 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iv/2/a/solution

m \ddot{r} = - Φ^{'} (r) \frac{r}{r} .

Differentiating the cross product gives

\dot{n} = \dot{r} \land \dot{r} + r \land \ddot{r} = 0,

since the acceleration is parallel to

r

. Hence

n = n (0)

is constant. Since

r \cdot n = 0

, the entire motion lies in the fixed plane through the origin with nonzero normal vector

n (0)

Choose polar coordinates in that plane, orienting

θ

so that its positive normal vector points along

n (0)

. The velocity is

\dot{r} = \overset{r}{˙} e_{r} + r \dot{θ} e_{θ}, r \land \dot{r} = r^{2} \dot{θ} e_{z} .

Therefore

r^{2} \dot{θ} = h = ∣ n (0) ∣ > 0.

m

identifies this with conservation of angular momentum; geometrically, the swept area per unit time is

h /2

\frac{d}{d t} (\frac{m}{2} ∣ \dot{r} ∣^{2} + Φ (r)) = m \dot{r} \cdot \ddot{r} + Φ^{'} (r) \overset{r}{˙} = 0.

Thus

\frac{m}{2} (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}) + Φ (r) = E .

Finally, for

u = 1/ r

and a prime denoting a derivative with respect to

θ

, the chain rule gives

\dot{θ} = h u^{2}, \overset{r}{˙} = - u^{- 2} u^{'} \dot{θ} = - h u^{'}, r \dot{θ} = h u .

Substitution into the conserved energy yields

\frac{m h ^{2}}{2} [(\frac{d u}{d θ})^{2} + u^{2}] + Φ (1/ u) = E .

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