1. The averaged temperature of the satellite.
2. Temperature variations.
1.
Accurate calculation of radiation fluxes from the Earth and the Sun is the "looking for fleas".
The most significant influence on the temperature of the satellite have coefficient of absorption of sunlight and coefficient of radiation in infrared.
Metals with a smooth and clean surface have a low coefficient of radiation in infrared.
Example:
Aluminium 0.03 ... 0.06
Steel 0.07
But steel with coarse finish surface 0.24
Other materials (mineral and organic) have a large coefficient of radiation in infrared.
examples:
rusty iron 0.7
fireclay bricks 0.85
paper 0.8 ... 0.9
varnish and paint 0.85 ... 0.95.
Exception — aluminum paint 0.28.
Coefficient of absorption of sunlight is clearly visible to the naked eye.
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The thermal radiation of the earth = 239 W/m2 at zero altitude.
If the height of the satellite's orbit = 350 km, the Earth takes up 42 percent of the area of the sky, not 50%.
Then the flow from the Earth would be = 200 W/m2.
If a satellite has a spherical shape, then:
its surface, which absorbs sunlight is equal to pi * R^2 (cross section);
its surface, which emits infrared is equal to 4 * pi * R^2 (full surface);
its surface, which absorbs infrared of the Earth, is equal to 42% * 4 * pi * R^2 (part of surface).
I do not consider the flow of the visible light reflected from the Earth (nealy 100 W/m2). It can simply be added to the solar constant, which is equal to 1400 W/m2.
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If the satellite is constantly illuminated by the sun and the absolute black (k_visible = 1, k_infrared = 1), the balance of power would be:
k_visible * 1400 * pi * R^2 + k_infrared * 239 * 0.42 * 4 * pi * R^2 = k_infrared * 4 * pi * R^2 * 0.0000000574 * T^4
where 0.0000000574 - Boltzmann constant.
T^4 = (k_visible * 1400 * pi * R ^ 2 + k_infrared * 239 * 0.42 * 4 * pi * R ^ 2) / (k_infrared * 4 * pi * R^2 * 0.0000000574)
T = sqrt4(k_visible * 1400 + k_infrared * 239 * 0.42 * 4) / (k_infrared * 4 * 0.0000000574) = sqrt4((1400 + 400) / 0.0000002296) = 298 K = 25 C
25 C — it is room temperature.
If a satellite was constantly in the shade, its temperature would be:
T = sqrt4(0 + k_infrared * 239 * 0.42 * 4) / (k_infrared * 4 * 0.0000000574) = sqrt4((400) / 0.0000002296) = 204 K = —69 C
But "constantly in the shade" is not real. Maximum for real orbit is 40% in the shade.
So, for 60% illuminated by the sun:
T == sqrt4((0.6 * 1400 + 400) / 0.0000002296) = 271 K = —2 C
Also quite a decent temperature for electronics.
The same result would be for the satellite, colored by dark paint, because ratio k_visible / k_infrared = ~ 0.8 / 0.8 = ~ 1.
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If the satellite is the sphere from aluminium (k_visible = 0.18, k_infrared = 0.04), the temperature would be:
T = sqrt4(0.18 * 1.0 * 1400 + 0.04 * 239 * 0.42 * 4) / (0.04 * 4 * 0.0000000574) = 413 K = 140 C (100% sun)
T = sqrt4(0.18 * 0.6 * 1400 + 0.04 * 239 * 0.42 * 4) / (0.04 * 4 * 0.0000000574) = 367 K = 94 C (60% sun)
White paint (k_visible = 0.25, k_infrared = 0.85):
T = sqrt4(0.25 * 1.0 * 1400 + 0.85 * 239 * 0.42 * 4) / (0.85 * 4 * 0.0000000574) = 244 K = —29 C (100% sun)
T = sqrt4(0.25 * 0.6 * 1400 + 0.85 * 239 * 0.42 * 4) / (0.85 * 4 * 0.0000000574) = 231 K = —42 C (60% sun)
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So, there is a strong dependence of the temperature from the ratio k_visible / k_infrared.
The best result is obtained at about ratio = 1.
And easy to get the desired temperature, regulate ratio of these coefficients.
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2.
To be continued... ?