Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/5/84; site ssc-vax.UUCP Path: utzoo!watmath!clyde!burl!ulysses!mhuxr!mhuxj!houxm!vax135!cornell!uw-beaver!ssc-vax!eder From: eder@ssc-vax.UUCP (Dani Eder) Newsgroups: net.space,net.astro Subject: Re: L5 Message-ID: <356@ssc-vax.UUCP> Date: Mon, 21-Jan-85 13:53:51 EST Article-I.D.: ssc-vax.356 Posted: Mon Jan 21 13:53:51 1985 Date-Received: Thu, 24-Jan-85 05:53:21 EST References: <79@drivax.UUCP> Distribution: net Organization: Boeing Aerospace Co., Seattle, WA Lines: 33 Xref: watmath net.space:3529 net.astro:465 > A bunch of us have heard of the LeGrange points in space, and with > a minimum of effort can understand the stability of L1, L2, and L3. > However, we lack an adequate explaination of the stability of the > remaining two. Could someone help us out? > I'll give it a try. A stable point is where there is no acceleration relative to the Earth and Moon. This requires that forces be balanced so there is no net force to produce an acceleration. In the case of the L2 point this is easy to see, since it is between the bodies, there is somewhere where the attractions are equal and opposite. The complication that throws most folks off is the fact that we have a dynamic rotating system. The Earth and Moon orbit about the common center of gravity of the system. This introduces a fictional 'centrifugal' force when we switch from a non-rotating frame of reference to a rotating frame centered on the center of gravity. A piece of paper and sketch will help here. Put the Earth and Moon down, and mark a point near the earth along the Earth-Moon (E-M) line as the center of gravity C. The actual location is about 1000 mi below the Earth's surface, since the Earth is so much more massive, but put the point further away for clarity. Now, a spacecraft at the L5 point is orbiting the common center of gravity C, and in our rotating frame of reference sees a centrifugal force pointing from C to L5. The attraction of the Earth and Moon combined acts as if it comes from the center of gravity (that's the definition of center of gravity), and hence points from L5 to C. Now we have two opposing vectors, and it remains to find the distance at which they are equal. The calculation is complex, but it turns out to be at the equilateral point. Dani Eder / Boeing / ssc-vax!eder / Ad Astra! (To the Stars!)