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6 Restricted circular three-body problem

 This section illustrates the formalism we introduced in this tutorial and it provides an example with long history dating back to Poincaré.


 
Figure 3:  Restricted three-body problem
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In restricted three-body problem one of the three masses is negligible as compared to the other two masses. In the circular version we assume that the remaining two masses move along circular orbits about their center of mass (see Figure 3).

It is natural to introduce a coordinate system in which the two heavy masses are at rest, say at points A and B and that the center of mass is at 0. The potential energy in this coordinate system is

U(Q) = - $\displaystyle{\frac{\mu_A}{\Vert Q-A\Vert}}$ - $\displaystyle{\frac{\mu_B}{\Vert Q-B\Vert}}$ (51)

where $\mu_{A}^{}$ and $\mu_{B}^{}$ are suitable constants. We assume that the light mass is 1 .

Thus the equations of motion are

$\displaystyle\dot{Q}$ = P - $\displaystyle\omega$ x Q   
$\displaystyle\dot{P}$ = - $\displaystyle\nabla$U(Q) - $\displaystyle\omega$ x P (52)
where P,Q $\in$ $\reals^{3}_{}$ . Of course, we may perform the differentiations and obtain a completely explicit system of ODE with 4 free parameters. Indeed, we may assume that A and B are on the x -axis and A = (a,0,0) , B = (- b,0,0) . We must have $\mu_{1}^{}$a = $\mu_{2}^{}$b which is the only relation between the four parameters. We may also assume that $\omega$ = (0,0,c) and that c is another parameter.
next up previous
Next: About this document ... Up: Lagrangian and Hamiltonian mechanics Previous: 5 Motion of a
Marek Rychlik
9/2/1997