Wax On, Wane Off

After a new moon, the crescent appears to grow slowly at first. At some point, the moon will be one-sixth full by area, then one-quarter full, and so on. Eventually, it becomes a half-moon, at which point its growth begins to slow down. How many times faster is the area of the illuminated moon growing when it is a half-moon versus a one-sixth moon?

With the simplifying assumption that the sun is sufficiently far away and that the moon orbits Earth much faster than Earth orbits the Sun, we are left with the simplified system where the illuminated portion of the moon from the observer's viewpoint on Earth is a spherical segment of angle $\theta$ for $\theta \in [0, \pi]$ (when the Moon is waxing).

Let's set up coordinates such that the $y$-axis is pointing towards the observer on Earth, and the $x$- and $z$-axis in the plane bounding the hemisphere that is visible from Earth. If the angle between the plane normal to the ray from the Moon to the Sun and the plane normal to the ray from the Moon to the Earth is $\theta$, then the dividing line between darkness and light on the face of a presume perfectly spherical Moon would be parameterized as $$ \{ (R_\text{moon} \cos \theta \sin \phi, R_\text{moon} \sin \theta \sin \phi, R_\text{moon} \cos \phi) \,\, \mid \,\, 0 \leq \phi \leq \pi\},$$ where $\phi$ is the polar angle from the $z$-axis.

If we were just measuring how fast the illuminated surface area increases, then it wouldn't be much of a question since the surface area grows linearly in the angle $\theta$, so the speed that the illuminated surface area was growing at would be constant at both half- and one-sixth-moons. So instead, we need to project the curve on the spherical surface into the circle in the $xz$-plane. In particular, the boundary between the illuminated and dark regions would then be $$\{(R_\text{moon} \cos \theta \sin \phi, R_\text{moon} \cos \phi) \, \mid \, 0 \leq \phi \leq \pi \}.$$

Switching to rectilinear coordinates and assume that we change units of measurement such that $R_\text{moon} = 1,$ we can calculate the illuminated area as $$I(\theta) = (1- \cos \theta) \int_{-1}^1 \sqrt{1-y^2} \, dy = \frac{\pi}{2} (1 - \cos \theta).$$ The rate of change of the illuminated area is given by $$\frac{dI}{d\theta} = \frac{\pi}{2} \sin \theta.$$

The moon is one-sixth illuminated when $I(\theta) = \pi \frac{1 - \cos \theta}{2} = \frac{\pi}{6},$ or equivalently, $\theta_{1/6} = \arccos \frac{2}{3}.$ So the illuminated area is growing a rate of $$\frac{dI}{d\theta} (\theta_{1/6}) = \frac{\pi}{2} \sin \left( \arccos \frac{2}{3} \right) = \frac{ \pi \sqrt{5}}{6}.$$ The moon is half illuminated when $\theta_{1/2} = \frac{\pi}{2},$ where the rate of change is $$\frac{dI}{d\theta} (\theta_{1/2})= \frac{\pi}{2}.$$ Therefore, the illuminated area is growing $\frac{\frac{\pi}{2}}{\frac{\pi \sqrt{5}}{6}} = \frac{3}{\sqrt{5}} \approx 1.34164\dots$ times faster when half full than one-sixth full.