 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/section-a/5/solution

At closest approach the velocity is tangential. The conserved specific orbital energy and specific angular momentum are

ε = \frac{v _{p}^{2}}{2} - \frac{GM}{r _{p}} = - 2.1972 \times 1 0^{8} m^{2} s^{- 2}, ℓ = r_{p} v_{p} .

a = - \frac{GM}{2 ε} = 3.0376 \times 1 0^{11} m .

Since

r_{p} + r_{a} = 2 a

r_{a} = 2 a - r_{p} = 5.58 \times 1 0^{11} m .

This treats the star as fixed, appropriate to

m ≪ M

. Equivalently, equate the energy at the two turning points using

v_{a} = ℓ / r_{a}

. The planet mass cancels; a circular-orbit relation would be inappropriate here.

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