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View Full Version : Lunar contribution to Earth's insolation

Lord Jubjub
2015-Jun-20, 05:35 PM
This might be a rather odd question, I suppose, but how much energy is added to Earth's atmosphere by the reflection of sunlight off the moon? It is only 0.12 currently, but if the moon had an albedo of 1 (yeah, not possible this close to the sun), it seems to me that we could effectively have two suns.

I suppose the answer lies in the total albedo of the moon and not merely the visible spectrum.

WaxRubiks
2015-Jun-20, 05:52 PM
I don't think an albedo of 1 would be enough to have two suns. and albido of 1 would only be enough to give daylight to a circle, on Earth, the radius of the moon.

cjameshuff
2015-Jun-20, 05:57 PM
I don't think an albedo of 1 would be enough to have two suns. and albido of 1 would only be enough to give daylight to a circle, on Earth, the radius of the moon.

Right. The angular area of the sun and moon are close to equal, so you would need to make the moon's surface brightness equal to the surface brightness of the sun, not to a sheet of paper in sunlight on Earth.

Lord Jubjub
2015-Jun-20, 05:57 PM
Which means that we would not have days and nights based on the sun, but constantly changing days and nights depending on where the moon was. Considering that we can see the moon during daytime, just how bright would the day be if the albedo was 1?

What is the spectrum of reflected light from the moon. . .guess I'll have to look that up.

Hornblower
2015-Jun-20, 06:02 PM
Both of you are missing the fact that the light striking the Moon is scattered in all directions, and only a tiny fraction reaches the Earth. That fraction is about 1/500,000 of what we get from the Sun directly. Increasing the albedo to 100% will only increase that to about 1/50,000. To get the effect of a second Sun we would need a flat, polished mirror in place of the Moon, and it would need to be as big in diameter as the Earth.

Lord Jubjub
2015-Jun-20, 06:03 PM
Good point. I think you answered my question.

NEOWatcher
2015-Jun-20, 06:08 PM
I don't think an albedo of 1 would be enough to have two suns. and albido of 1 would only be enough to give daylight to a circle, on Earth, the radius of the moon.
Or, to further elaborate... If you picture the moon and earth as flat discs, you have an area of about 7.5% of that of Earth. So; either that 7.5% is lighting up 7.5% of the earth or it is spread out to give much less light over the whole Earth.

ETA: it looks like I delayed posting long enough for some other good explanations.

cjameshuff
2015-Jun-20, 07:52 PM
Both of you are missing the fact that the light striking the Moon is scattered in all directions, and only a tiny fraction reaches the Earth. That fraction is about 1/500,000 of what we get from the Sun directly. Increasing the albedo to 100% will only increase that to about 1/50,000. To get the effect of a second Sun we would need a flat, polished mirror in place of the Moon, and it would need to be as big in diameter as the Earth.

I phrased my answer in terms of surface brightness specifically to avoid this issue. For a given solid angle, the received illumination is proportional to the surface brightness, and the surface brightness is independent of distance. The sun and moon have effectively the same solid angle.

You don't need a giant mirror and all the tricky aiming issues that involves, you just need to heat the moon up to around 6000 K.

Hornblower
2015-Jun-20, 09:19 PM
I phrased my answer in terms of surface brightness specifically to avoid this issue. For a given solid angle, the received illumination is proportional to the surface brightness, and the surface brightness is independent of distance. The sun and moon have effectively the same solid angle.

You don't need a giant mirror and all the tricky aiming issues that involves, you just need to heat the moon up to around 6000 K.
I was addressing Lord Jubjub and Frog march, and I was unaware that you were posting while I was finishing up my first post. My apology for not checking for nearly simultaneous posts before posting mine.

WaxRubiks
2015-Jun-20, 09:31 PM
I was addressing Lord Jubjub and Frog march, and I was unaware that you were posting while I was finishing up my first post. My apology for not checking for nearly simultaneous posts before posting mine.

I was only saying that if the moon had an albedo of 1 then it would only have enough radiation to make daylight for a circle, on Earth, the radius of the moon; that is to say if it acted like a mirror, which I know it wouldn't.

plphy
2015-Jun-23, 10:48 PM
The calculation goes like this:

Since the Moon is practically at the same distance from the Sun as the Earth, the solar irradiance at the lunar surface is equal to the solar constant, Is = 1366 W/m^2. The mean albedo of the lunar soil is about a = 0.12, so the lunar surface reflects about 12 percent of the incident radiation. The radiant power emitted per square meter (the "radiant emittance") of the lunar surface is therefore M = a*Is.

For the next step we need the radiance L, i.e. the radiant power emitted in a given direction per square meter and per steradian. Assuming that the lunar soil reflects the light diffusely,(*) the reflected light is evenly dispersed in all directions, resulting in a factor 1/pi: L = M/pi = a*Is/pi.

What is the irradiance at the Earth's surface, caused by the lunar disc as a light source? For a light source of small angular extent, as is the case here, the irradiance produced by the light source is the radiance of the source, integrated over the solid angle subtended by the source.(**) The lunar disc has an angular diameter of about 0.518 degrees and thus subtends a solid angle of about w = 0.000064 steradian.(***)

If we assume that the lunar disc has everywhere the same radiance, the integral of the radiance over the solid angle is simply the radiance multiplied by the solid angle: I = L*w = a*Is*w/pi.

The ratio of the irradiances due to the Moon and the Sun at the Earth's surface is therefore I/Is = a*w/pi = 0.12*0.000064/pi = 1/400000.

Check: The apparent magnitude of the Sun is -26.7 m, that of the Moon is -12.7 m. The difference of 14 magnitudes corresponds to a factor 1/(2.512^14) = 1/400000.

If the albedo of the Moon were a=1, it would still be 1*w/pi = 50000 times fainter than the Sun.

(*) This is not true at all, as evidenced by the fact that we see the full moon as a flat, evenly illuminated disc and not as a sphere which is more dimly illuminated towards the limbs where the sunlight strikes it obliquely. However, if we treat the lunar soil as diffusely reflecting _and_ the moon as a flat disc instead of a sphere, the two errors cancel and our calculation comes out right.

(**) For light sources of larger extent, the fact that light from different parts of the source hits the receiver at different angles has to be taken into account. The solid angle must then be replaced by the appropriate view factor.

(***) w = 2 pi (1 - cos(angular diameter / 2)) = 2 pi (1 - cos(0.518°/2)) = 0.000064 sr

Regards,
Thomas