Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site utastro.UUCP Path: utzoo!linus!philabs!seismo!ut-sally!ut-ngp!utastro!bill From: bill@utastro.UUCP (William H. Jefferys) Newsgroups: net.physics Subject: Re: Stable Grav. points (reply to Roger Noe) Message-ID: <776@utastro.UUCP> Date: Tue, 1-Nov-83 15:58:12 EST Article-I.D.: utastro.776 Posted: Tue Nov 1 15:58:12 1983 Date-Received: Fri, 4-Nov-83 03:22:47 EST Organization: UTexas Astronomy Dept., Austin, Texas Lines: 156 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. -- Roger Noe ...ihnp4!ihlts!rjnoe ************************************************************************ Roger, I am sorry, but you are wrong. I do know what I am talking about, since I did my dissertation on the restricted problem of three bodies and among my publications are approximately 20 scholarly papers on it. I teach the subject nearly every year at the upper-division or graduate level. But you (and the rest on the net) have a right to demand more than my credentials as evidence that what I say is correct, so let me state my case. Possibly you are just confused by the notation. Depending upon whom you read, the terms L1, L2 and L3 can refer to any of the colinear libration points. Only one of them is between the two masses (and therefore can be at the barycenter). Let us call it L2, and let L1 be the one on the far side of the Moon, and L3 the one on the far side of the Earth (away from the Moon). Therefore it doesn't make sense to say as you do that L1 is at the center of mass iff the masses are equal, and then to say that L2 is always at the barycenter (center of mass). I say that only L2 *can* be at the barycenter, and then only in the case that the masses are equal. You are right that the Earth-Moon barycenter is about 1700 km beneath the Earth's surface. This in itself should be sufficient to convince you that it can't be a libration point. Remember, the L2 point is a place where it possible to place a test mass and have it remain in place (relative to the two massive bodies). If one were closer to the Moon than the L2 point (and still on the line joining the Earth and Moon), one would be accelerated towards the Moon. But that means that every time the Moon was overhead, we all ought to fly off the Earth and be accelerated to the Moon, which doesn't happen. Roger might object, that we are rotating with the Earth. I say that if he were correct, then to launch a plane into space, it would be sufficient merely to fly it at exactly the speed required to compensate for the Earth's rotation (about 1000 mph at the equator), in such a way as to keep the Moon directly overhead. Since he says that we are closer to the Moon than the L2 point, by his reckoning, the plane should be increasingly accelerated towards the Moon. That would be a whole lot cheaper than building a shuttle, wouldn't it? Unfortunately, it won't work, because the L2 point is nearer the Moon than the Earth. Now for the mathematics. I am copying from Peter van de Kamp's *Elements of Astrodynamics* (Freeman, 1964) with notational changes to make him conform to my choice of L1, L2 and L3. At the libration point L2 which is between the masses, the *gravitational* acceleration on the body is given by f = A*mu/(B + x2)**2 - B*mu/(A - x2)**2, where mu = G*(M + m), A = M/(M + m), the distance of the Moon to the barycenter, B = m/(M + m), the distance of the Earth center to the barycenter, G = constant of gravity, M = mass of Earth, m = mass of Moon. Units of distance have been chosen so that the distance from Earth to Moon, A + B = 1. The *centripetal* acceleration at L2 is given by fc = v**2/x2 where v = x2*V and V is the relative [angular] velocity of m and M for which the general expression for circular velocity takes the form V**2 = mu. [Kepler's third law for unit distance] Hence fc = x2*mu. The equilibrium contition is the equality of the gravitational and centripetal accelerations. Equating the two expressions yields A/(B+x2)**2 - B/(A-x2)**2 = x2. The total mass, and hence, mu, does not, of course, figure in this relation. The locations of the other two libration points, L1 and L3 follow from the equations A/(x1+B)**2 + B/(x1-A)**2 = x1, A/(x3-B)**2 + B/(x3+A)**2 = x3. Here, x1 and x3 are the distances from the center of mass of M and m to the libration points L1 and L3 respectively. [End of extract from van de Kamp.] Note the similarity of these equations. All are fifth order polynomials. Roger claims that the L2 point is at the barycenter, so that X2 = 0. If he were right, then for any A and B such that A>0, B>0, A+B=1, the relationship A/B**2 - B/A**2 = 0 would be satisfied. I claim on the contrary that the only solution to this equation under the stated circumstances is A = B = 0.5. For B<<1, A is almost 1 and we have (approximately) 1/x2**2 - B/(1-x2)**2 = x2. Let u = 1-x2, then this becomes (assuming u small, as is the case) 1 - B/u**2 = 1 - u or u**3 = B. Therefore, for small B the distance from the center of the Moon to L2 is of the order of the cube root of B. For the Earth-Moon system, this means that the L2 point is about 90000 km from the Moon, less than 25% of the distance between the Earth and Moon. I hope that this detailed explanation will clear this up. PLEASE FOLKS, I provided one reference in my previous posting and another one here. Both of them are by acknowledged experts in the field (one of the authors of the Astronomical Journal reference, Victor Szebehely, is "Mr. Restricted Problem", author of a monumental treatise on the subject and the winner of the American Astronomical Society's Brouwer Prize for his work in this field.) Please read them carefully before posting any more claims to the net that the barycenter is a libration point (except for the equal mass case). After doing so, if you have a *proof* that Celestial Mechanicians have been wrong about this for the past 200 years, post it and we will be happy to read it. Bill Jefferys 8-% Astronomy Dept, University of Texas, Austin TX 78712 (Snail) {ihnp4,kpno,ctvax}!ut-sally!utastro!bill (uucp) utastro!bill@utexas-20 (ARPANET) s