Talk:Lagrangian point

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[edit] Arbitrary section header

Perhaps someone could generate a diagram of the various points?

While I was staring at the lagrange points diagram, I realized that the diagram resembles a large "peace sign". Has anyone noticed this? I wonder if the concept of "balance" (of gravity) that the lagrange points represent was borrowed to also represent peace.

Does anyone have any knowledge in this area? What do you think about adding a short sentence or two on this resemblance?

- Jamie E (USA)

Jamie, your question is a common one but the resemblance between the Lagrange point diagram and the peace symbol is only coincidental. The peace symbol is a superposition of the flag semaphore symbols for 'N' and 'D' and form the acronym "ND" for *N*uclear *D*isarmament. That these two semaphore symbols for N and D were encircled is no mystery either as many sigils exploit the symbolism of the circle, and the peace symbol was certainly an embodiment of the peace movement as such. Cheers, Astrobayes 04:54, 15 July 2006 (UTC)


That would be lovely, but I'm restricted to ASCII here and I'm not about to start fiddling with slashes and backslashes and capital letter O's all afternoon :P -- Paul Drye

Not exactly what you had in mind, I know - but I wondered if this would be any good for the page (it's NASA so presumably public domain, I havent seen anything contradicting that posted anywhere) - http://map.gsfc.nasa.gov/m_ig/990529/990529b.jpg --Si42 01:10, 31 January 2006 (UTC)
Template:PD-USGov-NASA says it is okay, and I think it'd be a useful picture to have, for instance in the Stability section. -- Jitse Niesen (talk) 15:42, 31 January 2006 (UTC)

The paragraph

The Earth's companion object Cruithne is in a somewhat Trojan-like orbit around the Earth, but not in the same manner as a true Trojan. It has a regular solar orbit that is bumped at times by Earth. When the asteroid approaches Earth, the asteroid takes orbital energy from Earth and moves into a larger, higher energy orbit. When the asteroid (in a larger and slower orbit) is caught up by Earth, Earth takes the energy back and so the asteroid falls into a smaller, faster orbit and eventually catches Earth to begin the cycle anew. Epimetheus and Janus, satellites of Saturn, have a similar relationship, though they are of similar masses and so actually exchange orbits periodically.

is fascinating information, but has nothing to do with Lagrangian points or Trojan objects. Whither should it be moved?

Maybe to asteroid? --AN
Which class of asteroids is Cruithne in and do we have a page for it?
Put back in the "but differing" that Xaonon took out. It's a critical part of the definition!

Are you sure? I'm fairly certain that two equal masses orbiting each other would result in libration points as well -- a binary star system, for example. -- Xaonon

Well, technically you get them, but without the mass difference you lose the fundamental quality of an L-point: stability. Unless...
  • Mass A is "substantially" larger than mass B -- by about a factor of 30.
  • Mass C, at the libration point, has essentially no mass in comparison to both A and B.
...the points aren't linearly stable and can't hold anything. Basically, the centre of gravity of the system must be pretty close to A or it doesn't work. See J.M.A. Danby's "Fundamentals of Celestial Mechanics" (I think) where the ratio is discussed. -- Paul Drye

Take "differing" out! Even if the above is true (which I doubt, at least for L1, L2, & L3), the intro makes no sense with "differing" when you consider the two "slightly differing" cases. It would have you believe the L points vanish at the point mass equality is passed. It's just nonsense.

And the idea in the intro that two masses combine to form L points is just lousy English, which amounts to more nonsense if you don't read between the lines.

And the intro fails to mention the important point that bodies at the L points are not at all in equilibrium, unless they have a certain velocity. Such bodies must be inserted into their L orbits as any orbital body must be.

This is my first and last contribution to the Wikipedia, as I see below that contributions must be licensed under the GNU FDL, which has proprietary features that require me to be less liberal than I normally care to be.

---

Very good article, but I think the explanation as to why L1 L2 and L3 are unstable compared to L4 and L5 needs to be clearer. If you map the gradient fields for these points you'll notice that L1 L2 and L3 are at the top of hills but L4 and L5 are at the bottom of a depression, im not sure why that is but I think it is an expanation as to why objects would stay in their holes.--ShaunMacPherson 07:11, 10 Mar 2004 (UTC)

No, L1-3 are at saddle points in the pseudopotential field, while L4, L5 are at the tops of hills. (follow the external link to a pretty picture of the field.) Objects at 1-3 can just wander off, while staying at the same level. Objects at 4 and 5 fall down the hills, but then the Coriolis force kicks in, and keeps them in orbit around the Trojan points. –– wwoods 09:34, 25 Mar 2004 (UTC)
In Lagrangian mechanics, a Lagrangian point is…

I changed this opening sentence to In celestial mechanics…, basically because:

  1. The old definition suggested that Lagrange points emerge uniquely in Lagrangian mechanics. But Lagrange points are a physical phenomenon, independent from the theories or formalisms you use. They exist in Newtonian mechanics and in general relativity just as well.
  2. While not actually a tautology to the insider, it may look confusingly so to an outsider. The old definition might have been true, but didn't really explain anything.
  3. It makes sense to define a concept in the context of a wider, more generally known concept. Celestial mechanics meets that criterion better than Lagrangian mechanics does.

—Herbee 00:15, 2004 Mar 20 (UTC)

An asteroid was discovered to be in Neptune's L4 point. I was wondering if someone could work it into the part talking about similar systems? The asteroid's name is 2001 QR322, and I just created a page for it. --Patteroast 16:50, 15 Jun 2004 (UTC)

Can anyone answer a hypothetical question for me? If one had two super massive bodies of precisely equal mass orbiting about each other, would they generate Lagrange points as described here? I suspect that the positions of L1, L2 and L3 will be similar, but will L4 and L5 still be at the 60Deg Trojan points?

Thanks in advance, PBA

Yes, a system of two equal masses (they don't have to be "super massive") orbiting around their common center of mass will have all five Lagrange points. However the L4 and L5 points will be unstable. They will also be in a much higher orbit than the masses. If the masses are at a distance r from the CM, then the L4/5 points will be \sqrt{3}r \approx 1.7 r from the CM.
--wwoods 19:07, 23 Jun 2004 (UTC)

So how does the rest of the solar system fit in? I mean it's all very well to speak of a 3 body system, but for all practical purposes, the other planetary bodies are going to interact with as well. How does that affect the relative stability of, for examaple, the Terra-Sol Lagrange points, or the Luna-Terra L-points? Is this possible effect the reason why one hears L5 advanced as a site for a sizable space habitat? Or is that due to some literary influence of which I am ignorant?

Yes, the presence of other masses in the real Solar System perturbs bodies at the L4 and L5 points, but obviously not too much, as evidenced by the presence of objects at various L4&5 points around the system.
L5 was proposed as a habitat site because the stability reduces the need for station-keeping, and because it was close to the Moon in terms of delta v (~0.7 km/s).
—wwoods 17:44, 21 Aug 2004 (UTC)

[edit] Coriolis effect

I'm unhappy with invoking the Coriolis effect for explaining why a point is stable. The Coriolis effect is merely a device to explain the apparent deviation of a free moving object when viewed from a rotating frame of reference. It cannot really explain the stability of a Lagrange point. What says anyone? Paul Beardsell 23:46, 23 August 2005 (UTC)

Nuthin'. Paul Beardsell 00:53, 5 September 2005 (UTC)


I realize that this point is a bit belated, but classifying L4 and L5 as "unstable" is done by looking at an effective potential for the two body system in a rotating reference frame. However, this effective potential doesn't take into account the existance of the Coriolis forces in this reference frame. Thus, the Coriolis forces effectively cause these points to be stable, when we are looking at it in a rotating reference frame. Unfortunately, to look at this problem in an inertial reference frame would be technically quite difficult, although perhaps conceptually clearer, since you don't need to invoke centrifugal or Coriolis forces. If you did do the problem in the inertial frame, you would see that the Lagrange points (Lagrange orbits, if you will) would be stable, in the sense that a small perturbation from the Lagrange orbit would effectively cause epicyclic motion about the circular Lagrange orbit. Grokmoo 04:24, 1 March 2006 (UTC)

[edit] needs pronounciation

This article needs a sentence about how to pronounce it. Is the g hard or soft (for both of them)?

Thanks for your comment. The first g is pronounced as in go, the second as in judge. I added the pronunciation in IPA. -- Jitse Niesen (talk) 12:45, 4 September 2005 (UTC)

[edit] Wrong in L2 Examples?

I got another value for the example values on the L2 point. Sun-earth OK: 1.5*10^6km from earth Moon-earath : 65348 km

Maybe due to measuring from surface and center. Nor my source, nor wiki specifices. However centre would make more sense, don't you think.

I also get (distance from earth): L1 sun-earth: 1.49*10^6km L3 sun earth: 2.992*10^8km

I also get (distance from moon): L1 sun-earth: 57660km L3 sun earth: 764956km

[edit] wrong example?

"Earth–Moon L2 would be a good location for a communications satellite covering the Moon's far side."

no it wouldn't. how would you send information to it? isn't the moon in the way? I did not remove the sentence yet. —The preceding unsigned comment was added by 212.120.85.242 (talk • contribs) .

Satellites operating "at" collinear libration points (such as L2) are generally not actually located at the point, but rather in a halo or lissajous orbit around the point. If the amplitude of this periodic orbit is sufficiently large then the spacecraft will always have line-of-sight to the Earth. --Allan McInnes (talk) 05:22, 1 March 2006 (UTC)
Especially line-of-sight to a comms array in geostationary orbit around the earth.--Si42 19:45, 15 April 2006 (UTC)
Allan, you should add your explanation to the article, since I wondered the same thing when I read the example. --euyyn 00:43, 13 May 2006 (UTC)

[edit] Deleted false sentence

I deleted this sentence from the first picture's caption since it's nonsense to me:

An object in free-fall would trace out a contour (such as the Moon, shown).

Just look at the sharpness of the contours near L1: an object in free-fall could never do those movements, since the centre of the (fictional) centrifugal force isn't at L1, but at the centre of masses of the Sun and the Earth. In addition, free-falling objects can perfectly have elliptical orbits, thus crossing several contours. So the sentence is false in every posible way it could be. --euyyn 00:43, 13 May 2006 (UTC)

[edit] L4/L5 explanation using only gravity and 'centrifugal' force?

The examples given under L1/2/3 nicely explain, imho, the existence of these points from the simple addition (cancellation) of the gravitational fields and the centrifugal force. But I notice there is no such explanation under the L4/5 points. Can someone who understands these points add in a similar example/explanation? If it were possible for a body to sit at exactly the top of the potential hill at L4/5 , then it seems to me that one could remove the Earth and that body would stay in exacly the same position and same orbit. This implies that the presence of the Earth for a body at these points of irrelevent..but I don't understand how that can be as adding the additional force of the Earth would surely drag the body out of its circular orbit about the Sun. Anyway, I'm just looking for a clearer physical picture using only gravity/centrifugal force to explain the L4/5 points. Thanks!

You need the coriolis force to explain them.... This link http://map.gsfc.nasa.gov/m_mm/ob_techorbit1.html has some good info (most of this page seems to be sourced from it) & if you know some maths you could download the pdf & have a look at it. 203.97.255.167 00:14, 3 September 2006 (UTC)

[edit] Third pronunciation

I pronounce "Lagrangian" as [ləˈgɹe(ɪ)n.dʒiˌʊn]. Does anyone else?  Denelson83  00:28, 12 June 2006 (UTC)

Yeah. do that too. Is that a "mispronunciation"? AEuSoes1 06:39, 24 August 2006 (UTC)

[edit] Fictional References

I was delighted to finally read a clear explanation of the L4 and L5 points as they are extensively used in the Transhuman Space roleplaying game. This includes the Trojans and the L4 and L5 points around Earth. I suppose this goes under the mention made of hard science, but I thought I might as well mention it. - Philippe J (FR)

Just noticed that the Peter F. Hamilton, Nights Dawn, refernce is incorrect. This occurs in the first book, The Reality Dysfunction, as part of the battle for Lalonde. p 1172-73 and 1189-94 in the 1997 Pan Books paperback edition. The episode in the second book involves making a gas giant go nova to create a debris front that the Lady Mac would be in front of, and the pursuing starship would be caught by.

[edit] stability of L4/L5

Aren't L4 and L5 stable points in the Earth-Sun model? The blue arrows in de picture show them as instable, which I think is incorrect. --Mushlack 17:32, 2 August 2006 (UTC)

They are potential peaks, hence the blue downwards-slope arrows. They are, as you say, stable, even though the potential diagram does not show it. It requires slightly more sophisticated maths... 203.97.255.167 00:15, 3 September 2006 (UTC)

[edit] Kordylewski cloud at Earth-Moon L4/L5

There are supposed to be coulds of dust at both of this points. Does anybody have any specific details how much dust there is (or if there even is dust there) in stable halo orbit? Since this orbit is the size of the earth this could have large combined mass (enough to collect and use?) or could pose a problem to any object in it's halo orbit. --Taho s 18:49, 24 February 2007 (UTC)

[edit] Lagrange points over time

Over a period of time, do Lagrange points change as the objects, such as earth, continue in their orbits?

And also, the equations and the points seem to be based on a circular orbit, wouldnt the Lagrangian points be different using the elliptical orbits that planets truly move in? Xlegiofalco 06:34, 22 November 2006 (UTC)


[edit] Please, re-check info on the page

This page and the one on the Advanced Composition Explorer contradict each other about the satellite's orbit. This page informs that ACE is kept on Lissajous orbit and ACE's page tells about an Halo orbit.

Can someone correct either?

--Dr. Pnz 04:11, 21 January 2007 (UTC)

[edit] "In orbit around" L2?

Is it correct to describe the satellite as being "in orbit around" the L2 point here:

"The Wilkinson Microwave Anisotropy Probe is already in orbit around the Sun–Earth L2."

Maybe they're two separate statements: "The probe is already in orbit" and "The probe is around (near) L2".--Nonpareility 22:43, 30 January 2007 (UTC)
It certainly is correct; see Lagrangian point#Stability. –EdC 23:17, 30 January 2007 (UTC)

[edit] Lagrangian points and spacecraft

The article speaks of multiple existing and planned spacecraft operating at a single Lagrangian point. Of course, it's impossible for multiple objects to occupy the same point in space, so I'm assuming that the spacecraft are/will be within a certain radius that's "close enough". Could some text be added to explain this?--Nonpareility 22:43, 30 January 2007 (UTC)

Again, Lagrangian point#Stability explains how quasi-stable, quasi-periodic orbits around the L-points can exist. Where relevant, "at" should be amended to "in orbit around". –EdC 23:19, 30 January 2007 (UTC)

[edit] Science Fiction Counter-Earth Lagrangian?

Is a science fiction "counter-Earth" actually a Lagrangian concept when it's not an object of "negligible" mass -- and the two Earths would seem to be just counter-balancing each other? (Well I mean "Earth's Evil Twin" -- same mass.)

[edit] Asteroid(s) orbiting Earth at L4/5?

Wasn't there a discovery in the last few years of one or two asteroids at/orbiting the Earth-Sun L4 and/or L5 points? —The preceding unsigned comment was added by 67.121.242.84 (talk) 23:53, 30 January 2007 (UTC).

[edit] Exact position of L3, plus minor amendments

[edit] Beginning

The L-points are not necessarily in interplanetary space.

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

[edit] 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

[edit] Diagrams

Recreated contour plot with n=25
Recreated contour plot with n=25

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)

[edit] 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)

[edit] External Links

Should include the paper itself, via Gallica.

Which paper?
> the one for which the reference in the main page is missing ...
> Lagrange, Joseph-Louis, ESSAI SUR LE PROBLÈME DES TROIS CORPS
> via <a href="http://www.merlyn.demon.co.uk/gravity4.htm#Refs">.
> It's written in Maths, slightly diluted with French.
Added as a reference. –EdC 02:15, 5 February 2007 (UTC)

[edit] See ...

<a href="http://www.merlyn.demon.co.uk/gravity4.htm#GLP">The Geometry of the Lagrange Points</a>.

Looks useful.
Added as an external link. –EdC 02:16, 5 February 2007 (UTC)

82.163.24.100 23:05, 2 February 2007 (UTC)

Thanks for your comments. –EdC 04:49, 3 February 2007 (UTC)
> 82.163.24.100 15:52, 3 February 2007 (UTC)

[edit] Exact position of L4 and L5, plus minor amendment to figure

In subsection L4 and L5 :

Para 1 has "The L4 and L5 points lie at the third point of an equilateral triangle whose base is the line " no, "The L4 and L5 points lie at the third corners of two equilateral triangles whose common base is the line " or " ... whose bases are ...".

Well, it could be "The L4 and L5 points each lie at the third point...". Your wording is better, though. EdC 17:39, 23 February 2007 (UTC)

Para 1 has "whose base is the line between the two masses" - suggest '... between the centres of the two ...'. That does not change the natural meaning, but it does stress 'centres'.

Yes. EdC 17:39, 23 February 2007 (UTC)

Para 1 has "the smaller mass in its orbit around the larger mass". But both masses orbit the barycentre, and L4/L5 are outside the orbit of the secondary. Omit "in ... mass"?

I don't think "in its orbit" is intended to convey that they share the orbit; rather it is intended to qualify "ahead"/"behind". It could be misleading, though; I've changed "in" to "with regard to". EdC 17:39, 23 February 2007 (UTC)

Result could be "The L4 and L5 points lie at the third corners of two equilateral triangles whose common base is the line between the centres of the two masses, such that the point is ahead of (L4), or behind (L5), the smaller mass."

Changed, modulo the above. EdC 17:39, 23 February 2007 (UTC)

By the way, L4 is ahead of the secondary but behind the primary, in angle frem the barycentre. If mass were steadily transferred from primary to secondary, then as the secondary became the primary and vice versa, so would L4 and L5 exchange names.

True, but rather irrelevant, as L4 and L5 are only stable when there is a large disparity in mass between primary and secondary; binary star systems don't have trojan points. EdC 17:39, 23 February 2007 (UTC)
Well, we're dealing with Lagrange points, not Trojans. Pluto-Chiron, mass ratio about 10, still has five L-points but no T-points. The property is necessary for an exact solution. It should have a place in more mathematical pages.
82.163.24.100 22:41, 26 February 2007 (UTC)

Last para ends ", and asteroids there are named after characters from the respective sides of the war". Nowadays they are; but the convention was not established in the early days. See entries for 'Trojan asteroid', 'Trojan camp', 'Greek camp'. Given the detail elsewhere, I think the quoted bit can be omitted, or replaced by something like "; but there is a spy in each camp".

A little too humourous for an encyclopaedia, I think. A "mostly" qualifier will suffice; the reader can anyway follow the links to the detail, as you point out. EdC 17:39, 23 February 2007 (UTC)

In subsubsection Examples the page has "The Sun–Earth L4 and L5 points lie 60° ahead of and 60° behind the Earth in its orbit around the Sun." No, just outside its orbit. Maybe better to put it from another aspect - something like "As measured from the centre of the Sun, L4 and L5 are 60° ahead of and behind the Moon in the line of its orbit."

Again, I think "in its orbit" has a different intended meaning from how you're reading it. I've changed "in its orbit around the Sun" to "as it orbits the Sun". EdC 17:39, 23 February 2007 (UTC)

Figure

If the top-of-page diagram can again be edited, I suggest moving the blue triangles at L3 outwards a modicum, so that the grey circle of the Earth's orbit can be seen as it passes L3.

82.163.24.100 12:11, 23 February 2007 (UTC)

Doable, yes; unfortunately every time that image gets edited it loses quality (it's in JPEG). I'd prefer to develop Image:Lagrange points.svg to the degree that it can replace the NASA image.
By the way, Wikipedia is the encyclopedia that anyone can edit; your suggested edits would have been fine had you made them yourself. Be bold, I think the saying is. EdC 17:39, 23 February 2007 (UTC)
(1) not until I'm more familiar with page-editing convention, (2) A second opinion is valuable.
82.163.24.100 22:41, 26 February 2007 (UTC)

[edit] Correction to first diagram?

Shouldn't the L4/L5 arrows in the first diagram face inward and be red?--Michalchik 00:51, 10 March 2007 (UTC)Michalchik

Nope. Sorry. AFAIK the diagram is correct. I never did get my head around why it's the other way than you would expect, but I think it's something to do with coriolis effects closing the orbit.WolfKeeper 08:30, 10 March 2007 (UTC)

[edit] Euler and collinear libration points

In a lecture at CalTech in 2004, Shane Ross asserted that Euler first "discovered" the collinear libration points in the 3-body problem. Is there any written source to cite for this? Sdsds 21:39, 25 March 2007 (UTC)

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