Kvant Physics Problem 41

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Verdicts: SKIP + SKIP
Solve time: 1m11s
Source on kvant.digital

Problem

By what factor is the illuminance during a moonlit night at full moon smaller than during a sunny day, for the same altitude of the Moon and the Sun 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 system consists of the Sun, the Moon, and an observer on Earth. The unknown is the ratio of illuminance at Earth during a full moon night, $E_{\text{moon}}$, to the illuminance during daytime under direct sunlight, $E_{\text{sun}}$, for the same altitude of the Sun and Moon above the horizon.

The given parameters are the Moon radius $R = 2000~\text{km}$ and the Earth–Moon distance $d = 400{,}000~\text{km}$.

The Moon is modeled as a body that intercepts sunlight over its geometric cross section $\pi R^2$ and then scatters all absorbed energy uniformly into space, meaning isotropic re-emission over a sphere centered at the Moon. The solar irradiance at Earth is taken to be the same for both cases and cancels in the final ratio. Atmospheric effects, lunar phase dependence beyond full moon geometry, surface anisotropies, and multiple scattering are neglected.

Physical Principles

The power intercepted from a parallel beam of intensity $I_\odot$ by a сферical body of radius $R$ is

$$P_{\text{abs}} = I_\odot \pi R^2.$$

If this power is re-emitted isotropically, the radiated flux at distance $d$ is distributed over a sphere of area $4\pi d^2$, so the irradiance at Earth due to the Moon is

$$E_{\text{moon}} = \frac{P_{\text{abs}}}{4\pi d^2}.$$

The daytime illuminance for the same solar altitude equals the solar irradiance itself,

$$E_{\text{sun}} = I_\odot.$$

Derivation

Substituting the absorbed power into the expression for lunar illuminance gives

$$E_{\text{moon}} = \frac{I_\odot \pi R^2}{4\pi d^2}.$$

Cancellation of $\pi$ yields

$$E_{\text{moon}} = I_\odot \frac{R^2}{4d^2}.$$

The ratio of night to day illuminance becomes

$$\frac{E_{\text{moon}}}{E_{\text{sun}}} = \frac{I_\odot \frac{R^2}{4d^2}}{I_\odot} = \frac{R^2}{4d^2}.$$

Substituting $R = 2000~\text{km}$ and $d = 400{,}000~\text{km}$ gives

$$\frac{E_{\text{moon}}}{E_{\text{sun}}} = \frac{(2000~\text{km})^2}{4(400{,}000~\text{km})^2}.$$

Squaring the quantities,

$$(2000~\text{km})^2 = 4\times 10^6~\text{km}^2, \quad (400{,}000~\text{km})^2 = 1.6\times 10^{11}~\text{km}^2.$$

Thus,

$$\frac{E_{\text{moon}}}{E_{\text{sun}}} = \frac{4\times 10^6}{4 \cdot 1.6\times 10^{11}}.$$

The factor $4$ cancels,

$$\frac{E_{\text{moon}}}{E_{\text{sun}}} = \frac{10^6}{1.6\times 10^{11}} = \frac{1}{1.6\times 10^5} = 6.25\times 10^{-6}.$$

Result

The illuminance ratio is

$$\frac{E_{\text{moon}}}{E_{\text{sun}}} = 6.25\times 10^{-6}.$$

The Moonlit illuminance is therefore smaller than the daylight illuminance by the factor

$$\frac{E_{\text{sun}}}{E_{\text{moon}}} = \frac{1}{6.25\times 10^{-6}} = 1.6\times 10^5.$$

Sanity Checks

Dimensional consistency holds since the ratio $\frac{R^2}{d^2}$ is dimensionless and all units of kilometers cancel before final evaluation.

The dependence on $R^2$ reflects that a larger reflecting body intercepts more sunlight, while the $1/d^2$ dependence reflects geometric spreading of isotropic radiation, matching the inverse-square law.

The magnitude $10^{-5}$ to $10^{-6}$ corresponds to a very small fraction of solar illuminance, consistent with the Moon appearing many orders of magnitude dimmer than the Sun. The most sensitive step is the transition from reflected power to isotropic distribution over $4\pi d^2$, where missing the factor $4$ would change the final result by a factor of four.