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:


*

*Maxwell's stress tensor.

*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?
 A: $\vec S$ 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:
$$
\vec F = \frac{\pi R^2 \vec S}{c}
$$
A: So what's your answer? I used information from the web that said the radiation pressure from the sun is about $1000$th of a $\textrm{g/m}^{2}$. Not much until you consider just how many $\textrm{m}^{2}$ face the sun!
I then took the radius of earth at the poles (assuming the pressure to be less when its striking a glancing blow so to speak) which is $6356,752.3 \,\textrm{m}$, squared is $40,408,299,803,555.29$, multiplied by $3.14159265358979 = 126,946,417,806,903.1 \,\textrm{m}^2$, divide by $1,000,000,000$ to get pressure in elephants (tonnes) which is $126,946.4178069031$ tonnes!
Nearly $127$ thousand tonnes! Surely this pressure is gradually shifting Earth's orbit outwards? How accurate is my figure? How reflective is the earth? I read pressure on a perfect reflector is twice that of a back body (which is sooooo cool). I'm guessing we must be somewhere between a black body and perfect reflector but I wouldn't know where to start to get that info.
What got me thinking about this was an article about the radiation pressure on the death star which would send it hurtling away from the planet it was destroying at over $3 \%$ of the speed of light .
"What about ‘recoil’?"
One aspect of the weapon’s capabilities that certainly doesn’t seem to be factored into the film, or any other calculations, is recoil. Professor Alexander Barnett, from Dartmouth University, discusses its possible side effects:
“Momentum carried by that much light energy is $p = E/c$ ~ $10^{}$ newton seconds. The Death Star doesn't have that much mass - even if we assume it's $10 \%$ solid metal, it only has a mass of around $10^{17}$ kg. So, that means once it’s finished firing the beam, by conservation of momentum, the Death Star is now flying backwards at:
$$\frac{10^{24}}{10^{17}} \sim 10^{7}\, \textrm{m/s}$$
“That’s around $3 \%$ of the speed of light, meaning it would travel $100$ times its own diameter ($100 ~ \textrm{km}$) every second. I didn't notice any movement in the film, but maybe it shoots another beam behind it to balance itself out that we just don’t see!”
see https://www.ovoenergy.com/blog/energy/as-rogue-one-arrives-in-cinemas-we-estimate-the-cost-of-powering-the-death-star for a laugh...
