Stars: Binary Systems

Binary star systems allow the determination of stellar masses.

The orbital velocity of stars in a binary system reflect the stellar masses since, according to Kepler’s Law, the velocityof a star is inversely proportional to it’s mass.

Binary stars orbit each other such that the center of mass of the combined

system is located closest to the more massive star

Thus, the orbital radius is larger for the least massive star, and since v=rthe least massive star orbits faster (since = 2/P is the same for both stars.)

Kepler’s 3rd Law

Center of Mass

In a coordinate system centered on the C of M,

m1r1 = m2r2

Law of Gravity

The force of gravity, F, provides the centripetal acceleration that keeps the stars in circular orbits

Fgrav = Gm1m2/a2 where a = r1 + r2

Kepler’s 3rd Law (Continued)

The centripetal force for each star is,

F1 = m1 (v1)2 / r1 and F2 = m2 (v2)2 / r2

Now, F1 = F2 = Fgrav and these equations can be combined to obtain Kepler’s 3rd Law;

P2 = 42 a3/ G(m1 +m2)

Question: Prove Kepler’s 3rd Law.

The Sun-Earth System

P2 = 42 a3/ G(m1 +m2)

For the Sun-Earth system, P = 1 year, a = 1 AU, m1 = Mo

and m2 = Mearth. Substituting these values we obtain,

12 = 42 13/ G Mo(1 + Mearth/ Mo)

But since Mearth/ Mo ~ 0, we obtain the result that the constant42 / G Mo = 1, and hence a simpler version of Kepler’s Law

P2 = a3/ (m1 +m2) which is valid when using the correct units,

ie. P in years, a in AU, and the masses in solar units.

Getting the separation, a

Although we can measure the period, P, directly, we need to knowthe distance to get the separation, a.

Recall that, d(pc) = 1/where is in arc seconds, which is basedon the definition that 1 arc sec is the angular separation of the Earth-Sun system (1 AU ) at a distance of 1 pc. It can be shown,by similar triangles, that the angular separation of a binary starsystem, a, in arc seconds, divided by the parallax, in arc seconds,is equal to the linear separation, a, in units of pc, so that

a(AU) = a`` (arc sec)/ arc sec)

Question: Prove a (AU) = a`` (arc sec)/ arc sec)

Yet another version of Kepler’s 3rd Law

So, we can replace the a in

P2 = a3/ (m1 +m2)

With a (AU) = a`` (arc sec)/ arc sec)To obtain,

P2 = [ a`` ]3 where a and are now in arc sec _____________________

[ (m1 +m2)

The benefit of this equation is that all the variables are expressedin terms of “observed” quantities

There are basically 2 types of binary systems

Visual Binaries – are well separated visually on the sky

Spectroscopic Binaries – resolved only spectrally

Spectroscopic Binaries

Some more details on BinariesVisual Binaries

In the rare situation that the binary system is face-on, and the orbits are circular, then

measure P, a(``) and ``) which when appliedto Kepler’s 3rd law, yields m1 +m2.

Then measure r1 and r2 which yields m1/ m2.

Then solve for m1 & m2.

In actuality, it’s not so simple because the orbital plane is usually inclined to the line of sight which requires somegeometry to correct for projection effects.

More on Spectroscopic Binaries

Doppler Effect

The wavelength, (–o shift in the spectral lines is caused by the Doppler effect and can be used to deduce the radial velocity, vr, of the binary stars using,

o = (–oo= vr/c

Where is the observed wavelength, o is the rest (zero velocity) wavelength, vr is the radial component of the orbital velocity andc is the speed of light.

The wavelength shifts for each star are plotted on a radial velocitygraph.

Radial Velocity Graph

The two curves ( one for each star ) are sinusoidal and oscillatewith exactly opposite phase ( one star approaches as the otherrecedes ). The amplitude of each velocity curve yields r1 and r2.The star with the largest velocity amplitude has the largest radius,and hence the smallest mass.

Getting the masses

Measure the velocity amplitudes, v1 and v2, since v1/ v2 = r1/ r2 = m2/ m1.

Also, use the period and the velocities to calculate the radii r1 and r2 separately for each star which yields the separation a.

Then use the period and Kepler’s 3rd law to get m1 + m2

Then solve for m1 & m2 separately.

As usual, there are complicationsIf the orbital plane is inclined to the line of sight then the observedradial velocity is only a fraction of the actual radial velocity sincevobs = vr sin i, where i is the angle of inclination.

The angle of inclination, i, is measured from the line of sightto the normal of the orbital plane.

i = 90o would be an edge on system. i = 0o would be a face-on system which would not be very useful since sin i = 0 and the observed radial velocity for such a system would be zero.

You can still get the mass ratio since m2/ m1 = v1/ v2 since the sin i’scancel. But only a lower limit on the masses; m sin3i.

Getting a handle on the inclination with Eclipsing

BinariesDetermining the inclination is problematic. However, if the inclination is close to 90o, then the two stars may eclipse eachother which will manifest as a time variable change in thebrightness of the binary system.

Eclipses can also yield ;

a) the sizes of the stars, by measuring the duration of the eclipses and

b) the temperatures of the stars, by measuring the depth of the eclipses

Primary minimum Secondary minimum

Sizes of stars from the duration of eclipses

The time for onset of the primary eclipse, t1, tells you about the size of the hotter star (that passes behind the cooler star).

(vp + vs) t1 = 2 rs

and,

the duration of the primary eclipse, t2, tells you about the size of the cooler star;

(vp + vs) t2 = 2 rp – 2 rs

The temperatures of stars from the depth of the eclipses

During the primary eclipse, the hotter star is eclipsed by the coolerstar, so the luminosity of the system is lower by an amountequal to the luminosity of the hotter star, LH, so

MBol,p – MBol,o = 2.5 log [(LH + Lc)/ Lc] = 2.5 log [(rH

2TH4 + 1)]

[ rc2 Tc

4 ]

And, knowing the sizes of the stars already from the duration of the eclipses, one can find the ratio of the stellar temperatures.(There is a more complicated equation in the book that uses the primary and secondary eclipse depths to get the ratio of effectivetemperatures, by eliminating the surface areas of the stars.)

The Mass – Luminosity Relation

The primary reason for using binaries to measure stellar masses is to calculate the relationship between stellar mass and luminosity.

L M3

More on L M3

L = k M3 where k is the constant of proportionality.

Substitute some numbers for the Sun;

L = 1 Lo, M=1Mo so the constant k = 1 !

Thus,

L = M3 for masses and luminosities in solar units.

Back to the H-R Diagram

Now, we can see that thehotter and more luminousstars are also the most massive.

Understanding why is the key to understanding the physics of stars.