 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2013-resit/10/a/solution

Put

μ = GM

. At the turning point of a radial launch the speed vanishes, so conservation of energy gives

\frac{1}{2} v_{0}^{2} - \frac{μ}{R} = - \frac{μ}{r _{0}}, r_{0} = (\frac{1}{R} - \frac{v _{0}^{2}}{2 GM})^{- 1} = \frac{R}{1 - v _{0}^{2} / v _{esc}^{2}}, v_{esc} = \frac{2 GM}{R} .

This result is exact within the exterior Newtonian gravitational potential. The condition

r_{0} ≫ R

v_{0}

is just below the escape velocity:

1 - v_{0}^{2} / v_{esc}^{2} ≪ 1

. Replacing gravity by a constant would be inappropriate in this regime.

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