Visual binaries and the masses of stars

Builds on: Sections 8.1 and 4.3
Reading time: 35 min
Exercises: 3
Olympiad problems: 2

Sirius, the brightest star of the night sky, is only 2.64 pc away, and in a large telescope a faint point of light can be seen beside it. This is Sirius B, a white dwarf, and it goes round Sirius A once every 50.1 years. Following the pair for a century has given the mass of each star, about 2.1 solar masses for Sirius A and 1.0 for Sirius B, so a star with about the mass of the Sun has been squeezed into a ball the size of Earth. Visual binaries like this one are almost the only way to weigh a star directly, and the masses they give are the foundation for everything we know about how stars live. In this section we see how the weighing is done.

8.2.1Weighing the two stars

Before you readMake a guess first

The two stars of a binary each move on an ellipse round their centre of mass. Which star has the larger ellipse?

In a visual binary we measure three things. The period $P$ comes from watching the companion go round. The angular size of the orbit, its semi-major axis $α$ in arcseconds, comes from the positions of the companion relative to the bright star over the years. The parallax $p$ gives the distance, and dividing the angle $α$ by the parallax turns it into astronomical units, since an angle of 1″ at a distance of $1 / p$ parsecs covers $1 / p$ au. Kepler's third law ((8.1)) then gives the total mass.

To split the total between the two stars we also need the motion of each star separately, measured against the distant background stars. Each star moves on its own ellipse round the centre of mass, with angular semi-major axes $α_{1}$ and $α_{2}$ that add up to $α$ , and by (3.16) the sizes are in the inverse ratio of the masses.

ResultMasses from a visual binary(8.2)

\begin{aligned} M_{1} + M_{2} & = \frac{(α / p)^{3}}{P^{2}} \\ \frac{M_{1}}{M_{2}} & = \frac{α_{2}}{α_{1}}, α = α_{1} + α_{2} \end{aligned}

$M_{1}, M_{2}$	the masses of the two stars, in solar masses
$α$	the angular semi-major axis of one star's orbit relative to the other, in arcseconds
$α_{1}, α_{2}$	the angular semi-major axes of the two stars' orbits round the centre of mass
$p$	the parallax, in arcseconds
$P$	the period, in years

The angles must be the true semi-major axes of the orbits, which the next subsection recovers from a tilted orbit. Since

α / p

is cubed, an error of 1% in the parallax becomes an error of 3% in the masses, which is why good masses need good parallaxes.

Quick checkTotal mass from an orbit

A visual binary has a parallax of 0.100″, an orbit of angular semi-major axis 2.00″ and a period of 50.0 years. What is its total mass?

solar masses

Example 8.3Weighing Sirius A and B

Sirius B's orbit relative to Sirius A has a true angular semi-major axis of 7.50″ and a period of 50.13 years, and the parallax is 0.379″. Against the background stars, Sirius A moves on an ellipse of angular semi-major axis 2.48″. Find the two masses.

Solution

The orbit's semi-major axis is $7.50 / 0.379 = 19.8$ au, so by (8.2) the total mass is ${19.8}^{3} / {50.13}^{2} = 3.08$ solar masses. Sirius B's own ellipse is $α_{B} = 7.50 - 2.48 = 5.02 ″$ , and $M_{A} / M_{B} = 5.02 / 2.48 = 2.02$ .

AnswerSirius B has

3.08 / 3.02 = 1.02

solar masses and Sirius A has

2.06

solar masses.

8.2.2The orbit we see is tilted

Most orbits are tilted to the sky. We see the true orbit squashed along one direction, the apparent orbit. A tilted ellipse still looks like an ellipse, but the bright star is no longer at its focus, and that gives the tilt away. The line where the orbit's plane cuts the plane of the sky is the line of nodes. Distances along it are unchanged, while distances at right angles to it are shortened by $\cos i$ , where the inclination $i$ is the angle between the orbit's plane and the plane of the sky. Astronomers fit the true orbit, the inclination and the direction of the line of nodes all at once to positions measured over many years.

Figure 8.2The true and the apparent orbit

Eccentricity

e

Inclination

i

Periastron angle

ω

Watch

Apparent orbit
Bright star

The figure shows the orbit of a companion B round a bright star A. The true orbit, dotted, is drawn as it would look face on, with A at one focus. The heavy apparent orbit is what we see when the true orbit is tilted by the inclination

i

about the line of nodes. Its centre is C and its foci are F, and A lies off both of them unless the orbit is face on. The angle

ω

sets where the companion is closest to A, measured round the orbit from the line of nodes. The figure can be set to , , or .

A circular orbit seen at inclination $i$ looks like an ellipse with axes in the ratio $\cos i$ , with the bright star at its centre. For an elliptical orbit the reasoning is the same but the working longer. In the olympiad, problems on visual binaries either say that the orbit is face on, or give the true semi-major axis.

8.2.3Dynamical parallax

For most visual binaries the parallax is too small to measure well. The period and the angular size of the orbit can still give the distance, because stars on the main sequence follow a close relation between mass and luminosity, roughly $L \propto M^{3.5}$ . We guess the masses, find the parallax from Kepler's third law, use it to find the luminosities, read off new masses, and repeat. Because the masses enter as a cube root, a poor first guess still gives a fair parallax, and the repetitions settle quickly. This distance is called the dynamical parallax.

ResultThe dynamical parallax(8.3)

p = \frac{α}{P^{2 / 3} (M_{1} + M_{2})^{1 / 3}}

$p$	the parallax, in arcseconds
$α$	the angular semi-major axis of the relative orbit, in arcseconds
$P$	the period, in years
$M_{1} + M_{2}$	the total mass, in solar masses

This is (8.2) solved for

p

. The luminosities from each estimate of

p

come from the distance modulus ((4.4)), and each new estimate of the masses from the relation between mass and luminosity.

Example 8.4A wrong guess for Siriusfinish it

Suppose we did not know the parallax of Sirius and guessed that the two stars together have 2 solar masses. What parallax would the orbit, with $α = 7.50 ″$ and $P = 50.13$ years, give?

The start

By (8.3) we need $P^{2 / 3} = {50.13}^{2 / 3} = 13.60$ and $(M_{1} + M_{2})^{1 / 3} = 2^{1 / 3} = 1.260$ .

Your stepNow find

p

and compare it with the true 0.379″.

″

Common errorsSection 8.2

The apparent orbit of a tilted binary is smaller than the true orbit in one direction. Using its semi-major axis in Kepler's law gives too small a mass unless the orbit is face on.
An angle in arcseconds is turned into astronomical units by dividing by the parallax in arcseconds, not by multiplying.
The heavier star has the smaller orbit, so $M_{1} / M_{2} = α_{2} / α_{1}$ , with the indices crossed.
A small error in the parallax becomes three times as large in the masses, because the size of the orbit is cubed.

ExercisesSection 8.23 questions

8.2.1
The binary of the quick check, with a total of 3.2 solar masses, has star 1 moving on an ellipse of 0.80″ round the centre of mass and star 2 on one of 1.20″. What is the mass of star 1?
solar masses
8.2.2
A circular orbit is tilted by $i = 60 °$ . What is the ratio of the short axis to the long axis of the apparent orbit?
8.2.3
The parallax of a binary is measured 5% too small. How wrong is its total mass?

Olympiad problemsPast IOAA problems, restated2 problems

2010A binary seen face on10 points

Adapted from IOAA 2010 (Beijing), theory problem S4. We have restated it in our own words and kept the original data. The solution was last checked on 21 October 2026.

A binary star is 10 pc away. The largest angular separation of its two stars is 7.0″ and the smallest 1.0″. The period is 100 years and the orbit is face on, at right angles to the line of sight. The orbit of one star round the centre of mass has an angular semi-major axis of $α_{1} = 3.0 ″$ . Estimate the mass of each star, in solar masses.

I solved it without the solution

AnswerStar 1, with the larger orbit, has 1.6 solar masses and star 2 has 4.8 solar masses.

2008Weighing α Centauri A and B80 points

Adapted from IOAA 2008 (Bandung), data-analysis problem 2. We have restated it in our own words, kept the original data, and given the table in this course's sign of the bolometric correction. The first part, plotting the table by hand, is left out, so the points add up to 80 of the original 100. The solution was last checked on 21 October 2026.

α Centauri A and B form a visual binary on the main sequence. The semi-major axis of their relative orbit is 17.59″ and the period 79.24 years. Their visual magnitudes are $- 0.01$ and 1.34 and their colour indices $B - V$ are 0.65 and 0.85. For main-sequence stars, the bolometric correction $B C = M_{bol} - M_{V}$ is

$B - V$	0.50	0.60	0.70	0.80	0.90	1.00
$B C$	−0.03	−0.07	−0.12	−0.19	−0.28	−0.40

and the mass is related to the absolute bolometric magnitude by $M_{bol} = - 10.2 \log (M / M_{⊙}) + 4.9$ .

(a) Find the apparent bolometric magnitudes of the two stars. [40 points]

(b) Find the mass of each star, without using a measured parallax. [40 points]

I solved it without the solution

Answer(a) $m_{bol} = - 0.105$ for A and $1.105$ for B. (b) The parallax settles at 0.754″, and the masses at 1.15 solar masses for A and 0.88 for B. The official solution reads the bolometric corrections from a hand-drawn curve as 0.10 and 0.25 and reaches 1.15 and 0.88 too. Modern measurements give a parallax of about 0.75″ and masses of 1.08 and 0.91, so the simple mass relation does well.