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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 48.

Figure 48: The co-rotating frame.
\epsfysize =2.25in

According to Chapter 7, the equation of motion of mass $m_3$ in the rotating reference frame takes the form

\ddot{\bf r} + 2\,\mbox{\boldmath$\omega$}\times \dot{\bf r}...
...ldmath$\omega$}\times(\mbox{\boldmath$\omega$}\times {\bf r}),
\end{displaymath} (1050)

where $\mbox{\boldmath$\omega$}=(0,\,0,\,\omega)$, and
$\displaystyle \rho_1^{\,2}$ $\textstyle =$ $\displaystyle (x+\mu_2)^2+y^2 + z^2,$ (1051)
$\displaystyle \rho_2^{\,2}$ $\textstyle =$ $\displaystyle (x-\mu_1)^2+y^2 + z^2.$ (1052)

Here, the second term on the left-hand side of Equation (1050) is the Coriolis acceleration, whereas the final term on the right-hand side is the centrifugal acceleration. The components of Equation (1050) reduce to
$\displaystyle \ddot{x} - 2\,\omega\,\dot{y}$ $\textstyle =$ $\displaystyle - \frac{\mu_1\,(x+\mu_2)}{\rho_1^{\,3}}- \frac{\mu_2\,(x-\mu_1)}{\rho_2^{\,3}}
+ \omega^2\,x,$ (1053)
$\displaystyle \ddot{y} + 2\,\omega\,\dot{x}$ $\textstyle =$ $\displaystyle - \frac{\mu_1\,y}{\rho_1^{\,3}}- \frac{\mu_2\,y}{\rho_2^{\,3}}
+ \omega^2\,y,$ (1054)
$\displaystyle \ddot{z}$ $\textstyle =$ $\displaystyle - \frac{\mu_1\,z}{\rho_1^{\,3}}- \frac{\mu_2\,z}{\rho_2^{\,3}},$ (1055)

which yield
$\displaystyle \ddot{x} - 2\,\omega\,\dot{y}$ $\textstyle =$ $\displaystyle -\frac{\partial U}{\partial x},$ (1056)
$\displaystyle \ddot{y} + 2\,\omega\,\dot{x}$ $\textstyle =$ $\displaystyle -\frac{\partial U}{\partial y},$ (1057)
$\displaystyle \ddot{z}$ $\textstyle =$ $\displaystyle -\frac{\partial U}{\partial z},$ (1058)

U = - \frac{\mu_1}{\rho_1} - \frac{\mu_2}{\rho_2} - \frac{\omega^2}{2}\,(x^2+y^2)
\end{displaymath} (1059)

is the sum of the gravitational and centrifugal potentials.

Now, it follows from Equations (1056)-(1058) that

$\displaystyle \ddot{x}\,\dot{x} - 2\,\omega\,\dot{x}\,\dot{y}$ $\textstyle =$ $\displaystyle -\dot{x}\,\frac{\partial U}{\partial x},$ (1060)
$\displaystyle \ddot{y}\,\dot{y} + 2\,\omega\,\dot{x}\,\dot{y}$ $\textstyle =$ $\displaystyle -\dot{y}\,\frac{\partial U}{\partial y},$ (1061)
$\displaystyle \ddot{z}\,\dot{z}$ $\textstyle =$ $\displaystyle -\dot{z}\,\frac{\partial U}{\partial z}.$ (1062)

Summing the above three equations, we obtain
\frac{d}{dt}\left[\frac{1}{2}\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right) + U\right] = 0.
\end{displaymath} (1063)

In other words,
C = - 2\,U - v^2
\end{displaymath} (1064)

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 13.3 [it is easily demonstrated that Equations (1039) and (1064) are identical]. Note, finally, that the mass $m_3$ is restricted to regions in which
-2\,U \geq C,
\end{displaymath} (1065)

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

next up previous
Next: Lagrange Points Up: The Three-Body Problem Previous: Tisserand Criterion
Richard Fitzpatrick 2011-03-31