## Binary stars and circular orbits

In this section we will see how Newton's laws of motion and gravitation can be applied to binary stars in circular orbits. Circular orbits are not the most general case of orbital motion, but the analysis is most straightforward, and most of the basic points are clearly illustrated. In the next section we will go to the general case of elliptical orbits.

We consider two stars, of masses m1 and m2, orbiting their common center of mass at distances r1 and r2, respectively (Fig. 5.8). From the

Line of Sight

Line of Sight

Inclination of an orbit.The orbit is an ellipse, which lies in a plane.The plane makes some angle i with a plane of the sky.The plane of the sky is defined to be perpendicular to the line of sight.

Binary system with circular orbits. Both stars orbit the center of mass (CM).The more massive star is closer to the center of mass.The center of mass must always be between the two stars, so the stars lie on opposite sides of it.

Inclination of an orbit.The orbit is an ellipse, which lies in a plane.The plane makes some angle i with a plane of the sky.The plane of the sky is defined to be perpendicular to the line of sight.

Binary system with circular orbits. Both stars orbit the center of mass (CM).The more massive star is closer to the center of mass.The center of mass must always be between the two stars, so the stars lie on opposite sides of it.

definition of center of mass, these quantities are related by

Combining equations (5.13) and (5.14) gives m^i = m2r2

m1m2

The center of mass moves through space subject only to the external forces on the binary star system. The forces between the two stars do not affect the motion of the center of mass. We will therefore ignore the actual motion of the center of mass, and view the situation as it would be viewed by an observer sitting at the center of mass.

Since the center of mass must always be along the line joining the two stars, the stars must always be on opposite sides of the center of mass. This means that the stars orbit with the same orbital period P. In general, the period of the orbit is related to the radius, r, and the speed, v, by

Note that we can divide both sides by m1. If we also use equation (5.10) to relate v1 to P, we find that

G m2

Since the periods of the two stars must be the same, equation (5.9) tells us that rjvi = r2/v2 (5.11)

Combining these with equation (5.8) gives vi/v2 = ri/r2 = m2/mi (5.12)

(Note: We could have also obtained m1v1 = m2v2 directly from conservation of momentum. This is not surprising since the properties of the center of mass come from conservation of momentum.)

We now look at the gravitational forces. The distance between the two stars is r1 + r2 , so the force on either star is given by Newton's law of gravitation as

This force must provide the acceleration associated with the change of the direction of motion in circular motion, v2/r. For definiteness, we look at the force on star 1, so

This can be simplified if we introduce the total distance R between the two stars:

Using equation (5.12), this becomes R = r1(1 + m1/m2) (5.18)

Substituting this into equation (5.16) gives 4^R3/G = (m1 + m2) P2 (5.20)

Let's look at how equation (5.20) can be used to give us stellar masses. For any binary system, we can determine the period directly if we watch the system for long enough. If the star is a spec-troscopic binary, we can see how long it takes for the Doppler shifts to go through a full cycle. If it is an astrometric binary we can see how long it takes for the 'wobble' to go through a full cycle. If it is an eclipsing binary, we can see how long it takes the light curve to go through a full cycle. If we can see both stars, we can determine R. Once we know R and P, we can use equation (5.20) to determine the sum of the masses, (m1 + m2 ). We can also obtain the ratio of the masses m1/m2, either from r1/r2, if both stars can be seen, or v1/v2, if both Doppler shifts are observed. Once we know the sum of the masses and the ratio of the masses, the individual masses can be determined. The situation we have outlined here is the ideal one, however. Usually, we don't have all of these pieces of information (as we will see below).

Example 5.2 Mass of the Sun We can consider the Sun and Earth as a binary system, so we should be able to apply equation (5.20) to

0 0