Two points within the Arctic Circle are chosen at random, using a uniform distribution over the entire area. That is, any region of a given area is as likely as any other region with that area, to receive a given point.
What is the expected value of their great circle distance from each other?
Assume:
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The Earth is a perfect sphere with radius 3,958.8 miles.
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The Arctic Circle is located at 66.55° North.
Both calculus answers and simulation answers are welcome. As we are using approximations here, especially about the perfect sphericity of the Earth, the exactness of calculus is not really needed.
Part 2:
... and how about two points on the whole Earth?
Part 2 is actually the easy part. We can always rotate the Earth so that one of our two random points is the "new" North Pole. So part 2 is effectively asking "What is the expected great circle distance from any one random point on the earth to the North Pole?"
So then how to choose a random point on the surface of a sphere? I looked that up, and this is a parameterization for a random point on a sphere's surface in spherical coordinates (North Pole is the positive z-axis):
Theta = random (0,2pi)
Phi = arccos(random (-1,1))
The great circle distance from the North Pole is then just Phi times the Earth's radius.
So to get the expected distance over all points we only need to evaluate 3958.8 * integral {-1 to 1} (1/2)*arccos(x) dx.
The integral is a pretty standard Calculus 2 problem and evaluates to pi/2. Then 3958.8 * pi/2 = 6218.5 miles.
This also makes sense from a more casual view.
For any random pair of points we only really need to see how far apart two points are on their expected great circle.
With that idea in mind, the closest two point can be is effectively adjacent (~0 radians apart) and the furthest is to be antipodes (pi radians apart).
So the expected value is halfway inbetween (pi/2 radians apart). This leads to the same result 3958.8 * pi/2 = 6218.5 miles.
Part 2 Solution
