 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013/11/a/solution

Let

μ = G M_{*}

U = - μ M_{p} / r

, so

L = \frac{M _{p}}{2} (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}) + \frac{G M _{*} M _{p}}{r} .

A central force keeps the orbit in a plane. The star is treated as stationary, appropriate when

M_{p} ≪ M_{*}

; otherwise the two-body reduction uses the reduced mass and the relative separation.

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