# Why is the hot side of the James Webb Space Telescope expected to be 85 C?

According to multiple sources, the heat shield on JWST is expected to exceed 85 °C. The average temperature of the earth is only 15 °C, so this is about 70 °C higher at roughly the same distance from the sun. I assume the reason is that successive layers of the heat shield reflect heat back towards Layer 1, rather than allowing it to diffuse through the whole spacecraft, in addition to the heat generated by active components on the spacecraft itself.

That is to say, I assume that the 85 °C temperature cited reflects the fact that the sun side is a combination of energy collector, energy reflector, and thermal radiator (although I recognize that the shield is also designed to dump heat out the sides as well). However, none of the sources I have looked at says that is actually the case. Can anyone confirm this theory or give a correct explanation?

• "The average temperature of the earth is only 15 C" has a lot to do with sitting on top of a huge heat sink, under a blanket that weight 10 tons per square meter, and gently rotating this rotisserie to bake all sides evenly . Rather look at the midday temperature of the Moon, which can hit 130C. Moon is a bit darker than shiny spaceship, hence JWST is bit cooler despite the same-ish sun. Dec 5, 2021 at 6:51
• Just leave your garden hose a couple of hours in the sun and then measure the temperature of the water inside. This experiment doesn't take into account the cooling effect of the air vent around it. Dec 5, 2021 at 9:41

At a most basic level, the surface area of a sphere is four times its apparent illuminated area whereas the surface area of a flat sheet is twice its apparent illuminated area. Since the power radiated goes as the fourth power of temperature, one would expect a flat sheet to reach an absolute temperature higher by a factor of the fourth root of two, or a temperature of 70 °C.

The real answer is of course much more complex. Neither the Earth nor JWST are black bodies. So the temperature will depend not just on the geometry but also on the absorptivity/emissivity of the surfaces at the various relevant wavelengths - perhaps other contributors can provide more detail?

• I think you've nailed it; a one-sided black-body (backside insulated) would be $2^{1/4}$ hotter than a two-sided one. As far as more detail I'm confident you can add the solar and Stefan–Boltzmann constants and get verification of the 85 C
– uhoh
Dec 5, 2021 at 3:14
• For ballpark calculations, this is a fantastic answer, thanks! Dec 5, 2021 at 22:24

The accepted answer is correct in terms of the calculation for a flat sheet absorbing on one side while radiating from both (a factor of $$2^{1/4}$$ higher), but does not really answer the original question.

(BTW the numbers are not quite 70°C or 85°C: even if JWST was at the same distance from the Sun as Earth and thus experienced the same flux the temperature would only be 331K, and at its position at $$L_2$$ where the solar flux is 97% (1317 Wm$$^{-2}$$) of that at Earth it would be 328K, or 55°C.)

However, the answer does not take into account that the first layer of the shield reflects 90% of the incoming Solar radiation. Due to this, if the only source of heat was the Sun the hot side of JWST would be colder than Earth at 185K, not hotter.

The original question is correct that the additional heat source is the telescope itself (more specifically, the spacecraft bus with the computers, electronics, etc), and the sunshield is also acting as a cooling system [1], as well as collecting some heat energy from the Sun.

Even then though the total power budget for JWST is only 2kW [2], 10% of the incident solar radiation (200kW total over the area of the sunshield), not enough to raise the temperature by that much. The point is that the media reports are (somewhat) misleading: it is not the whole heat shield which will reach 85°C, but some “parts of Webb on the hot side” [3], those nearest the system bus and other components such as the radiators.

• They had to carefully model or measure in a vacuum chamber the temperature and reflectivity in order to get the momentum flap sizing right. Sep 1, 2022 at 20:31