 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2017/s2/1/2/solution

Circular edge-on motion gives

V_{s} = 2 π a_{s} / P

a_{s} = a M_{p} / (M_{s} + M_{p})

. Eliminate

a

using the preceding period relation:

\frac{M _{p}^{3}}{( M _{s} + M _{p} ) ^{2}} = \frac{P V _{s}^{3}}{2 π G} .

Thus

M_{p}^{3} ≃ P V_{s}^{3} M_{s}^{2} / (2 π G)

when

M_{p} ≪ M_{s}

. For a non-edge-on orbit the measured velocity instead determines

M_{p} sin i

to this accuracy; neglecting eccentricity is also part of the stipulated model.

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