# Mass of a potential black hole in a binary system

So I've been given the velocity curve, parallax and apparent magnitude of a star in a binary system with what is potentially a black hole. I've calculated from the apparent magnitude and parallax that the star is a type F5V, which puts the mass at about 1.4 Solar Masses. The velocity curve has an inclination of 90, and oscillates back and forth between +/- 75km/s. There is no data on the companion of this star, just the fact that it could be a black hole. I'm supposed to estimate the mass by numerically approximating a polynomial. So far I've used this equation

$$\frac{M^{3}}{(m+M)^{2}} = \frac{Pv^{2}}{2\pi G}$$

where M is the mass of the thing I don't know, m is the mass of the known companion (1.4 solar masses) P is the period (5.59 days) and v is of course the velocity (75km/s)

I got lazy and wrote $$\frac{Pv^{2}}{2\pi G}$$ as $$k$$ and arrived at

$$M^{3} - kM^{2} - km^{2} = 0$$

Using python's optimize library I found the mass of this unkown partner to be about 0.018 solar masses. My question here is where did I go wrong, and if I didn't go wrong anywhere is that a realistic mass for a super small black hole / other very small, dense and invisible object?

• Under the usual conventions, if the inclination is zero then there is no radial velocity (face-on system), are you sure that was what was stated in the original problem?
– user24157
Mar 9, 2019 at 23:15
• Yeah I'm sorry I meant 90 degrees, in that the radial velocity times the sine of the inclination can be ignored as it's 1. I'll make the edit in the original question. Mar 9, 2019 at 23:18
• That final equation is also wrong. $(m+M)^2 \neq m^2 + M^2$ Mar 10, 2019 at 7:37

Your first equation is incorrect. The left side has dimensions of mass, the right side has dimensions of mass × time × length-1. The velocity semi-amplitude (usually denoted $$K$$ rather than $$v$$) should be raised to the third power.
As noted by @PM-2Ring in the comments, your second equation is also incorrect as you didn't expand the $$(m+M)^2$$ term correctly.
• @TheNerdyCoder I calculate the mass to be $\approx 1.1743 M_\odot$ Mar 10, 2019 at 7:40