[Krakenup]

Originally posted by Batman:
If we take the output of photons to be constant, then the same amount of photons is spread over a spherical shell at a distance R from the sun. The area of this shell is A=4/3*Pi*R^3. A sail of area B captures B/A of the total photons emitted by the star. At double that distance from the star, the area of the shell is A2 = 4/3 * Pi * (2R)^3 = 8*4/3*Pi*R^3 = 8*A. So the same sized sail now captures only B/(8*A) of the total photons (because they are less dense). At double the distance, the sail is 1/8 as effective, and so on. So it should be big if it is to maintain effectiveness for the whole voyage.

That's the volume of a sphere. The surface area is A=4*Pi*R^2. At double the distance, the sail is 1/4 as effective. (Yeah, I know, but I'm a member of the pick-a-nit-a-day club).