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

For two isolated point masses in a Newtonian bound orbit, $\boxed{P^2=4{\pi }^2{a}^3/[G(M_s+M_p)]}$, where $a$ is the relative orbit's semimajor axis. For a negligible planet mass, using years, astronomical units and solar masses gives $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/(M_s/M_{\odot })$. The mass in the denominator is the total gravitating mass, not the planet's mass alone.

Circular edge-on motion gives $V_s=2\pi a_s/P$ and center-of-mass balance gives $a_s=a{M}_p/(M_s+M_p)$. Eliminate $a$ using the preceding period relation:Thus $\boxed{M_p^3\simeq P{V}_s^3{M}_s^2/(2\pi G)}$ when $M_p\ll M_s$. 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.

A luminosity does not determine a stellar mass without a mass-luminosity calibration. Using the usual approximate main-sequence law $L/L_{\odot }=(M_s/M_{\odot }{)}^{\alpha }$ gives $M_s=3^{1/\alpha }{M}_{\odot }$. With $\alpha =3.5$, the numerical results areThe angular radius follows directly from $a(\mathrm{AU})/d(\mathrm{pc})$. Taking $\alpha =4$ changes these estimates by only a few percent, but the calibration must be stated. Hubble's 2.4 m aperture has visible-light diffraction scale $1.22\lambda /D\simeq 0.05$ arcsec at 500 nm, hundreds of times larger. The separation is therefore unresolved even before the severe star/planet contrast is considered.