[TML] Planets in binary systems

[Timothy Little]

On Sun, Mar 02, 2008 at 01:49:34PM -0500, Jerry W Barrington wrote:
>  I've pretty much concluded from my research that L1 will always be
> closer than L2 or L3, so just being able to find that one would be
> sufficient.  Since it's to be done by computer, I don't mind the
> math being a little ugly.

OK :-)

The L1/2/3 points are where the centripetal acceleration at a given
angular velocity is equal to the gravity from the two bodies.

So let r1,r2 be the orbital radii of the stars from the barycenter,
m1,m2 be their (known) masses, w be the fixed angular velocity of the
system, and R be the (unknown) radius of the L1 point from the
barycenter.  Label the bodies so that m1 < m2 (just so we know which
side of the barycenter L1 is on).

w^2 r1 = G m2 / (r1 + r2)^2
w^2 r2 = G m1 / (r1 + r2)^2
w^2 R = G m2 / (r2 + R)^2 - G m1 / (r1 - R)^2

The first two equations allow us to determine that r2/r1 = m1/m2, and
to make the math easier, let k = 1 + m1/m2.  Then G m2 = w^2 r1^3 k^2
and G m1 = w^2 r1^3 k^2 (k-1).

We're only interested in ratios of distance, so let's make the
substitution R = (1 - u) r1, so that u is the ratio between L1
distance from the smaller body and that body's orbital radius.
Then simplifying the last equation:

1 - u = k^2 / (k - u)^2 - k^2 (k - 1) / u^2

This can be multiplied out into a polynomial (so you don't have to
worry about dividing by zero):

k^2 (u^2 - (k - 1) (k - u)^2) - (1 - u) u^2 (k - u)^2 = 0

Unfortunately degree-5 polynomials generally don't have any formula
for exact solution in terms of common functions.  For any given value
of k it should be fairly easy to solve numerically for u, though.


- Tim