Calculating the Cross Section of Planar Radiation Upon a Sphere and Finding the Average Radiation Intensity Over the Sphere

The incidence of planar radiation upon a sphere, such as that from the sun upon the Earth, is an important calculation for climate science, so I am going to demonstrate that calculation.  Some people have doubts about the common claim that the average incident intensity of solar insolation upon the Earth per unit area is one-quarter of the radiation intensity at that part of the Earth where the surface is at normal incidence to the solar radiation.

Let the amplitude of the planar radiation be P₀, which is commonly expressed as W/m².   The incident radiation on the sphere surface normal to the planar radiation will have an intensity of P₀.  As one examines a unit surface are of the sphere which is non-normal to the planar wave of radiation, the unit area of the sphere intercepts less and less of the planar radiation the greater the surface normal angle is with respect to the direction of the plane wave.  Let us call the angle from the center of the sphere to the point on the surface upon which the planar wave has normal incidence and the surface annulus of points equidistance from the normal incidence point θ.  The surface of the hemisphere facing the planar radiation then consists of a normal point and a series of annuli, one for each Δθ that we choose.  The annuli have increasing radii perpendicular to the planar wave of radiation.  The angle θ to describe these annuli varies from 0 to π/2 radians or 90̊.  We will use the radian units.  Let the radius of the sphere be R.

Each annulus has a circumference of 2πR sin θ and it intercepts an annulus of the planar radiation wave with the same circumference, but with a narrower width than the annulus on the sphere surface has for θ > 0.  The width of the planar wave annulus that projects onto the sphere surface is given by Δθ R cos θ.  Thus, at angle θ the area of the plane wave of radiation intercepted by the sphere at angle θ, is

A_I (θ) = 2πR² sin θ cos θ Δθ

Integrating over the range from 0 to π/2 for θ, we find the area of the planar wave of radiation which is intercepted by the sphere.  The integral is

A_I = 2πR² ∫₀^π/2 sin θ cos θ dθ

The value of the integral is ½ sin²θ, which for these limits of integration is ½.  Consequently,

A_I = πR²

The total incident planar wave intensity on the sphere (all on one hemisphere) is then

P_T = P₀ A_I = πR² P₀

The total area of the sphere, A_S, is 4πR², so we can rewrite this as

P_T = (1/4) A_SP₀

Averaging over the entire area of the sphere, the total incident plane wave intensity average per unit area is then P₀/4.