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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 that 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,\,0,\,0)$ and $ {\bf r}_2=(\mu_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 9.5.)

Figure 9.5: Co-rotating frame.
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According to Section 6.2, the equation of motion of mass $ m_3$ in the rotating reference frame takes the form

$\displaystyle \ddot{\bf r} + 2\,$$\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle \times \dot{\bf r}= - \mu_1\,\frac{({\bf r}-{\bf r}_1)}{\rho_1^{\,3}} - \mu_2\,\frac{({\bf r}-{\bf r}_2)}{\rho_2^{\,3}} -$   $\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle \times($$\displaystyle \mbox{\boldmath$\omega$}$$\displaystyle \times {\bf r}),$ (9.21)

where $ \omega$ $ =(0,\,0,\,\omega)$ , and

$\displaystyle \rho_1^{\,2}$ $\displaystyle = (x+\mu_2)^2+y^{\,2} + z^{\,2},$ (9.22)
$\displaystyle \rho_2^{\,2}$ $\displaystyle = (x-\mu_1)^2+y^{\,2} + z^{\,2}.$ (9.23)

Here, the second term on the left-hand side of Equation (9.21) is the Coriolis acceleration, whereas the final term on the right-hand side is the centrifugal acceleration. The components of Equation (9.21) reduce to

    $\displaystyle \skew{3}\ddot{x} - 2\,\omega\,\skew{3}\dot{y}$ $\displaystyle = - \frac{\mu_1\,(x+\mu_2)}{\rho_1^{\,3}}- \frac{\mu_2\,(x-\mu_1)}{\rho_2^{\,3}} + \omega^{\,2}\,x,$ (9.24)
    $\displaystyle \skew{3}\ddot{y} + 2\,\omega\,\skew{3}\dot{x}$ $\displaystyle = - \frac{\mu_1\,y}{\rho_1^{\,3}}- \frac{\mu_2\,y}{\rho_2^{\,3}} + \omega^{\,2}\,y,$ (9.25)
and   $\displaystyle \skew{3}\ddot{z}$ $\displaystyle = - \frac{\mu_1\,z}{\rho_1^{\,3}}- \frac{\mu_2\,z}{\rho_2^{\,3}},$     (9.26)

which yield

    $\displaystyle \skew{3}\ddot{x} - 2\,\omega\,\skew{3}\dot{y}$ $\displaystyle = -\frac{\partial U}{\partial x},$ (9.27)
    $\displaystyle \skew{3}\ddot{y} + 2\,\omega\,\skew{3}\dot{x}$ $\displaystyle = -\frac{\partial U}{\partial y},$ (9.28)
and   $\displaystyle \skew{3}\ddot{z}$ $\displaystyle = -\frac{\partial U}{\partial z},$ (9.29)

where

$\displaystyle U(x,y,z) = - \frac{\mu_1}{\rho_1} - \frac{\mu_2}{\rho_2} - \frac{\omega^{\,2}}{2}\,(x^{\,2}+y^{\,2})$ (9.30)

is the sum of the gravitational and centrifugal potentials.

It follows from Equations (9.27)-(9.29) that

    $\displaystyle \skew{3}\ddot{x}\,\skew{3}\dot{x} - 2\,\omega\,\skew{3}\dot{x}\,\skew{3}\dot{y}$ $\displaystyle = -\skew{3}\dot{x}\,\frac{\partial U}{\partial x},$ (9.31)
    $\displaystyle \skew{3}\ddot{y}\,\skew{3}\dot{y} + 2\,\omega\,\skew{3}\dot{x}\,\skew{3}\dot{y}$ $\displaystyle = -\skew{3}\dot{y}\,\frac{\partial U}{\partial y},$ (9.32)
and   $\displaystyle \skew{3}\ddot{z}\,\skew{3}\dot{z}$ $\displaystyle =-\skew{3}\dot{z}\,\frac{\partial U}{\partial z}.$     (9.33)

Summing the preceding three equations, we obtain

$\displaystyle \frac{d}{dt}\!\left[\frac{1}{2}\left(\skew{3}\dot{x}^{\,2}+\skew{3}\dot{y}^{\,2}+\skew{3}\dot{z}^{\,2}\right) + U\right] = 0.$ (9.34)

In other words,

$\displaystyle C = - 2\,U - \varv^{\,2}$ (9.35)

is a constant of the motion, where $ \varv^{\,2}=\skew{3}\dot{x}^{\,2}+\skew{3}\dot{y}^{\,2}+\skew{3}\dot{z}^{\,2}$ . In fact, $ C$ is the Jacobi integral introduced in Section 9.3 [it is easily demonstrated that Equations (9.10) and (9.35) are identical; see Exercise 4]. Note, finally, that the mass $ m_3$ is restricted to regions in which

$\displaystyle -2\,U \geq C,$ (9.36)

because $ \varv^{\,2}$ is a positive definite quantity.


next up previous
Next: Lagrange points Up: Three-body problem Previous: Tisserand criterion
Richard Fitzpatrick 2016-03-31