# Ratio of masses of double star problem

Your core question appears to be:

Why are the angular velocities equal?

For some reason, Wikipedia's articles about the two body problem do not clarify this important point. Here are a couple of diagrams from that article:

Two bodies with similar mass orbiting a common barycenter external to both bodies, with elliptic orbits—typical of binary stars.

Two bodies with a "slight" difference in mass orbiting a common barycenter. The sizes, and this type of orbit are similar to the Pluto–Charon system (in which the barycenter is external to both bodies), and to the Earth–Moon system—where the barycenter is internal to the larger body.

According to Newton's law of gravitation, the gravitational force between two bodies acts along the straight line connecting their centres of mass. This is the key to your question about angular velocities.

(It can also be shown that the gravity of a spherically symmetric body acts as if all of the body's mass were concentrated at its centre, so we can treat the body as a point particle).

So we have two stars, $S_1$ and $S_2$, exerting gravitational force on each other. The centre of mass of this system must be located on the line $S_1S_2$ that connects the two stars' centres. We can choose a reference frame with the centre of mass as its origin, like in the above diagrams. (As Rob Jeffries says, we can do this because of conservation of momentum). So I'll call the centre of mass $O$.

Now as the stars orbit around $O$ the only forces they exert on each other *always* act along the line
$S_1OS_2$, so the stars and the centre of mass *must* remain collinear, although the line $S_1OS_2$ rotates, and may change in length (as it does in the elliptical orbit example).

The only way for that to happen is for the angular velocities of the two stars,$\omega_1$ and $\omega_2$, to *always* be equal to each other. Otherwise, $S_1OS_2$ turns into a triangle, not a straight line.

And what happened to the negative sign?

That negative sign merely tells us that the position vectors of the two stars, $OS_1$ and $OS_2$ point in opposite directions. That is, the two stars are on opposite sides of $O$.

In the absence of external forces, the total momentum of the system is constant. That means the stars must arrange themselves so that the centre of mass stays in the same place.

The only way they can do this is if they orbit with the same period and hence the same angular velocity, as shown below.

If that were not the case, then the centre of mass would "wobble about", which is clearly unphysical.

That also means you can immediately write $$m_1 r_1 = m_2 r_2\ .$$