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Q49CE

Expert-verifiedFound in: Page 95

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

**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”.**

- Radiation pressure from the sunlight on the earth is 5x10
^{-6}N/m^{2}. - The total force experienced by the earth from the radiation pressure is 6.38x10
^{8}N.

**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,**

**${\mathit{P}}{\mathbf{=}}\frac{\left(\frac{dp}{dt}\right)}{\mathbf{A}}\phantom{\rule{0ex}{0ex}}{\mathit{P}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{A}}{\left(\frac{dp}{dt}\right)}{\mathbf{}}{\mathbf{\dots}}{\mathbf{\dots}}{\mathbf{\dots}}{\mathbf{\dots}}{\mathbf{\dots}}{\mathbf{.}}{\mathbf{\left(}}{\mathbf{1}}{\mathbf{\right)}}$**

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

$E=pc\phantom{\rule{0ex}{0ex}}p=\frac{E}{c}$

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,

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

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

Now, substitute I = 1500 W/m^{2} and C = 3x10^{8}m/s in the above expression and we will get,

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

$P=\frac{F}{A}\phantom{\rule{0ex}{0ex}}F=PA$

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 $6.38\times {10}^{8}N$

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