 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2022/8/b/solution

r = r_{0} > 0

. Both its initial radial velocity and radial acceleration must vanish, giving

r (0) = r_{0}, \overset{r}{˙} (0) = 0, V_{eff}^{'} (r_{0}) = V^{'} (r_{0}) - \frac{J ^{2}}{m r _{0}^{3}} = 0.

Equivalently, the initial tangential speed and angular velocity obey

v_{θ, 0}^{2} = \frac{r _{0} f ( r _{0} )}{m}, \dot{θ} (0) = \frac{J}{m r _{0}^{2}}, J^{2} = m r_{0}^{3} f (r_{0}) .

The initial polar angle is arbitrary, and the energy is fixed at

E = V_{eff} (r_{0})

. Either orientation is possible. These conditions give an exact circular orbit even if it is unstable; stability requires an additional positive second derivative of the effective potential.

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