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© J R Stockton, ≥ 2015-02-21

Gravity 4 : The Lagrange Points.

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History is now in a separate page : The History of the Lagrange Points. The history is not as has commonly been believed.

Three Non-Negligible Masses

For the equilateral configurations, one can do very much better than the traditional argument shown below by following Lagrange's initial direction. For both the collinear and the equilateral configurations, one can do much better by considering the fundamental requirements for pattern preservation. I am not aware of such calculations having been presented previously - they are certainly not widely known.

See in Gravity 6 : Constant-Pattern Motion and in The Lagrange Points Reconsidered.

Two Masses and a Particle

This major section deals in the usual manner (as opposed to Lagrange's manner) with the theory of the positions of the Lagrange Points.

Notation and Numbers

This major section mainly refers to the Earth-Moon system, but Sun-Earth and Sun-Jupiter Lagrange Points, for example, are also significant. The collinear points are named, as is now usual, in order (p = primary, s = secondary) L3 p L1 s L2. Astronomers have preferred L3 p L2 s L1.

Upright letters are used for geometry, and italic letters for physical quantities. Thus 'm' is the location of a secondary body, and 'm' is its mass.

Distances are average, and from the centre of the nearest body
Primary Radius6371696000696000km
Primary Mass5.9736e241.9891e301.9891e30kg
Secondary Radius1737637169911km
Secondary Mass7.3477e225.9736e241.8986e27kg
Separation, L4, L5384399149598261778547200km
Distance Variation5.51.64.5± %
Mass Ratio81.33329821048

The Solar System Barycenter shows the motion of the Solar System barycentre relative to the size of the Sun, for years 2000 to 2050. In it, the distance of the barycentre from the centre of the Sun varies from about 14% to 200% of the Sun's radius; the barycentre is outside the Sun for about 60% of the time.

The Five Lagrange Points

The five Lagrange Points are named after Joseph-Louis Lagrange (1736-1813), an Italian-French mathematician. Some Web writers use "Langrange"!

In a system where a primary body and a secondary body, such as Sun and Earth, are gravitationally bound in circular or elliptical orbits, and there are no other significant gravity gradients, the entire system orbits in a plane around its centre of mass ("centre of gravity"), the Barycentre.

In the case of three bodies all of significant mass, each body could be considered as being at a Lagrange Point of the other two. But the Lagrange Point notation is used only when one body has negligible mass.

There are five co-moving locations in the plane where a small particle would be in equilibrium. These are variously called the Lagrange Points or the Lagrangian Points or the Lagrangian Equilibrium Points, and termed L1 to L5. Only L4 and L5 can give stable equilibrium.

 L1 L2 L3 For a particle at a Lagrange point to orbit the barycentre of the primary and secondary masses with the same period as that of the masses, it must feel a net attracting field proportional to its distance from the barycentre (ω2r for circular orbits).
At L1, the fields of the masses subtract, and at L2 and L3, they add.
At L4 and L5, the fields of the masses combine to give a field directed towards the barycentre.
The Roche Lobes of primary and secondary meet at L1; their boundaries are surfaces of equal quasi-gravitational potential. L1, L2 and L3 seem to be saddle-points of potential.

Point L1 lies between the two bodies; points L2 and L3 are on extensions of a straight line between the bodies. They are always unstable positions for particles, but spacecraft can readily be maintained at or near them.

If the primary is substantially more massive than the secondary, as is usual, L1 and L2 are near to the secondary and approximately equidistant from it; and L3 and the secondary are approximately equidistant from the primary.

Let the distance Mm be d.
The field at m from M is M×G/d2
The field at M from m is m×G/d2.
Let the distance MB = R, Bm = r.
We have r = d×M/(M+m), R = d×m/(M+m)
so M = r×(M+m)/d, m = R×(M+m)/d ;
The fields at m and M are
G/d2×r(M+m)/d = G(M+m)/d3×r
and G/d2×R(M+m)/d = G(M+m)/d3×R.
That fits with the acceleration for a circular orbit of radius r being ω2r.

Points L4 and L5 are to the sides, respectively leading and trailing, each forming an exact equilateral triangle with the centres of the two massive bodies. If the secondary body is much lighter than the primary body, L4 and L5 are stable positions for particles – and asteroids and spacecraft.

Hill Radius

For a large mass ratio, L1 and L2 are approximately at the Hill Radius from the secondary body.


The ancients hypothesized a Counter-Earth, as a counter-balance in Earth's circular orbit and always behind the Sun. Earth's orbit is not circular; nevertheless, Sun-Earth L3 is always behind the Sun. But L3 is not a point where a body can be positioned stably.

Balance of Fields

Balance of Gravitational Fields

Ignore corrections for the Sun's field and the motion of the Earth-Moon barycentre, and treat orbits as circular. As shown later, L1 is about 15% of the way from Moon to Earth.

There is a point H between L1 and Moon at which there is, from Earth and Moon, no net gravitational force. It is at M / Q2 = m / q2 where Q+q is the Earth-Moon separation, D. As M ≈ 81×m, Q ≈ 9×q and H is about 10% of the way from Moon to Earth.

An unpowered body at H will not be changing its velocity but will be stationary or move in a straight line. But H itself moves in a circle, so the body and H cannot stay together. For a particle to be stable at H, it would be necessary for Earth and Moon to be non-moving, and therefore to be held in position by some means other than gravity.

Conversely, the net pull of the Earth and the Moon will, at L1, hold an unpowered body in a circular orbit with a period of one sidereal Month. The body will remain at L1.

There is no true "balance of gravitational fields" at any of the Lagrange Points.

Unless the primary and secondary bodies have similar masses, an object at any of the Lagrange Points is basically orbiting the primary body, but in a manner modified slightly - in distance and stability - by the gravitational field of the secondary body.

At the Lagrange Points, the gravitational fields of the primary and secondary, acting jointly, provide the centripetal force which is needed to maintain a particle in its path.

There is a balance, I believe, in rotating coordinates; but one of the fields involved is a non-gravitational pseudo-field.

For better descriptions of the Lagrange Points in general, see in Gravity 6 : Constant-Pattern Motion.

Circular Orbits

Approximate Geometry

When the secondary is considerably lighter than the primary, the barycentre can be approximated by the centre of the primary, and L3, L4 and L5 are then at about the same radius as the secondary.

 Diagram, as per text Page 126 of "The High Frontier", by Gerard K O'Neill, (c) 1976 (you don't own it? My copy was £0.95 new in 1978) has a clear diagram, equivalent to that here.

It shows L3, Earth, L1, Moon, L2 in that order in a straight line. L3, L4 and L5 are in the Moon's orbit, with L3 being opposite to the Moon and L4 and L5 being 60° away from it.

Generally, L4 leads, L5 trails; most other references agree.

(However, Henry Spencer has said that astronomers preferred the order L1, Earth, L2, Moon, L3; again L4 leads, L5 trails. Encyclopaedia Britannica 15th Edn shows L1, secondary, L2, primary, L3 and shows L4 lagging, L5 leading.)

As at 2007-02-02, Wikipedia had L3 further from the primary than the secondary is : that seemed wrong; to thar approximation, it should have been on the circle. Changed by 2007-02-07. It is interesting to look at corresponding diagrams in pages for different languages.

Actual Lagrange Point Geometry

 Diagram, as per text It seems to me that, in respect of L3, several other authors have misrepresented the results of calculations, putting it too far out. L3 is slightly further from the barycentre than the secondary is, but it is slightly nearer to the primary than the secondary is. Also, it is clear that L4 and L5 must lie outside the orbit of the secondary, but they are often said to lie on it.

The diagram on the left is to scale for the Earth-Moon system. The barycentre is taken as fixed and the coordinates are non-rotating.

In the following detailed scale diagram, the sizes of Earth (M) and Moon (m) are in mutual proportion; but their mass ratio M:m has been changed from 81:1 to 9:1 in order to increase the differences between the various circles (at ratio 9:1, L4 and L5 are unstable; but they still exist). The densities have been scaled to give the correct relative diameters and to put the barycentre (C, Centre of Mass) at the same relative depth as for the real Earth and Moon. The yellow cross marks the fixed position of the barycentre. D is used for the distance between the centres of the two bodies.

The small fixed red circle shows the orbit of the centre of the Earth, and the fixed green circle shows the orbit of the centre of the Moon. Both of those orbits are centred on C.

The moving blue circle is centred on the Earth's centre, M; it is the set of points currently at a distance D from M.

The Lagrange Points are shown by red plus signs; the red crosses mark named geometrical points X, Y, Z. Points x, y, z exist corresponding to X, Y, Z but with M and m interchanged.

The outer thin fixed black circle is the path of L3; the inner one of L4 and L5.

 Diagram, as per text

 Diagram, as per text

To the right is like the centre line of the large diagram, but is scaled at 100 km/px for the Earth-Moon system with the correct mass ratio and diameters.
Centre Line
scale diagrams
 D px

L1 and L2, on the Earth-Moon line, are at similar, but different, distances from the centre of the Moon. The paths of L1 and L2 are of little interest, and so are not shown.

L3, L4, and L5 are all much nearer to, but are all outside, the Moon's orbit.

Note that L3 is outside the green circle but inside the blue one. Its path, centred on C, is shown by the outer thin black circle.

Obviously, L2 is further from C than M is, and so is outside the Earth's orbit. Note the L2/L3 symmetry and the L4/L5 interchange as M/m → 1 → m/M.

L3 is further from C than m is, and so is outside the Moon's orbit. L3 is at a smaller distance from M than m is, and so is inside the circle through Moon, L4 and L5.

Measured from the Moon, L4 and L5 are 60° from the Earth.

Measured from the centre of the Earth, L4 and L5 are 60° from the centre of the Moon, and at the same distance D, thus forming two equilateral triangles, M4m and M5m.

The blue circle, on which L4 and L5 lie, is NOT the path of L4 and L5, though they remain on it. It is centred on the Earth's centre, and so is not fixed; it is epicyclic, and rolls around the green circle. Distances Mm and MX are equal.

The path of L4 and L5 is centred on C; it is shown by the inner thin black circle.

Argument for L3 in relation to the green circle

The field of M alone suffices to make m follow the green fixed circle around C. So a particle released with the same speed from Y, at the same distance from C but with more field from the nearer M plus some field from m, will move within the green circle. Therefore L3 is outside the path of the Moon.

Argument for L3 in relation to the blue circle

Mm = MX = D   ;   Earth mass M, Moon mass m
Gravitational field on Moon is GM÷Mm2 = GM/D2 balancing ω2×Cm
Gravitational field on Earth is Gm÷mM2 = Gm/D2 balancing ω2×CM
Gravitational field at X is GM÷MX2 + Gm÷mX2   =   GM/D2 + Gm/4R2
Centrifugal effect at X is ω2×CX   =   ω2×(CY+YX)   =   ω2×Cm + ω2×CM×2
  =   GM/D2 + 2Gm/D2   which is greater by 1.75Gm/D2.
Therefore L3 is nearer to the centre of the Earth than the Moon is.

Two Extreme Cases

If the mass of the Moon were to tend towards zero, the yellow dot would move onto the black one and the red circle would shrink to zero; all of the large circles would correspondingly merge, and L1 and L2 would move into the Moon. The five Points would occupy four locations on the combined circle.

For equal masses in orbits of diameter D, corresponding Points are : L1 is central, L2 and L3 are at radius 1.198×D, L4 and L5 are on the dividing plane at radius D×√3/2 = 0.866×D.

Non-Circular Motion

Customarily, the elliptical case is not proved, but may be assumed. Lagrange proved it, and there is proof in The Lagrange Points Reconsidered. Non-rotating systems are an extreme case of elliptical ones

Elliptical Orbits

The situation is essentially the same for elliptical orbits. Bodies and points move in orbits of different sizes but the same shape and period. The pattern changes in size as it rotates. Page Gravity 5 : A Lagrange Movie can show the motions either in fixed coordinates, or in coordinates which rotate at the same varying angular velocity as the bodies and the points. The diagram then breathes in and out, with the bodies and the Lagrange points all moving in straight lines.

Non-Rotating Systems

Non-rotating systems are a particular case, with eccentricity of one. The "orbits" are straight lines with an end at the barycentre. The bodies and the L-points gravitate to the barycentre in a fixed pattern. If the bodies and particles do not, because of slight sideways motion, collide, they will then return outwards.

Particles in general positions will not retain their pattern. In particular, if two unequal masses are allowed to fall together, a particle at the point where the forces initially balance will not initially accelerate, although the balance-point itself will do so.

Lagrange made no reference to this specific case, but his method is general enough to include it.

Lagrange Points L1, L2 and L3

The existence of L1, L2 and L3 is obvious enough; their locations are calculated below. Note that the intermediate Lagrange Point L1 is not the place where the attractions balance; one must remember, for all Points, that the rotation about the centre of gravity of the system requires a net centripetal force.

Points L1, L2 and L3 are not stable, radial unstability seeming evident. Halo orbits exist around them and are used. I am uncertain of the stability of those, but I've seen it said that stable halo orbits exist for L1 and L2 at least - and that they do not.

No natural bodies can be at any L1, L2 and L3 points, though they could perhaps exist in halo orbits. The characteristic time of escape from Sun-Earth L1 and L2 is about 23 days (0.2/π years). There can be no Counter-Earth at L3. Apart from the intrinsic instability with a characteristic time of about 150 years, Venus would have a large effect there.

Calculation of L1, L2 and L3 Locations

In Chapter 4 of "A Fall of Moondust", Arthur C Clarke put the inner Earth-Moon Lagrange Point at 50,000 km above the Moon. No doubt he would have been able to calculate it as accurately as the data available around 1960 would support; but he probably didn't feel any need to.

Since initially writing that and what follows, I have for the first time read Sir Arthur's "Ascent to Orbit", A Scientific Autobiography. Chapter 12, "Stationary Orbits", treats L1 and L2, and gets essentially the same quintics as I do below.
The implied exact quintic below is an instance of "Lagrange's Quintic Equation".

The angular velocity ω about the barycentre is the same for the Earth, the Moon and the five Lagrange Points. The centripetal force required for a mass m to circle at a radius D is 2D.

The Moon (mass m) orbits the Earth (mass M) at a mean distance D  ~ 384,403 km; and, approximately, M = 81.25×m. The gravitational force GmM/D2 is balanced by 2D, from which ω2 = GM/D3 .

Let the distance between a Lagrange Point and the nearest body be L; let the mass of the particle be o.

First Approximation

To the first order, we can ignore the motion of the Earth's centre of gravity.

For a particle at the inner Lagrange Point L1,
  2(D-L)   is balanced by   Go(M/(D-L)2 - m/L2) ;
substituting for ω2,   M(D-L)/D3 = M/(D-L)2 - m/L2 .
Changing to units where D = 1, m = 1, and multiplying by L2(1-L)2,
  ML2(1-L)3 = ML2 - (1-L)2
  ML2 - 3ML3 + 3ML4 - ML5 = ML2 - 1 + 2L - L2
  ML5 - 3ML4 + 3ML3 - L2 + 2L - 1 = 0   (Equation L1).
The solution for M = 81 is near L = 0.151272; that gives ~58,137 km from the lunar centre. Thus the inner Lagrange Point is somewhat further from the Moon than geostationary orbit is from Earth.

If M is large, L will be small, and equation L1 reduces to 3ML3 = 1.

For a particle at the Lagrange Point L2 behind the Moon,
  2(D+L)   is balanced by   Go(M/(D+L)2 + m/L2) ;
substituting for ω2,   M(D+L)/D3 = M/(D+L)2 + m/L2 .
Changing to units where D = 1, m = 1, and multiplying by L2(1-L)2,
  ML2(1+L)3 = ML2 + (1+L)2
  ML2 + 3ML3 + 3ML4 + ML5 = ML2 + 1 + 2L + L2
  ML5 + 3ML4 + 3ML3 - L2 - 2L - 1 = 0   (Equation L2).
The solution for M = 81 is near L = 0.168327; that gives ~64,692 km from the lunar centre. Thus the farthest Lagrange Point is somewhat further from the Moon than the inner one.

If M is large, L will be small, and equation L2 reduces to 3ML3 = 1.

For a particle at the Lagrange Point L3 behind the Earth,
  2L   is balanced by   Go(M/L2 + m/(D+L)2) ;
substituting for ω2,   ML/D3 = M/L2 + m/(D+L)2.
Changing to units where D = 1, m = 1, and multiplying by L2(1-L)2,
  ML3(1+L)2 = M(1+L)2 + L2
  ML3 + 2ML4 + ML5 = M + 2ML + ML2 + L2
  ML5 + 2ML4 + ML3 - (M+1)L2 - 2ML - M = 0   (Equation L3).
The solution for M = 81 is near L = 1.001029. Thus the third Lagrange Point, at ~384,716 km is ~395 km further away from the centre than the Moon is.

If M is large, L will be near 1.0, so set L = 1+λ, use (1+λ)n ≈ 1+nλ and equation L3 becomes 12Mλ = 1.

These need checking - both algebra and figures in km. Program "Lagrange" will do it.

Approximate Quasi-Period of L1 & L2

Considering Kepler's Third Law, with now M being the mass of the primary, D the distance to the secondary, and T the orbital period of the secondary : ω2D   =   (2π/T)2D   =   GM/D2   →   T2   =   (4π2/GM) D3

The approximation 3ML3 = 1 becomes, with m being the mass of the secondary, d the distance to L1 or L2, and t the orbital period that a particle at L1 or L2 would have if the primary were absent, 3(M/m)(d/D)3 = 1   →   (M/D3)/(m/d3) = 1/3   →   (t/T)2 = 1/3 , so that t is 1/√3 times the year of the secondary.

Sun-Earth L1, L2 and L3

Likewise, with masses of 1.984×1030 kg and 5.98×1024 kg, and distance of 149.46×106 km :-

LAGRANGE.PAS  www.merlyn.demon.co.uk  >=2000-03-11
 Major mass, Minor mass, Separation ???
  1.984E30 5.98E24 149.46E6
Iterate L1 :     0.009982     1491926 km
Iterate L2 :     0.010049     1501921 km
Iterate L3 :     1.000000   149460038 km

The latest version of that program is in lagrange.zip; it gives the same results.

Is L2 in the Umbra?

Point raised in mail received 2009-08-04.


The length U of the umbra is Dr/(R-r). For M/m large and therefore L/D small, U is approximately r/R in units where D = 1.

Let the density of m be ρ, in units such that the density of M is 1. With also m = 1, (R/r)3 = ρM.

For large M, from First Approximation above for L2, we have 3ML3 = 1 with D = 1.

Therefore, U-3 = ρM and L-3 = 3M, so that U = L for ρ = 3.

On this approximation, L2 is at the tip of the umbra if ρ = 3, and is within the umbra if the secondary is less than three times denser than the primary, in the limiting case of a large ratio between the masses.

Therefore, on the large mass-ratio approximation, Sun-Earth L2 is somewhat outside the solar umbra (ρ ≈ 5.515/1.408 = 3.917). But Earth-Moon L2 is well within the umbra of earthshine (ρ ≈ 3.346/5.515 = 0.607), so it is not visible from anywhere on Earth.

The exact calculator below determines L from M, in real units. From that is now determined, for any given mass ratio M, the value of the density ratio ρ for which U = L and therefore L2 is at the tip of the umbra. R/r = (L+1)/L   and   (R/r)3 = ρM   →   ρ = ((L+1)/L)3 / M

How near to L2 is the edge of the Penumbra?

With R and r small in comparison with D, the radius of the penumbra must by inspection be r at the planet m and 2r at the point of the umbra.

More pending?

L1, L2 and L3 More Accurately

This is expressed for Earth-moon L1, L2 and L3, but applies to all systems.

                                  L4  x
                                     / \
                                    /   \
                                   /     \
                                  /       \
                                 /         \
                                /     L5    \
                               /      |      \
                              /       V       \
                             /                 \
                            /                   \
                           /  CoG                \
  |<---------L3--------->|/    |            |<-L1->|<-L2->|
                         |<-c->|                   |
  o                      M                  o      m      o

Masses : M is Earth, m is moon, o are particles at L1 L2 L3.
CoG is the Centre of Gravity (barycentre) of the Earth-moon system,
the centre of all rotation.  L4 and L5 are at D from M and m.

   CoG:          c = mR/(M+m)
 Earth: Mω²c       = GMm/D²
 (Moon: mω²(D-c)   = GMm/D²)
    L1: oω²(D-c-L) = GMo/(D-L)² - Gmo/L²
    L2: oω²(D-c+L) = GMo/(D+L)² + Gmo/L²
    L3: oω²(L+c)   = GMo/L²     + Gmo/(D+L)²

These equations cannot be simplified to the extent that the earlier ones were; but they can easily be iterated to a solution :-

LAGRANGE.PAS  www.merlyn.demon.co.uk  >=2005-04-28
 Major mass, Minor mass, Separation ??? 81 1 1
 Using CofG as centre of motion - c = 0.012195 :
  Iterate L1 ...  0.151109
  Iterate L2 ...  0.168048
  Iterate L3 ...  0.992886             + c → 1.005081

And :-

Full Calculator

This includes results associated with L4 and L5, using the 60° geometry below.

Exact Lagrange Points Calculator
For X Z Y, near to L3, see the large picture above.
Entries can be expressions.
Primary Mass, M :- Secondary Mass, m :-
Separation, D :-
Mass Ratio, M/m :- Primary Orbit, CM :-
m1, m2, M3 :-
C3 :- Secondary Orbit, Cm :-
MX, MY, mH :-
C4, C5 :- Intersection, MZ :-
L2 at umbra tip : density ratio secondary/primary = ρ :-

Computation of L2 corrected, 2005-04-28.
Computation of density ratio ρ for L2 at umbra tip added, 2009-08-08, needs checking.

 M/m	1e0    1e1    1e2    1e3    1e4    1e5    1e6    1e7    ie8
 ρ	14.4   5.85   4.06   3.45   3.12   3.09   3.04   3.02   3.01

L1 and Gravity Balance

Consider a particle moving along a line between primary and secondary. When in rotating coordinates it passes the balance point of gravitational and centrifugal forces, at L1, it begins to feel an outwards force with respect to that point (which is itself accelerating inwards, being in orbit). Later, when it passes gravity balance (M/R^2 = m/r^2), it begins to feel a force outwards with respect to the Universe, and really does accelerate towards the secondary.

Lagrange Point Transfers

Points L1 and L2 can be used as "gateways" for very low energy transfer orbits – see in Baez, in Wikipedia Interplanetary Transport Network, http://www.esm.vt.edu/~sdross/books/ (16MB), and elsewhere – which can be better than Hohmann orbits.

Lagrange Points L4 and L5

To see that L4 and L5 exist for a particle, it is sufficient to consider, as below, the balance of centrifugal and net attractive forces in [rotating] barycentric coordinates. The 60° is not measured barycentrically; L4 and L5 form an equilateral triangle with the primary and secondary bodies, for any mass ratio.

Determining the stability of L4 and L5 is beyond my present scope. There is a page at NASA GSFC.

There are in general stable regions around L4 and L5.

I have read that L4 and L5 in the Earth-Moon system are actually not stable points, because of lunar eccentricity and solar perturbations. I have read that stable orbits around them should exist, and that dust there may have been observed (Kordylewski clouds).

Verification of L4 and L5 Locations

Further (and superior) verification is in Gravity 6 : Constant-Pattern Motion and The Lagrange Points Reconsidered.

The Earth, the Moon, and L4 (or L5) form an equilateral triangle rotating about the Earth-Moon centre of gravity. A reasonably simple geometrical argument shows that the attraction on a particle at L4 (or L5) is correctly directed for this, and that it is of the correct magnitude.

For the diagram, one needs to reduce the mass difference.

 Diagram, as per text

Draw the equilateral triangle MmL. Mark the centre of gravity of M and m on line Mm as C, so that M×CM=m×Cm. Draw LC. Through C, draw CX parallel to mL, with X on ML. X clearly divides ML in the same ratio as C divides Mm : MC=MX=CX, Cm=XL.

As ML=mL, the two field components on L, from M and m, are proportional to M and m. One sees that LXC is a triangle of fields, so that the net gravitational field on L from M and m is indeed directed towards C; and also that the fields on L, m, and M are proportional to LC, mC and MC, as is required to provide centripetal forces for the three bodies rotating in unison. Note that LC > mC >> MC.

Since the Moon orbits the barycentre C rather than the Earth's centre M, it is clear that L4 and L5 lie outside the orbit of the Moon.

The above argument appears as applicable to elliptical orbits as to circular; the orbit of L is at 60° from that of m.

The argument is independent of mass ratio; for two equal masses, a particle can orbit (unstably) at two points on the bisector, 60° from each mass, and √3 times as far from the centre.

Stability of Equilateral Configuration

For masses in circular orbits, a particle at L4 or L5 is stable only if the primary/secondary mass ratio exceeds about 24.96. It is also stable, though (I've heard) less so, for matching elliptical orbits.

The period of a particle orbiting around Sun/Earth L4/L5 is about 89 days (ESA). Is it exactly three synodic months? Presumably not; Wikipedia implies a factor of about twelve for Jupiter trojans.

Routh's Criterion

The following stability condition, for circular orbits, can be derived from what is called "Routh's criterion"; but Gascheau found it first : see Danby (May 1964) and Sicardy in References.

JH has written :- In order to be stable, the masses of the three bodies must obey the formula: 27*(m_1*m_2 + m_2*m_3 + m_3*m_1) < (m_1 + m_2 + m_3)^2; and, later, I got it from an article in Volume 5 of "What's Happening in the Mathematical Sciences" by Barry Cipra. See also Danby (March 1964, equation 28), Sicardy, Sosnitskii.

Danby (May 1964), Abstract, indicates that the above holds for non-circular orbits, but with a varying numerical constant; that was proved in Danby (March 1964).

Circular orbits are stable if   27(m1m2 + m2m3 + m3m1)   <   (m1 + m2 + m3)2

For equal masses, put m1 = m2 = m3 = 1, and the comparison reduces to 27×3 < 32 so the configuration is clearly unstable.

The pattern is clearly stable if two of the masses are negligible : 27×0 < 12.

For the border of stability when the third body is a particle, m3 = 0 :- m12 - 25 m1m2 + m22   =   0
m1/m2 = (25 ± √ (625-4)) / 2   =   24.9599 and 1/24.9599

For the border of stability when the smaller masses are equal, m3 = m2 :- 27(2m1m2 + m22)   =   (m1 + 2m2)2
m1/m2 = (-50 ± √ (2500+92)) / 46   =   0.019819 = 1/50.4558

If m2 and m3 are small, let their sum be m and ignore their product ; the margin is where m2 - 25mm1 + m12   =   0
m/m1 = (25 ± √(625-4))/2   =   24.9599
The border is approximately the line m2 + m3 = 0.04008.

m1 :   m2 :   m3 :
Eccentricity :   Stable :

The above tests all points on a 100×100 grid against the given stability expression.

See newsgroup thread.

The stability criterion can be given as (Orbital Motion,
A.E. Roy, p. 138):  mu = (1/2) - (23/108)^(1/2) , where mu is the
reduced mass of the secondary body; that's 0.03852 = 1/25.96,
i.e. primary mass is 24.96 times secondary mass.

For m_1=1, m_3=0 I get m_2 = 1/24.9599 for equality. For total mass 1 and  m_1=1-u  m_2=u  m_3=0  that expression becomes 27 u (1-u) < 1 from which u = 1/25.9599 for equality. The difference of reciprocals is exactly 1.0, and corresponds to a mere difference in terminology.

Stability of Dual Equilateral Configuration?

From the above, it is clear that, in a system with a massive primary and a much lighter secondary, particles of negligible mass can simultaneously reside stably at the two Lagrange points L4 and L5. But is any stability condition known for the four-body dual-equilateral system with non-negligible masses?

Trojan Points

Asteroids around Lagrange Points were first observed early in the 20th century, in the Sun-Jupiter system, and were named after Trojan and Greek heroes of Homer's Iliad. Those around L5 are named after Trojans, with one Greek among them. Those around L4 are named after Greeks, with one Trojan among them. The misplaced are Hector of Troy and Patroclus (slain by Hector) of Greece.

The IAU supports the naming convention, but seems to lack a definition of "Trojan asteroid".

The Lagrange Points L4 and L5 of any system are termed Trojan Points if the primary/secondary mass ratio is large enough for them to be positions of stable equilibrium for particles.

A Trojan asteroid can be a considerable distance from its Lagrange Point while still remaining bound to the Point. Asteroid 2010 TK7, JPL, at Sun-Earth L4, is one such case.

Lagrange Movie

An appendix, showing elliptical orbits and the changes as mass moves from one body to the other, is in Gravity 5 : A Lagrange Movie.

Known Objects at Lagrange Points

Objects residing at L1, L2, or L3 must necessarily be artificial. As yet, there are only natural objects at any L4 and L5 Points. For more, see List of objects at Lagrangian points and other Wikipedia pages. The Pluto-Charon L4 and L5 points are unstable (mass ratio about 10).

Some Known Objects at Lagrange Points
L1 L2SunEarthL1: Sunshade, against Global Warming
"""L1: LISA Pathfinder (NASA/ESA)
"""L1: KuaFu / Kuafu (China)
"""L2: James Webb Space Telescope (NASA)
"""L2: Euclid (ESA telescope)
"""L2: Plato (ESA planet finder)
"""L2: Athena (ESA X-ray observatory)
L4 L5EarthMoonColonies, G K O'Neill - position replanned
L1 L2SunEarth
L3 L4 L5
L1 L2SunEarthL1: SOHO solar observatory
"""L1: Advanced Composition Explorer (ACE)
"""L2: Gaia (ESA astrometry)
L4 L5EarthMoondust clouds around ?
"SunEarthL4: 2010 TK7
"""dust clouds ?
""MarsL4: 1999 UJ7
"""L5: asteroid 5261 Eureka, and others
""JupiterThe Trojan Asteroids (see above)
""UranusL4: 2011 QF99
""NeptuneL4: Some trojan asteroids
""NeptuneL5: 2008 LC18
"SaturnDioneL4: Helene
"""L5: Polydeuces (S/2004 S5)
""TethysL4: Telesto
"""L5: Calypso
L1 L2SunEarthL1: ISEE-3 - became comet explorer
"""L1: Genesis - moved to Utah for sample return
"""L2: WMAP - moved to graveyard
"""L2: Herschel
"""L2: Planck
"""L2: Chang'e 2 - 4179 Toutatis flyby
"EarthMoonL1 & L2 : Artemis - moved to lunar orbit
L3 L4 L5
L1 L2SunEarthL1: GRAIL A+B - via L1 to Moon orbit, became Ebb & Flow
L3 L4 L5

See also Hilda family asteroids, which approach Sun-Jupiter L3.

According to the beginning of A Search for Natural or Artificial Objects Located at the Earth-Moon Libration Points, Earth-Moon L4 and L5 are not stable, because of the influence of the Sun; but there are stable orbits around those points.

Multi-Body Configurations


Klemperer rosettes consist of many similar moons orbiting a planet or a point in one ring, or similar (site index).

The neat stability equation boxed above shows that three bodies of equal mass are not stable.

SF authors show various related beliefs.

Strange N-body Orbits

Other periodic multi-body configurations have been found, in which the configuration is not fixed, but does repeat :-

Errors in Web Sites, Press Releases, etc.

At the Lagrange points, the gravitational forces do not "balance out" (a much-propagated error, often by those who should know better). They jointly provide the inwards (centripetal) force necessary to maintain a particle in orbit at a constant relative position in the rotating system. If rotating coordinates are intended, they need to be explained. Sample errors : 1, 2, 3, 4, 5, 6, etc.

The point between primary and secondary at which the gravitational fields of the two bodies are equal and opposite is never as such a Lagrange Point, though it does coincide with L1 if and only if the two bodies have the same mass.

Some sites say that L1 and L2 are equidistant from the secondary body, which is only approximately true.

Many sites have put L3 at a greater distance from the centre of the primary than the secondary is, though actually it is at a slightly smaller distance. But it is correct to say that L3 lies outside the orbit of the secondary. Some draw L3 unreasonably near to the primary.

At least one site has said that L4 and L5 lie on the orbit of the secondary; but they actually lie a little outside it.

Another error commonly made by Web authors seems to be ignoring what Lagrange himself actually wrote.

Leonhard Euler can be credited with finding L1 & L2, but apparently not L3, and not L4/L5.


Note : descriptions of what Lagrange actually did are commonly wrong.

The Lagrange Points

Lagrange's life

Some Early References to the Lagrange Points


Out-of-date Taiwanese copy of this page, by Wen-Nung Tsai of NCTU - http://www.cs.nctu.edu.tw/~tsaiwn/

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