Problem — Radiation pressure at Earth's orbit

Problem

Estimate the radiation pressure exerted by sunlight at Earth’s orbit, knowing that the irradiance is $I = 1{,}4\;\text{kW/m}^2$ and assuming complete absorption. Compare it with atmospheric pressure ($\sim 10^5\;\text{Pa}$).

Solution

Setup. For a surface that completely absorbs the radiation, the radiation pressure is the ratio between the intensity and the speed of light:

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

Numerical substitution.

$P_\text{rad} = \frac{1400}{3\cdot 10^8} \approx 4{,}7\cdot 10^{-6}\;\text{Pa}$

$\ev{P_\text{rad} \approx 4{,}7\cdot 10^{-6}\;\text{Pa} = 4{,}7\;\mu\text{Pa}}$

Comparison with the atmosphere.

$\frac{P_\text{rad}}{P_\text{atm}} = \frac{4{,}7\cdot 10^{-6}}{10^5} \approx 5\cdot 10^{-11}$

The radiation pressure is about $50$ billion times smaller than atmospheric pressure: tiny on the everyday scale, but not negligible for solar sails and for the equilibrium of stars.

Links

Topics: Onde elettromagnetiche Concepts: Pressione di radiazione Skills: Stime di Fermi