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Libration (Lagrange) Points

  1. May 23, 2006 #1
    How does one go about computing the locations of the 5 largrange points? I know where they are but do not know how to derive the equilibrium equations. I know you will get a quintic polynomial which can be solved numerically depending on the masses of the two large bodies, but using forces, how do you write the equations of motion.

    I would assume that the sumF side would just be m(-u1/r^2+u2/r^2), but what is the expression on the right side (kinematics). I really don't understand what the motion of the libration points are exactly
  2. jcsd
  3. May 23, 2006 #2


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    First, chose coordinates. The simplest to use are probably co-rotating cartesian coordinates, where masses m1 and m2 always lie on the x-axis, and the whole coordinate system rotates with an angular frequency of [itex]\omega[/itex].

    You can write down the forces, or you can write down the Lagrangian.

    If you want to use the Lagrangian approach, it's helpful to know that the kinetic energy, T, can be reduced to the following expression in terms of the variables x and y. [itex]\dot{x}[/itex] is shorthand for dx/dt, and [itex]\dot{y}[/itex] is shorthand for dy/dt.

    1/2\,m \left( {{\dot{x}}}^{2}+{{\dot{y}}}^{2} \right) +1/2\,m{\omega}^{2}
    \left( {x}^{2}+{y}^{2} \right) -m\omega \left( x{\dot{y}}-y{\dot{x}}

    The Lagrangian, L is just given by T-V, where V is the potential energy.

    If you want to draw a free body diagram (which is what you stated a desire to do), because you are in a rotating coordinate system, you have to include fictitious forces. Because you are looking for the equilibrium point (and not doing a stability analysis), you can safely ignore the coriolis forces -- as all the bodies are at rest, the coriolis forces will be zero. This means you only nead to add in the centrifugal force comonents which are

    [tex]fx = \omega^2 x \hspace{.5 in} fy = \omega^2 y[/tex]

    To these fictitious forces, you add the "real" forces due to gravity, and demand that the particles be in equilibrium.
  4. May 23, 2006 #3


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    i think you also want the origin (which is on the x-axis) to be the common center of rotation of which both m1 and m2 revolve.

    don't they call this inclusion of certain fictitious forces, "d'Alembert's principle"? i thought that was the name for it. essentially it says that you can replace the accelerated frame of reference with an inertial one where there is an additional fictional force acting on the center of mass of each body which brings the acceleration to zero. in this rotational case, that pseudo-force is often called the "centrifugal force". it is the force you think you feel that is pulling you outward when you rear around a corner or curve in a car at high speed. anyway, you have to apply that pseudo-force to the test particle (which is much smaller than either m1 or m2) in the free body diagram along with the real gravitational forces from the large bodies m1 and m2.

    then find the places where the total force is zero.
  5. May 24, 2006 #4
    So L = T - V
    I know d/dt(dL/dxdot) - dL/dx = something and also
    I know d/dt(dL/dydot) - dL/dy = something

    On the right hand side is there where I place the ficticious forces (for each corresponding equation)?
  6. May 24, 2006 #5
    I get the following expression for my kinetic energy, there is a sign difference in the last term, did you maybe mean to include a plus on the last term?

    [tex]1/2\,m \left( {{\dot{x}}}^{2}+{{\dot{y}}}^{2} \right) +1/2\,m{\omega}^{2} \left( {x}^{2}+{y}^{2} \right) -m\omega \left( y{\dot{x}}-x{\dot{y}} \right) [/tex]
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