4 ms·
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 T
by svrocks 15y ago
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.
- ColinWright 15y agoThat 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 15y agoAh you're right, I missed a factor of 2 in the top integral. Will edit shortly.
- YuriNiyazov 15y agoThe 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'
- deleted 15y ago[deleted]
- tolo 15y agoI'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 15y ago2 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 15y agoAnd also the surface area of a spherical zone, as shown here: http://mathworld.wolfram.com/Zone.html http://mathworld.wolfram.com/Zone.html