Question

Radiant energy from the sum arrives at Earth with an intensity of$\mathbf{1}\mathbf{.}\mathbf{5}\mathit{k}\mathit{w}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$. Making the rough approximation that all photons are absorbed, find (a) the radiation pressure and (b) the total force experienced by Earth due to this “solar wind”.

Short answer

1. Radiation pressure from the sunlight on the earth is 5x10^-6N/m².
2. The total force experienced by the earth from the radiation pressure is 6.38x10⁸N.

Step-by-step solution

Derivation to find the formula of radiation pressure:-

The pressure exerted by the radiation,

$\mathit{P}\mathbf{=}\frac{\mathbf{F}}{\mathbf{A}}$

Substitute that force is the rate of change of momentum $\mathit{F}\mathbf{=}\frac{\mathbf{d}\mathbf{p}}{\mathbf{d}\mathbf{t}}$, therefore we can write the above expression,such that,

$$\begin{gathered} \mathit{P}\mathbf{=}\frac{\left({\displaystyle \frac{dp}{dt}}\right)}{\mathbf{A}} \\ \mathit{P}\mathbf{=}\frac{\mathbf{1}}{\mathbf{A}}\left(\frac{dp}{dt}\right)\mathbf{}\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{\dots }\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

Calculate pressure

Expression for the energy of the radiation if the momentum of light

$$\begin{gathered} E=pc \\ p=\frac{E}{c} \end{gathered}$$

Substitute in the equation (1), the value of p and we get,

$\begin{array}{rcl}P& =& \frac{1}{A}\frac{d}{dt}\left(\frac{E}{c}\right)\\ & =& \frac{1}{Ac}\frac{dE}{dt}\\ & =& \frac{1}{Ac}\left(power\right)\end{array}$

The intensity of the radiation,

$\begin{array}{rcl}I& =& \frac{P}{A}\\ P& =& IA\end{array}$

Step 3:- Substitute values in the formula

$\begin{array}{rcl}P& =& \frac{1}{Ac}\left(IA\right)\\ & =& \frac{I}{C}\end{array}$

Now, substitute I = 1500 W/m² and C = 3x10⁸m/s in the above expression and we will get,

$\begin{array}{rcl}I& =& \frac{1500W/m^2}{3\times {10}^{8}m/s}\\ & =& 5\times {10}^{-6}N/m^2\end{array}$

Step 4:- Substitute Calculation of total force of radiation pressure exerted by the radiation

$$\begin{gathered} P=\frac{F}{A} \\ F=PA \end{gathered}$$

A cross-sectional area of the earth is a circle will be $A=\pi R^2$and therefore,

$\begin{array}{rcl}F& =& P\times {\mathrm{\pi R}}^2\\ & =& \mathrm{\pi }\times \left(5\times {10}^{-6}\mathrm{n}/{\mathrm{m}}^2\right)\times {\left(6.371\times {10}^6\mathrm{m}\right)}^2\\ & =& 6.38\times {10}^8\mathrm{N}\end{array}$

Therefore, the radiation pressure from the sunlight on the earth is $5\times {10}^{-6}N/m^2$and the total force experienced by the earth from the radiation pressure is$6.38\times {10}^{8}N$