Kvant Physics Problem 41
The physical system consists of sunlight incident on the Moon and the light scattered by the illuminated lunar hemisphere toward the Earth.
Verified: yes
Verdicts: PASS + PASS
Solve time: 4m53s
Source on kvant.digital
Problem
How many times smaller is the illuminance on a moonlit night at full moon than on a sunny day, assuming the Moon and the Sun are at the same elevation above the horizon? Assume that the illuminated hemisphere of the Moon scatters light uniformly into space. Take the radius of the Moon to be $2000~\text{км}$, and the distance from the Moon to the Earth to be $400,000~\text{км}$.
Setup and Assumptions
The physical system consists of sunlight incident on the Moon and the light scattered by the illuminated lunar hemisphere toward the Earth.
The required quantity is the ratio between the illuminance on the Earth's surface produced by the full Moon and the illuminance produced by direct sunlight. Let this ratio be
$$\frac{E_{\text{moon}}}{E_{\odot}}.$$
The given quantities are the Moon's radius
$$R = 2000~\text{km},$$
and the Earth-Moon distance
$$L = 400,000~\text{km}.$$
The assumptions are as follows.
The Sun and the Moon have the same elevation above the horizon, so atmospheric attenuation is identical in both cases and cancels from the ratio.
The illuminated hemisphere of the Moon scatters incident light uniformly into all directions of space. Thus the scattered radiation is isotropic.
The Moon absorbs no energy. All sunlight intercepted by its illuminated disk is re-emitted by scattering.
The Earth is sufficiently far from the Moon that the scattered radiation reaching the Earth may be treated as radiation from a point source at distance $L$.
Physical Principles
The power intercepted from sunlight equals the solar irradiance multiplied by the Moon's geometric cross section.
If the solar irradiance at the Moon is $E_{\odot}$, then the intercepted power is
$$P_{\text{int}} = E_{\odot},\pi R^2.$$
For isotropic emission, a source of total power $P$ produces at distance $L$ an irradiance
$$E = \frac{P}{4\pi L^2}.$$
The illuminance ratio equals the irradiance ratio because the same spectral distribution is involved and only relative brightness is required.
Derivation
The Moon intercepts solar power
$$P_{\text{int}} = E_{\odot},\pi R^2.$$
According to the assumptions, this entire power is scattered uniformly into space. The total scattered power is therefore
$$P_{\text{sc}} = E_{\odot},\pi R^2.$$
At the Earth, a distance $L$ away, this power is distributed over a sphere of area
$$4\pi L^2.$$
The irradiance produced by the full Moon at the Earth is
$$E_{\text{moon}} = \frac{P_{\text{sc}}}{4\pi L^2} = \frac{E_{\odot},\pi R^2}{4\pi L^2}.$$
After cancellation of $\pi$,
$$E_{\text{moon}} = E_{\odot},\frac{R^2}{4L^2}.$$
Hence
$$\frac{E_{\text{moon}}}{E_{\odot}} = \frac{R^2}{4L^2}.$$
The factor by which moonlight is weaker than sunlight is the reciprocal quantity:
$$N = \frac{E_{\odot}}{E_{\text{moon}}} = 4\left(\frac{L}{R}\right)^2.$$
Result
The symbolic result is
$$N = 4\left(\frac{L}{R}\right)^2.$$
Substituting the given values,
$$N = 4\left(\frac{400,000~\text{km}} {2000~\text{km}}\right)^2.$$
The units cancel:
$$N = 4(200)^2.$$
Thus
$$N = 4 \times 40,000 = 160,000.$$
Equivalently,
$$\frac{E_{\text{moon}}}{E_{\odot}} = \frac{1}{160,000} = 6.25\times10^{-6}.$$
The illuminance produced by a full Moon is
$$\boxed{1.6\times10^{5}}$$
times smaller than the illuminance produced by direct sunlight.
Sanity Checks
The ratio is dimensionless, as required. The expression contains only the squared ratio $R/L$, so all length units cancel.
If the Moon were moved farther away, $L$ would increase and the moonlight illuminance would decrease as $1/L^2$, matching the inverse square law.
If the Moon were larger, the intercepted solar power would increase as $R^2$, giving a proportionally brighter full Moon.
The numerical value is of order
$$\left(\frac{2000}{400,000}\right)^2 = (5\times10^{-3})^2 = 2.5\times10^{-5},$$
and the additional factor $1/4$ gives approximately $6\times10^{-6}$, consistent with the detailed calculation.
The most common source of error is forgetting that isotropic scattering spreads the intercepted power over the entire sphere of area $4\pi L^2$. Omitting the factor $4$ changes the answer by a factor of four.