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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.

Dan
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  • Equivalent problem: What is the probability that the shortest distance $\Delta$ from the plane to the center of the Moon is lesser than or equal to $r$. – Tveltzel Sep 17 '25 at 14:37
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    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. – JonathanZ Sep 17 '25 at 14:42
  • BTW, although the physical picture becomes impossible when $r$ becomes on the order of $d$, you can still ask the mathematical question. Have you simulated that? Because it feels very possible that $r/d$ is just a valid approximation for $r\lt \lt d$. – JonathanZ Sep 17 '25 at 14:48
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    @JonathanZ When $d=2r$, the probability seems to be exactly $1/2$. (In two million trials, the proportion was $0.499901$.) – Dan Sep 17 '25 at 14:50
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    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 $d/2.$ – David K Sep 17 '25 at 15:34
  • @DavidK I assume that the the two spheres are disjoint, which implies that the radius of the Earth is no greater than $d-r$. – Dan Sep 17 '25 at 15:40
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    That is what I meant by "physically plausible." – David K Sep 17 '25 at 15:47
  • @JonathanZ That should follow from symmetry of the sphere. Suppose it is not uniform, then some point would be more likely than some other point but I can find an isometry of the sphere that maps any given point to any other given point. Contradiction. – quarague Sep 18 '25 at 06:23
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    "If you look up into the night sky, sitting beneath your s.o., and think: , you should study something normal. If however you find yourself wondering: , you should definitely study math". - paraphrasing a professor I've heard once – return true Sep 18 '25 at 08:15
  • @quarague - The problem as stated involves picking three points. Your argument would work if we could always find an isometry mapping any triple of points to any other triple of points (i.e. "isometries are transitive on triples"). But they're not. – JonathanZ Sep 18 '25 at 13:11

1 Answers1

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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 $d-r$.

Robert Israel
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    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! – David K Sep 17 '25 at 15:28
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    FWIW, Here's a daily plot (courtesy of JPL Horizons) of the Earth-Moon distance for 13 (anomalistic) lunar months. https://i.sstatic.net/T6xpo.png See https://astronomy.stackexchange.com/a/49267/16685 – PM 2Ring Sep 19 '25 at 02:59