1 Introduction

Lagrange points are points near two massive co-orbiting bodies, that remain fixed relative to the two massive bodies and a small mass kept at these points will experience no force. This is a consequence of the gravitational force due to the two large bodies and centrifugal force cancelling out at these points. It turns out that there are five such points, labelled L1 to L5, around any pair of massive bodies that orbit around their center of mass. For example, the Sun-Earth system has one set of five Lagrange points while the Earth-Moon system has a different set of Lagrange points. The L1, L2 and L3 points lie along the line passing through the centres of the two large bodies while the L4 and L5 points each lie at the third vertex of an equilateral triangle formed with the centres of the two large bodies. Thus all Lagrange points lie in the same 2D plane containing the two massive bodies. Lagrange points are named in honour of the French-Italian mathematician and physicist Joseph Louis Lagrange. Given below is an image showing the Lagrange points of the Sun-Earth system.

Figure 1: Lagrange points of the Sun-Earth system

1

2 The Problem

The problem to find Lagrange points around a system of two massive bodies is actually a three-body problem. The three body problem is not exactly solvable. However, if the mass of the third body is negligible compared to other two masses, we can ignore the reaction force of the third mass on the other two masses. Hence the problem of Lagrange points is a restricted three body problem. Leonhard Euler had discovered the existence Lagrange points L1, L2 and L3 a few years before Lagrange discovered the points L4 and L5. Lagrange studied the general three body problem extensively.

Consider two massive bodies of masses M1 and M2 (let M1 > M2) orbiting around their center of mass (assume circular orbits). Let their positions be r1 and r2 (I have used bold face to denote vectors and normal face to denote their magnitudes throughout) respectively with the origin of the coordinate system at the center of mass of the two bodies. Let m be a negligible mass compared to M1 and M2, located in the orbital plane of M1 and M2. Let r be its position vector. All these terms are shown in the figure below.

Figure 2: Layout of the problem

3 Finding Positions of the Lagrange points

3.1 Using force

We have to find general solutions for the position of Lagrange points. First, let’s find the general equation for total gravitational force at a point in the orbital plane where mass m is located. From Figure 2, we obtain the equation for total gravitational force which is given as

F = −GM1m

|r − r1|3(r − r1) −GM2m

|r − r2|3(r − r2) (1)

2

To find the Lagrange points, one can solve equation 1 by equating F = 0. However, r1 and r2 are changing with time because the two bodies are orbiting around their centre of mass. One can obtain solution for r1(t) and r2(t) by solving the two body problem and then look for solution of the equation

F(t) = md2r(t)

dt2

This will keep the relative positions of the three bodies fixed. The stationary solutions obtained by this method will be the Lagrange points.

However, the easiest method to solve for stationary solutions is to adopt a co-rotating frame of reference in which the masses M1 and M2 are at fixed positions with their centre of mass as the origin. The the co-rotating frame simulates the orbital motion of the two masses. As this frame of reference is rotating, it has some angular frequency Ω, which is same as the angular frequency of the orbiting bodies. Thus from Kepler’s third law we have

Ω2R3 = G(M1 + M2) (2)

where R is the distance between the two masses.

Since we have adopted a co-rotating frame of reference, which is a non-inertial frame of reference, we need to include pseudo-forces in the equation for total force. Thus equation (1) now becomes

|r − r1|3(r − r1) −GM2m 

Frot = −GM1m

|r − r2|3(r − r2) − 2m(Ω ×drdt) − m(Ω × Ω × r) 

∴ Frot = −GM1m

|r − r1|3(r − r1) −GM2m

|r − r2|3(r − r2) − 2m(Ω × r˙) − m(Ω × Ω × r)

(3)

where Ω is the angular velocity and r˙ is the velocity of mass m. The third term in the above equation represents Coriolis force and the last term represents centrifugal force. Note that Euler force (−mr×Ω˙ ) is not included as the angular velocity Ω is assumed to be constant due to circular orbits. Equation 3 can be written in terms of acceleration as

¨r = −GM1

|r − r1|3(r − r1) −GM2

|r − r2|3(r − r2) − 2Ω × r˙ − Ω × Ω × r (4)

From Lagrangian mechanics we know that the generalised potential U for rotating non-inertial frames is given by

U = −GM1

|r − r1|−GM2

|r − r2|− r˙.(Ω × r) −12(Ω × r).(Ω × r)

|r − r2|− 2Ω(xy˙ − yx˙) −Ω22(x2 + y2)

∴ U = −GM1

|r − r1|−GM2

(5)

The last term is called the “centrifugal potential” corresponding to centrifugal force. One may say that second last term represents “Coriolis potential” but it is not quite true as Coriolis force is non-conservative and cannot have a potential.

3

Let r1 = −r1ˆi and r2 = r2ˆi. Now let R = −r1 + r2 which implies R = (r1 + r2)ˆi. Thus R denotes the distance between the two massive bodies.

As the origin of the system is at center of mass of the two massive bodies, we have M1r1 + M2r2 = 0 (6)

Thus we have

r1 = −αR and r2 = βR

M1 + M2and β =M1

where α =M2

M1 + M2. Also note that α + β = 1. 

For convenience, we can write the distances between mass m and M1 and M2 respectively as d1 = |r − r1|=p(x + αR)2 + y2 + z2 and d2 = |r − r2|=p(x − βR)2 + y2 + z2

To find the positions of Lagrange points, we need to obtain solutions of Frot = 0 or equivalently ¨r = 0. We can split equation 4 into x, y and z components as follows.

x¨ = −GM1(x + αR)

d31−GM2(x − βR)

d32+ 2Ω ˙y + Ω2x (7)

y¨ = −GM1y

d31−GM2y

d32− 2Ω ˙x + Ω2y (8)

z¨ = −GM1z

d31−GM2z

d32(9)

As the equilibrium points lie in the orbital plane of the two masses, these points are independent of z. However, for sake of completeness, we need to include z for stability analysis in Section 4. One needs to set the above x and y components to zero and solve the resulting set of coupled, fourteenth order differential equations. However, this approach is very cumbersome. A better approach is to exploit the geometrical symmetries of the problem. By physical intuition, we know that at least one point must lie on the axis passing through two massive bodies and that the system is symmetric about x-axis. Thus we can solve the equation of x-component of acceleration by setting y = 0 and z = 0. Further, we can write general solution of the equation as x = R(u + α) so that u represents the distance of equilibrium point from M2 in terms of R. Substituting this general solution in the equation of x-component of acceleration, we arrive at solving three quintic equations in u. These equations have the general form

u2((1 − s1) + 3u + 3u2 + u3) = α(s0 + 2s0u + (1 + s0 − s1)u2 + 2u3 + u4) (10)

where s0 = sign(u) and s1 = sign(u + 1). The three equations arise from the three values of the ordered pair (s0, s1) which are (−1, 1),(1, 1) and (−1, −1) that correspond to the Lagrange points L1, L2 and L3 respectively. The case (1, −1) cannot occur. Each quintic equation has exactly one real root which corresponds to respective Lagrange point as mentioned above. We are unable to solve the quintic equations for arbitrary values of α and hence we find approximate solutions for the condition α << 1. The general positions of Lagrange points L1, L2 and L3 are

4

L1 = L2 =

R

R

1 −

1 +

α 3

α 3

1/3 , 0, 0 1/3 , 0, 0

L3 =

−R

1 +512α , 0, 0

From the above solutions, we see that L1 and L2 are equidistant from M2, on opposite sides of it. However, this is not quite true. The L2 point lies further away from M2 that L1.

Finding solutions for L4 and L5 points requires a little more effort. We need to balance forces for y 6= 0. Centrifugal force acts radially outward and needs to be balanced with the gravitational force exerted by the two masses. In the direction perpendicular to centrifugal force, we need to balance only the gravitational forces. This suggests that we should balance forces in directions parallel and perpendicular to r. The perpendicular projection yields 

F⊥rot = αβyΩ2R3

1 

((x − βR)2 + y2)3/2−1

((x + αR)2 + y2)3/2

Setting F⊥rot = 0 and y 6= 0 we get a solution x =12(βR −αR) i.e. the x-coordinate of the Lagrange point (if one exists) lies exactly halfway between the two masses. Using this result, the force parallel to r becomes 

Fkrot = Ω2(x2 + y2) 2

1 

R3−1

((x − βR)2 + y2)3/2

Setting Fkrot = 0 we get y = ±

√3

2R. Thus L4 is located at the third vertex of an equilateral triangle 

of side R, formed by the two masses and L5 is mirror reflection of L4 about the x-axis. Thus L4 and L5 are located at the following coordinates: 

L4 =

R

2

M1 − M2 M1 + M2

,

√3

2R, 0

!

L4 =

R

2

M1 − M2 M1 + M2

, −

√3

2R, 0

! 

Thus we found general coordinates of all five Lagrange points by solving equation of force. 5

3.2 Using potential

The net potential is given as

U = −GM1

|r − r1|−GM2

|r − r2|− r˙.(Ω × r) −12Ω2r2.

Note that in addition to position, the potential has a velocity dependent term and, although it has no effect on position of the Lagrange points, it must be taken into account when looking at their stability (Section 4).

As we have to find stationary solutions, let r˙ = 0. Thus the equation of potential becomes |r − r1|−GM2

U = −GM1

|r − r2|−12Ω2r2(11) 

Differentiating U with respect to position will give us force as gravitational and centrifugal forces are conservative. Thus if we take gradient of potential, then we can find find equilibrium points by setting ∇U = 0. However, as this is equivalent of F = 0, we find ourselves back at solving equations of force as done previously. So instead of finding analytical solutions, let’s find numerical solutions using Wolfram Mathematica.

First, we need to non-dimensionalize the equation of potential, for it to be used in Mathematica. We achieve this as follows:

From equations 2 and 11 we have

|r − r2|−12G(M1 + M2)r2

|r − r1|−GM2

U = −GM1

= −pGM1

R3

(x − βR)2 + y2−12G(M1 + M2)r2

(x + αR)2 + y2− pGM2

Dividing by G(M1 + M2)/R we get

UR

R3

p(x − βR)2 + y2−12r2

p(x + αR)2 + y2−αR

G(M1 + M2)= −βR

R2

Now for given values of M1, M2 and R, G(M1 +M2)/R is a constant and hence LHS can be written as a function of position. Thus, 

U(x, y) = −βR

p(x − βR)2 + y2−12r2

p(x + αR)2 + y2−αR

R2

R2. . . . . . [∵ r =px2 + y2] 

= −βR q

−αR q

−12x2 + y2

R

(xR + α)2 +y2 R2

R

(xR − β)2 +y2 R2

= −β

q

(xR + α)2 +y2

R2

−α

q

(xR − β)2 +y2

R2

−12x2 + y2 R2

−12x2 + y2

∴ U(x, y) = −β

−1 − β

q

q

R2

(xR + 1 − β)2 +y2

(xR − β)2 +y2

R2

R2

6

. . . . . . [∵ α + β = 1] (12) 

If we plot U(x, y) and set its gradient to zero, i.e. ∇U(x, y) = 0, then we will get the positions of five Lagrange points. For some value of β and R, the effective potential was plotted in Mathematica and its gradient was set to zero to obtain corresponding Lagrange points. Following are the contour and 3D plots of the same: 

(a) Contour plot of U(x, y) (b) Surface plot of U(x, y) 

(c) Cross-sectional view of (b) - 1 (d) Cross-sectional view of (b) - 2

Figure 3: Plots of effective potential for β = 0.7 and R = 1.25 with corresponding positions of Lagrange points

It is insightful to plot the net gravitational potential and centrifugal potential in equation 12. Given below are the plots for the same:

7

(a) (b)

Figure 4: Plots of gravitational (a) and centrifugal (b) potentials for β = 0.7 and R = 1.25

As expected, the gravitational potential tends to zero as distance from the two masses increases. Centrifugal potential on the other hand becomes more negative (increases negatively) as we move away from the system. The addition of these two potentials gives us the effective potential in rotating frame of reference, as shown in Figure 3(a), (b).

4 Stability Analysis

Now that we have established the existence of fixed points in the restricted three body problem, we also need to determine whether they are stable. The shape of effective potential as seen in Figure 3(b) gives a rough idea of their stability, but it is not sufficient as the potential also depends on velocity. Instead we must perform a linear stability analysis about each equilibrium point and look at small perturbations. To do so, we first separate the components dependent on velocity:

(x − βR)2 + y2 + z2−Ω22(x2 + y2) (13)

(x + αR)2 + y2 + z2− pGM2

U0 = U + 2Ω(xy˙ − yx˙) = −pGM1

Thus the components of total force as given in equations 7-9 can be written as x¨ = −∂U0

∂x + 2Ω ˙y (14)

y¨ = −∂U0

∂y − 2Ω ˙x (15)

z¨ = −∂U0

∂z (16)

8

Also, it is useful to write the net potential as a sum of partial derivatives. Using the Taylor series expansion of U0(upto second order) around a Lagrange point (x0, y0, z0) we get

U0 = U00 + U0x(x − x0) + U0y(y − y0) + U0z(z − z0)

+12 U0xx(x − x0)2 + U0yy(y − y0)2 + U0zz(z − z0)2

+ U0xy(x − x0)(y − y0) + U0xz(x − x0)(z − z0) + U0yz(y − y0)(z − z0) (17) 

where U0 = U0

 (x0,y0,z0)and U0X and U0XX0 represent single and double partial derivatives of U0for 

any variable X and X0(X = X0 or X 6= X0).

However, note that U0xz = U0yz = 0 and by definition, U0x = U0y = U0z = 0 for all Lagrange points (as net force is zero). Thus we have

U0 = U00 +12 U0xx(x − x0)2 + U0yy(y − y0)2 + U0zz(z − z0)2 + U0xy(x − x0)(y − y0) (18) General coordinates for some point very close to the Lagrange points are given as:

x = x0 + δx x˙ = δx˙ (19)

y = y0 + δy y˙ = δy˙ (20)

z = z0 + δz z˙ = δz˙ (21)

Plugging the general coordinates in equation 18 gives us

U0 = U00 +12 U0xx(δx) + U0yy(δy) + U0zz(δz) + U0xyδxδy (22) Now using equations 14-16 and 22 we get

δx¨ = −U0xx − U0xy + 2Ωδy˙ (23)

δy¨ = −U0yy − U0xy − 2Ωδx˙ (24)

δz¨ = −U0zzδz (25)

We see from the above equations that z is independent of x and y; more specifically, δz and δz˙ are independent of δx, δx˙, δy and δy˙ and vice versa. So we can prove that all Lagrange points are stable in the z-direction. We have the following linearised matrix equation: 

d

dt

δz δz˙

=

0 1 −U0zz 0

δz δz˙

(26) 

Since d1 and d2 are always positive, from equation 16 we have U0zz > 0. Now, the eigenvalues of the coefficient matrix in equation 26 are:

±ipU0zz

9

Since both eigenvalues are always imaginary, we know from theory of stability of fixed points that purely imaginary pair of eigenvalues correspond to stable fixed points. Thus we conclude that all Lagrange points are stable in the z-direction.

Advancing in stability analysis, from equations 19, 20, 23 and 24 we see that x and y are dependent on each other. Thus our linearised equations of motion for x and y coordinates take the following matrix form: 

d

dt

 

δx δy δx˙ δy˙



 =

 

0 0 1 0 0 0 0 1 −U0xx −U0xy 0 2Ω −U0xy −U0yy −2Ω 0

 

 

δx δy δx˙ δy˙



(27) 

To check for stability, we need to compute the partial double derivatives of U0 at each of the Lagrange points and find the eigenvalues of corresponding evolution matrices.

4.1 Stability of L1 and L2

Just by looking at the shape of effective potential around L1 and L2 we can say that they are unstable points; saddle points to be exact. The partial double derivatives of U0 at L1 and L2 turn out to be (tedious calculation is skipped)

Point U0xx U0yy U0xy

L1 −9Ω2 3Ω2 0

L2 9Ω2 −3Ω2 0

The evolution matrix for L1 has the following eigenvalues: 

q

λ± = ±Ω

1 + 2√7 σ± = ±iΩq2√7 − 1 

As one of the eigenvalues is real and positive and one real and negative, L1 is a saddle point. Similar procedure can be done for L2 to get the same conclusion. Hence we conclude that L1 and L2 are saddle points and thus dynamically unstable. Small departure from these points will grow exponentially and mass m will drift away from these points.

4.2 Stability of L3

We show that L3 is also a saddle point and hence unstable. The partial double derivatives of U0 at L3 are found to be

U0xx = −3Ω2 U0yy =7α

8βΩ2 U0xy = 0

The evolution matrix for L3 has the following eigenvalues:

r3β 

λ± = ±Ω

8ασ± = ±iΩ√7 10

The real, positive and negative eigenvalues make L3 a saddle point and hence dynamically unstable. Just as L1 and L2, small departure from L3 will grow exponentially. However, as the eigenvalue is much smaller than in case of L1 and L2, an object kept at L3 will drift away much more slowly than it would if displaced by the same amount from L1 or L2. The unstable nature of L3 is also evident from shape of effective potential around L3 as in Figure 3(b).

4.3 Stability of L4 and L5

Looking at the plot of effective potential, one would think that L4 and L5 are also unstable as they correspond to local maxima of the effective potential. Quite surprisingly, stability analysis shows that L4 and L5 are in fact stable fixed points. Stability of these points is due to the effect of Coriolis force. When mass m departs from the point, it gains speed and Coriolis force (which is velocity dependent) comes into play, which makes the mass m start orbiting the Lagrange point. The partial double derivatives of U0 at L4 and L5 are found to be

U0xx = −34Ω2 U0yy = −94Ω2 U0xy = −3√3

4κ±Ω2

where κ± = ±(M1 − M2)/(M1 + M2). The eigenvalues of the linearised evolution matrices are found to be

λ± = ±iΩ2r2 −q27κ2± − 23 σ± = ±iΩ2r2 + q27κ2± − 23

Thus, L4 and L5 would be stable only if the eigenvalues are purely imaginary. Otherwise, there will be at least one eigenvalue such that Re(λ) > 0, which will make the points unstable in nature and 

hence

r

2 −

q

27κ2± − 23 must be real. Note that since |κ±|≤ 1, the condition 2−

q

27κ2± − 23 ≥ 0 

is always satisfied which leaves us with the condition 27κ2± − 23 ≥ 0. This condition is satisfied 

when

M2≥1 +

q

23

=25 + 3√69 

M1

1 −

q

27

23 27

2≈ 24.9599. 

This gives us the condition for stability of L4 and L5 which is M1/M2 ≥ 24.9599. If this condition is not satisfied then even L4 and L5 become unstable. Thus, Sun-Earth (M /M⊕ = 3, 33, 000) and Earth-Moon (M⊕/M$ = 81) systems have stable L4 and L5 points. Same applies for all Sun-planet systems in the solar system.

5 Discussion and Conclusion

We proved the existence of five Lagrange points and solved the restricted three-body problem to get general coordinates of the Lagrange points. We also found the Lagrange points numerically by using potential and plotting it in Mathematica. In stability analysis we found that L1, L2 and L3 points are unstable in nature while L4 and L5 are stable points.

Lagrange points are great spots to put purpose specific artificial satellites. For example, NASA’s Solar and Heliospheric Observatory (SOHO) is located at the Sun-Earth L1 point as this point

11

provides uninterrupted view of the Sun. For same reason, observations of the sunlit hemisphere of the Earth can also be made from Sun-Earth L1 point. The Sun-Earth L2 point lying beyond Earth is a good spot for making space based observations. The soon-to-be-launched James Webb Space Telescope (JWST) is going to be deployed at the Sun-Earth L2 point. The ESA GAIA space observatory is also at Sun-Earth L2. The Sun-Earth L3 was a popular subject of science fiction writers who would place a “counter Earth” at this point. The Sun-Earth L3 point is unstable and could not contain a natural object, large or small for very long. This is because the gravitational forces of other planets are stronger than that of Earth, for example, Venus comes within 0.3 AU of this point every 20 months.

Missions to the Lagrange points orbit the points in Lissajous orbit orbits or halo orbits (for L1, L2 and L3), instead of occupying the actual point. The L1, L2 and L3 points being dynamically unstable require periodic orbit corrections for artificial satellites to remain in the desired orbit. Thus duration of missions going to L1 − L3 points is limited by the fuel reserves of the satellite.

The L4 and L5 points being stable tend to pull natural objects into orbits around these points. These objects are called ’Trojans’ or ’Trojan asteroids’. In fact, the L4 and L5 points of Sun-Jupiter system house the most number of trojans (more than a million) than any other pair of bodies in the solar system. The region of stability around L4 and L5 is in fact large due to the effect of Coriolis force. Due to this, some trojans are found to exist more than 5◦ off of 60◦, where L4 and L5 are located. The Earth-Moon L4 and L5 points are stable for millions of years, if not billions of years. This is because small perturbations by other planets in the solar system eventually make an object place at these points drift away.

References

Web articles

• “Lagrangian Points” by Dennis Westra.

• “The Lagrangian Points” by Neil P. Cornish, University of Montana.

• Lecture L18 - Exploring the Neighborhood: the Restricted Three-Body Problem by S. Wid nall, MIT OCW.

• Lagrange Point Wikipedia

Books

• Course of Theoretical Physics, volume 1 by E. M. Lifshitz and L. D. Landau, Elsevier Inc., 3rd edition, 2011.