[TML] Planets in binary systems

[Jerry W Barrington]

On 3/3/08 1:43 AM, "Timothy Little" <tim at little-possums.net> wrote:

> 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.

Well, I couldn't quite follow some of your substitutions.  You said "R = (1
- u) r1, so that u is the ratio between L1 distance from the smaller body
and that body's orbital radius", but that seems to call for R=u*r1.

But I did find a couple web pages using the formula
    (1-z)^2-kz^2-z^2(1-z)^2((1+k)z-k)=0
where k=m1/m2 and z=R/(r1+r2) [in your units].

So I put that formula into an Excel spreadsheet, with spots to put m1 and m2
and (r1+r2).  Using Goal Seek tool to get the formula (close) to 0 by
changing z gives me a result.  Since Goal Seek seems have a rather arbitrary
precision, I found that multiplying the entire formula by 100,000 made it
find the correct value for equal masses, .50..., instead of 0.50017933232343
or 0.499999986081074.

Putting in Earth and Sun masses and distance, it gets 1,491,555 km (L2 is
listed as 1.5 Gm, and approximately equal to L1, so that sounds good).  The
Hill sphere formula give 1,496,558 km, which is only 5000 more (1:300
parts).

For Earth/Moon, it gives 58,019 km from the Moon.  Hill formula gives
61,524, so that's less than 6% off.

I suspect since the Hill formula is only an approximation anyway, my values
are better.  And they're smaller, and I'm only going to use 2/3 to 1/2 of
*that* value, so it's all good.  :)