 50m years in the future but, if there are likely to be significant consequences from such an event, then the real question is, "Is there another moon amoungst the 203 that are orbiting the outer planets that may disintergrate in a similar manner but perhaps next week?"

[Chris Peterson]

The disintergration of Phobos raises the question "What will be the consequences for earth when this happens?" A rather academic question given that it is 50m years in the future but, if there are likely to be significant consequences from such an event, then the real question is, "Is there another moon amoungst the 203 that are orbiting the outer planets that may disintergrate in a similar manner but perhaps next week?"

There is no reason to think it will have any consequences for Earth at all. It will form a belt of debris around Mars which will probably decay onto the surface eventually.

Asteroids in this 10 km class break up from time to time, either because of collisions or tidal disruption by Jupiter. Some of the debris may occasionally cross Earth's path in the form of meteors, but the debris from Phobos is gravitationally bound to Mars.

[VictorBorun]

The disintergration of Phobos raises the question "What will be the consequences for earth when this happens?" A rather academic question given that it is 50m years in the future but, if there are likely to be significant consequences from such an event, then the real question is, "Is there another moon amoungst the 203 that are orbiting the outer planets that may disintergrate in a similar manner but perhaps next week?"

how can tidal disintergration of a satellite send a fragment away from the planet's orbit to the interplanetary space?
We're talking a few km/s here, and is it possible for a cracking rock to kick a pebble so fast?

Or can the falling parts kick some pebbles from the surface of the planet? We know an asteroid can, but an asteroid easily has an escape level velocity after falling into the planet's gravitational well. A shower from a ring would have a smaller velocity by the factor of 1.4, would it not?

[orin stepanek]

I think it is better for it to break up into debris than to come
crashing down on the planet in one lump! The debris would make a
ring around the planet until they crashed into Mars as meteorites! If
anyone is living there at that time they are gonna need a shelter!

[johnnydeep]

The disintergration of Phobos raises the question "What will be the consequences for earth when this happens?" A rather academic question given that it is 50m years in the future but, if there are likely to be significant consequences from such an event, then the real question is, "Is there another moon amoungst the 203 that are orbiting the outer planets that may disintergrate in a similar manner but perhaps next week?"

how can tidal disintergration of a satellite send a fragment away from the planet's orbit to the interplanetary space?
We're talking a few km/s here, and is it possible for a cracking rock to kick a pebble so fast?

Or can the falling parts kick some pebbles from the surface of the planet? We know an asteroid can, but an asteroid easily has an escape level velocity after falling into the planet's gravitational well. A shower from a ring would have a smaller velocity by the factor of 1.4, would it not?

Where are you getting the value 1.4 from? It's close to the square root of 2, but is that significant for some reason here?

[VictorBorun]

A shower from a ring would have a smaller velocity by the factor of 1.4, would it not?

Where are you getting the value 1.4 from? It's close to the square root of 2, but is that significant for some reason here?

For a planet of radius R and acceleration of an object in free fall at surface g we can [with Newton's Universal gravitation] express the force of gravitational attraction at a distance of R+h from the center as
am = mgR²(R+h)⁻²
Thus the potential energy of being faraway and able to fall down to the surface is
∫(h from ∞ to 0)dh am = ∫(R+h from ∞ to R)d(R+h) mgR²(R+h)⁻² =
−mgR²(R+h)⁻¹ (with R+h from ∞ to R) = mgR.

Now the kinetic energy at a low orbit with almost zero potential energy can be derived as follows. To balance the force of gravitational attraction at a low orbit and the centrifugal force of orbiting object's reference we know that velocity v must be such that
mg = mv²R⁻¹
Thus the kinetic energy for a fragment in the ring before entering the atmosphere is
2⁻¹mv² = 2⁻¹mgR

Therefore the kinetic energy at a low orbit is 2 times smaller than the escape energy; after an escape energy of a distant and slowly approaching asteroid is transferred to the collision kinetic energy, the velocity is 1.4 times larger than that at a low orbit.

[MarkBour]

I wonder, if, as it approaches its Roche limit, will it perhaps begin to rotate?

[VictorBorun]

I wonder, if, as it approaches its Roche limit, will it perhaps begin to rotate?

I think that strong tidal forces would tidally lock a satellite on a close orbit.
Any rotation left in the satellite's reference would be because of an eccentricity of the orbit and that wiggle would not be infinitely fast, it would have a period of the orbiting.

So no spinning up for a satellite at the brink of the tidal fission

[Chris Peterson]

I wonder, if, as it approaches its Roche limit, will it perhaps begin to rotate?

Why would it? Once it starts to break up the individual pieces will assume their own rotations, with the total angular momentum conserved, of course.

[johnnydeep]

A shower from a ring would have a smaller velocity by the factor of 1.4, would it not?

Where are you getting the value 1.4 from? It's close to the square root of 2, but is that significant for some reason here?

For a planet of radius R and acceleration of an object in free fall at surface g we can [with Newton's Universal gravitation] express the force of gravitational attraction at a distance of R+h from the center as
am = mgR²(R+h)⁻²
Thus the potential energy of being faraway and able to fall down to the surface is
∫(h from ∞ to 0)dh am = ∫(R+h from ∞ to R)d(R+h) mgR²(R+h)⁻² =
−mgR²(R+h)⁻¹ (with R+h from ∞ to R) = mgR.

Now the kinetic energy at a low orbit with almost zero potential energy can be derived as follows. To balance the force of gravitational attraction at a low orbit and the centrifugal force of orbiting object's reference we know that velocity v must be such that
mg = mv²R⁻¹
Thus the kinetic energy for a fragment in the ring before entering the atmosphere is
2⁻¹mv² = 2⁻¹mgR

Therefore the kinetic energy at a low orbit is 2 times smaller than the escape energy; after an escape energy of a distant and slowly approaching asteroid is transferred to the collision kinetic energy, the velocity is 1.4 times larger than that at a low orbit.

Thanks for the formatting effort it took to explain that, but I'm afraid I'll need to read it about a dozen more times while simultaneously referencing Wikipedia. It's been over 40 years since I took high school physics.

[Chris Peterson]

Where are you getting the value 1.4 from? It's close to the square root of 2, but is that significant for some reason here?

For a planet of radius R and acceleration of an object in free fall at surface g we can [with Newton's Universal gravitation] express the force of gravitational attraction at a distance of R+h from the center as
am = mgR²(R+h)⁻²
Thus the potential energy of being faraway and able to fall down to the surface is
∫(h from ∞ to 0)dh am = ∫(R+h from ∞ to R)d(R+h) mgR²(R+h)⁻² =
−mgR²(R+h)⁻¹ (with R+h from ∞ to R) = mgR.

Now the kinetic energy at a low orbit with almost zero potential energy can be derived as follows. To balance the force of gravitational attraction at a low orbit and the centrifugal force of orbiting object's reference we know that velocity v must be such that
mg = mv²R⁻¹
Thus the kinetic energy for a fragment in the ring before entering the atmosphere is
2⁻¹mv² = 2⁻¹mgR

Therefore the kinetic energy at a low orbit is 2 times smaller than the escape energy; after an escape energy of a distant and slowly approaching asteroid is transferred to the collision kinetic energy, the velocity is 1.4 times larger than that at a low orbit.

Thanks for the formatting effort it took to explain that, but I'm afraid I'll need to read it about a dozen more times while simultaneously referencing Wikipedia. It's been over 40 years since I took high school physics. :-(

Here's a different way of looking at it, less rigorous but maybe more intuitive. Escape velocity from the surface of Mars is about 5 km/s. That's the lowest speed a body that wasn't gravitationally bound to Mars could strike the surface (but most things would hit much faster). Escape velocity (with respect to Mars) at the height of Phobos is about 3 km/s. Basically, material from Phobos striking the surface will be impacting at a much lower speed. The speed of something already gravitationally bound to Mars. So it makes sense that there wouldn't be enough energy to eject material fast enough to exceed the much greater martian surface escape velocity.

[johnnydeep]

For a planet of radius R and acceleration of an object in free fall at surface g we can [with Newton's Universal gravitation] express the force of gravitational attraction at a distance of R+h from the center as
am = mgR²(R+h)⁻²
Thus the potential energy of being faraway and able to fall down to the surface is
∫(h from ∞ to 0)dh am = ∫(R+h from ∞ to R)d(R+h) mgR²(R+h)⁻² =
−mgR²(R+h)⁻¹ (with R+h from ∞ to R) = mgR.

Now the kinetic energy at a low orbit with almost zero potential energy can be derived as follows. To balance the force of gravitational attraction at a low orbit and the centrifugal force of orbiting object's reference we know that velocity v must be such that
mg = mv²R⁻¹
Thus the kinetic energy for a fragment in the ring before entering the atmosphere is
2⁻¹mv² = 2⁻¹mgR

Therefore the kinetic energy at a low orbit is 2 times smaller than the escape energy; after an escape energy of a distant and slowly approaching asteroid is transferred to the collision kinetic energy, the velocity is 1.4 times larger than that at a low orbit.

Thanks for the formatting effort it took to explain that, but I'm afraid I'll need to read it about a dozen more times while simultaneously referencing Wikipedia. It's been over 40 years since I took high school physics.

Here's a different way of looking at it, less rigorous but maybe more intuitive. Escape velocity from the surface of Mars is about 5 km/s. That's the lowest speed a body that wasn't gravitationally bound to Mars could strike the surface (but most things would hit much faster). Escape velocity (with respect to Mars) at the height of Phobos is about 3 km/s. Basically, material from Phobos striking the surface will be impacting at a much lower speed. The speed of something already gravitationally bound to Mars. So it makes sense that there wouldn't be enough energy to eject material fast enough to exceed the much greater martian surface escape velocity.

Thanks. That helps some, but the orbital speed of Phobos averages 2.14 km/s (per Wikipedia). Wouldn't that increase the impact speed (and energy) over the 3 km/s escape velocity?

[Fred the Cat]

On HAL, by coincidence, was an document that discussed monoliths on Mars and Phobos.

And you have to trust HAL.

[MarkBour]

I wonder, if, as it approaches its Roche limit, will it perhaps begin to rotate?

Why would it? Once it starts to break up the individual pieces will assume their own rotations, with the total angular momentum conserved, of course.

I'm just looking at this image from the Wikipedia "Roche Limit" page and thinking about it intuitively (https://en.wikipedia.org/wiki/Roche_limit)

As the object approaches the Roche limit, its inner chunks are wanting to orbit angularly more rapidly than the rest, and its outer chunks are wanting to orbit angularly more slowly than the rest. If there is a period of time in which this is appreciable, but during which the body can still "hold it together", then I think it would induce a rotational motion. You might even have some chunks breaking loose, but then rolling along the surface in that direction as well.

My intuition, which is certainly a poor guide here, is that for a fluid satellite, it would not happen. The fluid would just stretch into an ellipsoid and slightly rotate forward, but only less than a single visible rotation of the "body". and then it would begin to disperse (exactly like this image).

But for a rubble pile, I have even more trouble guessing.

I wonder if someone has recorded an experiment that has found a way to show the details at the end before disintegration, or if some numerical simulation has worked it out with satisfactory conclusiveness.

The only example observation I know of is Comet Shoemaker–Levy 9, also discussed in that same Wikipedia article, but not much could be concluded from that example in regard to the fate of Phobos. If someone had a lot of money, they could easily enough set this up orbiting a few small satellites around the Moon. Of course, I doubt anyone is going to do that as a purposeful experiment. But maybe, with lots of varied activity up there, some related observations will be part of life in cislunar space.

[johnnydeep]

I wonder, if, as it approaches its Roche limit, will it perhaps begin to rotate?

Why would it? Once it starts to break up the individual pieces will assume their own rotations, with the total angular momentum conserved, of course.

Capture1.png

I'm just looking at this image from the Wikipedia "Roche Limit" page and thinking about it intuitively (https://en.wikipedia.org/wiki/Roche_limit)

As the object approaches the Roche limit, its inner chunks are wanting to orbit angularly more rapidly than the rest, and its outer chunks are wanting to orbit angularly more slowly than the rest. If there is a period of time in which this is appreciable, but during which the body can still "hold it together", then I think it would induce a rotational motion. You might even have some chunks breaking loose, but then rolling along the surface in that direction as well.

My intuition, which is certainly a poor guide here, is that for a fluid satellite, it would not happen. The fluid would just stretch into an ellipsoid and slightly rotate forward, but only less than a single visible rotation of the "body". and then it would begin to disperse (exactly like this image).

But for a rubble pile, I have even more trouble guessing.

I wonder if someone has recorded an experiment that has found a way to show the details at the end before disintegration, or if some numerical simulation has worked it out with satisfactory conclusiveness.

The only example observation I know of is Comet Shoemaker–Levy 9, also discussed in that same Wikipedia article, but not much could be concluded from that example in regard to the fate of Phobos. If someone had a lot of money, they could easily enough set this up orbiting a few small satellites around the Moon. Of course, I doubt anyone is going to do that as a purposeful experiment. But maybe, with lots of varied activity up there, some related observations will be part of life in cislunar space.

Hmm. What would happen to a 50 km carbon nanotube rod (assumed to be strong enough to resist being torn apart) oriented perpendicularly pointing to Mars and starting out in orbit at the Roche limit? Would it just rotate to be tangential to the orbit, overshoot a bit, then rotate back, etc., eventually settling in to a stable tangential orientation?

[Chris Peterson]

Why would it? Once it starts to break up the individual pieces will assume their own rotations, with the total angular momentum conserved, of course.

Capture1.png

I'm just looking at this image from the Wikipedia "Roche Limit" page and thinking about it intuitively (https://en.wikipedia.org/wiki/Roche_limit)

As the object approaches the Roche limit, its inner chunks are wanting to orbit angularly more rapidly than the rest, and its outer chunks are wanting to orbit angularly more slowly than the rest. If there is a period of time in which this is appreciable, but during which the body can still "hold it together", then I think it would induce a rotational motion. You might even have some chunks breaking loose, but then rolling along the surface in that direction as well.

My intuition, which is certainly a poor guide here, is that for a fluid satellite, it would not happen. The fluid would just stretch into an ellipsoid and slightly rotate forward, but only less than a single visible rotation of the "body". and then it would begin to disperse (exactly like this image).

But for a rubble pile, I have even more trouble guessing.

I wonder if someone has recorded an experiment that has found a way to show the details at the end before disintegration, or if some numerical simulation has worked it out with satisfactory conclusiveness.

The only example observation I know of is Comet Shoemaker–Levy 9, also discussed in that same Wikipedia article, but not much could be concluded from that example in regard to the fate of Phobos. If someone had a lot of money, they could easily enough set this up orbiting a few small satellites around the Moon. Of course, I doubt anyone is going to do that as a purposeful experiment. But maybe, with lots of varied activity up there, some related observations will be part of life in cislunar space.

Hmm. What would happen to a 50 km carbon nanotube rod (assumed to be strong enough to resist being torn apart) oriented perpendicularly pointing to Mars and starting out in orbit at the Roche limit? Would it just rotate to be tangential to the orbit, overshoot a bit, then rotate back, etc., eventually settling in to a stable tangential orientation?

A rigid body that is held together by forces much stronger than gravity doesn't have a Roche limit. I'd expect your rod to orbit in the perpendicular orientation forever, that being where it is tidally locked.

[Chris Peterson]

I wonder, if, as it approaches its Roche limit, will it perhaps begin to rotate?

Why would it? Once it starts to break up the individual pieces will assume their own rotations, with the total angular momentum conserved, of course.

Capture1.png

I'm just looking at this image from the Wikipedia "Roche Limit" page and thinking about it intuitively (https://en.wikipedia.org/wiki/Roche_limit)

As the object approaches the Roche limit, its inner chunks are wanting to orbit angularly more rapidly than the rest, and its outer chunks are wanting to orbit angularly more slowly than the rest. If there is a period of time in which this is appreciable, but during which the body can still "hold it together", then I think it would induce a rotational motion. You might even have some chunks breaking loose, but then rolling along the surface in that direction as well.

My intuition, which is certainly a poor guide here, is that for a fluid satellite, it would not happen. The fluid would just stretch into an ellipsoid and slightly rotate forward, but only less than a single visible rotation of the "body". and then it would begin to disperse (exactly like this image).

But for a rubble pile, I have even more trouble guessing.

I wonder if someone has recorded an experiment that has found a way to show the details at the end before disintegration, or if some numerical simulation has worked it out with satisfactory conclusiveness.

The only example observation I know of is Comet Shoemaker–Levy 9, also discussed in that same Wikipedia article, but not much could be concluded from that example in regard to the fate of Phobos. If someone had a lot of money, they could easily enough set this up orbiting a few small satellites around the Moon. Of course, I doubt anyone is going to do that as a purposeful experiment. But maybe, with lots of varied activity up there, some related observations will be part of life in cislunar space.

A rubble pile is essentially a fluid. (This isn't too difficult to model... much easier and less expensive than a physical experiment!)

[Chris Peterson]

Thanks for the formatting effort it took to explain that, but I'm afraid I'll need to read it about a dozen more times while simultaneously referencing Wikipedia. It's been over 40 years since I took high school physics. :-(

Here's a different way of looking at it, less rigorous but maybe more intuitive. Escape velocity from the surface of Mars is about 5 km/s. That's the lowest speed a body that wasn't gravitationally bound to Mars could strike the surface (but most things would hit much faster). Escape velocity (with respect to Mars) at the height of Phobos is about 3 km/s. Basically, material from Phobos striking the surface will be impacting at a much lower speed. The speed of something already gravitationally bound to Mars. So it makes sense that there wouldn't be enough energy to eject material fast enough to exceed the much greater martian surface escape velocity.

Thanks. That helps some, but the orbital speed of Phobos averages 2.14 km/s (per Wikipedia). Wouldn't that increase the impact speed (and energy) over the 3 km/s escape velocity?

It's in free fall. Its orbital speed isn't a factor. Consider as well that this isn't a rigid body with high material strength. When it breaks up, it's going to be boulders and gravel. Which will mostly burn up without even reaching the ground, and all of which will be substantially reduced in speed by the atmosphere, even if the ground is reached.

[VictorBorun]

Capture1.png

I'm just looking at this image from the Wikipedia "Roche Limit" page and thinking about it intuitively (https://en.wikipedia.org/wiki/Roche_limit)
As the object approaches the Roche limit, its inner chunks are wanting to orbit angularly more rapidly than the rest, and its outer chunks are wanting to orbit angularly more slowly than the rest. If there is a period of time in which this is appreciable, but during which the body can still "hold it together", then I think it would induce a rotational motion. You might even have some chunks breaking loose, but then rolling along the surface in that direction as well.

Well, if there is a ring, with some differential rotation, and a large globe orbiting along the median circle of the ring, the globe does get some spinning up, though far from infinitely large. The friction will work just until the equatorial velocity of the globe catches up with the headwind/tailwind of the matter of the ring at orbits higher/lower by the radius of the globe.

[johnnydeep]

Capture1.png

I'm just looking at this image from the Wikipedia "Roche Limit" page and thinking about it intuitively (https://en.wikipedia.org/wiki/Roche_limit)

As the object approaches the Roche limit, its inner chunks are wanting to orbit angularly more rapidly than the rest, and its outer chunks are wanting to orbit angularly more slowly than the rest. If there is a period of time in which this is appreciable, but during which the body can still "hold it together", then I think it would induce a rotational motion. You might even have some chunks breaking loose, but then rolling along the surface in that direction as well.

My intuition, which is certainly a poor guide here, is that for a fluid satellite, it would not happen. The fluid would just stretch into an ellipsoid and slightly rotate forward, but only less than a single visible rotation of the "body". and then it would begin to disperse (exactly like this image).

But for a rubble pile, I have even more trouble guessing.

I wonder if someone has recorded an experiment that has found a way to show the details at the end before disintegration, or if some numerical simulation has worked it out with satisfactory conclusiveness.

The only example observation I know of is Comet Shoemaker–Levy 9, also discussed in that same Wikipedia article, but not much could be concluded from that example in regard to the fate of Phobos. If someone had a lot of money, they could easily enough set this up orbiting a few small satellites around the Moon. Of course, I doubt anyone is going to do that as a purposeful experiment. But maybe, with lots of varied activity up there, some related observations will be part of life in cislunar space.

Hmm. What would happen to a 50 km carbon nanotube rod (assumed to be strong enough to resist being torn apart) oriented perpendicularly pointing to Mars and starting out in orbit at the Roche limit? Would it just rotate to be tangential to the orbit, overshoot a bit, then rotate back, etc., eventually settling in to a stable tangential orientation?

A rigid body that is held together by forces much stronger than gravity doesn't have a Roche limit. I'd expect your rod to orbit in the perpendicular orientation forever, that being where it is tidally locked.

So the idea that "As the object approaches the Roche limit, its inner chunks are wanting to orbit angularly more rapidly than the rest, and its outer chunks are wanting to orbit angularly more slowly than the rest." is irrelevant in that case? I would have thought that any such "twisting" forces would still be acting on any orbiting object regardless of its composition.

[johnnydeep]

Here's a different way of looking at it, less rigorous but maybe more intuitive. Escape velocity from the surface of Mars is about 5 km/s. That's the lowest speed a body that wasn't gravitationally bound to Mars could strike the surface (but most things would hit much faster). Escape velocity (with respect to Mars) at the height of Phobos is about 3 km/s. Basically, material from Phobos striking the surface will be impacting at a much lower speed. The speed of something already gravitationally bound to Mars. So it makes sense that there wouldn't be enough energy to eject material fast enough to exceed the much greater martian surface escape velocity.

Thanks. That helps some, but the orbital speed of Phobos averages 2.14 km/s (per Wikipedia). Wouldn't that increase the impact speed (and energy) over the 3 km/s escape velocity?

It's in free fall. Its orbital speed isn't a factor. Consider as well that this isn't a rigid body with high material strength. When it breaks up, it's going to be boulders and gravel. Which will mostly burn up without even reaching the ground, and all of which will be substantially reduced in speed by the atmosphere, even if the ground is reached.

I understand the fact that all pieces will have their impact speed reduced by atmospheric friction, but I guess it's back to orbital physics school I will need to go in order to understand the other stuff going on.

[Chris Peterson]

Hmm. What would happen to a 50 km carbon nanotube rod (assumed to be strong enough to resist being torn apart) oriented perpendicularly pointing to Mars and starting out in orbit at the Roche limit? Would it just rotate to be tangential to the orbit, overshoot a bit, then rotate back, etc., eventually settling in to a stable tangential orientation?

A rigid body that is held together by forces much stronger than gravity doesn't have a Roche limit. I'd expect your rod to orbit in the perpendicular orientation forever, that being where it is tidally locked.

So the idea that "As the object approaches the Roche limit, its inner chunks are wanting to orbit angularly more rapidly than the rest, and its outer chunks are wanting to orbit angularly more slowly than the rest." is irrelevant in that case? I would have thought that any such "twisting" forces would still be acting on any orbiting object regardless of its composition.

The Roche limit only applies to objects that are held together by their own gravity. It isn't some clearly defined distance from the parent. It's related to the structure of the satellite, to its shape, to its material strength, to its rotation. The nature of the tidal forces on an object don't change with distance.

[johnnydeep]

A rigid body that is held together by forces much stronger than gravity doesn't have a Roche limit. I'd expect your rod to orbit in the perpendicular orientation forever, that being where it is tidally locked.

So the idea that "As the object approaches the Roche limit, its inner chunks are wanting to orbit angularly more rapidly than the rest, and its outer chunks are wanting to orbit angularly more slowly than the rest." is irrelevant in that case? I would have thought that any such "twisting" forces would still be acting on any orbiting object regardless of its composition.

The Roche limit only applies to objects that are held together by their own gravity. It isn't some clearly defined distance from the parent. It's related to the structure of the satellite, to its shape, to its material strength, to its rotation. The nature of the tidal forces on an object don't change with distance.

So I guess there's no "tendency" for the lower orbit parts of bodies - whatever their structure - to "want" go faster than the higher orbit parts? (The rationale being that bodies in lower orbit travel faster.)

[Chris Peterson]

So the idea that "As the object approaches the Roche limit, its inner chunks are wanting to orbit angularly more rapidly than the rest, and its outer chunks are wanting to orbit angularly more slowly than the rest." is irrelevant in that case? I would have thought that any such "twisting" forces would still be acting on any orbiting object regardless of its composition.

The Roche limit only applies to objects that are held together by their own gravity. It isn't some clearly defined distance from the parent. It's related to the structure of the satellite, to its shape, to its material strength, to its rotation. The nature of the tidal forces on an object don't change with distance.

So I guess there's no "tendency" for the lower orbit parts of bodies - whatever their structure - to "want" go faster than the higher orbit parts? (The rationale being that bodies in lower orbit travel faster.)

There is. But all that does is slightly alter the fixed orientation of a tidally locked body very slightly. It can't create any spin (except in a limited sense before the body becomes tidally locked).