The Lagrangian points are locations in space in the vicinity of two orbiting masses where the gravitational forces and the orbital motion balance each other to form a point at which a third body of negligible mass would be stationary relative to the two bodies.

There are a number of web resources describing the Lagrangian points, some of them offering a very good introduction and description of the 5 Lagrangian points, while others insist on the techical aspects of the problem. Nonetheless, in the following article we'll try to add to the non-mathematical and intuitive descriptions available on the web (like, for instance, the

ESA page related to the 5 lagrangian points or the

Wikipedia article) a detailed-mathematical view of the problem, starting from the technical defintion which states that the Lagrangian points are the stationary solutions of the circular restricted three-body problem.

If we consider the restricted 3 body problem:

_{}

Considering an equilibrium point P we have the condition:

_{}

Then the resultant force is:

_{}

_{}

_{}

If we introduce the center of mass of a system like being a specific point where the system’s mass behaves as the system is concentrated in that specific point we will have the following relation:

_{}

For the particular Sun-Earth system the barycenter is located as following:

_{}

_{}

_{}

In the case _{} we have _{} (449 km for the Sun-Earth system)

Now let’s introduce in the system the Coriolis force and the centrifugal force.

Coriolis force:

_{}

Centrifugal force:

_{}

Then the resultant force is:

_{}

Correspondingly, the generalized potential is:

_{}

_{}

The first term from the formula above _{} determines the position of the equilibrium points, and the second one _{} determines the stability of motion about the equilibrium points.

If we consider a Cartesian coordinates system around the center of mass of the system Sun-Earth as in the following picture:

_{}

Considering all above we obtain the resultant force:

_{}

_{}In order to determine the static equilibrium points we put the condition

_{}

Then the resultant force became:

_{}

_{}_{}

_{}

_{}

From the condition _{}we will find the position of the equilibrium points.

_{}

In particular, if we consider _{} we can find the first 3 Lagrange points (located in the line Sun-Earth).

_{}

Now, in order to simplify the calculation we consider:

_{}

_{}

_{}

Let’s replace in the above formula the fact, that _{}.

_{}

Now we have basically three possibilities (the fourth case is practically impossible):

_{}

Correspondingly to the above cases we will have three equations:

_{} (1)

_{} (2)

_{} (3)

Or if we want to express the equations in a general form we can write:

_{}

Then the three particular cases become:

_{} (1)

_{} (2)

_{} (3)

We have now to solve these 3 equations. We can do this either by using an approximation considering the particularity of the system Earth-Sun _{} or we can solve it numerically using the Newton method.

Let’s first consider the approximation _{} and try to find a series which approximate the real solution.

We then have in the first equation:

_{}

If we note _{}

_{}

For the second equation we have:

_{}

For the third equation:

_{}

In the second case can be proven that the three equations have all just one real solution the other 4 being complex. The real solutions can be found being in the interval:

_{}

Using a numerical program and replacing the constants in the particular case of the system Earth-Moon-Sun we will have:

* M1=1.98892E30;//mass of Sun [kg]*

* M2=5.9742E24;//mass of Earth [kg]*

* M3=7.36E22;//mass of Moon [kg]*

* **R=1.4959610E8;//1 AU=R [km];*

* al=(M2+M3)/(M1+M2+M3);//alpha*

* **be=M1/(M1+M2+M3);//beta*

Then the roots of the three equations (calculated with a precision of 1e-14) are:

x_{1}=-0.0100113196805154

x_{2}=0.010078587510165

x_{3}=-0.0100332069481937

So the physical coordinates of the first three Lagrange points of the system Earth-Moon-Sun are:

L_{1x}=148097990.737622

L_{1y}=0

L_{2x}=151103362.50271

L_{2y}=0

L_{3x}=148094716.487738

L_{3y}=0

expressed in km from the center of mass of the system.

In the particular case of the L2 point which will focus on we found it at 1507717.38502939 km from Earth on the line which lies the Sun and the Earth.

Let’s find now the solutions for the other 2 Lagrange points.

We have the resultant force:

_{}

Considering the two directions (parallel _{}and perpendicular_{}) we obtain the projection on the perpendicular direction_{}:

_{}

_{}

If _{} and _{}

_{}

_{}

Let’s consider now the projection on the parallel direction:

_{}Replacing the previous solution _{}

_{}

we have:

_{}

_{}

From the condition _{}

_{}

So the two solutions are_{}.

Then the last two Lagrange points are located as:

_{}

Physically the solutions are found replacing the constants for the system Earth-Moon-Sun

_{}

expressed in kilometers from the center of mass of the system.

**Lissajous orbits around L2 point**

We have to find the general solution of the system of equations around L2 point:

_{}

with:

_{}

The general solution of the above system of equations can be proven to be of the form:

_{}

where,

_{}

where _{}is the solution of the equation which gives us the L2 point.

_{}

Based on the initial conditions the coefficients can be found as:

_{}

where:

_{}

We can simplify the system of equations:

_{}

If we put the supplementary condition _{}we obtain a Lissajous orbit in the liniarized restricted three body problem:

_{}

with _{}.

If _{}we obtain a Halo orbit around L2.

The following pictures show a typical Lissajous orbit with the conditions:

_{}

The following pictures show a general Lissajous orbit:

As can be seen, the exponential part of the equations results in a very fast grow of the orbit and, as consequence, the satellite will escape from the desired orbit. In conclusion the satellite should be kept in a liniarised Lissajous orbit such that the exponent is null.