(February 10, 2016 at 8:52 am)Alex K Wrote:(February 10, 2016 at 8:32 am)SteelCurtain Wrote: Wouldn't it have to be? Otherwise the surface of the moon would either store thermal energy or run out, yeah?
The energy captured by the moon rock by absorbing sunlight is reemitted by the rock as infrared thermal radiation which is added to the reflected light, and this way the moon is keeping the balance (*). This thermal radiation of the moon has the same low temperature as the rock and very different properties compared to the sunlight that is reflected directly, doesn't it. At first glance, it seems to me that it is with this IR thermal radiation (which is in th. equilibrium with the moon) that RMs argument would work, not with the reflected sunlight (which is not in th. equilibrium with the moon).
To elaborate, assume that the moon, for the sake of simplicity, absorbs half the sunlight and reflects half, then its thermal radiation and the reflected sunlight will have roughly the same power. HOWEVER, the emitted thermal radiation will be low-frequency (low-temperature) IR, whereas the reflected sunlight will simply be *dim* but higher-frequency light which, if you look at its spectrum, actually belongs to a temperature of thousands of degrees, but is greatly attenuated in intensity because the sun is far away.
(*) The power emitted by a body goes with the temperature to the fourth power, so the moon will heat up until the ~ T^4 thermal radiation carries away the same radiation power that the absorbed sunlight deposits in the rock.
As someone on the experimental park... I say we gather on a full moon night and test it! Anyone has a big lens?
