Point and circle on sphere probability

[BedrockGeometry]

Homework Statement

Given a randomly drawn circle on a sphere, calculate the probability that it will pass within a defined distance of a set point. To make it clear, imagine the example of the the Earth and Mt Everest. What is the probability that a randomly drawn circle will come within, say 3km of Everest?

Homework Equations

Surface area of sphere: 4*pi*r^2
Circumference of a circle: 2*pi*r

The Attempt at a Solution

Let the defined distance (3km in above example) be d. I think that you can approximate the probability for relatively small d (say 1/400th the circumference of a sphere) by calculating the ratio of (the surface area of a strip 2*d units high by the circumference of an averaged sized circle on a sphere) to that of (the entire sphere).

I assume that an arbitrarily drawn circle will on average be 1/2 of the great circle circumference of our sphere, or pi*r so the surface area of the strip would be 2*d*pi*r.

This would make the probability 2*d*pi*r/4*pi*r^2 or d/2*r for a ballpark figure?

It seems very straightforward for a good estimate but I want to bounce it off you all.
Thank you for taking a look.

 

[HallsofIvy]

My first guess would be the area of a circle of radius d divided by the surface area of the sphere.

 

[BvU]

Hello BG, welcome to PF !

I have some trouble interpreting your problem statement.
is it

(i) randomly drawn circle = circle with randomly chosen radius and randomly chosen center​

or is it

(ii) randomly drawn circle = circle with a fixed radius R and randomly chosen center​

The first is rather impossible: there must in all likelihood be an upper limit to this radius.


And even so: your

" I assume that an arbitrarily drawn circle will on average be 1/2 of the great circle circumference of our sphere, or pi*r so the surface area of the strip would be 2*d*pi*r. "​

What is this assumption based on ?

Then the criterion: "pass within a defined distance of a set point" .
Again, there must be an upper limit to d.
And "within" ? Then why start off with a strip ?


For all I understand, you pick a point, e.g. the North pole. Points within a distance d form a circle with area ##< \pi d^2## (you are on a sphere, remember!).
In case ii (fixed R) your criterion comes down to: center of circle less than R + d from the North pole. Still pretty complicated, because you are on a sphere.

Just assume that I'm not stupid, but somehow built like a computer: you have to be complete, specific and exact in order to get what you intended.

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