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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.

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    $\begingroup$ 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. $\endgroup$ – Puffin Jul 30 '16 at 20:05
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    $\begingroup$ 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. $\endgroup$ – Puffin Jul 30 '16 at 20:05
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    $\begingroup$ 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. $\endgroup$ – SDsolar Mar 3 '17 at 2:14
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    $\begingroup$ @TomSpilker I only have a Gedankenbolometer at the moment. $\endgroup$ – uhoh May 31 '18 at 20:43
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    $\begingroup$ @uhoh Oh ... hard to calibrate! $\endgroup$ – Tom Spilker May 31 '18 at 22:19
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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.

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