The Lagrangian

The Lagrangian

The Lagrangian of the system can be derived from simply writing down the kinetic and potential energies, as the Lagrangian is simply,

\[L(q,\dot{q}) = T - U(q).\]

Ultimately we want the Lagrangian to be written in the rotating frame’s coordinates such that \(q = (x,y,z)\) and \(\dot{q} = (\dot{x}, \dot{y}, \dot{z})\). The expression will end up in rotating coordinates as follows.

\[L(x,y,z,\dot{x},\dot{y},\dot{z}) = \frac{1}{2}\Big((\dot{x} - y)^2 + (\dot{y} + x)^2 + \dot{z}^2\Big) + \frac{1-\mu}{r_1(x,y,z)} + \frac{\mu}{r_2(x,y,z)} + \frac{1}{2}\mu(1-\mu)\]

Kinetic Energy

In inertial coordinates, the kinetic energy \(T\) is the usual \(v^2/2\) expression seen across many other dynamical systems.

\[T = \frac{1}{2}\left(\dot{X}^2 + \dot{Y}^2 + \dot{Z}^2\right)\]

Now recall the transformation between the inertial and rotating frames.

\[\begin{split}\dot{X} &= (\dot{x} - y)\cos t - (\dot{y} + x)\sin t \\ \dot{Y} &= (\dot{x} - y)\sin t + (\dot{y} + x)\cos t \\ \dot{Z} &= \dot{z}\end{split}\]

Therefore we can express the kinetic energy in terms of the rotating frame’s coordinates.

\[\begin{split}T = \frac{1}{2}\Big((\dot{x} - y)^2 + (\dot{y} + x)^2 + \dot{z}^2\Big) \\\end{split}\]

Proof

\[\begin{split}T &= \frac{1}{2}\left(\dot{X}^2 + \dot{Y}^2 + \dot{Z}^2\right) \\ &= \frac{1}{2}\Big(\big[(\dot{x} - y)\cos t - (\dot{y} + x)\sin t\big]^2 + \big[(\dot{x} - y)\sin t + (\dot{y} + x)\cos t\big]^2 + \dot{z}^2\Big) \\ &= \frac{1}{2}\Big(\big[(\dot{x} - y)^2\cos^2 t - 2(\dot{x} - y)(\dot{y} + x)\sin t\cos t + (\dot{y} + x)^2\sin^2 t\big] \\&\qquad\quad + \big[(\dot{x} - y)^2\sin^2 t + 2(\dot{x} - y)(\dot{y} + x)\sin t \cos t + (\dot{y} + x)^2\cos^2 t\big] \\&\qquad\quad + \dot{z}^2\Big) \\ &= \frac{1}{2}\Big((\dot{x} - y)^2\underbrace{(\cos^2 t + \sin^2 t)}_{1} + (\dot{y} + x)^2\underbrace{(\sin^2 t + \cos^2 t)}_{1} + \dot{z}^2\Big) \\ &= \frac{1}{2}\Big((\dot{x} - y)^2 + (\dot{y} + x)^2 + \dot{z}^2\Big)\end{split}\]

Potential Energy

The gravitational potential due to the two bodies is directly taken from the Newtonian potential

\[-\frac{GM}{r}\]

but applied to both bodies in nondimensional coordinates where \(G=1\).

The total gravitational potential is

\[-\frac{1-\mu}{r_1} - \frac{\mu}{r_2}.\]

• Before this is useful, we need to decide which coordinates to use

• Notice that it only depends on distance

Because the potential only depends on distance, we can actually write it directly in terms of rotating coordinates,

\[-\frac{1-\mu}{r_1(x,y,z)} - \frac{\mu}{r_2(x,y,z)}\]

where \(r_1\) and \(r_2\) are the distances of the particle from the primary and secondary, respectively.

\[\begin{split}r_1(x,y,z) &= \sqrt{(x + \mu)^2 + y^2 + z^2} \\ r_2(x,y,z) &= \sqrt{(x + \mu - 1)^2 + y^2 + z^2}\end{split}\]

Fig. 7 Distances from the primary and secondary in the rotating frame

As a finishing touch, we adjust the potential’s datum by adding a constant, finally arriving at the expression we’ll use.

\[U(x,y,z) = -\frac{1-\mu}{r_1(x,y,z)} - \frac{\mu}{r_2(x,y,z)} - \frac{1}{2}\mu(1-\mu)\]

Note

Adding in \(-\mu(1-\mu)/2\) to the potential is a standard approach to make some expressions that will come later behave more nicely across all values of \(\mu\) that may be studied.