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

The impactors' number density per diameter is $KD_{\rm im}^{-\alpha}/V$. With gravitational focusing neglected, the <geometric collision cross-section> is $\pi(D+D_{\rm im})^2/4$. The <planetesimal collision rate> is consequently
$$
R_{\rm col}(D,D_{\rm im})dD_{\rm im}
=\frac{K D_{\rm im}^{-\alpha}}{V}
\frac{\pi(D+D_{\rm im})^2}{4}v_{\rm rel}\,dD_{\rm im}
=A D_{\rm im}^{2-\alpha}\left(1+\frac D{D_{\rm im}}\right)^2dD_{\rm im},
$$
where
$$
\boxed{A=\frac{\pi K v_{\rm rel}}{4V}}.
$$

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

Two spheres of diameters $D_1,D_2$ collide geometrically when their projected centre separation is at most $(D_1+D_2)/2$, giving cross-section $\sigma=\pi(D_1+D_2)^2/4$.

= Geometric cross-section
{synonym}


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 Incoming links: Geometric collision cross-section