Eclipses justify an approximately <edge-on orbit>, and equal <velocity> <Amplitude>[amplitudes] imply equal masses. The relative <speed> is 200 km s$^{-1}$, so $a=(v_1+v_2)P/(2\pi)\simeq1.50\times10^{10}$ m, or $\boxed{a\simeq0.100\ \mathrm{AU}}$. Substitution in the binary period law gives $M_1+M_2\simeq4.52M_\odot$, hence $\boxed{M_1=M_2\simeq2.26M_\odot}$. Inclination corrections scale the dynamical masses as $\sin^{-3}i$ if exact edge-on viewing is relaxed.

Solution

Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S2 / 1 / 2 / Solution

 Solution (source code)