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)

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

$$x = x_0 + \delta x,$$ (9.68)

$$y =y_0 + \delta y,$$ (9.69)

$$z = 0,$$ (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

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

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

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

$$\delta\skew{3}\ddot{y} + 2\,\delta\skew{3}\dot{x} \simeq - U_{xy}\,\delta x -U_{yy}\,\delta y,$$ (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

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

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

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

It is convenient to define

$$A = \frac{\mu_1}{\rho_1^{\,3}} + \frac{\mu_2}{\rho_2^{\,3}},$$ (9.77)

$$B = 3\left[ \frac{\mu_1}{\rho_1^{\,5}} + \frac{\mu_2}{\rho_2^{\,5}}\right]y^{\,2},$$ (9.78)

$$C = 3\left[\frac{\mu_1\,(x+\mu_2)}{\rho_1^{\,5}}+\frac{\mu_2\,(x-\mu_1)}{\rho_2^{\,5}}\right]y,$$ (9.79)

$$D = 3\left[\frac{\mu_1\,(x+\mu_2)^2}{\rho_1^{\,5}}+\frac{\mu_2\,(x-\mu_1)^2}{\rho_2^{\,5}}\right],$$ (9.80)

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

$$U_{xx} = A - D - 1,$$ (9.81)

$$U_{yy} = A - B - 1,$$ (9.82)

$$U_{xy} = - C.$$ (9.83)

Figure: 9.13 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.

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

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

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

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

$$\frac{8}{9}\leq A \leq 1.$$ (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: 9.14 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.

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

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

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

In unnormalized units, this criterion becomes

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

Figure: 9.15 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.

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.