# Why are RTGs different colors?

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

This one was for one left on the Moon:

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

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

• 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. – called2voyage Sep 7 '16 at 17:48
• 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. – corsiKa Sep 7 '16 at 20:47
• I suppose that's possible. Hmmmm... – PearsonArtPhoto Sep 7 '16 at 21:13
• 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) – amI Sep 7 '16 at 23:09
• @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. – Drew Hall May 8 '18 at 6:23

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.

• That makes sense. I knew there was something I had to be missing! – PearsonArtPhoto Sep 7 '16 at 18:00
• 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. – uhoh Sep 8 '16 at 13:26
• 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. – uhoh Sep 8 '16 at 13:45
• 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. – Tristan Sep 8 '16 at 14:53
• @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) – uhoh Sep 9 '16 at 3:36

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)
0.93

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).

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%).

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.

• 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. – Uwe Jun 26 at 8:14
• @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. – David Hammen Jun 26 at 13:54
• @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... – uhoh Jun 26 at 14:38
• @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. – uhoh Jun 26 at 14:45
• @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. – David Hammen Jun 26 at 17:02