###### By Joshua N. Winn, Princeton University

## If we want to understand stars, we really need to know their mass. Mass determines a star’s gravitational pull, its internal pressure, and the amount of fuel that’s available to keep shining.

### Algol: An Eclipsing Binary

To understand mass, we need to consider a blue star in Perseus called Algol. Every three days, Algol drops in flux by 30%, and stays that way for 10 hours before going back to normal because it is actually a pair of stars that orbit each other and periodically eclipse one another.

Eclipsing binaries like Algol allow us to measure stellar masses. Using Kepler’s third law we can write: *a*^{3} = *GM* over 4π*r²* **×** *P*^{2}. Now, back in the planetary context, *a* was the semimajor axis of the planet’s orbit, *P* was the period, and *M* was the mass of the Sun.

### Two Stars in Motion

Following Newton’s third law: every reaction is accompanied by an equal and opposite reaction, if the Sun pulls on a planet, the planet must be pulling back on the Sun with equal force. But since acceleration equals force over mass, and the Sun is so much more massive than the planets, the Sun’s acceleration is much smaller than the planets’ and so we neglect the Sun’s motion.

But we with two stars of comparable mass pulling on each other, we find them travel in elliptical orbits, with a focal point at the center of mass of the system.

For 2 stars, with vector positions *r*1 and *r*2, the center of mass is located at *m*1 *r*1 plus *m*2 *r*2, divided by the total mass. With two stars obeying Newton’s second law, *F* equals *ma*. For star 2, m2 times the 2nd time-derivative of *r*2 equals *F*, which is the gravitational force exerted by star 1.

We can write a similar equation for star 1, except this time it’s minus *F*, because of Newton’s third law: the forces must be equal and opposite. The right side is zero, and the left side can be re-written as the second derivative of *m*1 *r*1 plus *m*2 *r*2. That’s proportional to the center of mass vector. So, we’ve just proven that the location of the center of mass doesn’t accelerate. The two stars pull each other around, but the center of mass moves in a straight line at a constant speed.

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### Heavier Star Moves Slowly

Each star moves in an ellipse, and the two ellipses have a common focus: the center of mass. The size of each ellipse is proportional to the other star’s mass. So, the ratio of semimajor axes, *a*2 over *a*1 equals *m*1 over *m*2.

As they move, the stars are always on opposite sides of the center of mass. They also have equal and opposite momenta. We can see that by taking the derivative of our center-of-mass equation: that gives *m*1 *v*1 = −*m*2 *v*2. So, their velocities are always in opposite directions, and the ratio of their speeds also tells us the mass ratio: *v*1 over *v*2 equals *m*2 over *m*1. The heavier star moves more slowly.

If we can measure the relative sizes of the orbits, or the orbital speeds, then we’ll learn the mass ratio. And if we also measure the orbital period, we can use Kepler’s third law to learn the total mass.

### Measuring the Size of the Orbits

We use the Doppler spectroscopy to measure the sizes of the orbits and the orbital period.

The Doppler effect is the shift in wavelength that you observe whenever the source of the waves is moving. The fractional shift in wavelength is equal to *vr* over *c*, where *c* is the speed of light and *vr* is the so-called radial velocity—the component of the velocity along the line between the star and us.

Radial here comes from the concept of the sky as a celestial sphere centered on the Earth. The motion towards or away from the Earth is the radial direction in that spherical coordinate system.

Even though the two stars of Algol are blended together in the image, the spectrum of that single point of light reveals two different sets of absorption lines, and they’re shifted in wavelength with respect to each other, because the stars are moving at different speeds. And as the stars go around, we can watch those lines shift back and forth.

### Figuring out the Masses

Suppose we monitor the radial velocities of both stars for 100 days. They vary sinusoidally, in opposite directions, as we expect. A full sine wave takes 40 days, so that’s the orbital period. And the amplitudes tell us the orbital speeds: 30 kilometers per second for star 1, and 60 kilometers per second, for star 2.

Now let’s use these data to figure out the stellar masses. We can determine the mass ratio, right away: *m*1 over *m*2 equals *v*2 over *v*1, which is 2. So, star 1 is twice as massive as star 2.

Next, we’ll use Kepler’s third law to figure out the total mass. We already know *P* is 40 days, but *a* is the separation between the stars, it’s the sum of *a*1 and *a*2. We can calculate *a*1 by remembering that over one period, star 1 travels the full circumference of its orbit, so *v*1 times *P* has to equal 2π*a*1. Likewise, *v*2 *P *equals 2π*a*2. Adding them together, and solve for *a*, we get *v*1 plus *v*2 times *P* over 2π. That’s 90 kilometers per second times 40 days over 2π, and when you convert all the units it comes out to be a third of an AU.

Now, we solve for *M*, and we write it as a scaling relation. For *P*, we insert 40 days, which is 0.11 years, and for *a*, we insert a third of an AU. That gives a total mass of 3 solar masses. So, we’ve got two stars with masses in a ratio of 2 : 1, and a total of 3. So, *m*1 is 2 solar masses, and *m*2 is one solar mass.

So, that’s how we measure stellar masses, using eclipsing binaries.

### Common Questions about Measuring Stellar Masses Using Eclipsing Binaries

**Q: What is a stellar mass?**

Mass determines a star’s gravitational pull, its internal pressure, and the amount of fuel that’s available to keep shining.

**Q: What is Algol?**

Algol, a blue star in Perseus, is actually a pair of stars that orbit each other and periodically eclipse one another.

**Q: How can we measure the orbital size?**

The orbital size can be measured using Doppler spectroscopy.