Fitting a single-lined spectroscopic orbit to a star of mass gives

PHYSICS · Subjective · Fermi University Physics · Astrophysics & Cosmology · Celestial mechanics and planetary systems · Celestial mechanics and planetary systems

Stars from the Inside, Ch Binary stars: fitted orbits, modelled eclipses and the Roche geometry

Problem

Fitting a single-lined spectroscopic orbit to a star of mass m1=1.0 M⊙m_1 = 1.0\,M_\odot gives P=18.0 dP=18.0\,\mathrm{d}, K1=45.0 km s−1K_1=45.0\,\mathrm{km\,s^{-1}} and e=0.60e=0.60. Infrared imaging finds a nearby star of mass 0.75 M⊙0.75\,M_\odot, a candidate for the companion. Someone argues: "The mass function needs only PP and K1K_1, so f(m)=1.036×10−7 K13P M⊙=1.036×10−7×45.03×18.0 M⊙=0.170 M⊙\begin{aligned} f(m) &= 1.036\times10^{-7}\,K_1^3P\ M_\odot \\ &= 1.036\times10^{-7}\times45.0^3\times18.0\ M_\odot = 0.170\,M_\odot \end{aligned} (with K1K_1 in km/s and PP in days). The largest mass function a 0.75 M⊙0.75\,M_\odot companion of a 1.0 M⊙1.0\,M_\odot star can produce, seen edge-on, is 0.753/1.752=0.138 M⊙0.75^3/1.75^2 = 0.138\,M_\odot. Since 0.170 M⊙0.170\,M_\odot exceeds this, the imaged star cannot be the companion, so the system must be a hierarchical triple." Find the error in this reasoning, give the corrected mass function, and decide whether the imaged star can be the companion. If it can, find the inclination it requires.

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