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Math Help - Average distance of two points on a sphere

  1. #1
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    Post Average distance of two points on a sphere

    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 by Zamadatix; May 25th 2010 at 12:40 PM.
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  2. #2
    Senior Member roninpro's Avatar
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    The only way I can see to do this involves calculus. Is that fair game for you?
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  3. #3
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    unless by this
    Quote Originally Posted by Zamadatix View Post
    "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
    \frac {\pi}{2} R


    if the question meant the distance in straight line, yes calculus.
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  4. #4
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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

    But if anyone wouldn't mind, how do you find out the straight line way using calculus?
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  5. #5
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    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 r?

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

    Let the other point be P.

    Consider the great circle through A and 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 P can be anywhere on the arc \overline{APB}.

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

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

    . . just as Haytham predicted . . .

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  6. #6
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    Thanks for the amazing answer Soroban, it is indeed correct , but I am afraid I didn't mean what I originally asked

    Quote Originally Posted by Haytham View Post
    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.
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  7. #7
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    Quote Originally Posted by Zamadatix View Post
    Thanks for the amazing answer Soroban, it is indeed correct , 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 1 and e^{i\theta} when \theta\in[0,\pi] (we will have to multiply the final answer by r).

    This equals: \int_0^\pi |1-e^{i\theta}|\frac{d\theta}{\pi}. However, 1-e^{i\theta}=-2i e^{i\theta/2}\sin\frac\theta2, so that |1-e^{i\theta}|=2\sin\frac\theta 2 and so the final answer is the value of \int_0^\pi 2\sin\frac\theta2\frac{d\theta}{\pi}, which I guess you know how to compute.
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  8. #8
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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 \pi  \theta  \text{Sin}\frac{\theta }{2} for any circle.
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  9. #9
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    Quote Originally Posted by Zamadatix View Post
    Ah thank you very much putting it on the complex plane was the thing to grasp in this.

    Integrating this you get \pi  \theta  \text{Sin}\frac{\theta }{2} for any circle.
    No, after integrating you get \frac{4}{\pi} (by the way, what would this \theta even mean?), so the average distance is \frac{4r}{\pi}.
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  10. #10
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    oops it is pi/4 but i have no clue what that theta meant, lowering the level of math to \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.
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