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From: rjnoe@ihlts.UUCP
Newsgroups: net.physics
Subject: Re: Stable Grav. points
Message-ID: <251@ihlts.UUCP>
Date: Mon, 31-Oct-83 09:44:58 EST
Article-I.D.: ihlts.251
Posted: Mon Oct 31 09:44:58 1983
Date-Received: Thu, 3-Nov-83 04:02:04 EST
References: <764@utastro.UUCP>
Organization: AT&T Bell Labs, Naperville, Il
Lines: 26

Yes, the center of mass of a two-body system is Lagrange libration point
L1 if and only if the masses are equal.  But the barycenter in any
event is L2.  Let m1 >= m2, u = m2/(m1+m2), and let the constant separation
of the masses be 1.  Choose cartesian coordinates such that the masses are
always in the x-y plane and let this be a rotating reference frame so that
m1 is always on the +X axis and m2 is always on the -X axis.  Remembering
that 0 < u <= .5, m1 is at (u,0) and m2 is at (u-1,0).  The Lagrange libration
points are:
	L1		(x1, 0)
	L2		(0, 0)
	L3		(x3, 0)
	L4		(u-.5, +sqrt(3)/2)
	L5		(u-.5, -sqrt(3)/2)
where x1 and x3 (x1 < u-1 < 0 < u < x3) are roots of a fifth order polynomial
with coefficients linear in u.  x3 is easy to numerically approximate but x1
is difficult for small u.  It doesn't really matter, since L1, L2, and L3 are
always unstable libration points.  L4 and L5 are stable iff 27u(1-u) < 1, i.e.
iff u < 0.03852 (approximately).  For the earth-moon system we have u = 0.01215
(approximately), so L4 and L5 are the stable Lagrange libration points, located
equidistant from both the earth and moon (centers), which is also the earth -
moon separation (these are equilateral triangles m1, m2, L4, and L5 make).  L1
is 64.5e3 km from the moon's center, L3 is 381.7e3 km from the earth's center,
and the barycenter L2 is only 4.7e3 km from the earth's center (toward the
moon, this time) or about 1700 km beneath the earth's surface.
-- 
		Roger Noe		...ihnp4!ihlts!rjnoe