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Were there any temperature measurements experiments done on the Moon where the temperature was measured ABOVE the surface for any period of time?

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    $\begingroup$ Where the temperature of what was measured? $\endgroup$ – brhans Oct 26 '19 at 1:04
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In air, on Earth, we can talk about the temperature of the air 2 meters above the surface.

Of course if it's the top of your hat, then the temperature of that is affected both by the air temperature, and by the speed of the wind and how much sunlight is hitting it and other things like how much infrared your hat is radiating up into the sky.

But 2 meters above the surface of the Moon is the vacuum of space. There is no air, there's essentially nothing to take the temperature of.

The temperature of the top of your helmet standing on the Moon is affected mostly by how much it is warmed by the Sun, and how much it radiates infrared back into space.

If it's pure white, it will absorb little sunlight, and if it also happens to be dark in infrared light at the same time, it will radiate and become very cold.

On the other hand, if it's black in visible light it could end up really quite hot.


So there is no way to answer

What would be the temperature range measured 2 meters above the Moon's surface?

as asked. You'd have to specify the details of an object, it's color in visible light, its albedo or emissivity in thermal infrared, the angle that the surface makes with respect to incident sunlight, and several other things.

Of course if it's night on the Moon at the time, then the object could be extremely cold. It might be warmed somewhat by infrared and visible light from the Earth if that is visible, but it would be quite extremely in comparison to the effect of the Sun.

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    $\begingroup$ Measuring the temperature of air on Earth is not that easy as it seems. The effects of sunlight, the ground below and the buildings nearby should be minimized. But to measure the temperature of a gas that is not present above the moon is impossible. $\endgroup$ – Uwe Oct 26 '19 at 13:03
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I'm not aware of any measurements made more than a couple of meters from the surface, but with some assumptions to be described below, there should be no observable difference over the altitude range you mention.

Ther are many factors that can influence a temperature measurement on an airless body. Yeah, yeah, some people talk about the moon's atmosphere. It's actually an exosphere, too thin to yield any significan heat transfer by convection. Also, any "thermometer" worth a ... darn ... wouldn't be heated or cooled significantly by conduction: it would be fully isolated from any support hardware. Thus the only thing that affects this thermometer, and therefore the temperature it indicates, is radiative heating by its surroundings. That would include: 1) heat radiated by the moon's surface; 2) heat from the sun; 3) heat radiated from Earth; and 4) heat from the cosmic background, ~2.7 K.

If the temperature of the thermometer is greater than any of these (such as the cosmic background), the heat radiated by the thermometer is greater than the heat received from that source, and it becomes a source of cooling of the thermometer.

Let's assume the thermometer absorbs and radiates heat equally over $4\pi$ steradians, i.e. from every direction.

If the moon's surface isn't a nice, uniform, smooth surface, the radiation from it will be very non-uniform. This is especially true if the "surface" where you start is inside a crater that is mostly shadowed, or near the edge of a mountain's shadow. Temperatures of those shadowed surfaces will be very low so not much heat will be received from them. The heat radiated from a surface is proportional to temperature to the fourth power, so warmer surfaces radiate a lot more heat than colder surfaces.

As you ascend from this shadowed region, first you start getting heat from the sun — a lot of it! — and as you rise above a crater's rim you start seeing the moon's average surface, much warmer than the shadowed crater interior, or the shadow of the mountain. The thermometer's indicated temperature zooms!

So let's assume that we're over a part of the moon that, most un-representatively, is completely smooth. If we ever found a spot like that, covering maybe 1 km by 1 km of the moon's surface, it would be like finding TMA-1 ("2001: a Space Odyssey" reference). But let's assume the smooth surface.

If you're right next to the surface, the moon's surface, at some temperature, fills $2\pi$ steradians (half of all directions), and that temperature varies with the angle between the local vertical and the direction to the sun; the sun provides a certain amount of heat; Earth provides a bit more; and cold space (the cosmic background) covers almost $2\pi$ steradians.

If you ascend 10 m, the relationship with the sun, Earth, and cold sky essentially haven't changed, they're so far away. And given the size of the moon, neither has the moon's. It still subtends about $2\pi$ steradians, within a tiny fraction of a percent. So the temperature indicated by the thermometer essentially doesn't change. The same is true 50 m above the surface. You have to get high enough that the moon no longer fills half your field of view — and that's really high, many km — and then the indicated temperature starts to come down.

As far as the temperature range goes, it would be the range of the surface temperature (~100-300 K, depending on the time of the lunar day), increased a bit by solar radiation. Permanently shadowed regions would be significantly cooler.

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    $\begingroup$ I think you haven't stressed that the temperature that a thermometer reads 2 m above the Moon's surface is nothing more than the temperature of the thermometer itself. It is not the temperature above the Moon. Remove the thermometer, and there is nothing there that is that temperature. The temperature of one thing 2m above the Moon doesn't tell you much about the temperature of another thing 2m above the Moon next to it; they are virtually unrelated unless the two objects happen to be identical. $\endgroup$ – uhoh Oct 26 '19 at 1:27
  • $\begingroup$ Yeah, I didn't want to get too far into radiative transfer. Not too many people have read Chandrasekhar. But this is why I stressed uniform $4\pi$ steradian response. $\endgroup$ – Tom Spilker Oct 26 '19 at 1:35
  • $\begingroup$ Related to the Moon, I've just added a bounty to this question but (fyi) I won't grant it to ten paragraph long essays unless they contain specific answers that state clearly and prominently what will actually happen to the air; where it will end up. $\endgroup$ – uhoh Oct 26 '19 at 2:46

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