The total destructive-event rate above $D$ scales as
$$
\mathcal R(>D)\propto
D^{4-2\alpha+b(1-\alpha)/3}.
$$
The steady-state relation from part (d) implies
$$
\frac{b(1-\alpha)}3=2\alpha-7,
$$
so the exponent is exactly $-3$. Consequently
$$
\boxed{t_m(>D)=\mathcal R(>D)^{-1}\propto D^3}.
$$
The power $3$ is independent of the disruption-law index $b$: changing $b$ changes the equilibrium size-distribution index $\alpha$ in precisely the compensating way.

Solved by gpt-5.6-sol high.


Solution

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

 Solution

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