Why Do You Divide the Solar Constant by 4 to Find Average Incoming Intensity on Earth?

You divide the solar constant by 4 because Earth intercepts solar radiation over a circular cross-section, but this energy is averaged over Earth’s entire spherical surface. The ratio of these areas is $\frac{\pi R^2}{4\pi R^2}=\frac{1}{4}$.

The Geometrical Reason

The solar constant $S$ is the solar power received per unit area by a surface perpendicular to the Sun’s rays at Earth’s orbital distance. In the IB Physics data booklet, $S=1.36\times10^3\ \text{W m}^{-2}$.

Earth intercepts sunlight as if it were a flat disk of radius $R$. Therefore, the total incoming solar power is

$$P_{\text{in}}=S\pi R^2.$$

However, Earth rotates, and its climate is considered across the whole surface. The surface area of a sphere is

$$A_{\text{Earth}}=4\pi R^2.$$

The average incoming intensity is therefore

$$I_{\text{average}}=\frac{P_{\text{in}}}{A_{\text{Earth}}}=\frac{S\pi R^2}{4\pi R^2}=\frac{S}{4}.$$

Numerically,

$$I_{\text{average}}=\frac{1.36\times10^3}{4}=340\ \text{W m}^{-2}.$$

This averaging assumes the solar radiation arriving at Earth is approximately parallel and that the value is averaged over both location and time. At any particular point, the instantaneous intensity depends on the angle between the rays and the local surface normal. The factor of 4 therefore does not mean every location actually receives $340\ \text{W m}^{-2}$.

A common misconception is that the factor of 4 comes only from day and night. Day and night provide a factor of 2, but the other factor of 2 comes from the curved surface receiving rays at different angles.

The value $S/4$ is the incoming average before reflection. If Earth’s albedo is $\alpha$, the average absorbed intensity is $(1-\alpha)S/4$.

For example, if Earth’s albedo is 0.30, then the absorbed average is $0.70(340)=238\ \text{W m}^{-2}$. This absorbed value is the quantity compared with outgoing infrared intensity in a simple Earth energy-balance model. It is lower than 340 because 30% of incident solar power is reflected.

IB Exam Technique

This is SL and HL content in B.2 Greenhouse effect. In an Explain question, state both areas, form their ratio, and distinguish incoming intensity $S/4$ from absorbed intensity $(1-\alpha)S/4$.