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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/ii/paper-3/13c/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 3 / 13C / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-05

With constant masses, differentiate the moment of inertia twice:

\frac{1}{2} I^{''} = \sum_{i} m_{i} \overset{r}{˙}_{i}^{2} + \sum_{i} m_{i} r_{i} \cdot \overset{r}{¨}_{i} = 2 K + \sum_{i} r_{i} \cdot F_{i} .

For each pair,

F_{ij} = - G m_{i} m_{j} (r_{i} - r_{j}) /∣ r_{i} - r_{j} ∣^{3}

and

F_{ji} = - F_{ij}

. Hence its contribution is

(r_{i} - r_{j}) \cdot F_{ij} = - G m_{i} m_{j} /∣ r_{i} - r_{j} ∣ = V_{ij}

. Summing over unordered pairs gives

\sum r_{i} \cdot F_{i} = V

. Time-averaging over

[0, T]

yields

2 \overline{K}_{T} + \overline{V}_{T} = \frac{I ^{'} ( T ) - I ^{'} ( 0 )}{2 T} .

For a bound system with

I^{'} (T) = o (T)

, or a stationary equilibrium with zero average left-hand derivative, the limit is the virial theorem

2 ⟨ K ⟩ + ⟨ V ⟩ = 0.

Bounded spatial extent together with bounded velocities suffices for the derivative condition; possible gravitational collisions must be excluded or otherwise regularized.

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