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?

  • $\begingroup$ 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? $\endgroup$
    – user24157
    Mar 9, 2019 at 23:15
  • $\begingroup$ 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. $\endgroup$ Mar 9, 2019 at 23:18
  • 2
    $\begingroup$ That final equation is also wrong. $(m+M)^2 \neq m^2 + M^2$ $\endgroup$
    – PM 2Ring
    Mar 10, 2019 at 7:37

1 Answer 1


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.

  • $\begingroup$ Oh jeez thank you so much. I'm getting answers that make much more sense, about 1.1 solar masses now. $\endgroup$ Mar 9, 2019 at 23:35
  • $\begingroup$ @TheNerdyCoder I calculate the mass to be $\approx 1.1743 M_\odot$ $\endgroup$
    – PM 2Ring
    Mar 10, 2019 at 7:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .