# Finding the mean distance between n points evenly distributed in a disc of radius r

In reading this article about updated estimates for the number of exoplanets in the Milky Way, I am curious how to get an estimate of the mean distance between them. The Milky Way is ~50,000 light years in radius and an average of ~1000 ly in thickness, and the article estimates there to be ~500 million exoplanets in the habitable zone around their stars.

To simplify this greatly (grotesquely!), I think the galaxy can be thought of as a plane, since I want to know the method of doing this for a disc anyway, and there are so many estimates here, it shouldn't affect the results greatly. So, the area of this disc would be:

$\pi (50000ly)^2 \approx 7853981634 ly^2$

However, beyond this, to find an even distribution of 500,000,000 points, I'm not really sure how to go about that. Given this, I can later apply it to an oblate disk, but I'd like to know the general method of evenly distributing points in a circle, and possibly finding the distance between them if it's not straightforward.

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You might want to say disc instead of circle — a circle is a curve, and it's relatively easy to calculate the mean distance of two uniformly randomly distributed points on a circle. On a bounded disc it's more complicated, but if you're willing to settle for numerical answers, it's quite easy to write a program to estimate the mean distance. (I can provide details if you're interested.) –  Zhen Lin Feb 20 '11 at 14:40
I'm sure you'll get an answer here for the question about distributing points equally on a disk, but astronomically speaking, it's the wrong question to ask. Exoplanets are not evenly ditributed, they are mostly orbiting stars. The ones that have been discovered up to now certainly are. Then, these stars are not distributed uniformly either. Our galaxy has a central bulge and spiral arms, which kills the even distribution assumption. Finally, what are you trying to figure out anyway? –  Raskolnikov Feb 20 '11 at 14:49
@Zhen Lin: The details for that would be very helpful. –  Sdaz MacSkibbons Feb 20 '11 at 16:04
@Raskolnikov: Yes, I know it's a gross oversimplification and likely wouldn't be applicable to the real situation. The question just brought up the question of how to do point distributions, as I realized I didn't even have an idea how to do that. It's just become a matter of curiosity now. (The MW isn't even a pure spiral, it's a barred spiral with major and minor arms, which would make the real calculations even more complex ;) –  Sdaz MacSkibbons Feb 20 '11 at 16:06
No problem. It's a nice question in itself. ;) –  Raskolnikov Feb 20 '11 at 16:10

Insofar as the number of points is practically infinite it is enough to calculate the average distance $a(D)$ of two independently chosen and uniformly distributed points in the unit disc $D$. The result is $$a(D)= {128\over 45\pi}.$$ I found this on page 1 of the following document:

where also references are given.

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Ah, that's a great paper. Thank you for the reference. –  Sdaz MacSkibbons Feb 20 '11 at 16:37

It seems a rough approximation is useful here, as the distance from one planet to the nearest will vary considerably. I would just calculate the volume of interest (taking account of the bulge and arms as best you can), then divide by the number of planets to get the volume each one "occupies". The expected distance will then be something like 1-2 times the radius of a sphere of this volume. Very rough, but doable.

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That's straightforward, and I see why it would give a reasonable estimate (given all of the other fudge factors herein, anyway). Thank you. –  Sdaz MacSkibbons Feb 20 '11 at 16:36
And the volume of a sphere is easy to calculate as we learned that $\pi=4$ in another thread. –  Ross Millikan Feb 20 '11 at 16:58
Based on this, I came up with a radius of ~26.5, giving a range of 26.5-53 ly, which sounds perfectly cromulent for this calculation and is about what I was expecting. Thanks again. –  Sdaz MacSkibbons Feb 20 '11 at 16:58
Haha, I missed that thread. I wish I'd used that; it's sure a lot easier than the old transcendental value! ;) It kind of goes with all the other fudge factors I've used though, so I guess I shouldn't laugh.. –  Sdaz MacSkibbons Feb 20 '11 at 16:59