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Let's say we have 2 planets, one being 'a' (the bigger one) and one being 'b'. Planet a is completely hollow inside with planet b is inside of it. My question is how much gravitational force would planet a have to exert on b to turn b into a molten core.(if it is possible) planet b is about 2.12x10^23kgs. I don't care how much planet a weighs, that's up to you.

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    $\begingroup$ Newton's shell theorem, 1687: a hollow sphere exerts no gravitational force on objects inside it. In the setup described in the question, planet a would exert no gravitational force on planet b. $\endgroup$ – AlexP Jan 23 '18 at 0:22
  • $\begingroup$ This is not a duplicate, though the setup does appear to be the same. I strongly recommend rewriting your question, or even rethinking the premise. $\endgroup$ – rek Jan 23 '18 at 3:18
  • $\begingroup$ @AlexP, but planet B would have its own gravity though. Planet A would not exert gravity onto B; we got that. But B would exert its own gravity perhaps creating a pull on A(?) thus creating gravity within A(?). See where I'm going with this? I'm not really sure if this helps OP in any way, but its interesting... $\endgroup$ – Len Jan 24 '18 at 16:13
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TLDR : Not going to happen.

Despite the shell theorem there is a net force between the two.

Object (a) on the outside ("the shell") does not exert any net force on any part of object (b) ("the core") regardless of the relative positions as long as (b) is entirely enclosed in (a) and (a) is spherically symmetrical. That's an absolute result from the Shell Theorem.

But the core (b) does exert a net attraction on every part of the shell (a). The shell theorem does not apply outside an object.

So the core will attract the shell and, as there must be an equal and opposite reaction, the core will also feel a force balancing that.

But this is a very, very unstable configuration.

The instant the core's center shifts even a fraction off the exact center of the shell it will inexorably move off the center, most likely accelerating into a collision with the shell.

Likewise if either the core or the shell are not perfectly spherically symmetrical they will drift off and eventually most likely collide.

My question is how much gravitational force would planet a have to exert on b to turn b into a molten core.(if it is possible) planet b is about 2.12x10^23kgs.

Never going to happen.

The net force on every part of the core is not compressive (inward) but outward (an opposite reaction to the attractive force it exerts on the shell). So while there could be extreme sheering forces if they cease to be in a stable configuration (and they would eventually) there would no be compressive forces globally on the core, only perhaps (a maybe) on local parts of the core undergoing deformation as part of sheering. But that's unlikely to melt the core, so much as break it up.

Likewise the forces on the shell are all pulling it inward, but this would also require the shell to support itself. This would almost certainly be impossible - you're describing a Dyson Sphere, in fact.

If you want at least part of the core to be molten it would have to do it under it's own efforts. This either requires that :

  • The core was formed in such a way that it has not yet dissipated it's heat of formation - actually Earth's condition, hence our own molten core.
  • or the core had a very radioactive interior (unlikely to be significant, IMO)

The Earth would have been a molten ball early in it's formation. It has cooled gradually down. That's how it works.

I don't think your system is at all possible without artificial support, which would be a feat of staggering difficulty, beyond out current science and probably going to stay that way for many, many tens of thousands of years if at all.

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  • $\begingroup$ @lochlan-harrison It appears you have marked my answer as accepted. Thank you, but note that on SE we'd prefer you wait at least a day before accepting an answer so that other people have a chance to comment and provide perhaps better answers. Of course you can still mark an answer as accepted earlier, but in all fairness it's perhaps better to wait. Thank you. $\endgroup$ – StephenG Jan 23 '18 at 1:49
  • $\begingroup$ That's not quite accurate. If the core world is shoved off center, there is no restoring force, so it will continue moving until it crashes into the shell. In other words, it is not dynamically stable. But it won't accelerate into the shell, either. The force is symmetric. The inner world will produce compressive stress on the outer shell, but if the shell is strong enough to hold itself up under its own weight, purely gravitational interacts between the inner world and the shell will not cause either one to accelerate. $\endgroup$ – Logan R. Kearsley Jan 23 '18 at 1:59
  • $\begingroup$ @LoganR.Kearsley If there is a net gravitational force, the by definition there will be acceleration. There's no way to avoid this in two unconnected bodies. $\endgroup$ – StephenG Jan 23 '18 at 5:13
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    $\begingroup$ There is no net force. A body inside a uniform spherical shell will not accelerate due to the gravity of the shell, nor will the shell accelerate around that body. The central body does apply a force to the shell, which induces compressive stress, but the forces applied to each infinitesimal part of the shell cancel each other out. If the stress isn't balanced by the shell's mechanical strength, then the shell will clearly collapse inward--but even in that case, there is no net acceleration of the center of mass of the shell or the central body until something breaks the symmetry. $\endgroup$ – Logan R. Kearsley Jan 23 '18 at 5:33
  • $\begingroup$ @LoganR.Kearsley Your point is simply what I've already said in my answer. But the point you're ignoring is that such a configuration is inherently unstable - even the slightest disturbance will result in failure. Put it anywhere but in an empty universe and it will fail eventually. And it's impossible to imagine it developing by accident and it would surely never (because of it's inherent instability) be designed and manufactured by an advanced civilization. $\endgroup$ – StephenG Jan 23 '18 at 5:43
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The way your question is set up, it does not sound like planet B would become a molten core at all. If planet A surrounds B in the manner described, B will be gravitationally pulled outward by A, the gravity will reduce B's pressure inward, not raise it.

If you are thinking that A's gravity is so strong that B is fully pulled apart and is itself hollow, a hollow molten core pushing outward along the inner wall of the hollow planet A because the hollow cavity is large enough and A's crust thick enough for that much gravity, I don't think that would happen either, and even if it did then B would not be a separate planet anymore but rather would be part of planet A.

Explanation

The pressures within our planet which contribute to our molten core would not be possible if it were not for the fact that all the rest of the world is sitting on top of it. In the case of your hollow world, nothing will be weighing upon planet B. Planet B can be considered a normal planet just like any other, and therefore treated as any other with its own normal geology, with the exception that it has the inner wall of another planet overhead.

The following assumes A is roughly spherical with a roughly spherical cavity which itself is centrally located within planet A. Now remember, even though the center of mass of A is at the center of its hollow cavity, B is being attracted to the matter of planet A, so it is actually being attracted outward. For a sufficiently large planet A, this could even result in a planet B rolling around the inner cavity wall of A.

For a planet B stabilized at the center of mass of A such that it is suspended in the center of the cavity, increasing the gravity will not pull B apart. If you increase A's gravity sufficiently, it would merely pull B out of its stabilized position with the entirety of B "falling" out toward some random direction and leave you with something more like the previous paragraph before this one.

Caveat

That said, there is a catch to what I said, something that might still be close to what you are after, in a way. If B settles down (down is subjective here; I am meaning toward the surface of a sphere halfway between A's outer surface and inner cavity surface), and if A's crust is sufficiently thick that B can undergo similar geological activity within A just as our core does in Earth, then it could still become molten given the right conditions. However, there are now a few issues:

  • At this point, again, B is no longer a separate planet but is part of A

  • Once B becomes molten, if the material in A is the same all the way around, then B will likely become a molten sphere going around the entirety of A, a molten sphere sandwiched between solid material on the inside and out.

  • This all assumes that A can hold its hollow shape through the process. I am not sure if there is a natural planetary make-up that could, so I would hand-waive that problem away and suggest that A is an artificial planet. Or you could come up with some outlandish quasi-scientific rationale for the hollow planet that aided in the suspension of disbelief. "The planet is electrically charged, and the build-up of a static charge is causing the particles to repel from each other just like your hair does when charged, so it has developed a hollow cavity." That would not really work, and your science purists will role their eyes, but that should be sufficient to appease the laymen.

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    $\begingroup$ B would not be pulled outward. The gravitational force from a spherical shell inside the shell is zero. As far as B is concerned, A might as well not even be there, at least as far as gravity is concerned. $\endgroup$ – Logan R. Kearsley Jan 22 '18 at 23:34
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    $\begingroup$ @LoganR.Kearsley Only for a perfectly balanced system. The moment it is off balance, that ceases to be the case. All you need to do to convince yourself of this is to think of Ringworld, but extrapolate that out into a similarly sized Sphereworld where the opposite edge of the inner cavity is so far as to be gravitationally negligible. You could walk around with ease on the inner surface of the planet with your head pointing toward the center and your feet toward the surface, assuming proper crust width. Now, how far do you need to shrink this cavity before what you describe takes precedence? $\endgroup$ – Loduwijk Jan 23 '18 at 0:00
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    $\begingroup$ @TimothyAWiseman: A hollow sphere exerts no gravitational force on an object inside it whatever the position of the object. $\endgroup$ – AlexP Jan 23 '18 at 0:25
  • $\begingroup$ @AlexP I stand corrected. Thank you for pointing that out. $\endgroup$ – TimothyAWiseman Jan 23 '18 at 0:40
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    $\begingroup$ @Aaron Only for a perfectly balanced system. The moment it is off balance, that ceases to be the case. This is still wrong. Newton's shell theorem says otherwise. Quote from the wikipedia article: If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell. $\endgroup$ – kingledion Jan 23 '18 at 1:13

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