What fraction of the power radiated by the sun is intercepted by the planet Mercury?

Question

What fraction of the power radiated by the sun is intercepted by the planet Mercury? The radius of Mercury is 2.44 x 10^6 m, and its distance from the sun at the moment is 5.79 x 10^10 m. Assume that the sun radiates uniformly in all directions.

Answer 1

You can think of this problem as
what fraction of the surface of a sphere of the distance from the sun is taken up by mercury

this can be calculated by thinking of mercury as a circle of area
pi * r^2 = pi * (2.44 * 10^6)^2 = 1.87 * 10^13 m^2

The surface area of the sphere is
4 * pi * r^2 = 4 * pi * (5.79 * 10^10)^2 = 4.21 * 10^22

so the fraction intercepted is
(1.87 * 10^13) / (4.21 * 10^22)

=4.44 * 10^-10 of the radiation emitted

The actual amount will be slightly different as the surface of a sphere is curved, but is counted as flat in this answer (a circle is used instead of part of the curved surface of the sphere), so a slightly different (but pretty much negligible) amount of the sphere will be covered.

Answer 2

The surface area of the sphere of radius equal to the distance from the sun to the planet Mercury divided by the cross sectional area blocked by the planet Mercury is the percentage being requested.

Surface of a sphere is 4/3*Pi*R*R
1.40E+22
Cross sectional Surface is Pi*R*R
1.87E+13
Percentage is 0.00000013%