dalex@eleazar.dartmouth.edu (Dave Alexander) (01/19/89)
In article <1989Jan18.044744.18328@sq.uucp> msb@sq.com (Mark Brader) writes: > Lagrange found that there are exactly 5 positions that the tertiary > can occupy with respect to the primary and secondary. The three > bodies can be in a straight line: position L1 has the tertiary in the > middle, L2 the secondary, and L3 the primary. Or they can be in an > equilateral triangle, with the tertiary either leading or trailing the > secondary as they move around the primary; these positions are L4 and > L5. ... > The L in each of these positions stands for libration, as a body near > those positions may librate or oscillate around them, and not for > Lagrange. ... > This is as far as I have gotten on my own. Now, I observe that if a > body is placed near one of the L1, L2, L3 points but slightly off the > straight line, then there will be a sideways force on it, so it makes > sense that, as I have read, these are unstable equilibria. So if L1, L2, and L3 are loci of unstable equilibrium, how can we expect an object to librate or oscillate about any of them? -- Dave Alexander -- "Experience has proved that some people indeed know everything." -- Russell Baker
henry@utzoo.uucp (Henry Spencer) (01/21/89)
In article <11854@dartvax.Dartmouth.EDU> dalex@eleazar.dartmouth.edu (Dave Alexander) writes: >So if L1, L2, and L3 are loci of unstable equilibrium, how can we expect >an object to librate or oscillate about any of them? You can't, and it won't, without help. An object *precisely* at one of those points, with *no* perturbations, would stay there, but in the real world, that doesn't work. There are trajectories near those points -- so-called "halo orbits" -- which are *almost* stable, requiring only very slight corrections to maintain. ISEE-3 was in a halo orbit around the Earth-Sun L1 point for some years, to study the solar wind "upwind" of Earth. -- Allegedly heard aboard Mir: "A | Henry Spencer at U of Toronto Zoology toast to comrade Van Allen!!" | uunet!attcan!utzoo!henry henry@zoo.toronto.edu
dalex@eleazar.dartmouth.edu (Dave Alexander) (01/22/89)
In article <1989Jan20.180839.7800@utzoo.uucp> henry@utzoo.uucp (Henry Spencer) writes: > In article <11854@dartvax.Dartmouth.EDU> dalex@eleazar.dartmouth.edu (Dave Alexander) writes: >> In article <1989Jan18.044744.18328@sq.uucp> msb@sq.com (Mark Brader) writes: >>> The L in each of these positions stands for libration, as a body >>> near those positions may librate or oscillate around them, and not >>> for Lagrange. >> So if L1, L2, and L3 are loci of unstable equilibrium, how can we >> expect an object to librate or oscillate about any of them? > You can't, and it won't, without help. An object *precisely* at one > of those points, with *no* perturbations, would stay there, but in the > real world, that doesn't work. I understand that. The question that I was *really* asking was "How can you say that the `L' in L1-5 stands for `libration,' when 60% of such points do not exhibit that behavior?" In light of that, I question Mr. Brader's assertion that the `L' stands for `libration' and not `Lagrange.' -- Dave Alexander -- "Experience has proved that some people indeed know everything." -- Russell Baker
henry@utzoo.uucp (Henry Spencer) (01/23/89)
In article <1989Jan18.044744.18328@sq.uucp> msb@sq.com (Mark Brader) writes: >... I don't even see why L4 and L5 exist. >I hope someone can explain it to me so that I do.... >... now can someone explain to me why the Trojan points are stable >equilibria, or even why they are equilibria at all? The same condition >of sideways force would seem to apply. Clearly (I think) the equilateral- >triangle position cannot be exact; the tertiary would then be in an orbit >exactly like the secondary's despite the secondary's perturbing gravity. >So where are the true L4 and L5 positions? And why? Odd though it sounds, the equilateral triangles are indeed exact. The key observation is that the tertiary is not in an orbit around the primary, it is in an orbit around the center of mass of the primary-secondary system. The secondary's "perturbing" gravity alters the net force vector on the tertiary just enough to point it at the center of mass. That's not a sufficient condition, but it's necessary. Beyond that, one pretty much has to resort to math. In particular, I know of no good intuitive way of explaining why L4 and L5 are stable and L1-3 aren't; the key question is not whether a nearby body feels a side force, but whether a perturbation (of velocity or position) remains bounded (body remains near the point) or grows (more or less) unboundedly. The L1-3 points themselves are unstable; carefully-chosen "orbits" around them are theoretically stable but in practice the conditions are too fussy for true stability; L4 and L5 are honestly stable, with slight gravity wells around them (subject to some conditions, see below). >Also, has anyone investigated situations where the masses are not so >unequal? I remember reading that the Trojan positions are stable if the >some ratio exceeds 27; I think it was the secondary/tertiary mass ratio. Not correct; the key requirement is that the primary/secondary ratio exceed approximately 25. (My favorite astrodynamics books, Archie Roy's "Foundations of Astrodynamics", derives it as a requirement that the ratio of secondary mass to total mass be under 1/2 - sqrt(23/108).) Again, as far as I know, the question "why?" cannot be answered without mathematics. -- Allegedly heard aboard Mir: "A | Henry Spencer at U of Toronto Zoology toast to comrade Van Allen!!" | uunet!attcan!utzoo!henry henry@zoo.toronto.edu
joe@cit-vax.Caltech.Edu (Joe Beckenbach) (01/24/89)
In his article dalex@eleazar.dartmouth.edu (Dave Alexander) writes: >The question that I was *really* asking was "How can >you say that the `L' in L1-5 stands for `libration,' when 60% of such >points do not exhibit that behavior?" In light of that, I question Mr. >Brader's assertion that the `L' stands for `libration' and not >`Lagrange.' I too thought that the 'L' was for the describer of the points, Lagrange. And I do remember a bumper sticker from several years back, in a space-enthusiast-humorous context: "FREE THE LAGRANGE FIVE!!" Whether he had it right, respondent saith not. :-) :-) :-) -- Joe Beckenbach joe@csvax.caltech.edu Caltech 256-80, Pasadena CA 91125 Users I'd like to see (: camera%observer@commsat.mars.solar-system postmaster@link1.L5.edu daemon@probe.titan.solar-system
msb@sq.uucp (Mark Brader) (01/25/89)
Last week I wrote: > I did figure out how to calculate L1, L2, and L3, but I don't even see > why L4 and L5 exist. I hope someone can explain it to me so that I do. I received 6 mail messages, from Bob Ayers, Alan Paeth, Marc Ringuette, and Steve Willner. Both Marc and Steve gave me the critical hint that I had missed, and I was able to complete the derivation. Then before I finished writing this followup, Henry Spencer posted a clearer version of the same hint, and now I can't prove I had it before he helped. :-( It turns out that a completely different approach is needed for the L4 and L5 points from what I used for the others: as Henry said, it is not possible to say, just because the mass of the secondary is much smaller than that of the primary, that you'll treat the whole system as revolving around the center of the primary. Accordingly, let us put the center of the one body at point P, and of another at point S. Somewhere on the line segment between these points is the center of gravity of the two bodies; call that C. Then the ratio SC/PC is equal to the ratio of the masses of the body at P and that at S. Now put the third body at point T, off to one side of the line PS. Choose a point B beyond T on the line ST, such that ST/BT is also equal to that ratio of masses, i.e., ST/BT = SC/PC. Then CTS and PBS are similar triangles, and so TC and BP are parallel. The gravitational force on the body at T will be the resultant of two forces, one toward P (that is in the direction TP), and one toward S (that is in the direction TS, or what is the same thing, direction BT). The ratio of these two forces will be the mass-ratio of P and S divided by the square of the distance-ratio from T to P and to S. That is, it will be (ST/BT)(ST/TP)2, where 2 denotes squared as in my earlier article. Now, so far nothing has been assumed about the relative distances of the bodies. Now add to the above considerations one more assumption: that T is equidistant from P and S. Replacing ST by TP in the above formula gives the ratio of forces as simply TP/BT. But the forces are respectively in the directions TP and BT. Then the resultant force must be in the direction BP, which, as shown above, is also the direction TC. That is, if one body is equidistant from two others, then the resultant gravitational force on it is directly toward the center of mass of the two others. But the center of mass of a three-body system is on a line between any one of the bodies and the center of mass of the other two. Therefore, in a three-body system, if two of the bodies are equidistant from a third, then the net force on the third is also directly toward the center of mass of the system. And so if all three bodies are equidistant -- forming an equilateral triangle -- then the net force on each one is directly toward the center of mass of the system, and so, no "sideways force" exists to move the bodies out of the equilateral configuration. And notice that this is true no matter how equal or unequal the masses are. Now, this does NOT prove that an equilibrium exists. It only shows that IF there is an equilibrium of the kind described, then it is exactly the equilateral triangle formation. For an equilibrium to exist, it must also be true that when the three bodies are set revolving around their common center of gravity, each has the same orbital period. I'll leave the proof of this as an exercise for some reader more energetic than I am. I'm content to see it proved that the equilateral triangle is in fact an exact one. A second exercise is to correct the equations I gave in my earlier article to allow for the fact that the orbits are really around the center of mass; at least allow for the secondary's contribution to that. In particular, consider that I computed the Earth-Sun L3 point as being 46.5 miles off the Earth's orbit (treated as circular). The center of mass of the Earth-Sun system is of course 280 miles from the center of the Sun. How does this affect the L3 point? I'm getting lazy. You do it. Now for a couple of other details. I also wrote: > Since r is 93,000,000 miles, the L1 and L2 points are each 936,000 miles > from the Earth. ... Since the ratio of the Sun's and Earth's > diameters is about 108, the Earth exactly eclipses the Sun's disk at > a point only 861,000 miles beyond the Earth ... It's a curious > coincidence how close L2 is to the exact-eclipse distance. One of my correspondents suggested that that was no coincidence, but happens because the Sun and Earth have similar densities. Actually they don't, but he was close. If the Earth was exactly 3 times as dense as the Sun, then the L2 point would be at the exact-eclipse distance... or almost exactly so, anyway. This number 3 is the factor of 3 in my equation [10] last week, but that equation was only approximate. And finally, Dave Alexander wondered how I could justify saying that > The L in each of these positions stands for libration, as a body > near those positions may librate or oscillate around them, and not > for Lagrange. ... when 3 of the 5 positions are unstable. The answer is that this is what I have read in books that have described the points. I don't remember which ones offhand, though one was probably Clarke's "The Promise of Space". Perhaps Lagrange named the points himself and was being modest but wanted to keep his initial in there! In any case, Henry tells us that the unstable libration points have some (barely) stable orbits around them, and a body in such an orbit might be considered to be librating, so the name is not really all that bad. One of the messages from Alan Paeth pointed out that the gravity wells around the L4 and L5 points have a curious comma-like shape, and they can even extend around the "back" of the primary and join up! So a body orbiting near one of these points need not stay near it all the time to remain captured. My thanks to all of my correspondents. It was fun. Mark Brader, Toronto "The singular of 'data' is not 'anecdote.'" utzoo!sq!msb, msb@sq.com -- Jeff Goldberg
dsmith@hplabsb.HP.COM (David Smith) (01/27/89)
In article <11854@dartvax.Dartmouth.EDU> dalex@eleazar.dartmouth.edu (Dave Alexander) writes: >So if L1, L2, and L3 are loci of unstable equilibrium, how can we expect >an object to librate or oscillate about any of them? I didn't work through the mathematics, but: In an article on ISEE, Aviation Week pointed out that L1 was at a saddle point. The spacecraft was attracted to the Earth-Sun line, orbiting it. In the direction of the line, position was unstable, but quite manageable to the fuel budget. -- David R. Smith, HP Labs dsmith@hplabs.hp.com (415) 857-7898