I'm trying to calculate the maximum heat that an object on the moon can reach with the energy provided from Sunlight. I've got a total power output of the sun per square meter at the distance of the Earth at 1366.1 W/m^2. With 54% of that power being in Infrared, that gives me 737.7 W/m^2 of Infrared radiation.

The object I was calculating for has a reflectivity of 95%, so that gives around 36.9 W/m^2.

I substituted this value into j in a rearranged Stephan-Boltzmann law ( ⁴√(j/ϵ𝝈) )to give me a temperature of T, and T to be equal to 64.73 degrees C.

Am I correct in thinking that the emissivity in that calculation would cancel out the effects of the reflectivity in calculating the input power, so no matter the reflectivity of the object, it would still reach the same total temperature?

  • 1
    $\begingroup$ Not sure why you are only counting the IR part of the solar spectrum. Visible and UV light get converted into heat just as well. The emissivity of the object does matter somewhat because it is not in thermodynamic equilibrium with the radiation that heats it. Moreover, on the moon the geometry of the surrounding environment matters, just like it does on Earth and that would require much more complex optical calculations to account for the specifics of the location. $\endgroup$ – CuriousOne May 21 '15 at 7:22
  • $\begingroup$ Oh, I figured it was just the IR spectrum that got turned into heat. Is it as prevalent as IR, or should I factor in the whole 1366.1 W in my calculations? Can I get away with just assuming direct sunlight from above, or would the geometry of the moon actually play a large part? $\endgroup$ – JamEngulfer May 21 '15 at 7:26
  • $\begingroup$ That geometry in a reflective environment plays a nontrivial part can be seen in the example of the focus of a convex mirror (or the focal point of a lens, if you want), which is at the extreme end of reflective environments that can get things much hotter than the simple radiation equilibrium assumption would predict. A more realistic assumption is probably that all of the incoming radiation gets converted and then radiated away into a $2\pi$ solid angle, rather than $4\pi$ for the case that the lunar surface wouldn't be there. This should get you well above 100 degrees Celsius. $\endgroup$ – CuriousOne May 21 '15 at 7:33
  • $\begingroup$ I wonder if there is any direct data on how hot the cameras got on the Apollo 11 mission? I've seen it said that they got up to 120C, but there wasn't a source for it. EDIT: By the way, that's what I'm trying to calculate the temperature of. $\endgroup$ – JamEngulfer May 21 '15 at 9:21
  • $\begingroup$ How did you manage to use Stephan-Boltz to convert absorbed power into a temperature? That's not how things work at all! At the very least you need the specific heat (Joules/gram/K) of the absorbing material and its thermal conductivity if you want to estimate its temperature. OH, and also: there's no such thing as "Maximum Heat" $\endgroup$ – Carl Witthoft May 21 '15 at 13:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.