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I need to find the thermal energy received by a satellite in LEO from the Earth, on the night side.

The thermal energy received from the Sun by a satellite is easy to calculate, because the Sun is a point source: $$ \Phi=\frac{L_\odot}{4\pi r^2} \approx 1367 \ \mathrm{W/m^2} $$ But a satellite in LEO has almost half its sky filled by a sphere with an average temperature of 283 K - so what is the flux from the Earth? Stefan-Boltzmann gives me a flux emitted from the Earth of $$ \Phi=\sigma T^4=364 \ \mathrm{W/m^2} $$ but how do I integrate that up? For a fast approximation, I calculated the size of the satellite's footprint, which gives me a radius of ~2200 km for an orbital altitude of 400 km. Multiplying the area of that circle with the flux above gives me $$ P_{Earth}=364 \ \mathrm{W/m^2} \times (2 200 000 \ \mathrm{m})^2 = 5.5 \cdot 10^{15} \ \mathrm{W} $$ which seems like a mind-bogglingly large number.

But now the next question arises: How does one calculate the flux at a distance $h$ from a sphere of this surface? I don't suppose I can use the inverse-square law, since the Earth isn't a point source.

So - can anyone help me with this calculation, or actually just provide me with a number: What is the flux at orbital altitude $h$ from the Earth on the night side?

OZ1SEJ
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    You seem to be aware of the basic error: that you need to calculate the irradiance at the satellite's altitude as a function of an extended source. If you treat the Earth as a Lambertian reflector/radiator, then the contributions fall off with distance from the perpendicular line up to the satellite (inverse cosine^2 or something like that) – Carl Witthoft Mar 10 '21 at 15:04
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    Possibly helpful: http://eodg.atm.ox.ac.uk/user/grainger/research/book/protected/Chapter3.pdf , – Carl Witthoft Mar 10 '21 at 15:11
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    Simple view: Calculate what percentage of the satellite's view is "earth". multiply that by 364 w/m2. at 400km, about 40%. Thus 145w... if course, thats under MANY assumptions. like: Earth radiates same in all directions. Earth radiates same at night as day!! Does not matter if loking at surface through 60km of atmosphere, or 80km, at extreme limb. etc, etc. – CuteKItty_pleaseStopBArking Mar 11 '21 at 12:55

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That mind-bogglingly large number is the total power emitted by the footprint of the earth in all directions, not the power absorbed by the satellite. As a quick and dirty approximation, you need to use the area of the satellite that can "see" the earth, and it needs to be scaled by the solid angle subtended by the earth's surface.

A more accurate answer would be done by integrating over the surface of the satellite to account for varying view factors, but this will get you "close enough"

Tristan
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    So would $$\sigma \varepsilon_E T_E^4 \frac{\Delta \Omega}{4 \pi}$$ do it, where subscript $E$ is Earth and $\Delta \Omega/4 \pi$ is the fraction of the sphere around Earth at the spacecraft distance that the satellite intercepts? And would we need an additional $\varepsilon_S$ for the fraction that's absorbed by the spacecraft and not reflected? – uhoh Mar 11 '21 at 03:11
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    @uhoh Pretty much. If you wanted to get really fancy, you'd also integrate over the electromagnetic spectrum to account for different emissivities/absorptivities at different wavelengths, but that may be needlessly overcomplicating things – Tristan Mar 11 '21 at 15:50