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Leaving this comment to make sure that others don't get confused by your response. 1. As you mentioned, the moon acting as greybody absorbs part of the radiati
by skolos 8y ago
Leaving this comment to make sure that others don't get confused by your response.
1. As you mentioned, the moon acting as greybody absorbs part of the radiation and scatters the rest. In the second sentence you mention that the moon's temperature is enough to describe this radiation. In the third sentence you mention that you don't need moon's temperature to do that. So which is which?
Correct answer - to describe re-radiated energy we need moon's temperature, but to describe scattered we don't. We can ignore re-radiated, since it is not visible light and not hot enough. We can use scattered, since it is visible and hot enough.
2. Not sure why it is "No" if you are agreeing with me. Also not sure what you mean by temperature represented by total radiation (as you mentioned in part 1 there are two parts - re-radiated - at temperature of the moon, and scattered - at temperature of Sun).
To see that there is a problem with your argument - first note that the spectrum of black-body radiation dominates spectrum of grey-body radiation [1] (i.e. if black-body radiation is not in visible spectrum, the grey body will not be in visible as well). Then think that at 100 degrees C there would be no visible radiation. So the light that we see from moon cannot be this cold. Its coming from sun and it is hot enough (since it is in visible spectrum).
[1] https://claesjohnsonmathscience.wordpress.com/2012/04/21/grey-vs-black-bodies/ https://claesjohnsonmathscience.wordpress.com/2012/04/21/gre...
- ummonk 8y ago>Correct answer - to describe re-radiated energy we need moon's temperature, but to describe scattered we don't. We can ignore re-radiated, since it is visible light and not hot enough. We can use scattered, since it is visible and hot enough. When dealing with a gray-body, the equillibrium temperature of the body will be equal to the effective temperature of the incoming light, which will be equal to the effective temperature of the re-radiated + scattered light, since at equillbrium energy out equals energy in. So, assuming the moon is a graybody (and most objects tend to be roughly graybodies), we can use the surface temperature of the moon in our calculations instead of the effective temperature of the light that falls on it.
- skolos 8y agoSuppose you are right. The effective temperature of the moon is 100 degrees Celsius. It is a grey-body, so its spectrum is dominated by black-body - e.g. it would be similar to spectrum of black-body with temperature lower than 100 degrees Celsius. So your argument is that the light we see from Moon is same as light we would see from a black body heated to less than 100 degrees C? Try putting a charcoal in boiling water and please report to us if you see it lighted up as white as the moon.
- ummonk 8y agoI think you're mixing up spectral temperature with effective temperature. Effective temperature is about total power (per the Stefan Boltzmann law), not about the color of the light (per Wien's law). Compare https://en.wikipedia.org/wiki/Effective_temperature https://en.wikipedia.org/wiki/Effective_temperature with https://en.wikipedia.org/wiki/Color_temperature https://en.wikipedia.org/wiki/Color_temperature
- fpoling 8y agoThe moon surface is not in equilibrium as there is thermal flow towards moon interior.