Comment by mapt | Hacker News Reader

Comment by mapt | Hacker News Reader

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The blackbody equilibrium temperature at 1AU from the sun is about 278K. All satellite materials are fine at 278K because that's within the expected range of storage and atmospheric launch temperatures.

The equilibrium temperature of a polished aluminum surface at 1AU from the sun is 416K, hot enough to melt polyethylene and at least weaken many of the relevant aerospace plastics like the PET in mylar film.

Painting polished aluminum black drastically raises emissivity along with the lowered reflectivity, and brings its behavior closer to a blackbody.

So does allowing aluminum to oxidize, which it does almost instantaneously in atmosphere. So it's not like it's going to change anything drastically.

The reason this seems like it should change things a lot is that you're used to convectively cooled matte surfaces on Earth, where emissivity and radiative cooling is a less relevant factor and the only significant effect of painting something black is primarily that it absorbs more energy.

"The blackbody equilibrium temperature at 1AU from the sun is about 278K. All satellite materials are fine at 278K because that's within the expected range of storage and atmospheric launch temperatures."

But that is NOT the temperature of something in LEO. You're ignoring everything else that adds energy to the system. Friction from collisions with atomic oxygen, down to heating up to temperatures as hot as 530 Kelvin just entirely dependent upon orientation to the sun.

My impression is that friction is negligible once firmly in LEO. Orientation of a sub-component surface definitely matters, as does Earth's presence in half of the viewshed. But my point is that surface finish also makes a big deal, and paint is not necessarily a bad thing, because of the emissivity increases.

(Though I'd rather anodize the thing black imperfectly if it helps avoid paint flecks becoming orbital debris)

How does this happen? Why would something which reflects light on the outside get hotter than one which absorbs light? This makes no intuitive sense to me.

The radiative equilibrium temperature is a function of (watts in - watts out), with a fourth power law shoved in there somewhere. A blackbody at radiative equilibrium absorbs whatever visible light you throw at it, and then spits it back out according to a distribution law that mostly places it in the thermal infrared bands (at temperatures we've familiar with, anyway).

Remove convection/conduction as heat transfer methods, and you end up with two numbers dictating radiative balance:

Percent reflectivity in the bands it's exposed to

Percent emissivity in the bands it's emitting

The balance between these dictates temperature, and they're generally inversely correlated. Mirrors are good reflectors, but very poor emitters.

Ah right, it reflects in both directions of course. Thanks.

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