 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2018/s7/3/d/solution

In the equatorial plane,

\overset{r}{˙} = p_{r} / m, \dot{ϕ} = p_{ϕ} / (m r^{2}), \overset{p}{˙}_{r} = p_{ϕ}^{2} / (m r^{3}) - k / r^{2}, \overset{p}{˙}_{ϕ} = 0.

Differentiating the given combination produces a cosine term

- \frac{p _{ϕ}}{m} (\frac{p _{ϕ}^{2}}{m r ^{3}} - \frac{k}{r ^{2}}) cos ϕ + (\frac{p _{ϕ}^{2}}{m r} - k) \frac{p _{ϕ}}{m r ^{2}} cos ϕ = 0

and a sine term

p_{r} p_{ϕ} \dot{ϕ} sin ϕ / m - p_{ϕ}^{2} \overset{r}{˙} sin ϕ / (m r^{2}) = 0

. Thus

\dot{G} = 0

. The expression is a Cartesian component of the Laplace-Runge-Lenz vector with a conventional overall normalization, reflecting the additional symmetry of the inverse-square-force problem.

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