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Co-Rotating Frame

Let us transform to a non-inertial frame of reference rotating with angular velocity $\omega$ about an axis normal to the orbital plane of masses $m_1$ and $m_2$, and passing through their center of mass. It follows that masses $m_1$ and $m_2$ appear stationary in this new reference frame. Let us define a Cartesian coordinate system $(x,\,y,\,z)$ in the rotating frame of reference which is such that masses $m_1$ and $m_2$ always lie on the $x$-axis, and the $z$-axis is parallel to the previously defined $\zeta$-axis. It follows that masses $m_1$ and $m_2$ have the fixed position vectors ${\bf r}_1=\mu_2\,(-1,\,0,\,0)$ and ${\bf r}_2=\mu_1\,(1,\,0,\,0)$ in our new coordinate system. Finally, let the position vector of mass $m_3$ be ${\bf r}=(x,\,y,\,z)$--see Figure 63.

Figure 63: The co-rotating frame.
\begin{figure}
\epsfysize =2.25in
\centerline{\epsffile{Chapter14/fig14.02.eps}}
\end{figure}

According to Chapter 8, the equation of motion of mass $m_3$ in the rotating reference frame takes the form

\begin{displaymath}
\ddot{\bf r} + 2\,\mbox{\boldmath$\omega$}\times \dot{\bf r}...
...ldmath$\omega$}\times(\mbox{\boldmath$\omega$}\times {\bf r}),
\end{displaymath} (1113)

where $\mbox{\boldmath$\omega$}=(0,\,0,\,\omega)$, and
$\displaystyle \rho_1^{\,2}$ $\textstyle =$ $\displaystyle (x+\mu_2)^2+y^2 + z^2,$ (1114)
$\displaystyle \rho_2^{\,2}$ $\textstyle =$ $\displaystyle (x-\mu_1)^2+y^2 + z^2.$ (1115)

Here, the second term on the left-hand side of Equation (1113) is the Coriolis acceleration, whereas the final term on the right-hand side is the centrifugal acceleration. The components of Equation (1113) reduce to
$\displaystyle \ddot{x} - 2\,\omega\,\dot{y}$ $\textstyle =$ $\displaystyle - \frac{\mu_1\,(x+\mu_2)}{\rho_1^{\,3}}- \frac{\mu_2\,(x-\mu_1)}{\rho_2^{\,3}}
+ \omega^2\,x,$ (1116)
$\displaystyle \ddot{y} + 2\,\omega\,\dot{x}$ $\textstyle =$ $\displaystyle - \frac{\mu_1\,y}{\rho_1^{\,3}}- \frac{\mu_2\,y}{\rho_2^{\,3}}
+ \omega^2\,y,$ (1117)
$\displaystyle \ddot{z}$ $\textstyle =$ $\displaystyle - \frac{\mu_1\,z}{\rho_1^{\,3}}- \frac{\mu_2\,z}{\rho_2^{\,3}},$ (1118)

which yield
$\displaystyle \ddot{x} - 2\,\omega\,\dot{y}$ $\textstyle =$ $\displaystyle -\frac{\partial U}{\partial x},$ (1119)
$\displaystyle \ddot{y} + 2\,\omega\,\dot{x}$ $\textstyle =$ $\displaystyle -\frac{\partial U}{\partial y},$ (1120)
$\displaystyle \ddot{z}$ $\textstyle =$ $\displaystyle -\frac{\partial U}{\partial z},$ (1121)

where
\begin{displaymath}
U = - \frac{\mu_1}{\rho_1} - \frac{\mu_2}{\rho_2} - \frac{\omega^2}{2}\,(x^2+y^2)
\end{displaymath} (1122)

is the sum of the gravitational and centrifugal potentials.

Now, it follows from Equations (1119)-(1121) that

$\displaystyle \ddot{x}\,\dot{x} - 2\,\omega\,\dot{x}\,\dot{y}$ $\textstyle =$ $\displaystyle -\dot{x}\,\frac{\partial U}{\partial x},$ (1123)
$\displaystyle \ddot{y}\,\dot{y} + 2\,\omega\,\dot{x}\,\dot{y}$ $\textstyle =$ $\displaystyle -\dot{y}\,\frac{\partial U}{\partial y},$ (1124)
$\displaystyle \ddot{z}\,\dot{z}$ $\textstyle =$ $\displaystyle -\dot{z}\,\frac{\partial U}{\partial z}.$ (1125)

Summing the above three equations, we obtain
\begin{displaymath}
\frac{d}{dt}\left[\frac{1}{2}\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right) + U\right] = 0.
\end{displaymath} (1126)

In other words,
\begin{displaymath}
C = - 2\,U - v^2
\end{displaymath} (1127)

is a constant of the motion, where $v^2=\dot{x}^2+\dot{y}^2+\dot{z}^2$. In fact, $C$ is the Jacobi integral introduced in Section 14.3 [it is easily demonstrated that Equations (1102) and (1127) are identical]. Note, finally, that the mass $m_3$ is restricted to regions in which
\begin{displaymath}
-2\,U \geq C,
\end{displaymath} (1128)

since $v^2$ is a positive definite quantity.


next up previous
Next: Lagrange Points Up: The Three-Body Problem Previous: Tisserand Criterion
Richard Fitzpatrick 2009-07-28