We now come to the binary stars which, because they comprise two stars in a dynamically associated system, can be analyzed using the same principles of Kepler's 3rd law we've used for the planets. Binaries are important in their own right, because they are exactly the systems needed for us to find and corroborate the masses of stars, and hence also test basic theories at the foundation of astrophysics, such as stellar evolution.

The diagram shows the simplest type of system, with binary components A (of mass m(A)) and B (of mass m(B)). This visual binary system, it's important to note, indicates the apparent relative orbit of the 2 stars - since ordinarily the plane of the real orbit will not lie in the plane of the sky (that is, perpendicular to the line of sight as portrayed). Usually, what we do is observe the motions of the fainter member about the brighter over a period of time sufficient to determine the orbit period, and then obtain the apparent relative orbit.

Geometrically (though this is beyond the scope of this blog and more in line with advanced astronomy problems) it is straightforward to show that an ellipse in one plane when projected onto another plane (say oblique to it) will yield another ellipse - but of different eccentricity, e = c/a. Most importantly, the foci of the original ellipse do not project onto the foci of the projected one. This means the primary (brightest star) though it is located at one focus of the true relative orbit, is not at the focus of the apparent relative orbit.

But it is this circumstance that makes it possible to determine the inclination of the true orbit to the plane of the sky. Basically, the problem reduces to finding the angle at which the true relative orbit must be projected in order to account for the amount of displacement of the primary from the focus of the apparent relative orbit.

Now, if the semi-major axis of the true relative orbit (e.g. the one it would have if displayed face-on) has an angular distance of a" (seconds of arc) and if the system is at a distance d parsecs, then the semi-major axis in astronomical units is:

a = (a" x d)

Then the sum of the masses of the two stars is given by Kepler's law:

m(A) + m(B) = (d a")^3/ P^2

where P is their period.

Of course, obtaining the total mass is only the first step. One then wishes to obtain the individual masses for each star. This is done by analyzing the motions of each member with respect to the center of mass of the system which ordinarily will be apart from either member. For example, if m(A) = m(B) then the c.m. will be exactly midway between them.

Example Problem:

Consider the visual binary system, Sirius A and B. The semi-major axis of the true relative orbit is 7½" and the distance from the Sun to Sirius is 2.67 pc. If the period of the Sirius binary system is 50 years and the component B is found to be twice as far from the center of mass as component A, then find the total mass of the Sirius system and the masses of each component.

Solution:

We first obtain the mass total:

m(A) + m(B) = (d a")^3/ P^2 = (2.67 pc x 7.5")/(50 yr)^2 = 3.2 solar masses

So:

m(A) + m(B) = 3.2 M_s

But in terms of the center of mass:

A O------------x cm-------------------------o B

where: xB = 2 (xA)

Then: xB/xA = 2, and: m(A)/m(B) = xB/xA = 2/1

so: m(B) = ½ m(A)

(since the more massive star is always closer to the center of mass)

Thus, m(A) ~ 2.13 M_s, and m(B) ~ 1.06 M_s

Spectroscopic binaries are also of much interest and derive their name because spectrsocopic analysis is needed to obtain the radial velocity curve for the system and hence the relative orbital velocity, V, for the pair. The distance around the relative orbit, its circumference, is just the relative orbital velocity V (deduced from the radial velocity profile) multiplied by the period. Then the distance between the stars a, is just:

a = (V x P)/ 2π

If, for instance, the relative velocity is a lower limit, then the separation we obtain is a lower limit for the system. If this is then applied to Kepler's 3rd law one can obtain a lower limit to the sum of the masses of the components:

m1 + m2 = a^3/P^2

Example Problem:

A spectroscopic binary system is found to have a relative velocity of 100 km/sec and a period of 17.5 days. Obtain a lower limit for the separation of the components, a, and thence a lower limit to the sum of the masses. If the spectroscopic analysis shows component (1) is 3 times the mass of component (2), find the lower limits on the masses of the components.

Solution:

First convert 100 km/sec to AU/yr.

Over one year: t = 3.156 x 10^7 s

total distance covered: d = v x t = 100 km/s (3.156 x 10^7 s) = 3.156 x 10^9 km

But 1 AU = 1.495 x 10^8 km

Then, the AU in this total distance:

d/AU = (3.156 x 10^9 km)/ (1.495 x 10^8 km) = 21.1 AU

The period in yrs. for 17.5 days:

P = 17.5/ (365.25) = 0.048 yr.

Then:

a = (21.1 x 0.048)/ 2π = 0.161 AU

The lower limit to the masses is therefore:

m1 + m2 = (0.161)^3/ (0.048)^2 = 1.8 solar masses

since star m1 has 3x the mass of star m2, then: m1 = 3m2

The diagram shows the simplest type of system, with binary components A (of mass m(A)) and B (of mass m(B)). This visual binary system, it's important to note, indicates the apparent relative orbit of the 2 stars - since ordinarily the plane of the real orbit will not lie in the plane of the sky (that is, perpendicular to the line of sight as portrayed). Usually, what we do is observe the motions of the fainter member about the brighter over a period of time sufficient to determine the orbit period, and then obtain the apparent relative orbit.

Geometrically (though this is beyond the scope of this blog and more in line with advanced astronomy problems) it is straightforward to show that an ellipse in one plane when projected onto another plane (say oblique to it) will yield another ellipse - but of different eccentricity, e = c/a. Most importantly, the foci of the original ellipse do not project onto the foci of the projected one. This means the primary (brightest star) though it is located at one focus of the true relative orbit, is not at the focus of the apparent relative orbit.

But it is this circumstance that makes it possible to determine the inclination of the true orbit to the plane of the sky. Basically, the problem reduces to finding the angle at which the true relative orbit must be projected in order to account for the amount of displacement of the primary from the focus of the apparent relative orbit.

Now, if the semi-major axis of the true relative orbit (e.g. the one it would have if displayed face-on) has an angular distance of a" (seconds of arc) and if the system is at a distance d parsecs, then the semi-major axis in astronomical units is:

a = (a" x d)

Then the sum of the masses of the two stars is given by Kepler's law:

m(A) + m(B) = (d a")^3/ P^2

where P is their period.

Of course, obtaining the total mass is only the first step. One then wishes to obtain the individual masses for each star. This is done by analyzing the motions of each member with respect to the center of mass of the system which ordinarily will be apart from either member. For example, if m(A) = m(B) then the c.m. will be exactly midway between them.

Example Problem:

Consider the visual binary system, Sirius A and B. The semi-major axis of the true relative orbit is 7½" and the distance from the Sun to Sirius is 2.67 pc. If the period of the Sirius binary system is 50 years and the component B is found to be twice as far from the center of mass as component A, then find the total mass of the Sirius system and the masses of each component.

Solution:

We first obtain the mass total:

m(A) + m(B) = (d a")^3/ P^2 = (2.67 pc x 7.5")/(50 yr)^2 = 3.2 solar masses

So:

m(A) + m(B) = 3.2 M_s

But in terms of the center of mass:

A O------------x cm-------------------------o B

where: xB = 2 (xA)

Then: xB/xA = 2, and: m(A)/m(B) = xB/xA = 2/1

so: m(B) = ½ m(A)

(since the more massive star is always closer to the center of mass)

Thus, m(A) ~ 2.13 M_s, and m(B) ~ 1.06 M_s

Spectroscopic binaries are also of much interest and derive their name because spectrsocopic analysis is needed to obtain the radial velocity curve for the system and hence the relative orbital velocity, V, for the pair. The distance around the relative orbit, its circumference, is just the relative orbital velocity V (deduced from the radial velocity profile) multiplied by the period. Then the distance between the stars a, is just:

a = (V x P)/ 2π

If, for instance, the relative velocity is a lower limit, then the separation we obtain is a lower limit for the system. If this is then applied to Kepler's 3rd law one can obtain a lower limit to the sum of the masses of the components:

m1 + m2 = a^3/P^2

Example Problem:

A spectroscopic binary system is found to have a relative velocity of 100 km/sec and a period of 17.5 days. Obtain a lower limit for the separation of the components, a, and thence a lower limit to the sum of the masses. If the spectroscopic analysis shows component (1) is 3 times the mass of component (2), find the lower limits on the masses of the components.

Solution:

First convert 100 km/sec to AU/yr.

Over one year: t = 3.156 x 10^7 s

total distance covered: d = v x t = 100 km/s (3.156 x 10^7 s) = 3.156 x 10^9 km

But 1 AU = 1.495 x 10^8 km

Then, the AU in this total distance:

d/AU = (3.156 x 10^9 km)/ (1.495 x 10^8 km) = 21.1 AU

The period in yrs. for 17.5 days:

P = 17.5/ (365.25) = 0.048 yr.

Then:

a = (21.1 x 0.048)/ 2π = 0.161 AU

The lower limit to the masses is therefore:

m1 + m2 = (0.161)^3/ (0.048)^2 = 1.8 solar masses

since star m1 has 3x the mass of star m2, then: m1 = 3m2

And: m2 + 3m2 = 1.8 or 4m2 = 1.8

so: m2 = 1.8/4 = 0.45 solar masses, and m1 = 3(0.45) = 1.35 solar masses

Other Problems:

(1) Find, approximately, the periods of revolution of the following binary star systems in which each star has the same mass as the Sun, and in which the semi-major axis of the relative orbits has the value:

(a) 1 AU

(b) 6 AU

(c) 100 AU

(2) For each of the systems in (1), at what distance would the two stars appear to have an angular separation of 1"?

(3) The true relative orbit of Epsilon Ursae Majoris has a semi-major axis of 2½" and the parallax of the system is 0."127. If its period is 60 years, find the sum of the components in solar mass units.

(4) A hypothetical spectroscopic-eclipsing binary system is observed and its period is 3 years. The maximum radial velocities with respect to the center of mass of the system are:

Star A: 4π/3 AU/yr

Star B: 2π/3 AU/yr

(a) Find the ratio of the masses of the components.

(b) Find the mass of each star in solar units.

(Assume the eclipses are central)

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