Transcript Stars: Binary Systems

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=rw, the
least massive star orbits faster (since w =
2p/P is the same for both stars.)
Kepler’s
rd
3
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 = r 1 + r2
Kepler’s
rd
3
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 = 4p2 a3/ G(m1 +m2)
Question: Prove Kepler’s 3rd Law.
The Sun-Earth System
P2 = 4p2 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 = 4p2 13/ G Mo(1 + Mearth/ Mo)
But since Mearth/ Mo ~ 0, we obtain the result that the constant
4p2 / 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 know
the distance to get the separation, a.
Recall that, d(pc) = 1/P, where P 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, P, in arc seconds,
is equal to the linear separation, a, in units of pc, so that
a(AU) = a`` (arc sec)/ P ( arc sec)
Question: Prove a (AU) = a`` (arc sec)/ P ( 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)/ P ( arc sec)
To obtain,
P2 =
[ a`` ]3
where a and P are now in arc sec
_____________________
[ P]3 (m1 +m2)
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 P (``) which when applied
to 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 some
geometry to correct for projection effects.
More on Spectroscopic Binaries
Doppler Effect
The wavelength, (l – lo), 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,
Dl/lo = (l – lo)/lo = vr/c
Where l is the observed wavelength, lo is the rest (zero velocity)
wavelength, vr 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 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 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
vobs = vr 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 = 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’s
cancel. But only a lower limit on the masses; m sin3i.
Getting a handle on the
inclination with Eclipsing
Binaries
Determining the inclination is problematic. However, if the
inclination is close to 90o, 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, 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 cooler
star, so the luminosity of the system is lower by an amount
equal to the luminosity of the hotter star, LH, so
MBol,p – MBol,o = 2.5 log [(LH + Lc)/ Lc]
= 2.5 log [(rH2TH4 + 1)]
[ rc2 Tc4
]
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 a M3
More on L a 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 the
hotter and more luminous
stars are also the most massive.
Understanding why is the
key to understanding the
physics of stars.