## How long does it take to fall through a ball of ice?

Planet FrictionlessIceball has a straight, narrow tunnel connecting two points on its surface. How long does it take to slide through the tunnel?

## Simplifying assumptions

As with most story problems, we have to make some simplifying assumptions. But I think they’re fairly reasonable.

We assume the faller starts out motionless, at one end of the tunnel.

We assume we don’t need an exact answer down to the nanosecond. We assume Newton’s laws of gravity and motion are good enough. We assume the faller can be modeled as a point, and the tunnel as a line segment.

The word “planet” implies a fairly large object, large enough to in hydrostatic equilibrium. To mathematicians, the word “ball” means a filled sphere. For a real planet to be close to spherical, it must not be rotating very quickly; so we assume this planet is not rotating.

From the word “ice”, we assume the planet is made of water ice, and that it has the usual density of the usual kind of water ice. I admit that this would not be true in reality. In reality, the ice would get at least a little denser toward the center, and for a planet of substantial size it would mostly be an exotic kind of ice.

If my calculations are correct, the answer is about 1 hour and 43 minutes.

Solving it involves some slightly difficult math and physics calculations, and I’m not going to attempt to derive the formula myself. I think the interesting thing is not how to solve it, but the fact that such an under-specified problem has a solution at all.

This puzzle is not original to me. I just wanted to see how succinctly I could formulate it. What’s usually asked is “How long would it take to fall through the Earth?”, which has a different answer.

## Outline of solution

We don’t know how big the planet is, and even if we did, we still wouldn’t know how long the tunnel is. But, surprisingly, it doesn’t matter. The fall/slide always takes 1 hour and 43 minutes. The shorter the tunnel, the slower you move, and the complete trip always takes the same amount of time.

I’m not sure how much of a coincidence this is, but it turns out that the answer is the same time it would take to orbit just outside the surface, from one pole to the other. The formula for the full period of such an orbit, based on the planet’s density ($\rho$), is actually quite simple, and does not depend on the planet’s radius at all:

$\displaystyle T = \sqrt{\frac{3\pi}{G \rho} }$

(Where $G$ is the gravitational constant.) Since we’re making a one-way trip, we want one half of this duration, or:

$\displaystyle T = \sqrt{\frac{3\pi}{4 G \rho} }$

Here’s a graph of this function:

That writing the formula in terms of density makes it simpler reminds me of my previous post on Hill Spheres.

I’ll arbitrarily use units of kilograms, meters, and seconds. For the density of ice, I’ll use 916.8 kg per cubic meter. In the proper units, $G$ is about $6.674 \times 10^{-11}$.

When I crunch the numbers, I get 6205.45 seconds, which is 1 hour, 43 minutes, 25.45 seconds.