 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2021/paper-1/section-b/b6/c/solution

Eliminate

v_{p}, v_{a}

using the angular momentum expressions from part (b):

\frac{J ^{2}}{2 m r _{p}^{2}} - \frac{GM m}{r _{p}} = \frac{J ^{2}}{2 m r _{a}^{2}} - \frac{GM m}{r _{a}} .

Hence

\frac{J ^{2}}{2 m} (\frac{1}{r _{p}^{2}} - \frac{1}{r _{a}^{2}}) = GM m (\frac{1}{r _{p}} - \frac{1}{r _{a}}) .

For distinct turning radii, factor the difference of squares and cancel the common factor to obtain

J^{2} = \frac{2 GM m ^{2}}{1/ r _{p} + 1/ r _{a}} = \frac{2 GM m ^{2} r _{p} r _{a}}{r _{p} + r _{a}} .

The circular Kepler orbit case is recovered continuously by setting

r_{p} = r_{a} = r

J^{2} = GM m^{2} r

. Combining this with the energy expression also gives

E = - GM m / (r_{p} + r_{a})

for the transfer ellipse.

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