Moonlight

If the sky were covered in moons it'd be almost a bright as day!

Walking to the pub through the Yorkshire Dales on a particularly brilliant moonlit evening, when everything was clearly visible, I wondered just how much less light there was than during daytime?  It turns out to be quite easy to estimate an upper bound - all you need to do is measure the angle subtended by the moon!

Let $A_e$ and $A_m$ be the cross sectional areas of the Earth and moon, $d$ be the distance to the moon, and $r$ be the distance to the Sun.  Now, suppose the Sun releases some energy $E_s$ then the amounts $E_{se}$ and $E_{sm}$ which land on the Earth and the moon are given by:
$$
\begin{align}
E_{se} &= \frac{A_e E_s}{4\pi r^2}\\
E_{sm} &\approx \frac{A_m E_s}{4\pi r^2}
\end{align}
$$
On a full moon, let's assume for the sake of calculating an upper bound that the moon reflects all of $E_{sm}$ equally in all hemispheric directions.  Then the energy reflected to the Earth is given by:
$$
\begin{align}
E_{me} &\approx \frac{A_e E_{sm}}{2\pi d^2}\\
&\approx \frac{A_e A_m E_s}{2 \pi d^2 \times 4\pi r^2}\\
&= \frac{A_m}{2\pi d^2} E_{se}
\end{align}
$$

So the ratio $E_{me}/E_{se}$ is one divided by the number of times you could fit the moon's cross-sectional area into a hemisphere of the same radius.  A clunkier way of putting this is that it's the solid angle subtended by the moon divided by $2\pi$.  Since the solid angle is quite small it's approximately $\pi \theta^2$ where $\theta$ is half the angle subtended.  If you hold your little finger up at arm's length the angle subtended by its width is about 1 degree.  Using this as a measure you can check that $\theta \approx 0.25^\circ = 0.25 \pi /180$, giving us an estimated upper bound of
$$
\begin{align}
\frac{E_{me}}{E_{se}} &\approx \frac{\pi 0.25^2 \pi^2 / 180^2}{2 \pi} \\
&\approx \frac{1}{105,000}
\end{align}
$$

POSTSCRIPT

I have tried validating this figure by looking for the actual data, but it's been quite hard to find.  This is because sources often quote values without specifying whether it is averaged over the whole month, day and night, or the top or bottom of the atmosphere.  According to Wikipedia the proportion of sunlight reflected by the moon (its albedo) is about 14%.  This link claims the power from the sun at a surface perpendicular to the Sun's rays at sea level on a clear day is about 1000 $W/m^2$.  It was a lot harder to find a reliable source for the power of moonlight at an equivalent surface.  The best I found was post 5 on this page which claims it's about 0.0015 $W/m^2$.  Setting $E_{se}$ to 1000 $W/m^2$ gives us an upper bound for $E_{me}$ of  1000/105000 = 0.0095 $W/m^2$ and multiplying this by the moon's albedo gives 0.0013 $W/m^2$ which is pretty close to the value the post claims has been measured.

POST-POSTSCRIPT

I've just realized that Sun Winter is Moon Summer!  My moonlit walk was on the 22nd of December - nearly the shortest day - and at 54 degrees north.  This explains why the moonlight appeared so brilliant that night. The moonlight/sunlight ratio, at this time of year and at this location, is greater than the value our calculation predicts by a factor of $cos(54^\circ-23^\circ)/cos(54^\circ+23^\circ)$, i.e. about 3.8.

POST-POST-POSTSCRIPT

I used the assumption light is reflected equally likely in every direction in the visible hemisphere.  This led to the simple result that the moonlight to sunlight ratio is $A_m / 2 \pi d^2$.  This isn't bad for a first stab but it is clearly a crude approximation.  If it were accurate then a half moon would be exactly half as bright as a full moon, not a mere 8% as bright as claimed here.  I played a bit with a different model where the angle of incidence always equals the angle of reflection.  This gave a moonlight to sunlight ratio of $A_m / 4 \pi d^2$.  But it also predicted that the reduction in brightness at half moon would be $cos(\pi/4) = 1/\sqrt{2}\approx 71\%$ which is even less accurate than the original model.  The moon obviously isn't a mirror ball - but if it were a collection of mirror balls then half would be in shadow during a half moon, which would bring the ratio down to about 30%.  Better, but still no biscuit.  However, the mental image led me to a eureka moment: the surface of the moon is covered in rocks!  This makes little difference to light raining down almost vertically, but light which is reflected at an acute angle is much more likely to hit a rock and never make it to Earth.  The slighter the angle the greater this effect is.  In fact anywhere on the moon in Earthlight-shadow is a place from which moonlight is blocked.  This neatly explains the mystery, but unfortunately means there is no neat mathematical formula for reflected light is scattered.  For that we need to just use measurements of moonlight intensity vs phase of moon.