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Mathematics 7 Online
OpenStudy (anonymous):

Assuming that the earth is a perfect sphere with radius 6378 kilometers, what is the expected straight line distance through the earth (in km) between 2 points that are chosen uniformly on the surface of the earth?

OpenStudy (anonymous):

I feel as if this question is really simple, but I know it's not.

OpenStudy (kainui):

I think this is the same problem as what the distance is between any two points on a circle.

OpenStudy (kainui):

Wait, is this a probability question?

OpenStudy (anonymous):

I actually have no idea

OpenStudy (anonymous):

Use expected value I suppose.

OpenStudy (anonymous):

All I know is, it'll involve some geometry lol

OpenStudy (anonymous):

Well, first come up with the probability distribution.

OpenStudy (anonymous):

We can have the first point at the top, and then all that matters is the second point. Even if the first point were not at the top, we would be able to rotate the sphere in a way tat it is at the top, without changing distances.

OpenStudy (richyw):

can't you use some symmetry to say that the points will be separated by an angle of pi/2?

OpenStudy (richyw):

like wrt to the centre of the earth?

OpenStudy (richyw):

so just would be \(\sqrt{2}R_e\)

OpenStudy (anonymous):

This problem is actually somewhat difficult.

OpenStudy (anonymous):

^

OpenStudy (anonymous):

I am thinking that spherical coords might be the best way to create a random uniform probability distribution for this...

OpenStudy (richyw):

oh wait. I have done this before. It's not what I said I don't think

OpenStudy (anonymous):

I found a solution someone wrote

OpenStudy (anonymous):

Oh?

OpenStudy (anonymous):

Anyway, If you have \(\phi \) and \(\theta\), then the straight line distance would be....

OpenStudy (anonymous):

"Fix a point to find the average distance of the other points from it. Take the point as A and centre O. Consider another point B on the surface. Let <AOB= theta. AB = 2sin(theta/2) with band radius sin theta."

OpenStudy (kainui):

I don't really know much about probability integrals but. |dw:1395643948332:dw| there's sort of this cone of distance that has an equal probability of being picked. Which is really just a circumference following the outside of the sphere.

OpenStudy (anonymous):

|dw:1395644054673:dw|

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