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

For the two-body problem, the centre of mass lies a distance $a{M}_J/(M_{\odot }+M_J)=a/1048$ from the Sun. With $a=5.2\mathrm{AU}$,The relative circular-orbit period is $P=2\pi \sqrt{a^3/[G(M_{\odot }+M_J)]}\simeq 11.85$ years. The Sun's orbital speed about that centre of mass is thereforeThis is the space-speed amplitude; the observed radial velocity amplitude also includes the orbital inclination.

Kepler's third law is $P^2=4{\pi }^2{a}^3/[G(M_S+M_P)]$, with $a$ the semimajor axis of the relative orbit. For a circular edge-on orbit, the star's speed is $V_S=2\pi a_S/P$, and the centre of mass relation gives $a_S=a{M}_P/(M_S+M_P)$. Cubing the speed expression and eliminating $a$ yieldsFor $M_P\ll M_S$, this reduces toWithout the edge-on assumption, a factor ${\sin }^{3}i$ multiplies the numerator, explaining the minimum-mass degeneracy of the radial-velocity method.

Converting luminosity to mass needs a main-sequence mass-luminosity relation. Using the near-solar approximation $L/L_{\odot }\simeq (M/M_{\odot }{)}^4$ gives $M_S\simeq 0.946M_{\odot }$. ThenFor $P=11.6$ days and $V_S=10\mathrm{m}{\mathrm{s}}^{-1}$, these give approximatelyThese estimates retain the circular, edge-on and low-companion-mass assumptions. Adopting a mass-luminosity exponent $3.5$ instead changes the quoted values by about one percent; luminosity alone does not determine a unique stellar mass without such a calibration.