Problem — Radiation pressure at ground level

Problem

Estimate the radiation pressure exerted by sunlight on the Earth’s surface, knowing that the average intensity of the radiation at ground level is $I \sim 1300\;\text{W/m}^2$. Compare the result with atmospheric pressure ($p_\text{atm} \approx 10^5\;\text{N/m}^2$).

Solution

Setting up. For a surface that completely absorbs the radiation, the radiation pressure is the ratio of the wave’s intensity to the speed of light:

$P_\text{rad} = \frac{I}{c}$

(For a perfectly reflecting surface it would be double; for the purposes of the estimate the order of magnitude does not change.)

Calculation.

$P_\text{rad} = \frac{1300}{3\cdot 10^8} \approx 4{,}3\cdot 10^{-6}\;\text{N/m}^2 \sim 10^{-6}\;\text{N/m}^2$

$\ev{P_\text{rad} \approx 4{,}3\cdot 10^{-6}\;\mathrm{N/m^2}}$

Comparison with the atmosphere.

$\frac{p_\text{atm}}{P_\text{rad}} \approx \frac{10^5}{10^{-6}} = 10^{11}$

The radiation pressure of sunlight is about one hundred billion times smaller than atmospheric pressure: this is why, in everyday life, the “push” of light is entirely negligible (it only becomes significant on astronomical scales, such as comet tails or solar sails).

Connections

Topics: Electromagnetic waves Concepts: Radiation pressure Skills: Fermi estimates