Write the <power-law size distribution> as $n(D)=kD^{-\alpha}$. The total geometric cross-section of the optically thin <debris disk> is
$$
\sigma_{\rm tot}=\int_{D_{\min}}^{D_{\max}}
\frac{\pi D^2}{4}kD^{-\alpha}\,dD
\simeq\frac{\pi k}{4(\alpha-3)}D_{\min}^{3-\alpha}.
$$
Because blackbody grains at radius $r$ intercept the fraction $\sigma_{\rm tot}/(4\pi r^2)$ of the stellar luminosity, the <fractional luminosity of a debris disk> gives
$$
k=16r^2f(\alpha-3)D_{\min}^{\alpha-3}.
$$
Another <integral> then yields
$$
\boxed{n(>D)=\frac{16r^2f(\alpha-3)}{\alpha-1}
D_{\min}^{\alpha-3}D^{1-\alpha}},
$$
for $D_{\min}\ll D\ll D_{\max}$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 316 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-316/3/a/solution