 Solution

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

The Lagrangian has no explicit time dependence, so its energy function is constant. With total speed

v^{2} = \overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2}

M_{p}

gives

ϵ = \frac{1}{2} v^{2} - \frac{μ}{r} = constant .

Consequently, between the initial point

(r_{i}, v_{i})

and any other point,

v^{2} (r) = v_{i}^{2} + 2 G M_{*} (\frac{1}{r} - \frac{1}{r _{i}}) .

This energy relation holds for nonradial motion too. Radially outward motion is the stipulated approximation for the subsequent probe calculation; angular momentum is then zero.

 New to topics? Read the docs here!