3 ms·
>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
by 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.