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While the moon is certainly not a good reflector of solar radiation, surely the radiation it reflects back heats the Earth (even if it is a terribly small amount).

How would one go about calculating (or estimating) this heating contribution on a night with a Full Moon?

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The Earth receives approximately $6.8\text{mW/m}^2$ of reflected sunlight from the moon (see below for details of how I calculated that).

However, the sunlight is also absorbed by the moon and this raise the surface temperature. So the moon also emits thermal radiation towards the Earth (assuming the highest day time temperature of 400K, see comments below for more information), $\epsilon_{\text{moon}}(1-A)\sigma (400K)^4 = 89\text{mW/m}^2$

So the total power received from the moon (reflected + thermal) is 10,438 times weaker than sunlight, i.e.

$$ \frac{6.8\text{mW/m}^2 + 89\text{mW/m}^2}{1000\text{W/m}^2} = \frac{1}{10438} $$

To answer your question about how much that heats the Earth, let's assume that the average daytime temperature of the Earth is 20$^\circ$C and the average nighttime temperature is 10$^\circ$C (these estimates could be improved, but it doesn't really change the answer significantly).

Therefore the incident solar energy causes a temperature difference $\Delta T=10^\circ$C between night and day. So we know that 1000 $\text{W/m}^2$ (solar irradiance on the Earth surface) cause a temperature increase of around $10^\circ$C. Let's assume that moonlight will also cause a temperature difference but one that is scaled proportionally by its intensity. Moonlight is 10,438 (reflected and thermal energy) times weaker than sunlight, the change in temperature of the earth from absorbing moonlight is,

$$ \frac{10^\circ C}{10,438} = 958 \mu K $$

Good luck measuring that...

Assumptions and method

  1. Solar irradiance is 1000 $\text{W/m}^2$ at the surface of the moon and the earth.
  2. The reflectivity of the moon is about $A=$10%.
  3. The solid angle of subtended by the moon in the sky is the same as that subtended by the sun $\epsilon_{\text{moon}} = 6.8\times10^{-5} \text{Sr}$. I say this because during an eclipse they appear to the same size so it's probably quite a good assumption.

From 1 and 2 we know that $100\text{W/m}^2$ is reflected at the surface of the moon. From 3, let's multiply that by the solid angle subtended by the moon as viewed from the Earth as this will give us the amount of the reflected energy that hits the Earth. So, $100\text{W/m}^2 \times 6.5\times10^{-5} = 6.5\text{mW/m}^2$.

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    $\begingroup$ Besides the reflected light from the sun of the moon, shouldn't you also take into account the thermal radiation? $\endgroup$ – fibonatic Dec 6 '13 at 10:50
  • $\begingroup$ The sun is a blackbody of approx. 6000K (this is the temperature of the light reflected), but the moons thermal spectrum is a blackbody between 300K to 400K. The total about of emitted energy is $\propto T^4$ so the thermal spectrum carries much less energy than the reflected light. $\endgroup$ – boyfarrell Dec 6 '13 at 11:49
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    $\begingroup$ But you also have to keep in mind that the Moon will scatter the reflected sunlight. And if you calculate the total Lunar irradiance you will see that it does matter significantly. For the reflection of sunlight of the moon I will use the Bond albedo ($A=0.123$), which means that its emission coefficient would be $\epsilon=1-A$. The average total solar irradiance at $1 AU$ is equal to $TSI=1366 \frac{W}{m^2}$. So the total irradiance of the Moon would be: $$ \frac{W}{m^2}=A TSI+(1-A)\sigma T^4 $$ Which is equal to $~1120 \frac{W}{m^2}$. $\endgroup$ – fibonatic Dec 6 '13 at 12:45
  • $\begingroup$ Yeah I agree with you, thanks for useful comments. I edited above to include thermal spectrum. $\endgroup$ – boyfarrell Dec 6 '13 at 13:47
  • $\begingroup$ Interesting approach. I would never thought of doing it this way. +1 $\endgroup$ – Jani Kovacs Apr 30 '14 at 11:49
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Assume the moon is at roughly the same distance from the Sun as the Earth and so receives the same amount of solar energy / area.

Find the area of the moon facing the Earth (hint, it's roughly the area of the moon's disk). Multiply by the reflectivity of the moon (about 12%).

But that power is reflected from the moon in all directions, so you need to consider a hemisphere at the distance of the Earth. Work out what fraction of this hemisphere corresponds to the Earth's disk - this is the fraction of the 12% reflected energy that hits Earth.

Compare that to the energy arriving at the Earth's disk from the sun

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protected by Qmechanic Sep 26 '15 at 17:50

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