 Solution

ID: past-exam-of-the-short-options-paper-shared-by-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2019/s2/1/b/solution

Assume a circular orbit and an orbital inclination

i ≃ 9 0^{\circ}

, as appropriate for a simple eclipsing binary estimate. At quadrature the two radial velocities are opposite and equal to the orbital speeds. The more massive component has the smaller amplitude. With the centre of mass as origin,

a_{1} = \frac{K _{1} P}{2 π}, a_{2} = \frac{K _{2} P}{2 π}, a = a_{1} + a_{2}, \frac{M _{1}}{M _{2}} = \frac{K _{2}}{K _{1}} = 2.

Thus

a_{1} = 0.1103 AU

a_{2} = 0.2206 AU

a = 0.3309 AU, M_{1} = 8.06 M_{⊙}, M_{2} = 4.03 M_{⊙} .

The orbital plane contains the observer's line of sight; at quadrature the two velocities point towards and away from that observer, as labelled in the original schematic below. For an elliptic orbit, the measured amplitudes instead give

a sin i = P (K_{1} + K_{2}) 1 - e^{2} / (2 π)

; the numerical answer requires the circular assumption.

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