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I was asked this question (in essence) 4 years ago in the context of an undergraduate introductory astronomy course.

Given a spherical moon with radius r, and periodic impact craters formed on the surface of that moon with radius i every t years, what is the expected time elapsed for the entire surface of that moon to be completely covered with impact craters? Assume that the locations of impact are uniformly and independently distributed across the surface of the moon, at time 0 there are no impact craters on the surface, and the impacts will surely occur at times t, 2t, 3t, ...

Also assume i is smaller than r in order to be somewhat realistic; think of the size of the visible impact craters on our own Moon.

According to my TA the correct answer was the surface area of the moon divided by the area of the crater multiplied by t. Of course that is incorrect; only if there are no overlapping impacts will that answer be correct.

My TA wrote "unnecessary" on my assignment when I tried an unsuccessful probabilistic approach to solve the question as it was posed, and his lack of effort and dismissal of my effort was hurtful enough to make me remember the question from time to time ever since. However I sympathize with a busy and overworked TA.

James
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  • This looks like an interesting and more complicated version of the Coupon Collector problem. You could try modeling the craters as the intersections of the large sphere with smaller spheres whose centers lie on the surface of the large one, but it's probably a nontrivial solution regardless. – platty Aug 07 '17 at 19:18
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    This is quite a hard problem even in the 2D case of rain droplets falling on a table! The solution in that case is asymptotic to $\mu\log\mu$, where $\mu$ is the ratio of the areas of the table and the raindrops. –  Aug 07 '17 at 19:23
  • Wow, I briefly read user joriki's answer and was very surprised at the complexity of this sort of problem in the 2D case. I can only assume the 3D case is just as complex if not more so. His idea to focus on the intersections among the circles is still relevant in this case. However his formulation 1−π(R−r)2/(πR2)=2μ−1(1+O(μ−1)) cannot be used here since we have no "edge" of the sphere to speak of. Perhaps if we split the sphere into equal halves, we can then have two surfaces with edges which joriki's approach can be applied to. – James Aug 07 '17 at 20:18
  • The cited answer for the 2D case looks pretty daunting. But what's the matter with this kind of argument: call the total surface area of the sphere $A$ and the surface area of a crater $a$, defining $\mu={a\over{A}}$. Now take as a sample a small area of the sphere $dA\ll{a}$. After $N$ craters the probability that $dA$ is not covered is $(1-\mu)^N$. For example, using $r_{moon}=1702$ km and $r_{crater}=1$ km, then $dA$ has a 98% chance of being covered only after about 11.3 million craters. – Joe Knapp Aug 09 '17 at 00:04
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    @JoeKnapp On that line of reasoning we could compute the expectation $\sum_{k=1}^{\infty} k\mu(1-\mu)^{k-1}$. But that only gives us the expected time elapsed to cover a specific point on the surface. – James Aug 09 '17 at 23:25
  • I was thinking that since there is nothing special about the $dA$ chosen, the same holds for all such $dA$ on the sphere. I guess where it breaks down is that if a given point is covered, there's a relatively high probability that its immediate neighbors are covered, since $dA\ll{a}$, the size of the crater. So it's hard to figure out how to add up the elements. – Joe Knapp Aug 09 '17 at 23:48

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