# Average distance between 2 points on surface of sphere?

How can I find an average distance between two points lying on surface of a sphere of a certain radius?

More importantly : can knowing the average distance between two points on surface of a disk ( this question has already an answer on MSE) be useful to answer the question about average distance between two points on surface of sphere? Or there are no immediate obvious relationship/generalization between the two question?

• If you mean what I think you mean, then you just want to consider a fixed point and average the distances along a great circle passing through that point. At that point, you can just drop down to the 2D problem of the average distance along a circle. Apr 27 '15 at 14:17
• @CameronWilliams : How can I generalise that to spheres of n dimesion? That is the motivation for asking this question. Apr 27 '15 at 14:28
• I could be wrong but if you make a slice along a great circle of a sphere in $\Bbb R^n$, I think you get a sphere in $\Bbb R^{n-1}.$ So, I think you can just keep reducing the problem all the way down to the case of the circle. I could be wrong though. Apr 27 '15 at 14:35
• I don't think the fixed-point-and-single-great-circle method generalizes well to even a sphere in three-dimensional space. If the points are uniformly distributed on the sphere, the probability density of the angular separation is greater at $\pi/2$ than at $0$ or $\pi$. But if both points are uniformly distributed, then one fixed point and one uniformly distributed point works. I'd try to find $dA/dr$ where $A$ is the measure of the portion of the sphere within $r$ units of the fixed point. Apr 27 '15 at 14:37
• By distance, I assume you mean the geodesic distance on the sphere. Consider the case the fixed point is the north pole. If you have a point at a distance $r$ from the north pole, its mirror image wrt to the $xy$-plane is at a distance $\pi - r$. Since the reflection wrt $xy$-plane is an isometry, .... Apr 27 '15 at 14:50

Without loss of generality, assume the first point is at the "north pole"; also without loss of generality, assume the second point is along the "prime meridian." Then the probability of being at "latitude" $x$ degrees north is equal to the probability of being at "latitude" $x$ degrees south (and is proportional to $\cos x$). Therefore, the average latitude is at the "equator," and the average distance is $\pi r/2$, as stated by Henry and achille hui.