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If you take a simple infrared "thermometer" which is actually a lot like a bolometer and point it at a clear sky, it says "cold!" because in the wavelength range that it's using (roughly 5 to 15 microns) the Earth's atmosphere is partially transparent and not really an effective blackbody radiator, so the "thermometer" is really somewhat exposed to the cold of space. Point it at a cloud, and it will register "not as cold". It's (roughly) why clear nights are colder than cloudy nights.

Since the Martial atmosphere is fairly thin (though probably has very roughly similar amount of CO2 as Earth's), if I put a flat chunk of 'black body' on top of an insulator and expose it to the Martian sky at night, is there some simple expression for the effective temperature of the sky - what is the radiative heating/cooling going to be? Is it looking at 100K?

There will be of course thermal contact (convective heat transfer) with the atmosphere, and that can be calculated separately. This question is just about radiative heating/cooling from exposure to to the $2\pi$ SR half-sphere above horizontal.

uhoh
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  • I guess one way to go about this would be to find the coldest surface temperature ever recorded anywhere on Mars. Another way would be to look for bolometers or infrared thermometers pointed UP on Mars landers or rovers. – uhoh Jul 30 '16 at 09:24
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    Interesting question, this must have been looked at on many occasions. I actually have a vague memory of doing some thermal analysis for a small equipment item on a Mars lander 15 years or so ago but have no memory of the detail. Generically, the thermal analyst will have to have assumed values for the three temperature boundaries - convection, radiative (all around, including sky) and conduction ( through leg pads from the surface) and these conditions, together with solar illumination, should be specified for seasonal hot and cold cases. – Puffin Jul 30 '16 at 20:05
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    I'd guess someone must have published an analysis so I'd suggest a search based on mission names e.g. "xyz lander thermal analysis" to see what crops up. – Puffin Jul 30 '16 at 20:05
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    I do know that the straight-up temp at night in the Arizona desert can be about what a jetliner experiences... Roughly -40 or so. – SDsolar Mar 03 '17 at 02:14
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    @uhoh Have you tried pointing the IR thermometer at the moon? It its field of view isn't too wide, by going from having the moon centered in its field of view to off-pointed just enough to get the "cold" display, you can get an idea of the beamwidth of that field of view. If the field of view is too wide the moon's radiant energy is too diluted by the sky around it and you'll still get "cold", even pointed directly at the moon. But mine sees the moon just fine. Also try at different lunar phases: full, half, quarter, etc. Also waxing quarter vs waning quarter. – Tom Spilker May 31 '18 at 20:21
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    @TomSpilker I only have a Gedankenbolometer at the moment. – uhoh May 31 '18 at 20:43
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    @uhoh Oh ... hard to calibrate! – Tom Spilker May 31 '18 at 22:19

1 Answers1

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You've asked a question that is very difficult to answer accurately without in situ measurements, which apparently we don't have. The short answer: We don't know closer than ~100K!

There was an experiment planned for the Mars Surveyor Lander, "MTERC" (Mars Thermal Environment and Radiator Characterization), that would have made those measurements. But that lander was a copy of the Mars Polar Lander that failed in December of 1999, a failure attributed to design flaws, so the Surveyor Lander mission was cancelled.

It appears that those measurements have still not been made. I called David Brinza of JPL, the 2nd author of the MTERC paper (the 1st author, Ken Johnson, is no longer at JPL), and he told me there has been no attempt to recover that investigation, despite the flight-ready hardware sitting in mothballs at JSC. No other Mars landers, NASA or not, have attempted those measurements either.

Calculating the effective sky temperature from "first principles" is a very difficult task and requires some detailed knowledge of the Martian atmosphere. It is essentially a radiative transfer problem through a real atmosphere, not a simplified model. While in the Science Division at JPL, one of my tasks was radiative transfer modeling of planetary atmospheres, trying to interpret the data from radio astronomical observations. To do a reasonable radiative transfer calculation for an atmosphere you need:

  • vertical profiles of temperature and pressure from the surface to space
  • vertical profiles of atmospheric absorptivity, at all wavelengths for which there is any significant energy involved, and from the surface to space
  • vertical profiles of the atmospheric dust load, and the scattering characteristics of that dust

The second bullet above is not easy. Not only is an atmosphere "not really an effective blackbody radiator" as you mention in the question, an atmosphere is wildly not an effective blackbody radiator! As an example, take a look at the at the transmissivity of Earth's atmosphere from radio to UV. There are absorption lines (which are also emission lines) all over the place, due to O2, CO2, O3, CH4 ... a whole long list of things complicating the spectrum. Mars will be somewhat similar. The lines will be narrower due to the lower surface pressure, but CO2 will be prominent, at some of the wavelengths where the sun is trying to radiatively heat things and Mars is trying to radiatively get rid of heat. Other molecules such as water and CO will modify the spectrum, but we're not sure how much because their mixing ratios vary a lot.

The third bullet above is also not easy. Scattering characteristics are a strong function of particle size and shape. Martian dust is a mix of particle sizes and shapes, so it's really difficult to model its scattering characteristics accurately.

On top of those, all three bulleted characteristics vary with time, so you can't calculate the effective sky temperature for one atmosphere model and have the problem solved. In 1970 Joseph Wachter did some radiative transfer calculations for various models of the Martian atmosphere (unfortunately behind a paywall—c'mon, gang, the paper is 48 years old! Release it!) and got effective sky temperatures all the way from ~80K to ~170K, as echoed in the MTERC paper. I don't have descriptions of the models Wachter used, so I don't know if the 80K and 170K are lower and upper limits, or are just "representative" of what you'd expect to see.

The net result is that we won't have any reliable knowledge about this until someone actually makes the measurements at Mars.

Tom Spilker
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    I like when someone can authoritatively state "we don't know." OTOH it surprises me no mission included such an instrument. I got my IR thermometer off Aliexpress for like $5 and it's the size of an AA battery, display and batteries included. Sure "for space" is not $5, but even at 10,000x the price it would still be peanuts by space standards, and not a noticeable hit to weight and energy budget either. – SF. Jun 02 '18 at 01:25
  • I've just asked When will NASA PubSpace really make publicly available most of NASA funded research papers? – uhoh Jun 02 '18 at 01:39
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    There's a section on this and Figure 4.4 even shows a plot of "...longwave irradiance at the surface, expressed as sky temperature..." in Mars Global Reference Atmospheric Model 2001 Version (Mars-GRAM 2001): Users Guide which shows mid-latitude values circa 150 to 200K. – uhoh Jun 02 '18 at 02:21
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    fyi I've just asked What is the science behind the variation of Mars' effective sky temperature with latitude and longitude? Why the “hot spot”? – uhoh Jun 02 '18 at 02:22
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    @uhoh The high sky temperatures in Gere Justus's MarsGRAM 2001, as you point out in your question "...Why the 'hot spot;?", indicate to me that the models Wachter ran did not include an upper limit case, since they apparently didn't include atmospheric heating from a significant dust load near solar noon. – Tom Spilker Jun 02 '18 at 21:40
  • @SF. The loss of interest in doing those measurements is probably related to the politics of NASA and its government overlords, the executive branch (NASA reports to the Vice President of the US). In the late 1990's there was a lot of discussion at the highest levels of NASA concerning future habitation at Mars, and the infrastructure needed for that, such as in-situ resource utilization (ISRU). That led to interest in rejecting heat from resource-utilizing machines, and thus to radiators and sky temperature. That died off for a while (15-20 years!) but ISRU might be making a comeback. – Tom Spilker Jun 02 '18 at 21:55
  • @TomSpilker Indeed. I asked this question a few weeks after posting this answer to the question Will suits worn on Mars lose kilograms of “expendable water” each time they are used? I suggested "ice batteries" (i.e. cold packs) left outside could recharge themselves via "...some simple insulation flaps that passively open at night so they can radiate to space..." I asked this question about sky temperature when I started thinking about "recharging" during the day as well. – uhoh Jun 03 '18 at 02:01
  • @TomSpilker: in one hand, maybe. On the other hand, strapping a simple IR thermometer to Cutiosity' arm wouldn't be hard, and it could check temperatures of rocks, use it for self-diagnostics, or point at the sky. – SF. Jun 03 '18 at 03:19
  • @SF. You might be surprised at how much being on an interplanetary flight mission can run up the cost of even a simple instrument. The requirements for reliability in the space environment means that inexpensive components have to be replaced by ones that are tested and documented, radiation-tolerant, etc., so a 50-buck instrument in a consumer electronics shop quickly becomes a 500k (or more) instrument. And that's before all the work & redesign involved with instrument/spacecraft compatibility: no electromagnetic interference, untoward mechanical or thermal loads,etc. Adds up to millions! – Tom Spilker Jun 03 '18 at 19:54
  • @SF. Often it's not hard to get an instrument on a mission if the total cost will be, say, 3 megabucks. But if the Project Scientist and Project Manager really want to fly a 22 Mbuck instrument for the mission's top science priority, and flying the 3 Mbuck instrument (that addresses a lower science priority) means they'd have to settle for a significantly inferior 19 Mbuck instrument for the top priority objectives, then they'll happily toss the 3 Mbuck instrument. Sometimes it comes down to what the science priorities du jour are. – Tom Spilker Jun 03 '18 at 20:03
  • I've linked this page to Could Ingenuity stay warmer at night by landing on (or near) Perseverance's RTG?. In order to have confidence that the helicopter will not drain its battery before morning in order to stay warm, they either had to know do some careful modeling and maybe used some data from Curiosity's thermometers, or simply over-design the insulation around the helicopter. But that might make it get too hot when flying. Hmm... I'm going to think about this further; I think there are several new questions here. – uhoh Aug 01 '20 at 03:57