Average distance of two points on a sphere

Nov 2009
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Question: What is the average distance between two points on the surface of a sphere with a radius r?

I came up with this question while thinking "What is the average distance of any point on earth from me" and cannot find out HOW I would go about figuring this out. Any guidance is greatly appreciated :)

EDIT: Revise the question above to "What is the average distance between two points on the surface of a sphere with a radius R by traveling directly in a straight line from point A to B" I apologize for the original wording.
 
Last edited:
Nov 2009
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The only way I can see to do this involves calculus. Is that fair game for you?
 
Jul 2009
68
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unless by this
"What is the average distance of any point on earth from me"
he means the distance around the sphere surface
in that case the answer can be seen by the sphere symmetry and its simply
\(\displaystyle \frac {\pi}{2} R\)


if the question meant the distance in straight line, yes calculus.
 
Nov 2009
11
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Wow, I was trying to figure out the straight line problem even when my question had "on the surface" in it. Thanks for clearing me up on my own question (Speechless)

But if anyone wouldn't mind, how do you find out the straight line way using calculus?
 

Soroban

MHF Hall of Honor
May 2006
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Lexington, MA (USA)
Hello, Zamadatix!

I think I have an answer . . .


What is the average distance between two points
on the surface of a sphere with a radius \(\displaystyle r\)?

With no loss of generalization, let one point \(\displaystyle A\) be at the North Pole of the sphere.
. . Let \(\displaystyle B\) be the South Pole.

Let the other point be \(\displaystyle P.\)

Consider the great circle through \(\displaystyle A\) and \(\displaystyle P.\)
It could pass through Greenwich Mean Time
. . (discovered by the Armenian explorer, Prime Meridian).
Code:
                A
              * o *
          *           *  P
        *               o
       *                 *
     
      *                   *
      *         * - - - - * E
      *         O    r    *
     
       *                 *
        *               *
          *     o     *
              * * *
                B

Point \(\displaystyle P\) can be anywhere on the arc \(\displaystyle \overline{APB}.\)

Since half the points are above the Equator \(\displaystyle OE\) and half are below,
. . the average distance would be: .\(\displaystyle \text{arc}\overline{AE}\)

Therefore, average distance .\(\displaystyle = \;\frac{1}{4}(2\pi r) \;=\;\frac{1}{2}\pi r\)

. . just as Haytham predicted . . .

 
Nov 2009
11
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Thanks for the amazing answer Soroban, it is indeed correct (Nod), but I am afraid I didn't mean what I originally asked

if the question meant the distance in straight line, yes calculus.
Is what my original question was actauly supposed to be about but I slipped up.
 

Laurent

MHF Hall of Honor
Aug 2008
1,174
769
Paris, France
Thanks for the amazing answer Soroban, it is indeed correct (Nod), but I am afraid I didn't mean what I originally asked



Is what my original question was actauly supposed to be about but I slipped up.
Let's do it in a soft way. As before, it suffices to consider the average along one great circle.

Let's say the circle is the unit circle in the complex plane, and we want the average distance between \(\displaystyle 1\) and \(\displaystyle e^{i\theta}\) when \(\displaystyle \theta\in[0,\pi]\) (we will have to multiply the final answer by \(\displaystyle r\)).

This equals: \(\displaystyle \int_0^\pi |1-e^{i\theta}|\frac{d\theta}{\pi}\). However, \(\displaystyle 1-e^{i\theta}=-2i e^{i\theta/2}\sin\frac\theta2\), so that \(\displaystyle |1-e^{i\theta}|=2\sin\frac\theta 2\) and so the final answer is the value of \(\displaystyle \int_0^\pi 2\sin\frac\theta2\frac{d\theta}{\pi}\), which I guess you know how to compute.
 
Nov 2009
11
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Ah thank you very much :) putting it on the complex plane was the thing to grasp in this.

Integrating this you get \(\displaystyle \pi \theta \text{Sin}\frac{\theta }{2}\) for any circle.
 

Laurent

MHF Hall of Honor
Aug 2008
1,174
769
Paris, France
Ah thank you very much :) putting it on the complex plane was the thing to grasp in this.

Integrating this you get \(\displaystyle \pi \theta \text{Sin}\frac{\theta }{2}\) for any circle.
No, after integrating you get \(\displaystyle \frac{4}{\pi}\) (by the way, what would this \(\displaystyle \theta\) even mean?), so the average distance is \(\displaystyle \frac{4r}{\pi}\).
 
Nov 2009
11
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oops it is pi/4 but i have no clue what that theta meant, lowering the level of math to\(\displaystyle \int_0^2 \sqrt{4-x^2} \, dx\)=pi/4 using the formula for a circle from algebra 2 to understand it instead of the complex plane.