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
.