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Looking up at the night sky, I wonder:

What is the probability that a plane through three random points on the surface of the Earth intersects the Moon? (The points are independent and uniform.)

Let:

  • $d=$ distance between centre of the Earth and centre of the Moon
  • $r=$ radius of the Moon

Earth and Moon

(Figure not to scale.)

Simulations suggest that the probability is simply $r/d$, for any values of $r$ and $d$, assuming that the two spheres are disjoint. (It seems that the radius of the Earth does not matter.) If this is true, then plugging in $r=1737$ km and $d=384400$ km, the probability is approximately $\frac{1}{221}$.

Using sphere point picking, each random point has two variables, so there are a total of six variables. It seems that such a simple result should be provable without integrating over six variables, but I cannot find a way.

The $2D$ case is not elegant.

Fun fact

The probability is indeed $r/d$. If we replace "moon" with "sun", the probability is approximately $\frac{7\times 10^5}{1.496\times 10^8}\approx\frac{1}{214}$. So there is a greater chance that the plane will intersect the sun, than the moon.

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    $\begingroup$ A geometric argument that the independence and uniformity of picking the three points implies that the unit normal to the plane is also uniformity distributed over the surface of a sphere should make this easy, but I don't have an argument for that. $\endgroup$ Commented Sep 17 at 14:42
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    $\begingroup$ @JonathanZ When $d=2r$, the probability seems to be exactly $1/2$. (In two million trials, the proportion was $0.499901$.) $\endgroup$ Commented Sep 17 at 14:50
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    $\begingroup$ As pointed out in the first posted answer, the radius of the Earth is also relevant, but only insofar as you need it to be physically plausible. The radius of the Earth has to be no greater than $d - r.$ I assume in your $d=2r$ simulation the radius of your Earth was not greater than $d/2.$ $\endgroup$ Commented Sep 17 at 15:34
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    $\begingroup$ That is what I meant by "physically plausible." $\endgroup$ Commented Sep 17 at 15:47
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    $\begingroup$ "If you look up into the night sky, sitting beneath your s.o., and think: <oh how wonderful>, you should study something normal. If however you find yourself wondering: <what is the probability that a random plane through the earth intersects the moon>, you should definitely study math". - paraphrasing a professor I've heard once $\endgroup$ Commented Sep 18 at 8:15

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First let's do an apparently easier problem: take a uniformly-distributed random plane with closest distance $h$ to the centre of the Earth. What is the probability it intersects the Moon? Since everything is rotationally invariant, rather than treating the Moon as fixed and the plane as random, we might as well transform so that the plane is fixed (let's say $z=h$) and the centre of the Moon is a random point $(X,Y,Z)$ on the sphere $x^2 + y^2 + z^2 = d^2$. The plane intersects the Moon iff $h-r < Z < h+r$. Since the area element on the sphere is $\cos(\theta) \; d\theta d\phi$ in spherical coordinates, where $z = d \sin(\theta)$, we see that as long as $|h| < d - r$ the probability is exactly $r/d$.

Of course, for a plane through points on the Earth's surface, $|h|$ is less than the radius of the Earth, and certainly less than $d-r$.

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    $\begingroup$ I was stuck thinking that we needed to know the distribution of $h$ -- but as you point out, as long as it is bounded by $\pm(d-r),$ we don't, because the distribution of $Z$ is uniform on $[-d,d].$ Wow, does that simplify the problem! $\endgroup$ Commented Sep 17 at 15:28
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    $\begingroup$ FWIW, Here's a daily plot (courtesy of JPL Horizons) of the Earth-Moon distance for 13 (anomalistic) lunar months. i.sstatic.net/T6xpo.png See astronomy.stackexchange.com/a/49267/16685 $\endgroup$ Commented Sep 19 at 2:59

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