# Talk:Lagrangian point

## How about an explanation for the common man?

This article is clear as mud. — Preceding unsigned comment added by 158.61.0.254 (talk) 21:37, 18 July 2012 (UTC)

I think that at least the first sentence should be changed. "The Lagrangian points are the five positions in an orbital configuration where a small object affected only by gravity can theoretically be part of a constant-shape pattern with two larger objects." is horrible. The problem is the "be part of a constant-shape pattern with two larger objects" bit. This might be the best general description of the phenomenon but it's difficult to understand. It should either be replaced with something that doesn't include the term constant-shape pattern, or immediately afterword there should be a paragraph explaining it. I would suggest something like: "The speed at which an object orbits around another is related to how close it is. The closer the two objects get, the faster the orbiting object must be. This means that distance between two objects orbiting the same celestial body at different altitudes can not remain the same with the exception of five points provided one of the orbiting objects is of negligible mass. At these five Lagrange points the gravitational attraction of the larger orbiting object on the object of negligible mass either reduces or increases the speed required to orbit the celestial body at a certain altitude by exactly the amount needed so the smaller object can have the same orbital period as the larger orbiting object at a different altitude." Now I realize that of course all three objects are orbiting a common center but to include that would make it unclear. Also I might be a bit wrong because I'm not a physicist. Especially suspect is the use of equidistance as a goal, which is probably only valid for circular orbits (I don't know if that is the case). If it isn't valid for elliptical orbits, perhaps relative angles should be used or circular orbits should be specified. I came here hoping to understand the concept and then clicked back and went to the ESA webpage on the subject because this article starts too abstract and then just gets very technical. I haven't made the modifications directly, because I could be partially or totally wrong. I would kindly request that someone with more knowledge of orbital mechanics check my paragraph and include it or a modified version of it. Kaanatakan (talk) 16:13, 22 October 2013 (UTC)

## Add M1 M2 to diagram

The article discussed M1 M2 etc., but these symbols do not appear in the diagram. One would have to read the whole article to discover the definition of M1 and M2. This is quite frustrating. —Preceding unsigned comment added by 75.53.54.121 (talk) 18:32, 28 February 2011 (UTC)

## Exact position of L3, plus minor amendments

#### Beginning

The L-points are not necessarily in interplanetary space.

Corrected to "in orbital configuration".
> OK ("in an orbital ..." ??)

#### History and concepts

His name was hyphenated : Joseph-Louis Lagrange.

Good catch.

It has "It took hundreds of years before his mathematical theory was observed". His theory was published around 1772; Trojans were observed around 1905. Thet's not "hundreds of years" later.

"Over a hundred years"?
> OK

#### Diagrams

The first, "... contour plot ...", diagram shows Earth, L3, L4 & L5 on a Sun-centred circle, and L1 & L2 reasonably close to Earth. That's satisfactory.

Actually, it shows L3 just outside the circle. It may not be all that clear.
> Agreed, agreed.
The problem is that the contour plot clearly shows a system where the ratio of masses primary:secondary is of the order of 10:1-50:1. In that case L3, L4 and L5 will be visibly off the secondary's orbit. I recreated the diagram here. –EdC 00:36, 5 February 2007 (UTC)

The second, "... far more massive ...", diagram shows L3 outside the circle. But if Earth, L4 & L5 all lie (as far as can be seen) on a primary-centred circle, then L3 should be similarly on that circle; not outside it.

The circle should be the orbital path of the secondary (centred on the barycentre), in which case L3 lies outside it. The diagrams should be fixed by moving the primary away from the barycentre, and L4 and L5 outside the circle.
> Doubt. Could be better to have "very much more massive" with Moon L3 L4 L5 on a circle centred on Earth, and L1 L2 very near Moon, AND also "considerably more massive" with everything properly shown. If the latter is a bit bigger, it will serve also for the L4 L5 geometrical srgument.
Yes, that could work. In that case the "very much more massive" diagram should have L1 and L2 pulled in as far as practicable. –EdC 00:36, 5 February 2007 (UTC)
Done, now we need to decide how to fix the contour plot. –EdC 02:16, 5 February 2007 (UTC)

The blue triangles (showing the gradient to be downhill going away from the points) indicate that L4 and L5 are unstable equilibria, whereas they are actually stable equilibria. —Preceding unsigned comment added by 208.71.200.91 (talk) 05:26, 24 February 2010 (UTC)

Yes L4 and L5 are the stable equilibria. But they really are the maxima of the pseudopotential; it takes the Coriolis force to keep objects from falling away from those points.
—WWoods (talk) 09:21, 24 February 2010 (UTC)
The triangles are equilateral and do not act effectively as arrows. It is therefore difficult if not impossible to be certain which direction they are intended to be pointing.
—thereaverofdarkness (talk) 20:47, 10 August 2015 (UTC)

#### Section "L3"

The page says : "L3 in the Sun-Earth system exists on the opposite side of the Sun, a little farther away from the Sun than the Earth is" - my italics. That wording will naturally be taken as saying that L3 is further from the centre of the Sun than the centre of the Earth is.

The better calculations measure distances from the barycentre, and show that L3 is a little further from the barycentre than the centre of the Earth is. But it seems that L3 is a little nearer to the centre of the Sun than the centre of the Earth is.

Hm. Yes, it is, isn't it?
> Not a lot of people know that, though.
Fixed - I hope. –EdC 01:05, 5 February 2007 (UTC)

New point: the article says "Example: L3 in the Sun–Earth system exists on the opposite side of the Sun, a little outside the Earth's orbit but slightly closer to the Sun than the Earth is." But how can it be OUTSIDE the Earth's orbit but CLOSER to the sun??

Because the Sun also orbits the barycenter – hence the Sun is closer to the far side of the Earth's orbit (if we ignore eccentricity, perturbation from other planets, and possibly a bunch of other things I forgot). :) — the Sidhekin (talk) 21:48, 24 May 2008 (UTC)

## Contradiction

The issue of whether L4 and L5 are stable has been raised several times above. Could this be sorted out please. At the moment there's a straight contradiction between the contour plot, showing blue arrows leading "downhill" from L4 and L5, and the statement "the triangular points (L4 and L5) are stable equilibria ...". I can't fix it myself because I don't know which one is right. Occultations (talk) 11:37, 24 February 2008 (UTC)

I came to this Talk page for the same reason. L4 and L5 would have to be low points in the potential in order for them to be stable (according to this NASA page) and have objects orbit those points. The fact that it's stable only if conditions regarding the M1/M2 ratio should come later as, presumably, the "islands" of stability just get smaller as the ratio decreases until they eventually disappear. bcwhite (talk) 12:09, 31 May 2012 (UTC)
The points are dynamically unstable (an object perturbed from L4 will continue to move away) but form stable equilibria (Coriolis force will curve an object's path back to L4). EdC (talk) 15:46, 24 February 2008 (UTC)
The plot's caption used to say, "Counterintuitively, the L4 and L5 points are the high points of the potential." Maybe something about the stability of L4 and L5 should be added to the introduction?
—WWoods (talk) 17:23, 24 February 2008 (UTC)
Stability of L4 & L5 depends on the mass ratio of the primary and secondary bodies. For Earth/Moon they have been analytically proved to be stable (ca ~1980 I think; in the sense that a small test mass near either L4 or L5, moving at low speed with respect to the Lagrange point, will remain in its vicinity), and I believe this result also applies to Sun/Earth & Sun/Jupiter, as ratio m1/m2 is even larger (~80 for Earth/Moon). However, the stability depends on the Coriolis force, as an object a small distance from the L4 (say) point will at first move away, but then move back towards it, looping I think in a rosette sort of path. Keith Symon's old textbook Mechanics (2nd edition, 1960, Addison Wesley) discusses this problem fairly extensively at an advanced undergraduate level, although in 1960 the long-term question was still open. I am not certain that the Earth/Moon proof actually applies to the real physical situation, with a somewhat (~0.05) eccentric orbit, perturbed by the Sun, but I believe long accurate numerical integrations suggest they are stable. Wwheaton (talk) 16:11, 6 May 2010 (UTC)

I'm still confused. This article doesn't mention "Coriolis" anywhere, yet that appears to be critical to stability. Obviously, L4 and L5 look unstable in terms of effective potential. Understanding that earth is orbiting the sun and therefore that L4 and L5 are orbiting the sun, I can see the potential for orbiting-like behavior around those points, but I haven't worked out the details. Also, I've never seen "effective potential" before. It looks like effective potential drops off at a distance as centrifugal force takes over. In a two-body problem, I think the effective potential for a given angular momentum would be an annular trough, the bottom of which would indicate the the radius of the stable circular orbit (and the trough shape indicating stability).
I think this article needs a description of the effects of perturbation of a mass near all Lagrange points, particularly L4 and L5. I'd like to see something like this: "Suppose a small object is placed at L4 with the same angular velocity as earth. Were earth not there, it would be in orbit around the sun, but with earth there, it slows down (falls back toward the earth), causing it to drop to a lower orbit, speed up angularly, move forward (away from the earth), etc., orbiting L4." Is that even right? —Ben FrantzDale (talk) 14:49, 4 March 2015 (UTC)

All we need is a simple analogy. L4 and L5 are stable. Like a marble in the centre of a valley. If it is moved away from the centre a tiny bit, it rolls back to the centre. The other points are unstable. Like a marble on the tiny flat spot on the top of a hill. If it is moved away, even a tiny bit, it will then roll away from the hill. — Preceding unsigned comment added by 121.212.53.120 (talk) 03:47, 1 February 2018 (UTC)

I added the following clarification to the stability section, which I'm hoping resolves this issue: "Although the L4 and L5 points are found at the top of a "hill", as in the effective potential contour plot above, they are nonetheless stable, as such a diagram ignores Coriolis acceleration (which depends on the velocity of an orbiting object and cannot be modeled as a contour map)." expensivehat (talk) 22:11, 5 September 2018 (UTC)

## Changes made to the Intuitive Explanation

The Intuitive Explanation as presented was simply wrong. The outward force sensed by the hand twirling the string is a real physical force, namely the tension. It is not the centrifugal force. Further, the fact that when released the revolving mass travels on a straight tangential trajectory has nothing to do with the centrifugal force being 'fictitious'. When the string is cut, there is no longer any centrifugal nor centripetal force, so we are in an entirely new dynamical situation which is unrelated to the previous state. In addition to fixing these problems I have made the section less verbose and replaced the word 'weight' (which actually means 'gravity force') by a specific item, the stone. PlantTrees (talk) 20:26, 12 March 2008 (UTC)

### Possibly related question

Is there a stable orbit which runs through L4 L5, but into the 3rd dimension in the diagram? Or any pair of L1 L2 L3 ? Or any other pair? Just a mental rambling on my part, L4 L5 looks fairly obvious, but I am not able to do the math. —Preceding unsigned comment added by 78.32.144.39 (talk) 18:26, 10 April 2009 (UTC)

If you are still here 9 years later, then Yes and Yes. There are a number of excellent YouTube videos that show actual orbits of the Trojan (at the L4 & L5 points) and Hilda (at the L3 point) asteroids around Jupiter. It shows them in a sort of a "perspective" projection, which means you get some idea of "vertical" motion. You can see all combinations. Some asteroids stay at L4 and L5, some move between L4 and L5, some move through all three points. Be aware that the "easiest to understand" visualisations are shown as rotating with Jupiter, so Jupiter itself appears stationary. Search YouTube for "Trojan asteroids". As part of "background", these usually show the normal asteroid belt in green. — Preceding unsigned comment added by 58.166.224.167 (talk) 00:19, 3 March 2018 (UTC)

Here it is: https://www.youtube.com/watch?v=yt1qPCiOq-8 — Preceding unsigned comment added by 121.212.147.109 (talk) 03:26, 3 March 2018 (UTC)

## History and concepts

In the final paragraph of the section, the word "area" appears inappropriate. It seems to me that, in rotating co-ordinates, the system of two bodies and five points maintains a constant shape, varying in size. Then, at any instant, each L-point has a point location, which moves in and out in rotating co-ordinates and on an elliptical track in fixed co-ordinates. 82.163.24.100 (talk) 16:40, 23 June 2008 (UTC)

http://www.merlyn.demon.co.uk/gravity5.htm can now show the Lagrange points for elliptical orbits as seen by a co-rotating observer, who sees that each of the five points moves in a straight line. The observer's angular velocity is of course not constant. The path traced out by the points, in these coordinates, does not enclose an area. 94.30.84.71 (talk) 22:37, 12 June 2011 (UTC)

## Proposal to add section "Proposed objects"

It might worth noting somewhere the proposals (Space sunshade and Solar shade) to place a large "sunshade" at the L1 point to counteract the effects of global warming. (Give these proposals some though before dismissing them out-of-hand.) There are sections for existing and fictional objects at the Lx points, but there is no section for proposed objects. 220.76.15.253 (talk) 18:12, 27 October 2008 (UTC)

## Earth-Moon L2 Point

The article doesn't mention go into detail on the Lagrangian point on the other side of the moon; for example this orbit is an option for Constellation Program Moon relay satellites. MithrasPriest (talk) 18:46, 5 November 2008 (UTC)

## Rubber sheet analogy

If the contour map were replaced with a rubber-sheet visualization (i.e., the kind that are popular for depicting gravitational distortions of space-time), would it be accurate to say that the L-points are the highest points on the ridges between the astronomical bodies? If that's the case, it would be fairly obvious that these locations are not stable, and that if an obejct at such a point is nudged away from that point, it will "fall into" the gravity well of one of the bodies, right? | Loadmaster (talk) 17:43, 3 December 2008 (UTC)

L-4 and L-5 are the highest points of the pseudopotential, yes. Picture a volcano, with a deep central crater, and a smaller crater breaking its rim.
If an object is released near one of those points, it falls away from them, but the Coriolis force pushes it sideways, driving it along the contour lines which are closed around the points, so it never leaves the vicinity.
By contrast, L-1, -2, and -3 are saddle points. An object released there can wander about the system
—WWoods (talk) 20:24, 3 December 2008 (UTC)

## Kidney bean-shaped orbit round L4 and L5?

The paper by Cornish cited in the article shows that orbits around L4 and L5 are characterised by two frequencies. Since these are not in general harmonically related, it seems unlikely that these orbits will have a simple closed form.

Suppose that we rotate the coordinate axes in Cornish's paper so that the x-axis is parallel to the 'ridge' in the potential 'hill'. The effect of this on the evolution matrix is to diagonalise the 2x2 submatrix forming its bottom left-hand corner. It then becomes

${\begin{bmatrix}0&0&1&0\\0&0&0&1\\{\frac {3}{4}}\left(2-{\sqrt {3\kappa ^{2}+1}}\right)\Omega ^{2}&0&0&2\Omega \\0&{\frac {3}{4}}\left(2+{\sqrt {3\kappa ^{2}+1}}\right)\Omega ^{2}&-2\Omega &0\\\end{bmatrix}}.$

The eigenvalues (which indicate the characteristic frequencies of the system when the orbit is stable) are of course unaffected by this rotation. If $\epsilon$  is an eigenvalue, it can be shown that

${\frac {\delta y}{\delta x}}={\frac {2\Omega \epsilon }{{\frac {3}{4}}\left(2+{\sqrt {3\kappa ^{2}+1}}\right)\Omega ^{2}-\epsilon ^{2}}}$

Substituting the eigenvalues in turn into this, we obtain

${\frac {\delta y}{\delta x}}=\pm i{\frac {\sqrt {2-{\sqrt {27\kappa ^{2}-23}}}}{2-{\frac {1}{4}}{\sqrt {27\kappa ^{2}-23}}+{\frac {3}{4}}{\sqrt {3\kappa ^{2}+1}}}}$

and

${\frac {\delta y}{\delta x}}=\pm i{\frac {\sqrt {2+{\sqrt {27\kappa ^{2}-23}}}}{2+{\frac {1}{4}}{\sqrt {27\kappa ^{2}-23}}+{\frac {3}{4}}{\sqrt {3\kappa ^{2}+1}}}}$

That these are pure imaginary is a consequence of having diagonalised part of the evolution matrix. It follows that the orbit has the general form

$\delta x(t)=A_{1}\sin \left({{\frac {1}{2}}{\sqrt {2-{\sqrt {27\kappa ^{2}-23}}}}\Omega t+\phi _{1}}\right)+A_{2}\sin \left({{\frac {1}{2}}{\sqrt {2+{\sqrt {27\kappa ^{2}-23}}}}\Omega t+\phi _{2}}\right)$
$\delta y(t)=A_{1}{\frac {\sqrt {2-{\sqrt {27\kappa ^{2}-23}}}}{2-{\frac {1}{4}}{\sqrt {27\kappa ^{2}-23}}+{\frac {3}{4}}{\sqrt {3\kappa ^{2}+1}}}}\cos \left({{\frac {1}{2}}{\sqrt {2-{\sqrt {27\kappa ^{2}-23}}}}\Omega t+\phi _{1}}\right)$
$+A_{2}{\frac {\sqrt {2+{\sqrt {27\kappa ^{2}-23}}}}{2+{\frac {1}{4}}{\sqrt {27\kappa ^{2}-23}}+{\frac {3}{4}}{\sqrt {3\kappa ^{2}+1}}}}\cos \left({{\frac {1}{2}}{\sqrt {2+{\sqrt {27\kappa ^{2}-23}}}}\Omega t+\phi _{2}}\right)$

where $A_{1}$ , $A_{2}$ , $\phi _{1}$  and $\phi _{2}$  are constants chosen to match a particular set of initial conditions. The orbit is thus a kind of epicycle in which the circular motions have been replaced by elliptical ones; it is not, however, a squashed epicycle since the eccentricities of the ellipses differ.

The relationships between the shapes of these elliptical components and that of an equipotential are shown in the above. The innermost (solid) ellipse is an equipotential while the middle (dashed) ellipse shows the shape of the slower of the two elliptical components and the outer (dotted) one that of the faster.

The above shows part of an orbit for a system with an earth-moon mass ratio. The periods of the elliptical components are about 3.35 and 1.05 times that of the two-mass system. It is fairly apparent from the picture how, as the body falls down the potential hill, the Coriolis force changes its direction of motion to bring it back up again.

Can the orbit ever be kidney bean-shaped? Cornish's paper - and thus the above - deal with approximations to orbits which lie 'close' to a Lagrangian point. Presumably, if (i) the size of the orbit became such that the curvature of the potential 'ridge' were large enough to 'bend' the ellipses and (ii) if we had a situation analogous to that in which either $A_{1}$  or $A_{2}$  were 0, we would obtain a kidney bean-shaped orbit rather than an elliptical one but the probability of the second condition occurring naturally would seem to be 0.

--IanHH (talk) 17:27, 16 March 2009 (UTC)

My impression is that when they talk about a "kidney bean" shaped orbit, they usually mean with respect to a frame co-rotating with the smaller mass. So if you revolve your Spirograph, then as you said, depending on your A1/A2/A3, you may well be able to get a "mutant" kidney shape.

Finally, a number of simulations shown on YouTube of specifically the Trojan asteroids, show several that visit around L3, L4 and L5. It's a bit hard to tell, though whether they are "kidney beaning". I would have made the asteroids different colours :D — Preceding unsigned comment added by 121.212.147.109 (talk) 03:31, 3 March 2018 (UTC)

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