Time to roll through subterranean chords

Question

Imagine a spherical airless body. It is small enough that central pressure allows a tunnel to be built from north pole to the south pole. I jump in the tunnel at the north pole and fall to the south pole. How long would it take me? What I believe to be the correct answer is the period of a circular orbit having the planet's radius, r. If I remember correctly, I heard this argument from Dr. John Stockton:

Break the acceleration vector of the orbiting body into orthogonal components. The orbiting body's acceleration vector would have the same north/south component as the object bobbing up and down the mohole. So the period of the bobbing object would be $2\pi \sqrt{r^3/\mu }$

But how long would it take for a subway car to roll from one end of the chord to the other if the chord doesn't pass through the body's center?

I'm imagining a science fiction setting where 12 subway stations are built on the surface of Ceres, each station corresponding to the vertices of an icosohedron.

Tunnels linking stations to their nearest neighbors are chords subtending 63.43º. These tunnels from the edges of an icosohedron.

Tunnels linking stations to their second nearest neighbors are chords subtending 116.56º. These tunnels form the edges of a small stellated dodecahedron.

And of course there'd be 6 tunnels linking stations 180º from one another. I think I have handle on these.

For the moment I'm not considering friction or the so called centrifugal force a spinning Ceres would induce.

Answer 1

One equates the centrifugal force with the gravitational, viz

$F = GM m / R^2 = 4\pi^2 m R/T^2$

This gives $\frac 43 \pi G \rho R^3 = 4\pi^2 R^3/T^2$ whence $3\pi = G\rho T^2$ where $T$ is the time taken. ie $T = \sqrt{3\pi / G \rho}$

For something like the earth, you have T = 84 minutes. As your graphic shows, you can travel in a tube one way in half this time (42 min), and this applies to all tubes, not just the one through the centre.

The time to travel along a chord representing an icosahedron-edge is then the same as for the longer chords, is the same as a diameter, 42 minutes. This is a known result involving trinometry and calculus, which is beyond my current skill.

[C:\Program Files\JPSoft\TCMD16x64]rxc sqrt(pi(3)/6.672e-11/5520)/60
 84:3745 5191,3420 E981,6312 4184,2699 V999,6987 4450
 84.311 488 287,445 946 571,778 196 492,255 355 903,447

Answer 2

This question was asked 8 years ago, but it's been a fun diversion for me, so I figured I'd answer it.

Falling through the center of the Earth

First let's go over the basic argument for falling through the center of the Earth.

As described by the question, the time to fall through the center of the earth and back out the other side is half the period of a low Earth orbit. This is because inside a spherical body of constant density, the acceleration due to gravity, $g$ is proportional to the distance from the center. For a body with constant density $\rho$, $$g = G\frac{M}{r^2} $$ $$ M = \frac 4 3 \pi \rho r^3$$ where $G$ is the gravitational constant, $r$ is the distance to the center, and $M$ is the mass of the planet "below" $r$.

Then $$g = G \frac{\frac 4 3 \pi \rho r^3}{r^2} $$ so $g\propto r$.

At the surface of the Earth, the acceleration due to gravity has constant magnitude, and the direction always points directly toward the center. Then by a basic vector decomposition, the acceleration along the tunnel axis is proportional to the distance from the center along the axis, so, the acceleration at every point is the same, and therefore the periods are the same.

Rolling along a subterranean chord

A surprising result: All frictionless subterranean chords take the same amount of time to roll through.

Again, this applies only to a ball of constant density. The argument is actually pretty simple. Consider any chord, and any point along the chord. As before, the gravitational force points toward the center, and is proportional to distance to the center. Then, by vector decomposition, the force along the tunnel is proportional to the distance to the center of the tunnel.

In other words, even though a non-diameter chord is shorter than falling through the center, it is also shallower, and the force along the chord is proportionally weaker.

Therefore, all chords (including the diameter) have the same period as each other, so the time to roll through any frictionless subterranean chord is the same as half the period of an orbit at the surface.