The Lagrange Points

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Document Created by Neil J. Cornish for WMAP Education and Outreach/ 1998

The Lagrange Points

There are ve equilibrium p oints to be found in the vicinity of two orbiting

masses. They are called Lagrange Points in honour of the French-Italian

mathematician Joseph Lagrange, who discovered them while studing the re-

stricted three-b o dy problem. The term \restricted" refers to the condition

that two of the masses are very much heavier than the third. Today we

know that the full three-b o dy problem is chaotic, and so cannot be solved

in closed form. Therefore, Lagrange had go o d reason to make some approx-

imations. Moreover, there are many examples in our solar system that can

be accurately describ ed by the restricted three-b o dy problem.

M 1 M 2

r1 r2

Figure 1: The restricted three-b o dy problem

The pro cedure for nding the Lagrange points is fairly straightforward:

We seek solutions to the equations of motion which maintain a constant

separation b etween the three b o dies. If M and M are the two masses, and

1 2

~r and r~ are their resp ective p ositions, then the total force exerted on a

1 2

third mass m, at a p osition ~r , will b e

GM m GM m

2 1

~r r~ ~r r~ : 1 F =

1 2

3 3

j~r r~ j j~r r~ j

1 2

The catch is that both ~r and ~r are functions of time since M and M

1 2 1 2

are orbiting each other. Undaunted, one may pro ceed and insert the orbital 1

Document Created by Neil J. Cornish for WMAP Education and Outreach/ 1998

solution for ~r t and ~r t obtained by solving the two-body problem for

1 2

M and M and lo ok solutions to the equation of motion

1 2

d ~r t

F t= m ; 2

that keep the relative p ositions of the three b o dies xed. It is these stationary

solutions that are know as Lagrange p oints.

The easiest way to nd the stationary solutions is to adopt a co-rotating

frame of reference in which the two large masses hold xed p ositions. The

new frame of reference has its origin at the center of mass, and an angular

frequency given by Kepler's law:

2 3

R = GM + M : 3

1 2

Here R is the distance b etween the two masses. The only drawback of using

a non-inertial frame of reference is that we have to app end various pseudo-

forces to the equation of motion. The e ective force in a frame rotating

~ ~

with angular velocity is related to the inertial force F according to the

transformation

d~r

~ ~ ~ ~ ~

m ~r : 4 F = F 2m

The rst correction is the coriolis force and the second is the centrifugal force.

The e ective force can b e derived from the generalised p otential

~ ~ ~

U = U ~v ~r + ~r ~r ; 5

as the generalised gradient

F = r U + r U : 6

~r ~v

The velocity dep endent terms in the e ective p otential do not in uence the

p ositions of the equilibrium points, but the are crucial in determining the

dynamical stability of motion ab out the equilibrium p oints. A plot of U

with ~v =0, M =10M =1 and R =10is shown in Figure 2. The extrema

1 2

of the generalised potential are lab eled L1 through L5. 2 Document Created by Neil J. Cornish for WMAP Education and Outreach/ 1998

L 4

L 3 L 2

L 1

L 5

Figure 2: A countour plot of the generalised p otential.

Cho osing a set of cartesian co ordinates originating from the center of

mass with the z axis alined with the angular velocity, we have

= k

~r = xt^ +y t^

~r = R ^

~r = R ^ 7

where

M M

2 1

= ; = : 8

M + M M + M

1 2 1 2

To nd the static equilibrium p oints we set the velo city ~v = d~r=dt to zero

and seek solutions to the equation F = 0, where

3 3

x RR x + RR

F = ^ x

2 2 3=2 2 2 3=2

x + R + y x R + y

3 3

y R yR

^ : 9 y

2 2 3=2 2 2 3=2

x + R + y x R + y 3

Document Created by Neil J. Cornish for WMAP Education and Outreach/ 1998

Here the mass m has b een set equal to unity without loss of generality. The

brute-force approach for nding the equilibrium p oints would be to set the

magnitude of each force comp onent to zero, and solve the resulting set of

coupled, fourteenth order equations for x and y . A more promising approach

is to think ab out the problem physically, and use the symmetries of the

system to guide us to the answer.

Since the system is re ection-symmetric ab out the x-axis, the y comp o-

nent of the force must vanish along this line. Setting y = 0 and writing

x = R u + so that u measures the distance from M in units of R , the

condition for the force to vanish along the x-axis reduces to nding solutions

to the three fth-order equations

2 2 3 2 3 4

u 1 s +3u +3u + u = s +2s u +1+s s u +2u + u ; 10

1 0 0 0 1

where s = signu and s = signu + 1. The three cases we need to solve

0 1

have s ;s equal to 1; 1, 1; 1 and 1; 1. The case 1; 1 cannot

0 1

o ccur. In each case there is one real ro ot to the quintic equation, giving

us the p ositions of the rst three Lagrange points. We are unable to nd

closed-form solutions to equation 10 for general values of , so instead we

seek approximate solutions valid in the limit 1. To lowest order in ,

we nd the rst three Lagrange p oints to b e p ositioned at

! "

1=3

; 0 ; L1: R 1

! "

1=3

; 0 ; L2: R 1+

L3: R 1+

; 0 : 11

6 8

For the earth-sun system 3 10 , R = 1AU 1:5 10 km, and

the rst and second Lagrange points are lo cated approximately 1.5 million

kilometers from the earth. The third Lagrange p oint - home of the mythical

planet X - orbits the sun just a fraction further out than the earth.

Identifying the remaining two Lagrange points requires a little more

thought. We need to balance the centrifugal force, which acts in a direction

radially outward from the center of mass, with the gravitational force exerted

by the two masses. Clearly, force balance in the direction p erp endicular to 4

Document Created by Neil J. Cornish for WMAP Education and Outreach/ 1998

centrifugal force will only involve gravitational forces. This suggests that we

should resolve the force into directions parallel and p erp endicular to ~r . The

appropriate pro jection vectors are x^ + y^ and y^ x^ . The p erp endicular

pro jection yields

1 1

? 2 3

: 12 F = y R

2 2 3=2 2 2 3=2

x R + y x + R + y

Setting F = 0 and y 6= 0 tells us that the equilibrium points must be

equidistant from the two masses. Using this fact, the parallel pro jection

simpli es to read

2 2

x + y 1 1

F = : 13

3 2 2 3=2

R R x R + y

Demanding that the parallel comp onent of the force vanish leads to the

condition that the equilibrium p oints are at a distance R from eachmass. In

other words, L4 is situated at the vertex of an equilateral triangle, with the

two masses forming the other vertices. L5 is obtained by a mirror re ection

of L4 ab out the x-axis. Explicitly, the fourth and fth Lagrange p oints have

co ordinates

R M M 3

1 2

L4: ; R ;

2 M + M 2

1 2

R M M 3

1 2

L5: ; R : 14

2 M + M 2

1 2

Stability Analysis

Having established that the restricted three-b o dy problem admits equilib-

rium p oints, our next task is to determine if they are stable. Usually it is

enough to lo ok at the shap e of the e ective potential and see if the equilib-

rium p oints o ccur at hills, valleys or saddles. However, this simple criterion

fails when wehaveavelo city dep endent p otential. Instead, wemust p erform

a linear stability analysis ab out each Lagrange p oint. This entails linearising

the equation of motion ab out each equilibrium solution and solving for small 5

Document Created by Neil J. Cornish for WMAP Education and Outreach/ 1998

departures from equilibrium. Writing

x = x + x ; v = v ;

i x x

x = y + y ; v = v ; 15

i y y

where x ;y isthe p osition of the i-th Lagrange p oint, the linearised equa-

i i

tions of motion b ecome

1 0

0 0 1 0

0 0 1 1

C B

x x

C B

B CB C B C

0 0 0 1

B CB C B C

y y

B CB C B C

2 2

B CB d U C = B d U C : 16

B CB C B C

dt 0 2

B CB C B C

v v

x x

dx dxdy

B CB C B C

C B

@ @ A A

C B

2 2

A @

d U d U

v v

y y

2 0

dy dx dy

Here the second derivatives of U are evaluated at ~r =x ;y .

i i

Stability of L1 and L2

The stability of the rst and second Lagrange p oints of the earth-sun system

is an imp ortant consideration for some NASA missions. Currently the solar

observatory SOHO is parked at L1, and NASA plans to send the Microwave

Anisotropy Prob e MAP out to L2. It has also b e suggested that the Next

Generation Space Telescop e NGST should b e p ositioned at L2.

The curvature of the e ective p otential near L1 and L2 reveals them to

be saddle points:

2 2 2 2

d U d U d U d U

2 2

= 9 ; = 3 ; = =0: 17

2 2

dx dy dxdy dy dx

Solving for the eigenvalues of the linearised evolution matrix we nd

q q

p p

= 1+2 7 and = i 2 7 1 : 18

The presence of a p ositive, real ro ot tells us that L1 and L2 are dynamically

unstable. Small departures from equilibrium will grow exp onentially with a 6

Document Created by Neil J. Cornish for WMAP Education and Outreach/ 1998

e-folding time of

2 1

= : 19

For the earth-sun system = 2 year and 23 days. In other words, a

satellite parked at L1 or L2 will wander o after a few months unless course

corrections are made.

Stability of L3

A p opular theme in early science ction stories was invasion by creatures

from Planet X. The requirement that Planet X remain hidden behind the

sun places it at the L3 point of the earth-sun system. Unfortunately for

our would-be invaders, the L3 p oint is a weak saddle point of the e ective

potential with curvature

2 2 2 2

d U d U 7M d U d U

2 2

= 3 ; = ; = =0: 20

2 2

dx dy 8M dxdy dy dx

To leading order in M =M , the eigenvalues of the linearised evolution matrix

2 1

are

= and = i 7 : 21

The real, p ositive eigenvalue sp ells disaster for Planet X. Its orbit is exp o-

nentially unstable, with an e-folding time of roughly = 150 years. While

there can be no Planet X, the long e-folding time makes L3 a go o d place to

park your invasion force while nal preparations are made...

Stability of L4 and L5

The stability analysis around L4 and L5 yields something of a suprise. While

these p oints corresp ond to lo cal maxima of the generalised p otential - which

usually implies a state of unstable equilibrium - they are in fact stable. Their

stability is due to the coriolis force. Initially a mass situated near L4 or L5

will tend to slide down the p otential, but as it do es so it picks up sp eed and

the coriolis force kicks in, sending it into an orbit around the Lagrange p oint.

The e ect is analogous to howahurricane forms on the surface of the earth:

as air rushes into a low pressure system it b egins to rotate because of the 7

Document Created by Neil J. Cornish for WMAP Education and Outreach/ 1998

coriolis force and a stable vortex is formed. Explicitly, the curvature of the

potential near L4 is given by

2 2 2 2

d U d U d U 3 d U 3 9 3

2 2 2

; ; = = ; 22 = =

2 2

dx 4 dy 4 dxdy dy dx 4

where =M M =M + M . The eigenvalues of the linearised evolution

1 2 1 2

matrix are found to equal

= i 2 27 23

= i 2+ 27 23 : 23

The L4 point will be stable if the eigenvalues are pure imaginary. This will

be true if

27 23 2 : 24 and

The second condition is always satis ed, while the rst requires

0 1

1+ 1 4=625

@ A

M 25M : 25

1 2

When the L4 and L5 p oints yield stable orbits they are refered to as Tro jan

points after the three Tro jan asteroids, Agamemnon, Achilles and Hector,

found at the L4 and L5 p oints of Jupiter's orbit. The mass ratios in the

earth-sun and earth-mo on system are easily large enough for their L4 and

L5 p oints to be home to Tro jan satellites, though none have been found.

Author

These notes were written by Neil J. Cornish with input from Jeremy Go o d-

man. 8