This is an image of the radioisotope thermoelectric generator (RTG) for Cassini:

enter image description here

This one was for one left on the Moon:

enter image description here

And this one is for the Multi-Mission RTG, used by Curiosity on Mars:

enter image description here

One is black, one grey, and one white. Why such a difference in the color of these RTGs?

  • $\begingroup$ I bet it has something to do with the thermal properties of the fins, but I haven't been able to find anything specific. I do know that some RTGs have beryllium housing/fins and some have aluminum housing/fins. $\endgroup$
    – called2voyage
    Commented Sep 7, 2016 at 17:48
  • $\begingroup$ By the looks of it, that grey one might not be grey - it might just be covered in something, shall we say, fine and powdery. $\endgroup$
    – corsiKa
    Commented Sep 7, 2016 at 20:47
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    $\begingroup$ The dust is on the pallet but not on the fins. The RTG and the 'soil' are both as black as asphalt. (photo is from Apollo 14) $\endgroup$
    – amI
    Commented Sep 7, 2016 at 23:09
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    $\begingroup$ @gerrit: That looks to be an engineering mock-up or test article, rather than flight hardware. No need to work in a cleanroom environment unnecessarily--it just makes everything more difficult. $\endgroup$
    – Drew Hall
    Commented May 8, 2018 at 6:23
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    $\begingroup$ Please reconsider which answer you accepted in this question. There is a newer answer which appears to be a better explanation. $\endgroup$
    – DrSheldon
    Commented May 21, 2020 at 16:28

3 Answers 3


Answer: Thermal radiating coating technology has improved, so they are no longer forced to be sub-optimally black in visible light. They can now be white and reflect incident sunlight to improve thermoelectric efficiency by staying cooler.

The color has nothing to do with the atmosphere. It has to do with sunshine!

Curiosity's MMRTG is producing about 2 kW of power constantly. A small amount of heat is dissipated by being converted to electricity and some is carried away by liquid in tubes to warm the rover's innards on cold nights.

The efficiency of the thermoelectric conversion depends upon the cooling fins remaining effective at dissipating heat. If the fins were black they would efficiently absorb sunlight and get hotter. The Multi-Mission Radioisotope Thermoelectric Generator (MMRTG) fact sheet gives the size as 64x66 cm, a black square that size could receive as much as 200 Watts of heating from sunlight on Mars, and that would be a serious efficiency hit.

So the reason that modern RTGs such as the MMRTG look white in visible light is so they don't get heated by sunlight.

It is true that the MMRTG has several design features related to working in various atmospheres (planets and moons) but the white visible color is to avoid getting hot in the Sun.

This diagram mentions Aptek 2711 coating.

From that link (hat tip to @DavidHammen's comment):

APTEK 2711 was developed for use as a thermally conductive coating where excellent resistance to intense UV light exposure is required.

Total normal emittance (ASTM E-408)

Solar absorption vs thickness (ASTM E-903)
               α    mils (0.001 inch)
             0.20        2
             0.185       3
             0.17        4

Here's something similar mentioned in comments (also found here).

AZ-93: http://www.aztechnology.com/materials-coatings-az-93.html

Application of AZ-93 creates a nonspecular white coating that provides superior thermal control / protection by allowing only 14-16% of the solar radiation impinging on the spacecraft external surface to be absorbed through to the interior systems while emitting 89-93% of the internal heat generated to the cold vacuum of space. By incorporating a highly stabilized pigment system with a silicate binder, this spacecraft / satellite paint forms a bendable ceramic coating that has been tested time and again and has proven itself stable in the harshness of the space environment. AZ-93 has been exposed by NASA to atomic oxygen (AO) fluence of 5.6 x 1022 atoms/cm2, charged particle radiation of 4.5 x 1015 e-/cm2, and vacuum ultraviolet (VUV) radiation (from 118 nm to 170 nm) of 701 equivalent solar hours with less than 4% deterioration in solar absorptance (α_s) and less than 1% change in thermal emittance (ε_t).

It mentions

Thermal Emittance (ε_t)    0.91 ± 0.02
Solar Absorptance (α_s)    0.15 ± 0.01 at ≥ 5.0 mils thickness
Use Temperature Range     -180 C to 1400 C

So these coatings are "white" in visible light with about 85% reflectance, but "black" in thermal infrared with an emissivity of about 0.91 (which also means it would only reflect about 9%).

From Radioisotope Power Systems for Space Exploration:

cutaway model of a Multi-Mission Radioisotope Thermoelectric Generator (MMRTG)

A cutaway model of a Multi-Mission Radioisotope Thermoelectric Generator (MMRTG). The vertical red blocks in the center are individual heat source modules and the white fins on either side are radiators.

  • $\begingroup$ The color may also depend on the exposure and lighting of the picture. A dark grey surface may be pictured as medium grey, very dark grey or even black. But the white RTG is definatly a different thing and will increase efficiency. Of course a white surface may be pictured as brilliant white or very light grey using different exposures. $\endgroup$
    – Uwe
    Commented Jun 26, 2019 at 8:14
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    $\begingroup$ @uhoh - Re I can't find the visible light absorbtivity versus thermal infrared emissivity for it ... The numbers are in the link you yourself provided on APTEK 2711. On the second page, under CURED PHYSICAL PROPERTIES, it specifies solar absorption as a function of thickness in mils (thousandths of an inch), ranging from $\alpha_s=0.20$ at 2 mils to 0.17 at 4 mils. It also specifies a total normal emittance of 0.93. $\endgroup$ Commented Jun 26, 2019 at 13:54
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    $\begingroup$ @OrganicMarble no, the best option that was available was black, which was not what you'd want for the color in visible light. It probably didn't matter for deep space when the Sun wasn't so bright, but it certainly wasn't optimal on the Moon. I've been thinking about asking "Why were the Apollo RTGs on the Moon black?" to bring this non-ideal situation out, but people will answer "so it can radiate" and then someone will say "but that's in the thermal IR not visible" and... $\endgroup$
    – uhoh
    Commented Jun 26, 2019 at 14:38
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    $\begingroup$ @OrganicMarble if it hadn't been invented yet, then it wasn't an option. It's hard to make something very black in IR and simultaneously white in visible and simultaneously good thermal conductivity and simultaneously good UV and radiation hardness and simultaneously good thermal stability and simultaneously proven in space. $\endgroup$
    – uhoh
    Commented Jun 26, 2019 at 14:45
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    $\begingroup$ @OrganicMarble - What you're possibly missing is that a good absorber is a good emitter, while a bad absorber is a bad emitter. Black bodies (objects that are completely black at all frequencies; such objects don't exist) perfectly absorb all incoming frequencies of electromagnetic radiation. Black bodies are also ideal thermal radiators. Black carbon is a very close to a black body, absorbing nearly 100% of incoming radiation and also emitting across a rather wide range of frequencies, from microwave to UV. $\endgroup$ Commented Jun 26, 2019 at 17:02

The big difference between the two darker RTG fins (Black and Grey) and the white RTG fins, is that the white fins were destined for use in an atmosphere (Mars). The presence of an atmosphere, even as diffuse as Martian air, would allow increased heat transfer from the RTG fins via convection and conduction, vs. the space based versions which would entirely rely on radiation to transfer heat.

Radiation heat transfer is affected by color and is likely the reason for the color differences.

  • $\begingroup$ That makes sense. I knew there was something I had to be missing! $\endgroup$
    – PearsonArtPhoto
    Commented Sep 7, 2016 at 18:00
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    $\begingroup$ I'd really like to see a technical link that explains this quantitatively. Ballparking: for convective heat transfer to (Earth) air, I see numbers like 5 W/m^2/K. If Mars is say 1% and it scales linearly with # of molecular collisions/time, then a 300K difference on a 0.5 m^2 area would be only about 10 Watts. $\endgroup$
    – uhoh
    Commented Sep 8, 2016 at 13:26
  • $\begingroup$ Further - why not make it black and pick up the extra power via radiation? The atmosphere goes through wide daily variations in temperature and air speed which will affect the efficiency of the system. Adding radiative cooling would moderate those fluctuations, especially in "warm", windless spells. After poking around the internet and seeing that in every one of Curiosity's "selfies" on Mars the heat sink is indeed white, it does suggest that unless it's some kind of exotic material that the IR emissivity will be low. $\endgroup$
    – uhoh
    Commented Sep 8, 2016 at 13:45
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    $\begingroup$ The white coating is probably something like Z93, which maintains high thermal emissivity while having low absorbtivity in much of the solar spectrum. It's used on the ISS radiators as one example. $\endgroup$
    – Tristan
    Commented Sep 8, 2016 at 14:53
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    $\begingroup$ @Tristan now that makes sense! "...allowing only 14-16% of the solar radiation impinging on the spacecraft external surface to be absorbed through to the interior systems while emitting 89-93% of the internal heat generated to the cold vacuum of space." You'd want it "white" in the visible to minimize solar loading, but "black" in the IR to maximize radiation. If it were indeed the case here, it would suggest that this answer - roughly speaking "the color doesn't matter because theres a little bit of air" - is incomplete. Also thanks for linking to the Z93 - that's cool stuff! (punintended) $\endgroup$
    – uhoh
    Commented Sep 9, 2016 at 3:36

I'll elaborate further on some of the points in other answers and provide some additional background.

To put it as simply as possible, a good white spacecraft paint such as AZ93 (ZnO pigment in a silicate binder) is very white in the visible where you eye responds. Also where most of the energy from the sun is located. So the white paint reflects sunlight while absorbing only a small amount (like 10 to 15 %). Now what is important is in the infrared wavelength range where you cannot see, the white paint is BLACK, like really black. It runs over 90% absorption in the IR and beyond (beyond a few microns). Body heat peaks at around 10 microns for reference. A 90% absorption also means in the IR the "white paint or coating" also emits radiation at around 90% of an ideal black body.

Sorry, but this can be confusing at first. Simply put, what energy goes in [at a given wavelength] must go out. At the same wavelength absorbance equals emittance; in other words you cannot create energy by heat transfer; that is unless you wonder into the Quantum Field and then all bets are off. Furthermore, the black body radiation is a function of the four power of temperature, therefore the visible white paint can dump a lot of heat in the infrared region. In general an ideal thermal control radiator would reflect all of the incident sunlight (low solar absorptance, high reflectance) and radiate in the IR (black body radiation at a very high rate). Spacecraft white paints are very good at this, also second surface mirrors (quartz side exposed), and silver Teflon (Teflon exposed). I have a lot of good schematics that help to explain this, but hard to use them here.

Note, if you are not exposed to sunlight, then a solid black coating will work, that is a good black coating/paint. Also must be stable in the space environment, like damage from solar UV, high energy particulate radiation (electrons and protons), and for LEO orbits (atomic oxygen).

  • $\begingroup$ Welcome to Stack Exchange! This is a very helpful explanation and while it overlaps somewhat with other answers it explains things in a different way which some readers may find much easier to understand. I added some paragraph breaks to make it a little easier to read, feel free to edit further, thanks! $\endgroup$
    – uhoh
    Commented Jul 25, 2020 at 3:43

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