# Is it true that we can estimate the mass of a star more accurately if it has a companion star?

I remember I hear a theory that if a star has a companion star, we can know the mass of the star more easily, is that true? if it is true, while I understand that by measuring the period of the companion star, we can know the mass ratio of 2 stars,but how can the companion star also help to measure the mass of parent star?

• I also think Astronomy SE is a much better place for this particular question. That said, if you can measure spectroscopic doppler shifts you get the components of the velocities in our direction; combine those with the period and you can calculate both masses if you know the inclination of the orbit also. If they eclipse each other, you are in luck.Otherwise, if you don't know the inclination - I'm not sure.
– uhoh
Aug 17 '16 at 10:14
• The inclination is, in theory, measurable by watching the orbits of the system. If the apparent center of mass of the system doesn't line up with the focus of the orbit, you can attribute that due to a projection effect caused by the inclination of the orbit. The degree to which they don't line up relates to the inclination. You can also just take a statistical approach and say $i \approx 42$°, which is the most probably inclination. Aug 18 '16 at 0:58
• @zephyr Wow that's interesting! I though I'd have a nice quiet morning with my coffee, but now I'm not going to rest until I can calculate that! Is it actually the most probable (something like $\partial \text{P} / \partial \theta$) or is it really the median (half are above, half are below)? Maybe it should be added here, 42 being a "notable number'.
– uhoh
Aug 19 '16 at 23:03

In general, masses can only be precisely estimated for stars in binary systems (asteroseismology and some other indirect techniques can yield estimates too).

There are two broad types of binary where this is possible - astrometric binaries, where the motion of both stars can be seen on the sky and where the distance to the binary is known (e.g. Sirius AB); or eclipsing binaries, where the stars eclipse each other and their motions are inferred from their line of sight velocities, measured by spectroscopy (e.g. YY Gem).

You are correct, that the relative motions of the stars only yields their mass ratio. What is needed to get the individual masses is something that fixes the absolute sizes of the orbits and the orbital period, which can then be used with Kepler's third law to give the total mass of both stars and hence combined with the mass ratio to give the masses of the individual components.

In astrometric binaries you can measure the angular size of the orbits and calculate an absolute size using the known distance. In close (unresolved) binaries, that size is inferred from the velocities and orbital periods, but also requires knowledge of the inclination of the orbital plane. This is known quite accurately (is close to 90 degrees) only for eclipsing binaries.

I could present the Maths if you really like, but these are quite standard results.

I'll just fill in the blanks with the math that Rob Jeffries alluded to. There's a variety of scenarios you can consider, but I'll just run through the basics of two binary stars, where both stars are distinctly visible. Such an example is Antares, a well known and rather bright star in the northern sky. Such a star system is usually referred to as a Visual Binary.

If you want to solve for the masses of both stars, you'll need two equations. If algebra has taught you anything, it's that you need as many equations as you have unknowns if you want your system to be completely determined.

### Mass Ratio

To start with, consider two binary stars of mass $$M_1$$ and $$M_2$$ which are located at positions $$\vec{r}_1$$ and $$\vec{r}_2$$, respectively. The center of mass for that system, $$\vec{R}$$, is defined to be

$$\vec{R} = \frac{\vec{r}_1 M_1 + \vec{r}_2 M_2}{\vec{r}_1 + \vec{r}_2}$$

Now, you'll notice I'm defining the positions of the stars without setting up a coordinate system. Currently $$\vec{r}_1$$ and $$\vec{r}_2$$ have no physical meaning. I'm going to choose a convenient coordinate system such that the origin is at the center of mass. This means that $$\vec{R}=0$$ and we find that

$$\frac{M_1}{M_2} = \frac{\vec{r}_2}{\vec{r}_1}$$

This basic concept gives you the mass ratio, if only you can measure the ratio of the physical distance each star is from the center of mass. This is easily achieved by noting that a physical distance is related to the angular distance, $$\alpha$$, (something quite measurable with a telescope) by the relation $$r=\alpha d$$, where $$d$$ is the distance to the binary system. This allows us to say that

$$\frac{M_1}{M_2} = \frac{\alpha_2}{\alpha_1}$$

As stated, $$\alpha$$ is an easily measurable parameter (more or less) by simply imaging the binary system with a telescope over a long enough time period. There's a bit more involved in this since I haven't talked at all about the inclinations of the orbits, but this is the basic premise.

### Mass Sum

Now that we have a way of getting the mass ratio, the other equation we'll look for is the sum of the masses. Anytime you hear sum of masses, you should immediately think Kepler's third law.

$$P^2 = \frac{4\pi^2}{G(M_1+M_2)}a^3$$

In this equation $$P$$ is the orbital period, $$G$$ is the gravitational constant, and $$a = r_1 + r_2$$ is the semi-major axis of the reduced mass system (there's some more math and assumptions behind this, but let's roll with it for now). This problem is as simple as solving for the sum of the masses. However, we'll also note that $$a = \alpha d$$, where $$\alpha = \alpha_1 + \alpha_2$$. Solving for the sum of masses gives us

$$M_1 + M_2 = \frac{4\pi^2}{GP^2}\alpha^3d^3$$

As stated in the mass ratio section, it is a bit more involved than this because I haven't said anything about the inclination of the system. If the system has an inclination, it will tend to muck up the math and make it more difficult, but the principle concepts will be the same.

### Putting it together

You now have two equations and two unknowns. Let's solve for the masses.

$$M_1 = \frac{4\pi^2}{GP^2}d^3\alpha^2\alpha_2$$

$$M_2 = \frac{4\pi^2}{GP^2}d^3\alpha^2\alpha_1$$