[svrocks]

Assumptions: 1) Earth is a sphere with radius r 2) "Sky" is a hollow spherical shell with radius R

The surface area of the entire spherical sky is: 4\piR^2

This can be represented by a spherical integral that I'm not sure I can write cleanly here.

We just need to change the bounds of that integral to find the area of the observable part of that shell. The Intersecting Chord Theorem along with some trigonometry can be used to find these bounds.

The answer I get is:

(1 - cos(x)) / 2 where x = Arctan(sqrt(r(R-r)) / r)

This seems to have the correct asymptotic behavior (as r approaches 0, cos(x) approaches cos(pi/2) = 0, and the answer approaches 50%

EDIT: My previous answer assumed the shapes were cones instead of spheres. Sorry about the confusion.

[YuriNiyazov]

The way the question is posed, it seems to me that the radius R of hollow spherical shell should not figure in the calculation. Assume there is no Earth, and you are hanging out in the center of a large hollow sphere. In that case, you can see 100% of that sphere, no matter how large it is. Now, let's say that earth magically appears, and now r' varies from 0 to r. The question asks to find percentage of the sky visible as a function of r'

[ColinWright]

That can't be right - it doesn't have the right behavior as R->oo. As R->oo you need to get an answer of 0.5, but this gives an answer of 0.25.

[svrocks]

Ah you're right, I missed a factor of 2 in the top integral. Will edit shortly.

[tolo]

I'm not sure if I'm missing something, but I get a much simpler answer. Surface area of a spherical cap is 2 pi R h, with h = R - r. Divided by the total surface area (4 pi R^2), gives (R-r)/(2 R)

[svrocks]

2 pi R h gives you the surface area of the side of a cylinder with height h, since you are taking the circumference of a circle and multiplying it by h

[tolo]

And also the surface area of a spherical zone, as shown here: http://mathworld.wolfram.com/Zone.html

[svrocks]

After choosing a different way to find the angle, my answer matches yours. Kudos for having a much simpler answer.

[svrocks]

wow that's an interesting and counterintuitive result. I need to figure out why my integral doesn't line up with that