Newsgroups: sci.astro
Subject: Re: Binary stars, lifezones. HELP!
From: bankst@kauri.vuw.ac.nz (Timothy Banks)
Date: Fri, 4 Sep 1992 01:36:54 GMT
Organization: Victoria University of Wellington

>Scott writes:
>
>I have been asked the following questions and am appealing to my 
>fellow netters who may be more familiar with the field.  A caller to 
>the planetarium has asked how one might determine the separations of 
>stars in a binary (or multiple) star system if you know the masses.  I 
>recall something along the conservation of angular momentum giving the 
>ratio of the masses proportional to the ration of the individual 
>distances and the sum of the masses proportional to the total 
>separation(?).  But as I recall, one also needs the inclination angle 
>of the system to the Earth-system line (if I'm not way off base here), 
>which he does not seem to have.  Are there approximations that can be 
>used and a set of equations one can pursue?  The non-mathematical 
>astronomy texts we have here at the planetarium are no help in this 
>regard, and I will gladly hunt any references provided.
>


OK, firstly lets assume a circular orbit of radius r and period T. The
second object is orbiting the larger first one at speed v resulting in
the centripetal acceleration

	a = v^2 / r = ( 4 pi^2 r ) / T^2

Assuming both are point masses, the force exerted by the first body on
the second is towards the first, and of magnitude

	( G M1 M2 ) / r^2

where G is the gravitational constant. M1 and M2 are the masses of the two
bodies. The acceleration of the second body, by Newton's second, is then

	a = F / M2 = G M1 / r^2

which leads to

	a = ( G M1 / r^2 ) = v^2 /r = 4 pi^2 r / T^2

that rearranges to

	T^2 = r^3 ( 4 pi^2 / G M1 )

While this has been derived for circular orbits, it will also apply to any
elliptical orbit if we treat r as the mean distance of the bodies apart. This
wil work fine for things like satellites and the planets. So, if we know the
period and the mass, we can get the distance.

We've cheated a little in the above, in that we assumed that the mass of
the first object is large compared to the second. Lets now have them
orbiting about their common centre of mass properly, i.e

	T^2 = r^3 ( 4 pi^2 / G (M1 + M2 ) ) 		eqn *

(lets continue to assume lotsa things!)


If you now look at _Astronomy - the Structure of the Universe_ by
A.E. Roy and D. Clarke (Adam Hilger, 1977, recently reprinted though)
on page 118 they do the following:

	The corresponding formula for the Earth's orbit about the Sun is

	T(e) = 2 pi ( sqrt ( R^3 / G (M(s) + M(e) ) ) )		eqn **

	which is simply equation * above rearranged with T(e) the year,
	M(s) the Sun's mass, and M(e) the Earth's mass. 

	R is the radius of our orbit. They then put T
	into units of a year, and consider that the Sun's mass is much
	greater than the Earth's. This causes the equation immediately
	above to reduce to

	1 = 2 pi ( sqrt ( R^3 / ( G M(s) ) ) )

	which rearranges to

	sqrt (G) = 2 pi ( sqrt ( R^3 / M(s) ) )

	they substitute this into * giving

	T = sqrt ( ( a^3 M(s) ) / ( (M1 + M2) R^3 ) )

	or
	
	M1 + M2 = ( r / R )^3 ( M(s) / T^2 )

	If we let the solar mass equal unity then we have

	M1 + M2 = ( r / R )^3 T^(-2)

	allowing us to determine the separation if we know the masses and
	the orbital period.

But how do we get to know the masses? Roy and Clarke then go to explain
the basics of light curve analysis for eclipsing binaries, and radial
velocity curves for spectroscopic binaries. As they point out the first
will give you the luminosity ratio, radii ratio relative to the orbit, 
the system's inclination, their shape, orbital eccentricity, and maybe
limb darkening (though you tend to assume an appropriate value. Binnendijk
explains why it is hard to get good limb darkening values). The second gives
you the product of the sine of the orbital inclination and the major axis
of the orbit, eccentricity, and the product of each mass and the sine of the
inclination, i.e

	( r sin (i) )^3 / T^2 = ( M2 sin i )^3 / ( M1 + M2 ) ^2

Hence, if you can get the inclination off the light curve then you can
obtain the absolute masses and radii of the components.

Incidentally, on the question raised by another netter on obtaining the
relative radii from a light curve, I think Forbes explained it well in his
1991 thesis  for a first approximation model:

	Consider a light curve with a limb (outside eclipse) intensity
	of B. b1 and b2 are the primary and secondary eclipse "peak"
	intensities. If the maximum surface area presented to an 
	observer is s1,s2 with flux f1,f2 for each component of the
	system then

		s1f1 + s2f2 = kB

	where k is a constant relating the system's luminosity from
	the stellar surfaces s1 and s2 to that seen by the observer.
	During the primary eclipse a fraction alpha (from which Kopal's
	alpha integrals come) of the first star's surface s1 is hidden
	by the secondary star, so

		( 1 - alpha ) s1 f1 + s2 f2 = k b1

	and for the secondary eclispe you have

		s1 f1 + ( s2 - alpha s1 )f2 = k b2

	Combining the above equations gives

		alpha s1 f1 = k b1
		alpha s2 f2 = k b2

	and so

		f1 / f2 = b1 / b2

	Rearranging the above will give

		s1 / s2 = b2 / ( alpha B - b1 )

	and as S1,2 is approx r1,2^2, then ratio of the radii will be
	
		r1 / r2 = sqrt ( b2 / ( alpha B - b1 ) )

	and as the flux is the luminosity per unit surface area L approx 
	F r^2 the luminosity ratio is

		L1/ L2 = b1 / ( alpha B - b1 )

-- 
  Timothy Banks, Physics Department, Victoria University of Wellington, NZ.
       bankst@kauri|rata|matai.vuw.ac.nz, banks@beagle.phys.vuw.ac.nz.
    "He's dead, Jim!" "OK, you take the tricorder, I'll take the wallet!"