AstronomyMeasuring the Stars

Binary Stars and Stellar Masses

Two stars orbiting each other reveal their masses, and the main sequence turns out to be a sequence of mass.

Not started Jump to practice

About half of the stars like the Sun have at least one companion star. Pairs bound by gravity, binary stars, are a gift: by weighing with orbits they give the masses of stars, the one property we cannot read from light alone. And when those masses are put on the H–R diagram, they explain the main sequence.

Why it works

The two stars orbit their common center of mass, the balance point between them. Like children on a seesaw, the heavier star sits closer: $M_1 r_1 = M_2 r_2$. Both go around in the same time, so each one's speed is proportional to its distance from the center: $\frac{M_1}{M_2} = \frac{r_2}{r_1} = \frac{v_2}{v_1}$. Newton's form of Kepler's third law gives their total mass from the size $a$ (in AU) and period $P$ (in years) of their orbit: $M_1 + M_2 = \frac{a^3}{P^2}$. With the sum and the ratio, each mass follows. The speeds come from Doppler shifts; the sizes from watching the stars move across the sky, or from eclipses when the orbit is edge-on.

Key idea

For a binary: $M_1 + M_2 = \dfrac{a^3}{P^2}$ (solar masses, AU, years) and $\dfrac{M_1}{M_2} = \dfrac{r_2}{r_1} = \dfrac{v_2}{v_1}$. Binaries show that the main sequence is a sequence of mass: luminosity rises roughly as $M^{3.5}$.
Two stars circling their center of mass. The heavier star moves on the smaller circle, more slowly, so its Doppler curve (gold) is smaller.

Worked examples

Worked example: Sirius A and B

Sirius B orbits Sirius A with $a = 19.8$ AU and $P = 50.1$ years. Find their total mass. Sirius A stays about half as far from the center of mass as Sirius B; find each mass.
  1. Total: $\frac{19.8^3}{50.1^2} = \frac{7762}{2510} = 3.09$ solar masses.
  2. Seesaw: $r_B = 2r_A$, so $M_A = 2M_B$.
  3. $M_A + M_B = 3M_B = 3.09$, so $M_B = 1.03$ and $M_A = 2.06$ solar masses.

Answer: About $3.1\,M_\odot$ in all: $2.06$ and $1.03$.

Worked example: From Doppler speeds

In a binary, star A moves at $20$ km/s and star B at $60$ km/s around the center of mass. How do their masses compare?
  1. $\frac{M_A}{M_B} = \frac{v_B}{v_A} = \frac{60}{20} = 3$.
  2. Star A, the slower one, is three times as massive.

Answer: Star A has three times the mass of star B.

Common mistake

It is tempting to read the main sequence as a life story, with stars sliding down it as they age. Binary masses show something else: along the main sequence, mass changes. A $10\,M_\odot$ star is thousands of times more luminous than the Sun, and a $0.2\,M_\odot$ red dwarf less than a hundredth as luminous. A star stays near one spot on the main sequence for most of its life. Check: in binaries, the brighter main-sequence star is always the more massive one.
Main-sequence luminosity against mass on log scales: roughly $L \propto M^{3.5}$.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

Loading practice…

Builds on

Leads to

Sources and further reading

  • OpenStax, Astronomy 2e, section 18.2, Measuring Stellar Masses
  • OpenStax, Astronomy 2e, section 18.4, The H–R Diagram
@teddyrookbook GitHub