# Force on Earth due to Sun's radiation pressure

I have been asked by my Classical Electrodynamics professor to calculate the force that the Sun exerts in the Earth's surface due to its radiation pressure supposing that all radiation is absorbed and a flat Earth, and knowing only that the magnitude of the Poynting vector in the surface is $\left\langle {\bar S} \right\rangle = 13000{\rm{[W/}}{{\rm{m}}^{\rm{2}}}{\rm{]}}$ using:

1. Maxwell's stress tensor.
2. The absorbed momentum.

Using Maxwell's stress tensor I get ${\rm{35.6}} \cdot {\rm{1}}{{\rm{0}}^8}{\rm{[N]}}$, which seems plausible since we consider a flat Earth and no radiation reflection. But I'm lost on how to obtain an answer using the variation of electromagnetic momentum.

I think I should start by writing

$$\vec F = \frac{d}{{dt}}{\vec p_{EM}} = \frac{d}{{dt}}\int\limits_V {{\varepsilon _0}{\mu _0}\left( {\vec E \times \vec H} \right)dV}$$

But, how do I take it from here?

• Your Poynting vector should be 1300 W/m$^2$. – Rob Jeffries Dec 9 '15 at 8:57

$\vec S$ is the flux, so you need an area integral of the surface of the earth.
The pressure $P$ you will have is force per area, $F/A$. The pressure is flux $S$ divided by speed of light, since you have a momentum of $hf/c$ in the photons.
Then you should integrate over the pressure (i. e. multiplying with the cross section of the earth) to get the force: $$\vec F = \frac{\pi R^2 \vec S}{c}$$
• I see... so I just simply integrate S/c in the surface of the Earth? But wouldn't I get $\left\langle F \right\rangle = \frac{{\pi {R^2}}}{c}\left\langle S \right\rangle$? – Mike Jan 23 '13 at 21:39
• Yes, the 4 should not be there (I was thinking surface of a sphere apparently), and I forgot the $c$. So your result is correct. – Martin Ueding Jan 25 '13 at 13:37