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.

  • 2
    $\begingroup$ I think this is very related to, but maybe not exactly a duplicate of, space.stackexchange.com/questions/39641/… $\endgroup$
    – Erin Anne
    Feb 15 at 9:06
  • $\begingroup$ @ErinAnne doesn't even come close to an answer $\endgroup$ Feb 15 at 12:28
  • 2
    $\begingroup$ 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. $\endgroup$
    – Hobbes
    Feb 15 at 15:31
  • 2
    $\begingroup$ @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 $\endgroup$ Feb 15 at 16:30
  • $\begingroup$ 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. $\endgroup$ Feb 15 at 19:01

1 Answer 1


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


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)

  • 1
    $\begingroup$ How on earth did those tiles glow brighter than the plasma? $\endgroup$ Feb 15 at 14:22
  • 16
    $\begingroup$ @Abdullah The plasma was optically thin. en.wikipedia.org/wiki/Optical_depth $\endgroup$
    – John Doty
    Feb 15 at 14:26
  • 7
    $\begingroup$ @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. $\endgroup$ Feb 15 at 22:15
  • 1
    $\begingroup$ @leftaroundabout - burning hydrogen is very good at emitting light, it is just in the UV not the visible. $\endgroup$
    – Jon Custer
    Feb 16 at 13:11
  • $\begingroup$ LRSI has a coating? I thought it was just the bare silica "foam" material. $\endgroup$
    – ikrase
    Feb 17 at 9:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.