Lagrangian Mechanics Made Simple Part 5 The Stable Lagrange Points and the Coriolis Force

Lagrangian Mechanics Made Simple
Part 5
The Stable Lagrange Points
and the Coriolis Force

Ron Steinke <>

Last month we discussed the motion of a small body in the gravitational field of two larger bodies. We showed that the Lagrangian was

L = \frac{1}{2} m \left({\bf\dot r} + \hbox{\setbox0=\hbox{$...
...2 \frac{R^2 R_1}{\left\vert{\bf r} - {\bf R}_2\right\vert} \,.
\end{displaymath} (1)

Here, $m$ is the mass of the smaller body. $\omega$ is the angular velocity of the two larger bodies in their orbits around each other. $R_1$ and $R_2$ are the distances from each of the larger bodies to their common center of rotation, given in vector form by

{\bf R}_1 = \left(-R_1, 0, 0\right) \,,
\kern 1 in
{\bf R}_2 = \left(R_2, 0, 0\right) \,.
\end{displaymath} (2)

Finally, $R = R_1 + R_2$ is the distance between the two bodies.

We described the five solutions to the equations of motion (known as the Lagrange points) where ${\bf r}$ is a constant. Two of these solutions,

{\bf r}_\pm = \left(\frac{R_2 - R_1}{2}, \pm \frac{\sqrt{3}}{2} R, 0\right) \,,
\end{displaymath} (3)

are known to be stable in certain cases. Our goal this month is to show this.

Perturbations About the Stability Points

In the last column, we briefly discussed the stability of the solutions ${\bf r}_\pm$ by perturbing them a small constant distance $\Delta{\bf r}$ . To do a full stability analysis, we examine the dynamics of these perturbations by allowing $\Delta{\bf r}$ to be a function of the time $t$ . That is, we define

\Delta{\bf r}(t) = {\bf r}(t) - {\bf r}_\pm \,.
\end{displaymath} (4)

In terms of $\Delta{\bf r}$ , the Lagrangian is

$\displaystyle L$ $\textstyle =$ $\displaystyle \frac{1}{2} m \omega^2 \left\vert{\bf r}_\pm\right\vert^2
- m {\b...
\kern-.035em\raise.0433em\box0}\times \Delta{\bf r}\right)^2$  
    $\displaystyle + m \omega^2 \frac{R^2 R_2}{\left\vert{\bf r}_\pm + \Delta{\bf r}...
...frac{R^2 R_1}{\left\vert{\bf r}_\pm + \Delta{\bf r} - {\bf R}_2\right\vert} \,,$ (5)

where we have used the fact that ${\bf r}_\pm$ and $\hbox{\setbox0=\hbox{$\omega$}\kern-.035em\copy0\kern-\wd0
\kern-.035em\raise.0433em\box0}$ are perpendicular to simplify the equation.

Since we are only interested in small perturbations, we expand the Lagrangian in powers of $\Delta{\bf r}$ . Using the fact that

\frac{1}{\sqrt{1 + x}} = 1 - \frac{1}{2} x + \frac{3}{8} x^2 + O(x^3) \,,
\end{displaymath} (6)

for small values of $x$ , we find that the Lagrangian is

$\displaystyle L$ $\textstyle =$ $\displaystyle m\omega^2 R^2 + \frac{1}{2} m\omega^2 \left\vert{\bf r}_\pm\right\vert^2
\mp \sqrt{3} m \omega R \Delta\dot r_x$  
    $\displaystyle + \frac{1}{2} m \left\vert\Delta{\bf\dot r}\right\vert^2
- m \omega \Delta\dot r_x \Delta r_y
+ m \omega \Delta\dot r_y \Delta r_x$  
    $\displaystyle + m \omega^2 \left[ \frac{3}{8} \Delta r_x^2
\pm \frac{3\sqrt{3}}...
...ta r_x \Delta r_y
+ \frac{9}{8} \Delta r_y^2 - \frac{1}{2} \Delta r_z^2 \right]$  
    $\displaystyle + O\left(\Delta{\bf r}^3\right) \,.$ (7)

Dropping terms of $O\left(\Delta{\bf r}^3\right)$ and higher from the Lagrangian gives linearized equations of motion for $\Delta{\bf r}$,

$\displaystyle \Delta\ddot r_x$ $\textstyle =$ $\displaystyle -2\omega\Delta\dot r_y + \frac{3}{4} \omega^2\Delta r_x
\pm \frac{3\sqrt{3}}{4} \frac{R_1 - R_2}{R} \omega^2\Delta r_y$  
$\displaystyle \Delta\ddot r_y$ $\textstyle =$ $\displaystyle 2\omega\Delta\dot r_x + \frac{9}{4} \omega^2\Delta r_y
\pm \frac{3\sqrt{3}}{4} \frac{R_1 - R_2}{R} \omega^2\Delta r_x$  
$\displaystyle \Delta\ddot r_z$ $\textstyle =$ $\displaystyle - \omega^2\Delta r_z \,,$ (8)

where the $\pm$ in the $x$ and $y$ equations corresponds to the choice of a solution $\Delta{\bf r}_\pm$.

Normal Mode Frequencies and Stability

Examining the $z$ equation first, we see that it has the solution

\Delta r_z(t) = A \cos(\omega t) + B \sin(\omega t) \,,
\end{displaymath} (9)

where $A$ and $B$ are some constants. This shows that a small perturbation in the $z$ direction will result in the particle oscillating about $\Delta r_z = 0$ with frequency $\omega$ . The solutions $\cos(\omega t)$ and $\sin(\omega t)$ are called normal modes of the perturbation, and $\omega$ is the normal mode frequency.

To see whether perturbations in the $x$ and $y$ directions result in the same stability as a perturbation in the $z$ direction, we will find the normal mode frequencies of the $x$ and $y$ modes. That is, we will look for solutions to the $x$ and $y$ equations of motion of the form

\Delta r_x(t) = e^{i\nu t} \Delta \bar r_x \,,
\kern 1 in
\Delta r_y(t) = e^{i\nu t} \Delta \bar r_y \,,
\end{displaymath} (10)

where $i = \sqrt{-1}$ , and the exponential has the usual definition

e^{i\nu t} = \cos(\nu t) + i \sin(\nu t) \,,
\end{displaymath} (11)

and the useful property

\frac{d}{dt} e^{i\nu t} = i \nu e^{i\nu t} \,.
\end{displaymath} (12)

The solutions $\cos(\omega t)$ and $\sin(\omega t)$ in the $z$ case are simply linear combinations of two normal modes with $\nu = \pm\omega$ . If $\nu$ is real (has no imaginary part proportional to $i$ ), the perturbation has an oscillatory normal mode, like the one in $z$ . An imaginary part in $\nu$ results in exponential growth of the perturbation, and will quickly drive us out of the region where our displacement from the stability point can be considered small. Thus, the Lagrange points are stable if all their normal mode frequencies are real.

Inserting the trial solution (10) in the $x$ and $y$ equations of (8) gives

e^{i\nu t}
\left( \matrix{
\nu^2 + \frac{3}{4} \omega^2 &
...\matrix{ \Delta \bar r_x \cr \Delta \bar r_y } \right)
= 0 \,.
\end{displaymath} (13)

For (13) to have a solution, the matrix must have a zero eigenvalue. This can happen if and only if its determinant is zero. Setting the determinant to zero gives an equation for $\nu^2$ ,

\left(\nu^2 + \frac{3}{4} \omega^2\right)
\left(\nu^2 + \fra...
...} \frac{\left(R_1 - R_2\right)^2}{R^2} \omega^4\right)
= 0 \,.
\end{displaymath} (14)

Notice that the $\pm$ has vanished from the equation. The fourth and fifth Lagrange points have the same normal frequencies, as should be expected from symmetry.

Eq. (14) is quadratic in $\nu^2$ , so we can solve it to obtain

\frac{\nu^2}{\omega^2} = \frac{1}{2} \pm \frac{1}{4}
\sqrt{27 \frac{\left(R_1 - R_2\right)^2}{R^2} - 23} \,.
\end{displaymath} (15)

There are two possible solutions for $\nu^2$ , giving four solutions for $\nu$ . This is to be expected, since we started with two second order differential equations. The critical question is whether all the values of $\nu$ are real. If the argument of the square root is negative, then $\nu^2$ is complex, and so is $\nu$ , leading to instability. If, on the other hand, the argument of the square root is positive, we know it is not more that $4$ , since $\vert R_1 - R_2\vert / R \le 1$ . This gives values of $\nu^2$ between $0$ and $\omega^2$ , and all values of $\nu$ are real. The condition for stability is

\frac{\left\vert R_1 - R_2\right\vert}{R} \ge \sqrt{\frac{23}{27}} \,.
\end{displaymath} (16)

This requires that the smaller of the masses $M_1,M_2$ of the two large bodies not be more than 3.7% of the larger mass.

This concludes this series of articles. It's been fun, and I hope some of the readers have found them useful.

Ron Steinke 2002-09-27