Lagrangian Mechanics Made Simple
Part 4
Orbital Motion and Kepler's Equation

Ron Steinke <rsteinke@w-link.net>

Last month we introduced the problem of a system two particles whose potential energy $U$ depended only on the distance $r$ between them. The dynamics of this system reduced to

\begin{displaymath} \frac{dr}{d\theta}= r \sqrt{ 2 \left(E - U(r)\right) \frac{\mu r^2}{l^2} - 1} \,, \end{displaymath} (1)

where the quantities $l$ and $E$ are the angular momentum and energy of the system, and
\begin{displaymath} \mu = \frac{m_1 m_2}{m_1 + m_2} \end{displaymath} (2)

is the reduced mass.

Kepler's Equation

For the force of gravity, 1the potential is

\begin{displaymath} U(r) = -\frac{k}{r} \,, \end{displaymath} (3)

where $k$ is a constant. The $1/r$ dependence of the potential reflects the $1/r^2$ dependence of the gravitational force. The potential is negative, since the particles have less potential energy when they are close together than when they are far apart. Inserting (3) in (1) gives
\begin{displaymath} \frac{dr}{d\theta} = \frac{r^2}{l} \sqrt{2E\mu + 2\frac{k\mu}{r} - \frac{l^2}{r^2}} \,. \end{displaymath} (4)

The solution to this equation is
\begin{displaymath} \frac{l^2}{\mu k} \frac{1}{r} = 1 + \cos\theta \sqrt{1 + \frac{2El^2}{\mu k^2}} \,. \end{displaymath} (5)

This solution is unique up to a constant shift in the angle $\theta$, $\theta \to \theta + \Delta\theta$.

Eq. (5) describes a conic section,

\begin{displaymath} \frac{\alpha}{r} = 1 + \epsilon \cos\theta \,. \end{displaymath} (6)

The quantity
\begin{displaymath} \epsilon = \sqrt{1 + \frac{2El^2}{\mu k^2}} \end{displaymath} (7)

is called the eccentricity of the orbit. An eccentricity of 0 gives a circular orbit, between 0 and 1 gives an ellipse, 1 gives a parabola, and greater than 1 gives a hyperbola. Notice that $\epsilon = 1$ corresponds to $E = 0$. A parabolic orbit has just enough energy to escape the system. Higher (positive) energies give hyperbolic orbits, and lower (negative) energies give circular or elliptical orbits, where the two particles are bound together by the gravitational force. Note also that for an orbit of zero angular momentum, (5) gives a constant value for $\theta$, and the two bodies fall straight into each other.

The Lagrange Stability Points

The three body problem is not solvable in the general case. However, some approximate solutions are quite interesting. Consider the case of a small body of mass $m$ orbiting two larger bodies of masses $M_1$ and $M_2$. For simplicity we will assume that the two large bodies orbit each other circularly with angular velocity $\omega$, and are unaffected by their attraction to the smaller body. If $R_1$ and $R_2$ are the distances of the two bodies from their common center of rotation, then

$\displaystyle R_1$ $\textstyle =$ $\displaystyle \frac{M_2}{M_1 + M_2} R$  
$\displaystyle R_2$ $\textstyle =$ $\displaystyle \frac{M_1}{M_1 + M_2} R \,,$ (8)

where $R$ is the distance between the two bodies,
\begin{displaymath} R = \left(\frac{G \left(M_1 + M_2\right)}{\omega^2}\right)^{1/3} \,. \end{displaymath} (9)

Here, $G$ is Newton's gravitational constant.

The two larger bodies lie at positions ${\bf R}_1 = (-R_1, 0, 0)$ and ${\bf R}_2 = (R_2, 0, 0)$ in a coordinate system with angular velocity $\hbox{\setbox0=\hbox{$\omega$}\kern-.035em\copy0\kern-\wd0\kern.07em\copy0\kern-\wd0\kern-.035em\raise.0433em\box0}= (0, 0, \omega)$. The Lagrangian for the lighter particle at position ${\bf r}(t)$ is

\begin{displaymath} L = \frac{1}{2} m \left({\bf\dot r} + \hbox{\setbox0=\hbox{$... ...+ \frac{G M_2 m}{\left\vert{\bf r} - {\bf R}_2\right\vert} \,, \end{displaymath} (10)

where we have used the Lagrangian for a rotating coordinate system, which we derived in Part 2 of this series. To simplify this a bit, we use the fact that (8) and (9) tell us that
$\displaystyle GM_1 = G(M_1 + M_2) \frac{R_2}{R}$ $\textstyle =$ $\displaystyle \omega^2 R^2 R_2$  
$\displaystyle GM_2 = G(M_1 + M_2) \frac{R_1}{R}$ $\textstyle =$ $\displaystyle \omega^2 R^2 R_1 \,.$ (11)

This substitution gives the Lagrangian the simpler form
\begin{displaymath} 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} (12)

Notice that all terms are now proportional to $m$. This constant factor will drop out of the equations of motion, and all dependence on the masses of the particles will be parameterized by the lengths $R_1$ and $R_2$.

To find metastable points, we look for solutions where ${\bf\dot r}$ is zero. We do this by examining the effective potential,

\begin{displaymath} U_{{\rm eff}}({\bf r}) = -\frac{1}{2} m \left(\hbox{\setbox0... ...2 \frac{R^2 R_1}{\left\vert{\bf r} - {\bf R}_2\right\vert} \,, \end{displaymath} (13)

which is just $-L$ with ${\bf\dot r}$ set to zero. Examining $U_{{\rm eff}}({\bf r})$, we see that when ${\bf r}$ is near ${\bf R}_1$ or ${\bf R}_2$, the potential is dominated by gravitational attraction, so the light particle will tend to fall towards one of the heavy ones. Similarly, for large ${\bf r}$, the potential is dominated by the centripetal term, and the light particle will tend to fly off to infinity.

There exist points for which the centripetal acceleration is balanced by the attraction of the larger bodies. At these points, the force,

\begin{displaymath} {\bf F} = -\nabla U_{{\rm eff}}({\bf r}) = - m \hbox{\setbox... ...} - {\bf R}_2}{\left\vert{\bf r} - {\bf R}_2\right\vert^3} \,, \end{displaymath} (14)

is zero. The $z$ component of ${\bf F}$ has no $\omega^2 r$ centripetal term, so it only vanishes when $r_z$ is zero. Setting the other two components of the force equal to zero gives us equations for $r_x$ and $r_y$,
$\displaystyle r_x$ $\textstyle =$ $\displaystyle R^2 R_2 \frac{r_x + R_1}{\left\vert(r_x + R_1)^2 + r_y^2\right\ve... ...}} + R^2 R_1 \frac{r_x - R_2}{\left\vert(r_x - R_2)^2 + r_y^2\right\vert^{3/2}}$  
$\displaystyle r_y$ $\textstyle =$ $\displaystyle R^2 R_2 \frac{r_y}{\left\vert(r_x + R_1)^2 + r_y^2\right\vert^{3/2}} + R^2 R_1 \frac{r_y}{\left\vert(r_x - R_2)^2 + r_y^2\right\vert^{3/2}} \,,$ (15)

where we have dropped the common factor of $m\omega^2$.

It is possible to solve the second equation in (15) by setting $r_y$ to zero. The value of $r_x$ is then found by solving the first equation. There are three of these solutions, one with $-R_1 < r_x < R_2$, between the two larger bodies, and one each with $r_x > R_2$ and $r_x < - R_1$, off to each side. These three points are saddle points. If you perturb the small mass $m$ off one of these points by a small amount along the direction of its orbit, it will tend to fall back into place. If you perturb it into a higher or lower orbit, however, it will tend to move away from the stability point, either towards one of the two heavier bodies, or pulled away by the centripetal component of the potential (13).

There are also two solutions where $r_y$ is not equal to zero. These are

\begin{displaymath} r_x = \frac{1}{2} \left(R_2 - R_1 \right) \,, \kern 1 in r_y = \pm \frac{\sqrt{3}}{2} R \,. \end{displaymath} (16)

These are what are commonly known as the fourth and fifth Lagrange points, and are known to be stable. However, a quick examination of the force (14) shows that they do not derive this stability from the effective potential. Let ${\bf r}_\pm$ be the two solutions (16), and examine the force at a point ${\bf r}_\pm + \Delta{\bf r}$, where $\Delta{\bf r}$ is small. The components of the force are
\begin{displaymath} \left( \matrix{ {\bf F}_x \cr {\bf F}_y \cr {\bf F}_z } \rig... ...matrix{ \Delta r_x \cr \Delta r_y \cr \Delta r_z } \right) \,. \end{displaymath} (17)

The force in the $z$ direction is proportional to $- \Delta r_z$, indicating that it is pulling against the displacement. Any small perturbation of a particle in the $z$ direction will result in a force which tends to pull it back to $z = 0$. The same is not true of perturbations in the $x$ and $y$ directions, however. The effect of $\Delta r_x$ and $\Delta r_y$ on the force are determined by a $2 \times 2$ portion of the matrix in (17). The eigenvalues of this submatrix are
\begin{displaymath} \lambda = \frac{3}{2} \left( 1 \pm \sqrt{1 - 3 \frac{R_1 R_2}{R^2}} \right) \,, \end{displaymath} (18)

which are both positive. Therefore, for small perturbations in the $x$ and $y$ directions, the force (14) tends to push particles away from the Lagrange points. The stability of orbits at these points is due to the Coriolis force, which we have neglected. We'll discuss how that works next month.

About this document ...

Lagrangian Mechanics Made Simple
Part 4
Orbital Motion and Kepler's Equation

Footnotes

... gravity, 1
This section leans heavily on J. Marion and S. Thornton, Classical Dynamics of Particles and Systems, 3rd ed., pp. 256-260
Ron Steinke 2002-08-11

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