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The traditional answer is that it's because black radiates heat better. The problem is that in order to get rid of heat at all, the surface must glow brighter than the surrounding plasma. And that seems doubtful to me.

Abdullah is not an Amalekite
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    I think this is very related to, but maybe not exactly a duplicate of, https://space.stackexchange.com/questions/39641/why-is-the-heat-shield-of-esas-solar-orbiter-black-in-colour?rq=1 – Erin Anne Feb 15 '24 at 09:06
  • @ErinAnne doesn't even come close to an answer – Abdullah is not an Amalekite Feb 15 '24 at 12:28
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    I'm not convinced that the surface must glow brighter than the surrounding plasma. The Stefan-Boltzmann law says that radiation is proportional to the 4th power of absolute temperature, not the temperature difference. – Hobbes Feb 15 '24 at 15:31
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    @Hobbes the ship would keep heating up until the radiated power equals irradiated power. Unless the plasma is not completely opaque, that will be at equal temperature – Abdullah is not an Amalekite Feb 15 '24 at 16:30
  • Not that it changes the basis of your question, but I don't think it's related so much to spaceplanes but instead to the choice to use refractory insulated tiles instead of ablative coatings. Although as it turns out spaceplanes (and Starships) tend to use tiles instead of ablative, mainly for reusability. Tiles in lower temperature areas of the Shuttle were white (later replaced by insulated blankets) and could withstand 1200° F, whereas tiles on higher temperature areas were coated with a borosilicate glass coating which gave them their black color, and could handle up to 2300° F. – Steve Pemberton Feb 15 '24 at 19:01
  • "They chose black because it represented the easiest development path, and despite the higher temperatures the front shield would experience. The white shield also mentioned in the study uses different materials: MLI (multilayer insulation, ed.) would use ABS (Alumina- Boria-Silica, ed.) as the front layer, which would have an a/ε of about 0.45 and high temperature resistant metallic foils such as titanium. ... This ABS MLI would be a new development, while carbon-carbon is well-known [and has a] a/e (absorption/emission constant, ed.) close to 1" = why the bottoms of spaceplanes are black. – Mazura Feb 16 '24 at 02:11
  • It's the answer below that doesn't make any sense w/o knowing what https://en.wikipedia.org/wiki/Emissivity is. "temperature emittance" – Mazura Feb 16 '24 at 02:12
  • @Mazura if you think that’s the case, then why not write an answer instead of using the comments? – fyrepenguin Feb 16 '24 at 02:32
  • @fyrepenguin Why is the Heat Shield of ESA's Solar Orbiter Black in colour? – Mazura Feb 16 '24 at 02:33
  • Another factor: Having a white face and a black face that you can choose which to point at the sun is useful for thermal regulation. – Loren Pechtel Feb 16 '24 at 07:18

1 Answers1

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I'm afraid that in this case, at least, the conventional wisdom is correct.

For the space shuttle at least:

A black version with good high-temperature emittance is used on the lower surface of the orbiter, and a white version with low solar absorption...is used on the upper surfaces of the orbiter.

Source: Space shuttle orbiter reusable surface insulation flight results

Also

The ceramic tiles that were developed for use on the Orbiter were divided into two categories: the High- temperature Reusable Surface Insulation (HRSI) and the Low-temperature Reusable Surface Insulation (LRSI). The primary difference in these two types of tiles is the color of the surface coating. The HRSI that were found primarily on the lower surface of the Orbiter were coated with a black borosilicate glass, while the LRSI tiles had a white glass coating. The HRSI tiles have a coating containing a black pigment for the proper high temperature emittance value (ϵ > 0.8) that is needed in the high-temperature applications on the Orbiter. The LRSI tiles have a white coating with the proper optical properties ( α/ϵ < 0.4) that is needed to maintain the proper on-orbit temperatures for vehicle thermal control purposes.

Source: Orbiter Thermal Protection System Lessons Learned (emphasis mine)

Organic Marble
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    How on earth did those tiles glow brighter than the plasma? – Abdullah is not an Amalekite Feb 15 '24 at 14:22
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    @Abdullah The plasma was optically thin. https://en.wikipedia.org/wiki/Optical_depth – John Doty Feb 15 '24 at 14:26
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    @Abdullah your intuition that it's strange that the tiles would glow brighter than the plasma is wrong. There is nothing surprising about that. For one thing, the plasma is extremely thin (specifically, the thinnest that's enough to brake the orbiter). For another, gas / plasma is generally not actually very good at absorbing / emitting light, c.f. the pale flames of hydrogen or a clean Bunsen burner. If a plasma is bright it's usually either because of strong currents knocking away electrons, or because of solid particles like soot. The entering orbit is a single big solid particle. – leftaroundabout Feb 15 '24 at 22:15
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    @leftaroundabout - burning hydrogen is very good at emitting light, it is just in the UV not the visible. – Jon Custer Feb 16 '24 at 13:11
  • LRSI has a coating? I thought it was just the bare silica "foam" material. – ikrase Feb 17 '24 at 09:42
  • @ikrase both colors had a reaction-cured glass coating http://www.nasa-klass.com/Curriculum/Get_Oriented%202/Space%20Shuttle%20Information/RDG_Space-Shuttle-Info-Additional/Orbiter%20Thermal%20Protection%20System.pdf page 3 – Organic Marble Feb 17 '24 at 12:10