Lagrange Points

Let us search for possible equilibrium points of the mass $m_3$ in the rotating reference frame. Such points are termed Lagrange points. Thus, in the rotating frame, the mass $m_3$ would remain at rest if placed at one of the Lagrange points. It is, thus, clear that these points are fixed in the rotating frame. Conversely, in the inertial reference frame, the Lagrange points rotate about the center of mass with angular velocity $\omega$, and the mass $m_3$ would consequently also rotate about the center of mass with angular velocity $\omega$ if placed at one of these points (with the appropriate velocity). In the following, we shall assumed, without loss of generality, that $m_1\geq m_2$.

The Lagrange points satisfy $\dot{\bf r} =\ddot{\bf r} = {\bf0}$ in the rotating frame. It thus follows, from Equations (1056)-(1058), that the Lagrange points are the solutions of

$$\frac{\partial U}{\partial x} = \frac{\partial U}{\partial y}= \frac{\partial U}{\partial z}=0.$$ (1066)

Now, it is easily seen that

$$\frac{\partial U}{\partial z} = \left(\frac{\mu_1}{\rho_1^{\,3}}+ \frac{\mu_2}{\rho_2^{\,3}}\right) z.$$ (1067)

Since the term in brackets is positive definite, we conclude that the only solution to the above equation is $z=0$. Hence, all of the Lagrange points lie in the $x$-$y$ plane.

If $z=0$ then it is readily demonstrated that

$$\mu_1\,\rho_1^{\,2} + \mu_2\,\rho_2^{\,2} = x^2+y^2+\mu_1\,\mu_2,$$ (1068)

where use has been made of the fact that $\mu_1+\mu_2=1$. Hence, Equation (1059) can also be written

$$U = - \mu_1\left(\frac{1}{\rho_1}+\frac{\rho_1^{\,2}}{2}\rig... ...rho_2}+\frac{\rho_2^{\,2}}{2}\right) + \frac{\mu_1\,\mu_2}{2}.$$ (1069)

The Lagrange points thus satisfy

$$\frac{\partial U}{\partial x} = \frac{\partial U}{\partial \rho_1}\,\frac{\partial \rho_1}{\partial x}+\frac{\partial U}{\partial \rho_2}\,\frac{\partial \rho_2}{\partial x}=0,$$ (1070)

$$\frac{\partial U}{\partial y} = \frac{\partial U}{\partial \rho_1}\,\frac{\partial \rho_1}{\partial y}+\frac{\partial U}{\partial \rho_2}\,\frac{\partial \rho_2}{\partial y}=0,$$ (1071)

which reduce to

$$\mu_1\left(\frac{1-\rho_1^{\,3}}{\rho_1^{\,2}}\right)\left(\frac{... ...t(\frac{1-\rho_2^{\,3}}{\rho_2^{\,2}}\right)\left(\frac{x-\mu_1}{\rho_2}\right) = 0,$$ (1072)

$$\mu_1\left(\frac{1-\rho_1^{\,3}}{\rho_1^{\,2}}\right)\left(\frac{... ..._2\left(\frac{1-\rho_2^{\,3}}{\rho_2^{\,2}}\right)\left(\frac{y}{\rho_2}\right) = 0.$$ (1073)

Now, one obvious solution of Equation (1073) is $y=0$, corresponding to a Lagrange point which lies on the $x$-axis. It turns out that there are three such points. $L_1$ lies between masses $m_1$ and $m_2$, $L_2$ lies to the right of mass $m_2$, and $L_3$ lies to the left of mass $m_1$--see Figure 48. At the $L_1$ point, we have $x = -\mu_2+\rho_1=\mu_1-\rho_2$ and $\rho_1=1-\rho_2$. Hence, from Equation (1072),

$$\frac{\mu_2}{3\,\mu_1} = \frac{\rho_2^{\,3}\,(1-\rho_2+\rho_2^{\,2}/3)}{(1+\rho_2+\rho_2^{\,2})\,(1-\rho_2)^3}.$$ (1074)

Assuming that $\rho_2\ll 1$, we can find an approximate solution of Equation (1074) by expanding in powers of $\rho_2$:

$$\alpha = \rho_2 + \frac{\rho_2^{\,2}}{3} + \frac{\rho_2^{\,3}}{3} + \frac{51\,\rho_2^{\,4}}{81}+ {\cal O}(\rho_2^{\,5}).$$ (1075)

This equation can be inverted to give

$$\rho_2 = \alpha - \frac{\alpha^2}{3} - \frac{\alpha^3}{9} -\frac{23\,\alpha^4}{81}+{\cal O}(\alpha^5),$$ (1076)

where

$$\alpha = \left(\frac{\mu_2}{3\,\mu_1}\right)^{1/3}.$$ (1077)

is assumed to be a small parameter.

At the $L_2$ point, we have $x = -\mu_2+\rho_1=\mu_1+\rho_2$ and $\rho_1=1+\rho_2$. Hence, from Equation (1072),

$$\frac{\mu_2}{3\,\mu_1} = \frac{\rho_2^{\,3}\,(1+\rho_2+\rho_2^{\,2}/3)}{(1+\rho_2)^2\,(1-\rho_2^{\,3})}.$$ (1078)

Again, expanding in powers of $\rho_2$, we obtain

$$\alpha = \rho_2 -\frac{\rho_2^{\,2}}{3} + \frac{\rho_2^{\,3}}{3} + \frac{\rho_2^{\,4}}{81}+ {\cal O}(\rho_2^{\,5}),$$ (1079)

$$\rho_2 = \alpha + \frac{\alpha^2}{3} - \frac{\alpha^3}{9} -\frac{31\,\alpha^4}{81}+{\cal O}(\alpha^5).$$ (1080)

Finally, at the $L_3$ point, we have $x = -\mu_2-\rho_1=\mu_1-\rho_2$ and $\rho_2=1+\rho_1$. Hence, from Equation (1072),

$$\frac{\mu_2}{\mu_1} = \frac{(1-\rho_1^{\,3})\,(1+\rho_1)^2}{\rho_1^{\,3}\,(\rho_1^{\,2}+3\,\rho_1+3)}.$$ (1081)

Let $\rho_1=1-\beta$. Expanding in powers of $\beta$, we obtain

$$\frac{\mu_2}{\mu_1} = \frac{12\,\beta}{7} + \frac{144\,\beta^2}{49} + \frac{1567\,\beta^3}{343} + {\cal O}(\beta^4),$$ (1082)

$$\beta = \frac{7}{12}\left(\frac{\mu_2}{\mu_1}\right) - \frac{7}{12}\left(... ...left(\frac{\mu_2}{\mu_1}\right)^3 + {\cal O}\left(\frac{\mu_2}{\mu_1}\right)^4,$$ (1083)

where $\mu_2/\mu_1$ is assumed to be a small parameter.

Let us now search for Lagrange points which do not lie on the $x$-axis. One obvious solution of Equations (1070) and (1071) is

$$\frac{\partial U}{\partial \rho_1} = \frac{\partial U}{\partial \rho_2} = 0,$$ (1084)

giving, from Equation (1069),

$$\rho_1 = \rho_2 =1,$$ (1085)

or

$$(x+\mu_2)^2 + y^2 = (x-1+\mu_2)^2 + y^2 = 1,$$ (1086)

since $\mu_1=1-\mu_2$. The two solutions of the above equation are

$$x = \frac{1}{2}-\mu_2,$$ (1087)

$$y = \pm \frac{\sqrt{3}}{2},$$ (1088)

and specify the positions of the Lagrange points designated $L_4$ and $L_5$. Note that point $L_4$ and the masses $m_1$ and $m_2$ lie at the apexes of an equilateral triangle. The same is true for point $L_5$. We have now found all of the possible Lagrange points.

Figure 49: The masses $m_1$ and $m_2$, and the five Lagrange points, $L_1$ to $L_5$, calculated for $\mu _2=0.1$.

Figure 49 shows the positions of the two masses, $m_1$ and $m_2$, and the five Lagrange points, $L_1$ to $L_5$, calculated for the case where $\mu _2=0.1$.