Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 316 / 1 / i

Solution

For the relative position

r = r_{1} - r_{2}

\ddot{r} = - \frac{μ}{r ^{3}} r, μ = G (M_{1} + M_{2}) .

v = \dot{r}

gives

\frac{d}{d t} (\frac{1}{2} v^{2}) = - \frac{μ}{r ^{3}} r \cdot \dot{r} = \frac{d}{d t} (\frac{μ}{r}) .

Integration therefore yields conservation of specific orbital energy:

\frac{1}{2} v^{2} - \frac{μ}{r} = C .

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