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 velocity of 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=r  the least massive star orbits faster (since  = 2  /P is the same for both stars.)

Kepler’s 3 rd Law Center of Mass In a coordinate system centered on the C of M, m 1 r 1 = m 2 r 2 Law of Gravity The force of gravity, F, provides the centripetal acceleration that keeps the stars in circular orbits F grav = Gm 1 m 2 /a 2 where a = r 1 + r 2

Kepler’s 3 rd Law (Continued) The centripetal force for each star is, F 1 = m 1 (v 1 ) 2 / r 1 and F 2 = m 2 (v 2 ) 2 / r 2 Now, F 1 = F 2 = F grav and these equations can be combined to obtain Kepler’s 3 rd Law; P 2 = 4  2 a 3 / G(m 1 +m 2 ) Question: Prove Kepler’s 3 rd Law.

The Sun-Earth System P 2 = 4  2 a 3 / G(m 1 +m 2 ) For the Sun-Earth system, P = 1 year, a = 1 AU, m 1 = M o and m 2 = M earth. Substituting these values we obtain, 1 2 = 4  / G M o (1 + M earth / M o ) But since M earth / M o ~ 0, we obtain the result that the constant 4  2 / G M o = 1, and hence a simpler version of Kepler’s Law P 2 = a 3 / (m 1 +m 2 ) 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 know the distance to get the separation, a. Recall that, d(pc) = 1/  where  is in arc seconds, which is based on 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 star system, 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 3 rd Law So, we can replace the a in P 2 = a 3 / (m 1 +m 2 ) With a (AU) = a`` (arc sec)/  arc sec) To obtain, P 2 = [ a`` ] 3 where a and  are now in arc sec _____________________ [    (m 1 +m 2 ) The benefit of this equation is that all the variables are expressed in 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 Binaries Visual Binaries In the rare situation that the binary system is face-on, and the orbits are circular, then measure P, a(``) and  ``) which when applied to Kepler’s 3 rd law, yields m 1 +m 2. Then measure r 1 and r 2 which yields m 1 / m 2. Then solve for m 1 & m 2. In actuality, it’s not so simple because the orbital plane is usually inclined to the line of sight which requires some geometry 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, v r, of the binary stars using,  o = (  –  o  o  = v r /c Where  is the observed wavelength, o is the rest (zero velocity) wavelength, v r is the radial component of the orbital velocity and c is the speed of light. The wavelength shifts for each star are plotted on a radial velocity graph.

Radial Velocity Graph The two curves ( one for each star ) are sinusoidal and oscillate with exactly opposite phase ( one star approaches as the other recedes ). The amplitude of each velocity curve yields r 1 and r 2. The star with the largest velocity amplitude has the largest radius, and hence the smallest mass.

Getting the masses Measure the velocity amplitudes, v 1 and v 2, since v 1 / v 2 = r 1 / r 2 = m 2 / m 1. Also, use the period and the velocities to calculate the radii r 1 and r 2 separately for each star which yields the separation a. Then use the period and Kepler’s 3 rd law to get m 1 + m 2 Then solve for m 1 & m 2 separately.

As usual, there are complications If the orbital plane is inclined to the line of sight then the observed radial velocity is only a fraction of the actual radial velocity since v obs = v r sin i, where i is the angle of inclination. The angle of inclination, i, is measured from the line of sight to the normal of the orbital plane. i = 90 o would be an edge on system. i = 0 o 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 m 2 / m 1 = v 1 / v 2 since the sin i’s cancel. But only a lower limit on the masses; m sin 3 i.

Getting a handle on the inclination with Eclipsing Binaries Determining the inclination is problematic. However, if the inclination is close to 90 o, then the two stars may eclipse each other which will manifest as a time variable change in the brightness 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,  t 1, tells you about the size of the hotter star (that passes behind the cooler star). (v p + v s )  t 1 = 2 r s and, the duration of the primary eclipse,  t 2, tells you about the size of the cooler star; (v p + v s )  t 2 = 2 r p – 2 r s

The temperatures of stars from the depth of the eclipses During the primary eclipse, the hotter star is eclipsed by the cooler star, so the luminosity of the system is lower by an amount equal to the luminosity of the hotter star, L H, so M Bol,p – M Bol,o = 2.5 log [(L H + L c )/ L c ] = 2.5 log [(r H 2 T H 4 + 1)] [ r c 2 T c 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 effective temperatures, 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  M 3

More on L  M 3 L = k M 3 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 = M 3 for masses and luminosities in solar units.

Back to the H-R Diagram Now, we can see that the hotter and more luminous stars are also the most massive. Understanding why is the key to understanding the physics of stars.