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Stability of Lagrange Points

We have seen that the five Lagrange points, $L_1$ to $L_5$, are the equilibrium points of mass $m_3$ in the co-rotating frame. Let us now determine whether or not these equilibrium points are stable to small displacements.

Now, the equations of motion of mass $m_3$ in the co-rotating frame are specified in Equations (1056)-(1058). Note that the motion in the $x$-$y$ plane is complicated by presence of the Coriolis acceleration. However, the motion parallel to the $z$-axis simply corresponds to motion in the potential $U$. Hence, the condition for the stability of the Lagrange points (which all lie at $z=0$) to small displacements parallel to the $z$-axis is simply (see Section 3.2)

\left(\frac{\partial^2 U}{\partial z^2}\right)_{z=0} = \frac{\mu_1}{\rho_1^{\,3}} + \frac{\mu_2}{\rho_2^{\,3}}>0.
\end{displaymath} (1096)

This condition is satisfied everywhere in the $x$-$y$ plane. Hence, the Lagrange points are all stable to small displacements parallel to the $z$-axis. It, thus, remains to investigate their stability to small displacements lying within the $x$-$y$ plane.

Suppose that a Lagrange point is situated in the $x$-$y$ plane at coordinates $(x_0,\,y_0,\,0)$. Let us consider small amplitude $x$-$y$ motion in the vicinity of this point by writing

$\displaystyle x$ $\textstyle =$ $\displaystyle x_0 + \delta x,$ (1097)
$\displaystyle y$ $\textstyle =$ $\displaystyle y_0 + \delta y,$ (1098)
$\displaystyle z$ $\textstyle =$ $\displaystyle 0,$ (1099)

where $\delta x$ and $\delta y$ are infinitesimal. Expanding $U(x,y,0)$ about the Lagrange point as a Taylor series, and retaining terms up to second-order in small quantities, we obtain
U = U_0 + U_x\,\delta x+ U_y\,\delta y + \frac{1}{2}\,U_{xx}...
..._{xy}\,\delta x\,\delta y
+ \frac{1}{2}\,U_{yy}\,(\delta y)^2,
\end{displaymath} (1100)

where $U_0=U(x_0,y_0,0)$, $U_x=\partial U(x_0,y_0,0)/\partial x$, $U_{xx}=\partial^2 U(x_0,y_0,0)/\partial x^2$, etc. However, by definition, $U_x=U_y=0$ at a Lagrange point, so the expansion simplifies to
U = U_0 + \frac{1}{2}\,U_{xx}\,(\delta x)^2+ U_{xy}\,\delta x\,\delta y
+ \frac{1}{2}\,U_{yy}\,(\delta y)^2.
\end{displaymath} (1101)

Finally, substitution of Equations (1097)-(1099), and (1101) into the equations of $x$-$y$ motion, (1056) and (1057), yields
$\displaystyle \delta\ddot{x} - 2\,\delta\dot{y}$ $\textstyle =$ $\displaystyle - U_{xx}\,\delta x -U_{xy}\,\delta y,$ (1102)
$\displaystyle \delta\ddot{y} + 2\,\delta\dot{x}$ $\textstyle =$ $\displaystyle - U_{xy}\,\delta x -U_{yy}\,\delta y,$ (1103)

since $\omega =1$.

Let us search for a solution of the above pair of equations of the form $\delta x(t) = \delta x_0\,\exp(\gamma\,t)$ and $\delta y(t) = \delta y_0\,\exp(\gamma\,t)$. We obtain

\gamma^2 + U_{xx}& -2\,\gamma+U_{xy...
... = \left(
0\\ [0.5ex]
\end{displaymath} (1104)

This equation only has a non-trivial solution if the determinant of the matrix is zero. Hence, we get
\gamma^4 + (4+U_{xx}+U_{yy})\,\gamma^2 + (U_{xx}\,U_{yy}-U_{xy}^{\,2}) = 0.
\end{displaymath} (1105)

Now, it is convenient to define
$\displaystyle A$ $\textstyle =$ $\displaystyle \frac{\mu_1}{\rho_1^{\,3}} + \frac{\mu_2}{\rho_2^{\,3}},$ (1106)
$\displaystyle B$ $\textstyle =$ $\displaystyle 3\left[ \frac{\mu_1}{\rho_1^{\,5}} + \frac{\mu_2}{\rho_2^{\,5}}\right]y^2,$ (1107)
$\displaystyle C$ $\textstyle =$ $\displaystyle 3\left[\frac{\mu_1\,(x+\mu_2)}{\rho_1^{\,5}}+\frac{\mu_2\,(x-\mu_1)}{\rho_2^{\,5}}\right]y,$ (1108)
$\displaystyle D$ $\textstyle =$ $\displaystyle 3\left[\frac{\mu_1\,(x+\mu_2)^2}{\rho_1^{\,3}}+\frac{\mu_2\,(x-\mu_1)^2}{\rho_2^{\,3}}\right],$ (1109)

where all terms are evaluated at the point $(x_0,\,y_0,\,0)$. It thus follows that
$\displaystyle U_{xx}$ $\textstyle =$ $\displaystyle A - D - 1,$ (1110)
$\displaystyle U_{yy}$ $\textstyle =$ $\displaystyle A - B - 1,$ (1111)
$\displaystyle U_{xy}$ $\textstyle =$ $\displaystyle - C.$ (1112)

Consider the co-linear Lagrange points, $L_1$, $L_2$, and $L_3$. These all lie on the $x$-axis, and are thus characterized by $y=0$, $\rho_1^{\,2} = (x+\mu_2)^2$, and $\rho_2^{\,2} = (x-\mu_1)^2$. It follows, from the above equations, that $B=C=0$ and $D=3\,A$. Hence, $U_{xx}=-1-2\,A$, $U_{yy} = A-1$, and $U_{xy}=0$. Equation (1105) thus yields

\Gamma^2 + (2-A)\,\Gamma + (1-A)\,(1+2\,A) = 0,
\end{displaymath} (1113)

where $\Gamma=\gamma^2$. Now, in order for a Lagrange point to be stable to small displacements, all four of the roots, $\gamma$, of Equation (1105) must be purely imaginary. This, in turn, implies that the two roots of the above equation,
\Gamma = \frac{A-2\pm\sqrt{A\,(9\,A-8)}}{2},
\end{displaymath} (1114)

must both be real and negative. Thus, the stability criterion is
\frac{8}{9}\leq A \leq 1.
\end{displaymath} (1115)

Figure 56 shows $A$ calculated at the three co-linear Lagrange points as a function of $\mu _2$, for all allowed values of this parameter (i.e., $0<\mu_2\leq 0.5$). It can be seen that $A$ is always greater than unity for all three points. Hence, we conclude that the co-linear Lagrange points, $L_1$, $L_2$, and $L_3$, are intrinsically unstable equilibrium points in the co-rotating frame.

Figure 56: The solid, short-dashed, and long-dashed curves show $A$ as a function of $\mu _2$ at the $L_1$, $L_2$, and $L_3$ Lagrange points.
\epsfysize =2.5in

Let us now consider the triangular Lagrange points, $L_4$ and $L_5$. These points are characterized by $\rho_1=\rho_2=1$. It follows that $A=1$, $B=9/4$, $C=\pm\sqrt{27/16}\,(1-2\,\mu_2)$, and $D=3/4$. Hence, $U_{xx} = -3/4$, $U_{yy}=-9/4$, and $U_{xy} = \mp\sqrt{27/16}\,(1-2\,\mu_2)$, where the upper/lower signs corresponds to $L_4$ and $L_5$, respectively. Equation (1105) thus yields

\Gamma^2 + \Gamma + \frac{27}{4}\,\mu_2\,(1-\mu_2) = 0
\end{displaymath} (1116)

for both points, where $\Gamma=\gamma^2$. As before, the stability criterion is that the two roots of the above equation must both be real and negative. This is the case provided that $1 > 27\,\mu_2\,(1-\mu_2)$, which yields the stability criterion
\mu_2 < \frac{1}{2}\left(1-\sqrt{\frac{23}{27}}\right) = 0.0385.
\end{displaymath} (1117)

In unnormalized units, this criterion becomes
\frac{m_2}{m_1+ m_2} < 0.0385.
\end{displaymath} (1118)

We thus conclude that the $L_4$ and $L_5$ Lagrange points are stable equilibrium points, in the co-rotating frame, provided that mass $m_2$ is less than about $4\%$ of mass $m_1$. If this is the case then mass $m_3$ can orbit around these points indefinitely. In the inertial frame, the mass will share the orbit of mass $m_2$ about mass $m_1$, but will stay approximately $60^\circ$ ahead of mass $m_2$, if it is orbiting the $L_4$ point, or $60^\circ$ behind, if it is orbiting the $L_5$ point--see Figure 55. This type of behavior has been observed in the Solar System. For instance, there is a sub-class of asteroids, known as the Trojan asteroids, which are trapped in the vicinity of the $L_4$ and $L_5$ points of the Sun-Jupiter system (which easily satisfies the above stability criterion), and consequently share Jupiter's orbit around the Sun, staying approximately $60^\circ$ ahead of, and $60^\circ$ behind, Jupiter, respectively. Furthermore, the $L_4$ and $L_5$ points of the Sun-Earth system are occupied by clouds of dust.

next up previous
Next: Lunar Motion Up: The Three-Body Problem Previous: Zero-Velocity Surfaces
Richard Fitzpatrick 2011-03-31