 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-iv-resit/4/a/solution

\dot{e}_{r} = \dot{θ} e_{θ}

and

\dot{e}_{θ} = - \dot{θ} e_{r}

. Applying the product rule twice to

r = r e_{r}

\dot{r} = \overset{r}{˙} e_{r} + r \dot{θ} e_{θ}, \ddot{r} = (\overset{r}{¨} - r \dot{θ}^{2}) e_{r} + (r \ddot{θ} + 2 \overset{r}{˙} \dot{θ}) e_{θ} .

m (\overset{r}{¨} - r \dot{θ}^{2}) = F (r), m (r \ddot{θ} + 2 \overset{r}{˙} \dot{θ}) = 0.

Multiplying the azimuthal force equation by

r / m

yields

\frac{d}{d t} (r^{2} \dot{θ}) = 0, r^{2} \dot{θ} = h .

This is conservation of specific angular momentum. Here

F (r)

is the signed radial component of the force; a negative value points inward.

When

h \neq = 0

θ

is a valid local parameter along the trajectory. Put

u = 1/ r

and let primes denote derivatives with respect to

θ

. The chain rule yields

\dot{θ} = h u^{2}, \overset{r}{˙} = - \frac{u ^{'}}{u ^{2}} \dot{θ} = - h u^{'}, \overset{r}{¨} = - h^{2} u^{2} u^{''}, r \dot{θ}^{2} = h^{2} u^{3} .

Substituting into the radial Newton's second law equation and dividing by

- m h^{2} u^{2}

u^{''} + u = - \frac{F ( 1/ u )}{m h ^{2} u ^{2}} .

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