Power-law size distribution

A differential power-law size distribution has $n(D)dD=k{D}^{-\alpha }dD$ objects with sizes between $D$ and $D+dD$.

For equal-density spherical bodies with differential number distribution $n(D)\propto D^{-\alpha }$, a fixed logarithmic bin has width $dD=Dd\log D$. Multiplying the body mass proportional to $D^3$ by $n(D)dD$ gives $m_{\log }\propto D^{4-\alpha }$. Distinguishing differential-number and logarithmic-mass slopes is essential in collisional cascade plots.

For spherical fragments of common mass density $\rho$, with $n(D)=K{D}^{-\alpha }$ between specified diameter cutoffs, mass normalization of a power-law size distribution and integration of the geometric cross-section giveIndeed $m=(\pi \rho K/6)\int D^{3-\alpha }dD$ and ${\sigma }_{\mathrm{tot}}=(\pi K/4)\int D^{2-\alpha }dD$; eliminating $K$ proves the formula. If both cutoff ratios raised to $\alpha -3$ and $4-\alpha$ are small, respectively, this reduces to $3m(4-\alpha )D_{\min }^{3-\alpha }{D}_{\max }^{\alpha -4}/[2\rho (\alpha -3)]$. Approaching either endpoint requires the corresponding logarithmic integral.

If a spherical parent and its fragments have the same mass density, substitute $m=\pi \rho D_c^3/6$ into the endpoint-dominated cross-section of a power-law fragment population. The diameter obeysThis inversion requires specified fragment cutoffs. If $D_{\max }=\eta D_c$ instead, solving that dependence changes the power of $D_c$ to $\alpha -1$. It is useful for estimating the parent comet needed to supply an observed debris disk, subject to the assumed fragment model.

For spherical planetesimals with mass density $\rho$ and total-number distribution $n(D)=K{D}^{-\alpha }$ between $D_{\min }$ and $D_{\max }$,Thus $K=6M_{\mathrm{tot}}(4-\alpha )/[\pi \rho (D_{\max }^{4-\alpha }-D_{\min }^{4-\alpha })]$ for $\alpha \ne 4$. At $\alpha =4$, use $K=6M_{\mathrm{tot}}/[\pi \rho \log (D_{\max }/D_{\min })]$. Divide $n(D)$ by belt volume to obtain the number-density distribution used in a planetesimal collision rate.