Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S2 / 1 / 2

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- Solution 2

Solution

Kepler's third law is

P^{2} = 4 π^{2} a^{3} / [G (M_{S} + M_{P})]

, with

a

the semimajor axis of the relative orbit. For a circular edge-on orbit, the star's speed is

V_{S} = 2 π a_{S} / P

, and the centre of mass relation gives

a_{S} = a M_{P} / (M_{S} + M_{P})

. Cubing the speed expression and eliminating

a

yields

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

For

M_{P} ≪ M_{S}

, this reduces to

M_{P}^{3} ≃ \frac{V _{S}^{3} P M _{S}^{2}}{2 π G} .

Without the edge-on assumption, a factor

sin^{3} i

multiplies the numerator, explaining the minimum-mass degeneracy of the radial-velocity method.

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