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I recently took interest in toroidal planets, not because they necessarily exist, but rather due to a more practical result (it can be easily represented with a rectangle). This got me curious about gravitation on such planets.

With spherical planets, in which the mass is a single clump of matter, we can use the shell theorem to show that there is no gravitational force inside the sphere, and it simplifies calculations for the force outside the sphere. However, I believe there is not a similar formula when dealing with a toroidal mass.

I am most interested about the motion of an object in the center of the toroid (in the hole). If an object is placed off-center in that hole, will it start oscillating inside the hole? If a person in the interior ring of the toroid (from a top-down view) jumps up, what would happen?

Venkataram Sivaram
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  • Related: https://physics.stackexchange.com/a/423422/123208 & https://physics.stackexchange.com/q/428986/123208 Also see http://www.aleph.se/andart/archives/2014/02/torusearth.html – PM 2Ring Feb 22 '21 at 16:38

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For the context of this answer let's assume no rotation at all and a uniform density throughout the torus. From symmetry considerations alone we can then deduce the gravity in the center of the hole. Because we are in the center, every part of the planet exerting gravity on us has an equivalent part right opposite and at the same distance. These forces necessarily all cancel (adding a vector to its negative always gives zero). So there is no net gravitational pull exactly in the center.

An object placed not bang in the middle of the hole, but offset within the plane of the torus will be closer to one side than the other. That side will exert a larger gravitational pull than the corresponding mass on the opposite side, as it is closer. Therefore an object placed there will fall to the closer surface. Similarly, a person standing on the inside of the torus and jumping up is pulled right back down (unless it jumps high enough to cross the center, then it will land on the opposite side).

An interesting case is when an object is placed right in the center of the hole, but offset along the axis going through the hole. There is now no force towards the surface of the torus, but one towards the center of the hole. Such an object would oscillate, but this motion is very unstable, as any offset from the central axis makes it drop to the surface.

noah
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    Are you sure about paragraph 2? I didn't calculate it and prove you wrong, but you'd expect it to work that way for spheres, and it doesn't... – user253751 Feb 22 '21 at 16:48
  • @user253751 How would the analogous expectation go for spheres? – noah Feb 22 '21 at 16:57
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    An object placed not bang in the middle of the hollow sphere, but offset within the sphere will be closer to one side than the other. That side will exert a larger gravitational pull than the corresponding mass on the opposite side, as it is closer. Therefore an object placed there will fall to the closer surface. BUT IT DOESN'T. – user253751 Feb 22 '21 at 17:00
  • Ah yes, a very good point. For the torus the forces won't cancel out because it's not spherically symmetric, but also not quite 2D (the shell theorem would hold for a 2D ring). But I agree that the formulation of my argument is a bit too handwavy. – noah Feb 22 '21 at 17:41