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Underground Kingdom ------------------- I've been enjoying reliving some of my childhood by reading through a choose-your-own-adventure book named "Underground Kingdom" which has been kindly hosted on the typed-hole.org [1] (also on gemini! [2]). The premise of the story is that the Earth is hollow due to the presence of a primordial black hole which has eaten out the interior. According to Professor Bruckner, who in the book was responsible for discovering this fact, "If you were to stand on the inner surface of the earth, like a fly on the inner shell of an enormous pumpkin, you would see the black hole directly overhead, like a black sun." A pretty neat symmetry! He proceeds to say, "The gravity of the earth's thick shell would hold you to the inner shell of the earth, though you would weigh much less than you would on the outer surface because the mass of the Black Sun would tend to pull you toward it." I immediately wondered whether this is the way things would actually be. It's been a while since I've studied physics, but I know that spheres are special in many ways. It turns out the answer is: no, the gravitational pull of the shell would not "stick" you to the inner surface at all. There's a very succinct explanation of why on NASA's website [3] which, for the convenience of curious gophers, I shall reproduce here: --- snip --- The gravitational force inside a hollow sphere shell of uniform areal mass density is everywhere equal to zero, and may be proved by the following argument: Let the sphere have a radius a. Place a point P inside the sphere at a distance r from the center where r < a; i.e., r is strictly less than a. Draw a line through P to intersect the sphere at two opposite points. Call these points α and β. Let the distance from P to α be r₁, and the distance from P to β be r₂. Now place a differential area dA_α at α, and project straight lines through P to acquire its image dA_β at β. These two areas subtend a solid angle dϕ at P. Let the sphere have areal mass density ρ (kg/m²). Then the net differential attraction dF of dA_α and dA_β at P directed toward α is just dF = ρ( dA_α /r₁² - dA_β/r₂²). But dA_α = r₁² dϕ, and dA_β = r₂² dϕ by definition of the solid angle. Thus, dF = ρ((r₁² dϕ)/r₁² - (r₂² dϕ)/r₂²) = 0. This result is true for all choices of dA_α and dA_β. The gravitational force within the sphere is everywhere equal to zero. --- snip --- So back to the hollow Earth of Professor Bruckner's theory, instead of sticking to the inside of the shell you'd instead float around inside of it, exactly as if the hollow Earth-shell wasn't there at all. Of course, this is without the presence of the black hole! You'd certainly feel the gravity of _that_, and would probably wind up being sucked into it. Furthermore, there'd be nothing to attract the black hole to the centre of the shell: the black hole would float around aimlessly as you would, probably just destroying the remainder of the Earth in the process. (Of course, mixing black holes and Newtonian gravitation is problematic, but whatever happened wouldn't be pretty.) All this is a bit of a shame for the story. Then again, as far as physics gaffes in fiction go, this isn't _too_ bad. :-) CORRECTION: Of course, for a _rotating_ sphere, things are slightly different. On the inside of the spherical Earth, assuming it is really thin so that the inner radius of the shell is almost the same as the outer radius, the acceleration due to the apparent centrifugal force at the equator would be R·ω² ≃ 0.033 m/s². This is super tiny, being only about 0.3% of the acceleration due to gravity on the outer surface, and probably wouldn't provide much protection against being sucked into the black hole... --- [1]: gopher://typed-hole.org:70/1/cyoa [2]: gemini://typed-hole.org/cyoa/underground [3]: https://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/grvtysp.htm