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

Table of contents
- Solution i

Solution

A body of diameter

D

has mass

m (D) = π ρ D^{3} /6

. Integrating the power-law size distribution therefore gives

M = \int_{D_{m i n}}^{D_{m a x}} m (D) K D^{- α} d D = \frac{π ρ K}{6 ( 4 - α )} (D_{m a x}^{4 - α} - D_{m i n}^{4 - α}) .

Hence

K = \frac{6 M ( 4 - α )}{π ρ ( D _{m a x}^{4 - α} - D _{m i n}^{4 - α} )} .

When

D_{m a x} ≫ D_{m i n}

and

3 < α < 4

, the belt mass is dominated by its largest bodies and

K ≃ 6 M (4 - α) / (π ρ D_{m a x}^{4 - α})

.

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