 Solution

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

v_{a} = \frac{r _{p}}{r _{a}} v_{p} = \frac{1}{3} \frac{3}{2} v_{0} = \frac{v _{0}}{6} .

v_{c} = GM / (3 r_{0}) = v_{0} / 3

. Thus

Δ v_{2} = v_{0} (\frac{1}{3} - \frac{1}{6}), f_{2} = \frac{Δ v _{2}}{v _{0}} = 0.1691 ≃ 16.9%

of the original fuel supply is used in the second burn. This is about

21.8%

of the fuel remaining after the first burn. The total transfer consumes

f_{1} + f_{2} ≃ 39.4%

of the original supply, leaving about

60.6%

. The two positive impulse magnitudes are added for the fuel budget, even though their directions at opposite ends of the orbit differ.

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