Let the belt contain $n(D)dD=CD^{-\alpha}dD$ bodies. Its total mass fixes
$$
\boxed{C=
\frac{6M(4-\alpha)}
{\pi\rho\left(D_{\max}^{4-\alpha}-D_{\min}^{4-\alpha}\right)}}
\simeq\frac{6M(4-\alpha)}{\pi\rho D_{\max}^{4-\alpha}}.
$$
For equal <mass density>, an impactor of diameter $D$ supplies <specific impact energy>
$$
Q=\frac12\left(\frac{D}{D_t}\right)^3v_{\rm rel}^2.
$$
Thus catastrophic disruption requires $D\geq D_{\rm cc}=X_cD_t$, where
$$
\boxed{X_c=\left(\frac{2Q_D^*}{v_{\rm rel}^2}\right)^{1/3}}.
$$
The impactor <number density> per diameter is $CD^{-\alpha}/V$. Neglecting <gravitational focusing>, the <geometric collision cross-section> is $\pi(D_t+D)^2/4$, so the exact <catastrophic planetesimal collision rate> in this model is
$$
\boxed{R_{\rm cc}(D_t)=
\frac{\pi Cv_{\rm rel}}{4V}
\int_{X_cD_t}^{D_{\max}}(D_t+D)^2D^{-\alpha}\,dD}.
$$
Equivalently, with $A=\pi Cv_{\rm rel}/(4V)$,
$$
R_{\rm cc}=A\left[
\frac{D^{3-\alpha}}{3-\alpha}
+\frac{2D_tD^{2-\alpha}}{2-\alpha}
+\frac{D_t^2D^{1-\alpha}}{1-\alpha}
\right]_{X_cD_t}^{D_{\max}}.
$$
If $D_{\min}\ll X_cD_t\ll D_t\ll D_{\max}$, the steep distribution makes impactors just above $X_cD_t$ dominate and their collision cross-section is approximately $\pi D_t^2/4$. Hence
$$
\boxed{R_{\rm cc}(D_t)\simeq
\frac{A}{\alpha-1}X_c^{1-\alpha}D_t^{3-\alpha}}.
$$
This approximation also assumes a common $v_{\rm rel}$ and size-independent <catastrophic disruption threshold> $Q_D^*$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 2 / iii / Solution

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Gravitational focusing increases a physical collision cross-section because mutual gravity bends initially non-colliding trajectories toward one another.


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 Incoming links: Gravitational focusing