A body of diameter $D$ has mass $m(D)=\pi\rho D^3/6$. Integrating the <power-law size distribution> therefore gives
$$
M=\int_{D_{\min}}^{D_{\max}}m(D)KD^{-\alpha}\,dD
=\frac{\pi\rho K}{6(4-\alpha)}
\left(D_{\max}^{4-\alpha}-D_{\min}^{4-\alpha}\right).
$$
Hence
$$
\boxed{K=\frac{6M(4-\alpha)}{\pi\rho\left(D_{\max}^{4-\alpha}-D_{\min}^{4-\alpha}\right)}}.
$$
When $D_{\max}\gg D_{\min}$ and $3<\alpha<4$, the belt mass is dominated by its largest bodies and $K\simeq6M(4-\alpha)/(\pi\rho D_{\max}^{4-\alpha})$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 316 / 1 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-316/1/i/solution