# The Lagrange points in the Earth-Moon system - Esa The Lagrange points in the Earth-Moon system - Esa

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The Lagrange points in the Earth-Moon system

Introduction:

Matthias Borchardt

Tannenbusch Secondary School, Bonn

borchardt. matthias@t-online.de

What forces are felt by a space probe that is travelling unpowered in the vicinity of the Earth and

the Moon?

First there are the attractive forces of the two celestial bodies that, according to the law of

gravitation, act upon the probe to a greater or lesser extent.

Added to this is a third force that only comes into play once you stop viewing the system of masses

as static, but instead take into account the dynamics of the rotation of the Earth and the Moon

around a shared centre of gravity. This movement creates a centrifugal force - a force that is

directed away from the centre of gravity.

These three forces at five points in space add up to zero - they are the five Lagrange points L1, L2,

L3, L4 and L5. They are also called the ‘parking lots’ of the Earth-Moon system because a space

probe that is positioned exactly at these positions remains stationary in relation to the two celestial

bodies.

Even small deviations from these special points quickly cause complicated behaviour: At L1, L2

and L3 we can see that the space probe is drifting away - although initially this happens at a very

low speed such that it should be possible to maintain the position for a longer period of time with

minimal course corrections.

A probe behaves differently in the vicinity of L4 and L5: it circles these points on looping paths,

and a course correction is not necessarily required as the curves always remain in the vicinity of the

Lagrange points.

Earth

Centre of gravity

Moon

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The centre of gravity:

The Earth and the Moon rotate around a common centre of gravity, the position of which can be

calculated as follows:

M-xM = m-xm

if r = xM + xm then

M-xM =m-(r-xM) or M-(r- xM ) =

M -xM = mr -m-xM M-r-M-xm =m-xm

(M + m)-xM = m-r M r = (m + M)-xm

If M = 5.97 • 10 24 kg , m = 7.35 • 10 22 kg and r = 384 400 km (half-way between the Earth and the

Moon) then XM = 4675 km.

This means that the centre of gravity S is still within the terrestrial globe.

For the purpose of the overview we will however always show S as being outside of it.

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The centrifugal force:

Let’s consider the centrifugal acceleration: , where x is the distance from the centre of

gravity S.

The angular speed of this rotation is:

( r - distance between M and m)

This formula should be derived:

Movement of the Earth around S Movement of the Moon around S

The gravitational force acts as radial force

if then if then

equal

Of course we must obtain the same value for the angular speed of the Earth as for the Moon.

Stated differently: both celestial bodies require the same period of time to orbit the common centre

of gravity.

The orbit period yields:

= 27.3 days. That is the sidereal month

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The gravitational forces:

The gravitational accelerations yield the following:

The Earth exerts an acceleration of on a body.

The Moon exerts an acceleration of on a body.

x is always the distance to the respective centre of mass.

The Lagrange point L1:

The origin of our coordinate system should always be the centre of gravity S:

With the aid of the drawing the three accelerations yield: (if: r = xM + xm)

At Lagrange point L1 the accelerations should add to zero.

Therefore aG1 = aG2 + az applies here.

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(Stated differently: In L1 the gravitation of the Earth cancels out the gravitation of the Moon and

the centrifugal force.)

The equation would have to be solved after x1 if you wanted to

calculate the position of the Langrange point.

This is mathematically very complex. Therefore we should take an easier route in order to

determine L1, that is, we should enlist the aid of a spreadsheet program.

This type of software is used to help calculate both the acceleration aG1 and the sum

applies

. The Lagrange point L1 is located at the position of the x1 value for which the following

This method with the aid of a spreadsheet delivers the following result for L1:

x1 = 321.688.900 m, which approximates to x1 = 321.689 km.

The Lagrange point L2:

With the aid of the drawing the three accelerations yield: (if: r = xM + xm)

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At Lagrange point L2 the accelerations should add to zero.

Therefore aG1 + aG2 = az applies here.

(Stated differently: In L2 the gravitational forces of the Earth and the Moon cancel out the

centrifugal force.)

The spreadsheet method delivers the following result for L2:

x2 = 444.260.800 m, which means approximately x2 = 444.261 km.

The Lagrange point L3:

The three accelerations yield: (if: r = xM + xm)

At Lagrange point L3 the accelerations should add to zero.

Therefore aG1 + aG2 = az applies here.

(Stated differently: In L3 the gravitational forces of the Earth and the Moon cancel out the

centrifugal force.)

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This method with the aid of a spreadsheet delivers the following result for L3:

x3 = -386.347.900 m, which means approximately x3 = -386.348 km

The Lagrange points L4 and L5:

At the Lagrange points L4 and L5 the relationships are somewhat more complicated as the forces

there are no longer parallel or anti-parallel, they must instead be added together according to the

Gravitational force of the Earth

Centrifugal force

Gravitational force of the Moon

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At these Lagrange points the two gravitational forces add to a resulting one that exactly corresponds

to the amount of the centrifugal force but which is directed against it. (Seen by an outside observer

one would have to say: the two gravitational forces together generate the radial force that an orbit

with the orbit period T=27.3 days around S generates.)

A complex mathematical analysis yields a surprising result for the position of the points L4 and

L5: The centres of mass of M and m and the Lagrange point L5 form an equilateral triangle with

the edge length r.

The coordinates of the Lagrange point L5 can therefore be obtained through simple geometric

examination (e.g. if is the height in an equilateral triangle with edge length r ):

and correspondingly for L4

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Now we want to prove that the three forces (accelerations) in L5 actually do add to zero if we

select L5 as a corner of the equilateral triangle described.

The cosine rule means:

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if

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Now we should calculate the centrifugal acceleration az :

, where d is the distance from L5 to the centre of gravity. This distance can be calculated

with the aid of Pythagoras’ theorem as the coordinates of L5 relative to the centre of gravity are

Thus if then:

That is the same value as the one resulting from aGi and aG2.

This shows that the three forces in L5 actually do cancel each other out.

The movement of a space probe in the vicinity of the Lagrange points:

How does a space probe behave when we position it in the vicinity of a Lagrange point?

We cannot carry out a mathematical analysis of the movements as it is exceedingly complex.

But using the simulation program ‘lagrange.exe’ important characteristics of the Lagrange points

can be investigated very easily.

The program calculates the two gravitational forces and the centrifugal force exerted on the location

of a spacecraft and thus determines the direction and amount of acceleration and speed of the craft

and hence its new position after a period of time Δt (Euler-Cauchy method).

However only those paths that lie on the plane of the Earth, centre of gravity and Moon are

calculated. It is therefore not possible to follow the path of a probe that has been placed above or

below this plane.

From a teaching point of view this reduction is especially advantageous as the relationships

between the forces are still able to be imagined to some degree. There is an additional advantage

due to the possibility of mapping the path of the space probe onto the surface of the gravitational

potential (‘potential surface’) and to draw the entire thing as an axonometric projection. This

creates a very vivid impression of the topology of the energetic relationships in the rotating Earth-

Moon system.

The coordinate system in which the space probe’s movements are calculated and drawn rotates

around the common centre of gravity together with the Earth-Moon system. This means that the

Earth and the Moon are resting in this reference system, which allows a clear representation of the

space probe’s path as we can see how it moves only in relation to the Earth and the Moon.

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The simulation shows that L1, L2 and L3 are apparently not stable points of equilibrium.

Topologically they are similar to a saddle.

In contrast, L4 and L5 are (very flat) hills and one could assume that a space probe would ‘slide

down’ these hills and would then drift away over time. Surprisingly this does not happen, however,

because as a result of sliding down the space probe gathers speed and, due to the rotation around the

centre of gravity, a Coriolis force is created that so strongly warps the path that looped paths are

created around the Lagrange point. In the process the space probe only moves away a certain

distance - spatially it thus remains almost stationary in relation to the Earth and the Moon. Course

corrections are normally not required.

Therefore L4 and l5 are ideal points for a space station that has the task of observing both the Earth

and Moon over a long periods of time.

The following screenshots show the path of a space probe that has been placed at various points on

the Earth-Moon plane at a speed of zero (in relation to the coordinate system with which it is

rotating).

A little to the right of L1 the space probe drifts towards the Moon, and a little to the left of L1

towards the Earth - the saddle-like topology of the surrounding area prevents the space probe from

moving forwards or backwards in spite of the Coriolis forces being exerted.

In the vicinity of L5 the situation looks completely different: Here we have a hill rather than a

saddle, and the space probe - driven by Coriolis forces - makes looped circuits over it.

Note: The saddle shape and the hill shape of the Lagrange points can hardly be seen on the

axonometric projection of the three dimensional image of the gravitational potential as it is only

very weak.

Matthias Borchardt

Tannenbusch Secondary School Bonn

borchardt. matthias@t-online.de

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Start at L1 (1km to the right of Lagrange point)

PATH

Launch position Launch speed

Start at L1 (1km to the left of Lagrange point)

PATH

Launch position Launch speed

slow

slow

Start

Start

Stop

Simulation

Stop

Simulation

POTENTIAL

Axonometric

projection

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Delete

fast

Delete

fast

Time

Time

Moon

orbits

Days

Moon

orbits

Days

Plane

with Moon

without Moon

Rotation

around the

x axis

y axis

z axis

Zoom

Info

Data

End

POTENTIAL

Axonometric

projection

Plane

with Moon

without Moon

Rotation

around the

x axis

y axis

z axis

Zoom

Info

Data

End

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Path in the vicinity of L5

PATH

Launch position Launch speed

PATH

Launch position Launch speed

slow

slow

Start

Start

Stop

Simulation

Stop

Simulation

POTENTIAL

Axonometric

projection

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Delete

fast

Delete

fast

Time

Time

Moon

orbits

Days

Moon

orbits

Days

POTENTIAL

Axonometric

projection

Plane

with Moon

without Moon

Rotation

around the

Plane

x axis

y axis

z axis

Zoom

Info

Data

End

with Moon

without Moon

Rotation

around the

x axis

y axis

z axis

Zoom

Info

Data

End

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