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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.

The equations of motion of mass $ m_3$ in the co-rotating frame are specified in Equations (9.27)-(9.29). 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 2.7)

$\displaystyle \left(\frac{\partial^{\,2} U}{\partial z^{\,2}}\right)_{z=0} = \frac{\mu_1}{\rho_1^{\,3}} + \frac{\mu_2}{\rho_2^{\,3}}>0.$ (9.67)

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

    $\displaystyle x$ $\displaystyle = x_0 + \delta x,$ (9.68)
    $\displaystyle y$ $\displaystyle =y_0 + \delta y,$ (9.69)
and   $\displaystyle z$ $\displaystyle = 0,$ (9.70)

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

$\displaystyle U \simeq U_0 + U_x\,\delta x+ U_y\,\delta y + \frac{1}{2}\,U_{xx}\,(\delta x)^2+ U_{xy}\,\delta x\,\delta y + \frac{1}{2}\,U_{yy}\,(\delta y)^2,$ (9.71)

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}$ , and so on. However, by definition, $ U_x=U_y=0$ at a Lagrange point, so the expansion simplifies to

$\displaystyle U \simeq U_0 + \frac{1}{2}\,U_{xx}\,(\delta x)^2+ U_{xy}\,\delta x\,\delta y + \frac{1}{2}\,U_{yy}\,(\delta y)^2.$ (9.72)

Finally, substituting Equations (9.68)-(9.70), and (9.72) into the equations of $ x$ -$ y$ motion, (9.27) and (9.28), and only retaining terms up to first order in small quantities, we get

    $\displaystyle \delta\skew{3}\ddot{x} - 2\,\delta\skew{3}\dot{y}$ $\displaystyle \simeq - U_{xx}\,\delta x -U_{xy}\,\delta y,$ (9.73)
and   $\displaystyle \delta\skew{3}\ddot{y} + 2\,\delta\skew{3}\dot{x}$ $\displaystyle \simeq - U_{xy}\,\delta x -U_{yy}\,\delta y,$     (9.74)

as $ \omega =1$ .

Let us search for a solution of the preceding 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

$\displaystyle \left( \begin{array}{cc} \gamma^{\,2} + U_{xx},& -2\,\gamma+U_{xy...
...end{array} \right) = \left( \begin{array}{c} 0\\ [0.5ex] 0 \end{array} \right).$ (9.75)

This equation only has a nontrivial solution if the determinant of the matrix is zero. Hence, we get

$\displaystyle \gamma^{\,4} + (4+U_{xx}+U_{yy})\,\gamma^{\,2} + (U_{xx}\,U_{yy}-U_{xy}^{\,2}) = 0.$ (9.76)

It is convenient to define

    $\displaystyle A$ $\displaystyle = \frac{\mu_1}{\rho_1^{\,3}} + \frac{\mu_2}{\rho_2^{\,3}},$ (9.77)
    $\displaystyle B$ $\displaystyle = 3\left[ \frac{\mu_1}{\rho_1^{\,5}} + \frac{\mu_2}{\rho_2^{\,5}}\right]y^{\,2},$ (9.78)
    $\displaystyle C$ $\displaystyle = 3\left[\frac{\mu_1\,(x+\mu_2)}{\rho_1^{\,5}}+\frac{\mu_2\,(x-\mu_1)}{\rho_2^{\,5}}\right]y,$ (9.79)
and   $\displaystyle D$ $\displaystyle = 3\left[\frac{\mu_1\,(x+\mu_2)^2}{\rho_1^{\,5}}+\frac{\mu_2\,(x-\mu_1)^2}{\rho_2^{\,5}}\right],$     (9.80)

where all terms are evaluated at the point $ (x_0,\,y_0,\,0)$ . It thus follows that

    $\displaystyle U_{xx}$ $\displaystyle = A - D - 1,$ (9.81)
    $\displaystyle U_{yy}$ $\displaystyle = A - B - 1,$ (9.82)
and   $\displaystyle U_{xy}$ $\displaystyle = - C.$     (9.83)

Figure: The solid, dashed, and dotted curves show $ A$ as a function of $ \mu _2$ at $ L_1$ , $ L_2$ , and $ L_3$ Lagrange points, respectively.
\begin{figure}
\epsfysize =3.5in
\centerline{\epsffile{Chapter08/fig8.13.eps}}
\end{figure}

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 preceding equations, that $ B=C=0$ and $ D=3\,A$ . Hence, $ U_{xx}=-1-2\,A$ , $ U_{yy} = A-1$ , and $ U_{xy}=0$ . Equation (9.76) thus yields

$\displaystyle {\mit\Gamma}^{\,2} + (2-A)\,{\mit\Gamma} + (1-A)\,(1+2\,A) = 0,$ (9.84)

where $ {\mit\Gamma}=\gamma^{\,2}$ . For a Lagrange point to be stable to small displacements, all four of the roots, $ \gamma$ , of Equation (9.76) must be purely imaginary. This, in turn, implies that the two roots of the preceding equation,

$\displaystyle {\mit\Gamma} = \frac{A-2\pm\sqrt{A\,(9\,A-8)}}{2},$ (9.85)

must both be real and negative. Thus, the stability criterion is

$\displaystyle \frac{8}{9}\leq A \leq 1.$ (9.86)

Figure 9.13 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: Positions of the Trojan asteroids (small circles) and Jupiter (large circle) projected onto ecliptic plane (viewed from north) at MJD 55600. The $ X$ -axis is directed toward vernal equinox. Raw data from JPL Small-Body Database.
\begin{figure}
\epsfysize =4in
\centerline{\epsffile{Chapter08/fig8.14.eps}}
\end{figure}

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 and lower signs corresponds to $ L_4$ and $ L_5$ , respectively. Equation (9.76) thus yields

$\displaystyle {\mit\Gamma}^{\,2} + {\mit\Gamma} + \frac{27}{4}\,\mu_2\,(1-\mu_2) = 0$ (9.87)

for both points, where $ {\mit\Gamma}=\gamma^{\,2}$ . As before, the stability criterion is that the two roots of the preceding 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

$\displaystyle \mu_2 < \frac{1}{2}\left(1-\sqrt{\frac{23}{27}}\right) = 0.0385.$ (9.88)

In unnormalized units, this criterion becomes

$\displaystyle \frac{m_2}{m_1+ m_2} < 0.0385.$ (9.89)

Figure: Positions of the Trojan asteroids (small circles) and Jupiter (large circle) at MJD 55600. $ Z$ is normal distance from the ecliptic plane. $ {\mit \Delta }\lambda $ is the difference in ecliptic longitude between the asteroids and Jupiter. Raw data from JPL Small-Body Database.
\begin{figure}
\epsfysize =4in
\centerline{\epsffile{Chapter08/fig8.15.eps}}
\end{figure}

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$ percent 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 it 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 9.12.) This type of behavior has been observed in the solar system. For instance, there is a subclass of asteroids, known as the Trojan asteroids, that are trapped in the vicinity of the $ L_4$ and $ L_5$ points of the Sun-Jupiter system [which easily satisfies the stability criterion in Equation (9.89)], and consequently share Jupiter's orbit around the Sun, staying approximately $ 60^\circ$ ahead of and $ 60^\circ$ behind, Jupiter, respectively. These asteroids are shown in Figures 9.14 and 9.15. The Sun-Jupiter system is not the only dynamical system in the solar system that possess Trojan asteroids trapped in the vicinity of its $ L_4$ and $ L_5$ points. In fact, the Sun-Neptune system has eight known Trojan asteroids, the Sun-Mars system has four, and the Sun-Earth system has one (designated 2010 TK7) trapped at the $ L_4$ point. The $ L_4$ and $ L_5$ points of the Sun-Earth system are also observed to trap clouds of interplanetary dust.


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Next: Exercises Up: Three-body problem Previous: Zero-velocity surfaces
Richard Fitzpatrick 2016-03-31