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>Moon surface temperature argument is incorrect To the extent that the moon acts as a greybody under sunlight, it is correct. And like most things, the moon wi

by ummonk 8y ago

>Moon surface temperature argument is incorrect To the extent that the moon acts as a greybody under sunlight, it is correct. And like most things, the moon will be close enough to a greybody that you could use the surface temperature as a first order approximation. It isn't necessary of course, since you can easily just directly estimate the amount of scattered / re-radiated energy from the amount of sunlight falling on it, without needing to use the temperature. >Hence the question is now - can we concentrate moon light enough so that intensity at the concentration point is higher than thermal loss into environment (only then we will be able to raise temperature in the concentration area enough for combustion - remember that light is "hot" enough for this)? No, because the body you're trying to heat up will act as an approximate greybody, so (neglecting conduction and convection which could further lower the temperature) its temperature has to max out at the Stefan-Boltzmann temperature represented by the total radiation.

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 ago
    Suppose 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 ago
      I 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 ago
    The moon surface is not in equilibrium as there is thermal flow towards moon interior.
smaddox 8y ago
> And like most things, the moon will be close enough to a greybody that you could use the surface temperature as a first order approximation. No, it won't, and no you can't. The OP already pointed out that the Moon is too cold for its blackbody radiation to reach the visible. All the visible light from the Moon is reflected from the Sun. The Sun's radiation's blackbody temperature is the ultimate limit, here, not the Moon's.

by ummonk 8y ago

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 ago
  Suppose 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 ago
    I 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 ago
  The moon surface is not in equilibrium as there is thermal flow towards moon interior.

smaddox 8y ago

> And like most things, the moon will be close enough to a greybody that you could use the surface temperature as a first order approximation. No, it won't, and no you can't. The OP already pointed out that the Moon is too cold for its blackbody radiation to reach the visible. All the visible light from the Moon is reflected from the Sun. The Sun's radiation's blackbody temperature is the ultimate limit, here, not the Moon's.