Two masses $m_1$ and $m_2$ orbit their common center of mass. A spacecraft that has a mass so small that it does not affect bodies $m_1$ and $m_2$ gravitationally, but its trajectory is determined by the two larger bodies.

We choose a reference frame that rotates such that the origin is at the center of mass, and bodies $m_1$ and $m_2$ are fixed on the x-axis.


Dimensionless coordinates parameter:

\begin{equation} \mu=\frac{m_1}{m_1+m_2}\end{equation}

Equations of Motion (EOM)

\begin{equation} f(\mathbf{x})=\begin{bmatrix} \ddot{x} \\\ddot{y} \\\ddot{z} \end{bmatrix}=\begin{bmatrix} x+2 \dot{y} -\frac{1-\mu}{r_1^3}(x+\mu)-\frac{\mu}{r_2^3}(x-1+\mu)    \\ y-2\dot{x} -\frac{1-\mu}{r_1^3}y-\frac{\mu}{r_2^3}y \\  -\frac{1-\mu}{r_1^3}z-\frac{\mu}{r_2^3}z \end{bmatrix}\end{equation}


\begin{equation} r_1= \sqrt{(x+\mu)^2+y^2+z^2}           \end{equation}


\begin{equation} r_2= \sqrt{(x-1+\mu)^2+y^2+z^2}           \end{equation}


Jacobi Integral C

The equations of motion imply that there is a conserved quantity. We call this quantity C, and it is called the Jacobi integral.

\begin{equation} C= x^2+y^2+2\frac{1-\mu}{r_1}+2\frac{\mu}{r_2}-(\dot{x}^2 +\dot{y}^2+\dot{z}^2 )       \end{equation}

It is equivalent to the total energy of a body on a ballistic trajectory in an inertial reference frame in a conservative force field. For any given location $(x,y,z)$ , the greater the velocity $(\dot{x},\dot{y},\dot{z})$ is, the smaller the value of C is because of the minus sign in front of the velocity term.

Zero velocity curves and Libration Points

In an inertial reference frame and a conservative force field, a body has kinetic and potential energy, and the total energy which is constant. The potential energy depends on the location only. When we throw a rock up in the air, it will stop at one point at which all the energy is in the potential energy.
Similarly in the CRTBP, there are locations where the velocity is zero, and all of the C quantity is contained in the location. These locations are either points (five libration points) or curves (zero velocity curves) which are closed curves that enclose one or two of the bodies (for values of C > C(L5), see below ). 

Libration Points

Since we don't have the values of C for the zero velocity points, we can't use the Jacobi integral equation to find these points. We use the EOM's such that

\begin{equation} f(\mathbf{x})=\mathbf{0}\end{equation}

The first three libration points $L_1$,$ L_2$, and $L_3$ lie on the x-axis:

\begin{equation} y=0, z=0, \dot{x}=0,\dot{y}=0,\dot{z}=0 \end{equation}

Only the first EOM gives a non-trivial condition for x. The distances reduce to

\begin{equation} r_1= |x+\mu|           \end{equation}


\begin{equation} r_2=|x-1+\mu|           \end{equation}

and the condition for x for $L_1$,$ L_2$, and $L_3$ is
\begin{equation} 0=x -\frac{1-\mu}{{|x+\mu|}^3}(x+\mu)-\frac{\mu}{{|x-1+\mu|}^3}(x-1+\mu)    \end{equation}
Librations points $L_4$ and $L_5$ lie on the corner of an equilateral triangle whose base is the line between the two bodies and has the length $l=1$.
The x-coordinates are in the middle between the two bodies:
\begin{equation} x_{4,5}= -\mu+\frac{1}{2}   \end{equation}
The y-coordinates are $y_{4,5}=\pm \frac{\sqrt{3}}{2} $ 
\begin{equation} L_4= (-\mu+\frac{1}{2},  \frac{\sqrt{3}}{2}) \end{equation}
\begin{equation} L_5= (-\mu+\frac{1}{2},  -\frac{\sqrt{3}}{2}) \end{equation}
Now that we have the coordinates of the libration points, we can compute the corrresponding Jacobi energy values $C_1$, $C_2$, $C_3$ ,$C_4$, and $C_5$. 
The distances are
\begin{equation} r_1= |L_1+\mu|           \end{equation}

\begin{equation} r_2=|L_1-1+\mu|           \end{equation}

and the energy level of $C_1$ is
\begin{equation} C_1=  (L_1)^2+2\frac{1-\mu}{r_1}+2\frac{\mu}{r_2}\end{equation}
(The equivalent equations for $C_2$, $C_3$ ,$C_4$, and $C_5$.)

Zero velocity curves


We obtain the zero velocity curves by setting a value for C and then finding the $(x,y,z)$ which solve the Jacobi equation with $(\dot{x},\dot{y},\dot{z})=(0,0,0)$.


Classes of Orbits and Trajctories


On the zero velocity curves, the velocity is zero which means that the spacecraft does not move. From here, the spacecraft could go both ways by a perturbation: back into the region or away from the region. A stability examination would reveal that a perturbation would move the spacecraft back into the region. Therefore, the zero-velocity curves present an impenetrable boundary for ballistic trajectories.

The energy level $C_{L_1}$ of the $L_1$ libration point is of particular interest. If the energy of the spacecraft is below that level, then it is caught in an orbit around either $m_1$ or $m_2$. If the energy level of the spacecraft is $C_{L_1}$, then the spacecraft can barely pass through $L_1$ and transit from an orbit around one of the two masses to an orbit around the other. If the energy of the spacecraft is above $C_{L_1}$, then the spacecraft can pass through points near $L_1$, in addition to $L_1$. Therefore, the energy level $C_{L_1}$ represents the minimum energy to travel from one body to the other.