B.2.6—Mean solar intensity

Why the factor is 1/4

A planet intercepts incoming sunlight over its projected disk, area $$\pi r^2$$. The intercepted power is then averaged over the planet’s whole spherical surface, area $$4\pi r^2$$.

Mean incoming intensity

If the solar constant is S,

$$\overline I=\frac{S\pi r^2}{4\pi r^2}=\frac S4$$

This is the mean intensity before accounting for reflection or atmospheric absorption.

Add albedo when required

If the planetary albedo is a, the globally averaged absorbed intensity is

$$I_{\mathrm{abs}}=(1-a)\frac S4$$

provided the problem’s model treats the planet as a uniform system.

Worked example from local Question Bank row 30218

For $S=1400\,\mathrm{W\,m^{-2}}$ and atmospheric albedo $a=0.30$, the transmitted incident intensity is

$$I=(1-a)S=(0.70)(1400)=980\,\mathrm{W\,m^{-2}}$$

Averaging the intercepted power over the full sphere gives

$$\overline I=\frac{980}{4}=245\,\mathrm{W\,m^{-2}}$$

This result combines reflection with geometry: $(1-a)S/4$.

Common trap

Do not divide S by 4 because sunlight is four times weaker at every point. The factor comes from intercepted disk area divided by total spherical area.