 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2012/paper-2/section-a/a3/solution

a_{P} / a_{E} = (P_{P} / P_{E})^{2/3}

. Therefore

a_{P} = 1.5 \times 1 0^{8} (248)^{2/3} ≃ 5.92 \times 1 0^{9} km

. The furthest point is apoapsis, so

r_{m a x} = a_{P} (1 + e) ≃ 7.40 \times 1 0^{9} km .

The supplied orbital period and eccentricity are sufficient; no present-day ephemeris is needed.

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