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Gravitational field, two body problem, lagrangian points

  1. May 6, 2005 #1
    I want to calculate the lagrangian points of the gravitational field in a 2-body problem. i want to superpose the two gravitational fields of the bodys and after that transform them into a rotating system which has the same angular velocity as the 2-body system does have. from this newscalarfield i want to calculate the stationary points to get the lagrangian points. Can somone tell me, if this would work? And How I can apply the rotation to the superposed gravitational field?
     
  2. jcsd
  3. May 6, 2005 #2

    pervect

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    The gravitational field isn't the tricky part at all. You determine the potential energy as -GM1m/R1 - G M2 M / R2 (m being the test mass, R1 and R2 the distance from the test mass to M1 and M2, respectively) - then you take the gradient of the potential. It doesn't matter to the gravitational field whether the system is rotating or not (at least for Newtonian mechanics).

    The tricky part is dealing with the centrifugal and coriolis pseudo-forces that arise when one switches to a rotating coordinate system.

    If you are familiar with Lagrangian mechanics it makes the job less error prone to use it - , otherwise you just have to manually add-in the centrifugal and coriolis pseudo-forces into the force equation.
     
  4. May 6, 2005 #3

    rbj

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    i presume the OP knows that the two large masses are revolving around their common center of mass. i think i know what you mean by "centrifugal" pseudo-force. it's the application of D'Alembert's principle principle to the accelerated frame of reference, right? but i dunno exactly what you mean by a "coriolis pseudo-force". do you mean the Coriolis effect? if so, how does that relate to the Lagrange points in the two-body problem? i just don't get that.

    r b-j
     
  5. May 6, 2005 #4
    What exactly are lagrange points?

    Coriolis effect = coriolis inertial force. If you are in a rotating frame and are moving relative to it with velocity v, then you feel a force of magnitude 2mwv perpendicular to your direction of motion, where w is the angular velocity of the frame.
     
  6. May 6, 2005 #5

    pervect

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    I'll post the derivation using Lagrangian mechanics, then point out how one can interpret the results in terms of the coriolis and centrifugal pseudo-forces.

    I'll use x' and y' as the non-rotating coordinates, and
    x and y as the rotating ones.

    In non-rotating coordinates, we have

    L = T - V = .5*m*((dx'/dt)^2 + (dy'/dt)^2) - V(x',y')

    here V(x',y') is the gravitational potential, which is equal to

    -G m M1 /sqrt((x'-x1)^2 + (y'-y1)^2) - G m M2 / sqrt((x'-x2)^2+(y'-y2)^2)

    x1 is the X coordiante of M1, y1 is the y coordiante of M1
    x2 is the x coordiante of M2, y2 is the y coordinate of M2

    To switch to rotating coordinates, we need to make the substitutiion

    x(t) = x'(t)*cos(wt)-y'(t)*sin(wt)
    y(t) = y'(t)*cos(wt)+x'(t)*sin(wt)

    w being the angular frequency of rotation of the system.

    The original expression for T was

    T = .5*m*((dx'/dt)^2 + (dy'/dt)^2)

    After much application of the chain rule, one gets the value of T in the new coordiante system

    eq #1
    T = .5*m*((dx/dt)^2+(dy/dt)^2 + 2*x*w*(dy/dt) - 2*y*w*(dx/dt)+w^2(x^2+y^2)

    A change in notation becomes convenient at this point, we write

    [tex]
    L = .5 m (\dot{x}^2 + \dot{y}^2 + 2 x \omega \dot{y} - 2 y \omega \dot{x} + w^2(x^2+y^2)) - V(x,y)
    [/tex]

    V(x,y) remains essentially unchanged from its expression in the non-rotating system V(x',y'), except that x1,x2,y1,and y2 are now constants as desired, rather than varying with time as they did before we switched to the rotating coordinates.

    Note that V(x,y) is a function of x and y only, it does not depend on dx/dt or dy/dt.

    Now we apply Lagrange's equation to get the equations of motion

    http://scienceworld.wolfram.com/physics/LagrangesEquations.html

    [tex]
    \frac{d}{dt} (\frac{\partial L}{\partial \dot{x}}) = \frac{\partial L}{\partial x}
    [/tex]
    [tex]
    \frac{d}{dt} (\frac{\partial L}{\partial \dot{y}}) = \frac{\partial L}{\partial y}
    [/tex]

    This gives us the resulting two equations of motions

    [tex]
    \ddot{x} - \omega \dot{y} = \omega^2 x + \omega \dot{y} - \frac{1}{m} \frac{\partial V}{\partial x}
    [/tex]
    [tex]
    \ddot{y} + \omega \dot{x} = \omega^2 y - \omega \dot{x} - \frac{1}{m}\frac{\partial V}{\partial y}
    [/tex]

    Let's look at the first equation. Moving the duplicate term so that it appears only on the right hand side of the equation, we can say that

    m d^2x/dt^2 aka [tex]m \ddot{x}[/tex] as the mass-acceleration term (in the rotating coordinate system x,y).

    2*m*w*dy/dt as the coriolis force
    m*w^2 * x as the centrifugal force
    -dV/dx as the gravitational force

    So we say that the acceleration is equal to the sum of the centrifugal force, the coriolis force, and the gravitational force in the rotating coordinate system.

    Similarly, the second equation says

    m*d^2 y/ dt^2 aka [tex]m \ddot{y}[/tex] is the mass-acceleration term
    -2*m*w*dx/dt is the coriolis force
    m*w^2*y is the centrifugal force
    -dV/dy as the gravitational force

    and again, acceleration is the sum of tthe centrifugal force, the coriolis force, and the gravitational force.

    The dervitaion on the webpage at

    http://scienceworld.wolfram.com/physics/LagrangePoints.html

    may or may not help you follow the one I did - it should at least serve as a double check for any typos which I may have missed in transcribe from my old archived worksheets on the problem to this post.

    Without Lagrangian mechanics, it's probably just as easy to look at the final result for the equations of motion and not bother with the formalism of the derivation. Where Lagrangian mechanics shines is in its ability to keep the signs straight, something that's very easy to mess up when one does the problem without it, adding the separate terms together "by hand".

    To further simplify the problem, when solving for the Lagrangian points, the coriolis terms drop out, because one assumes that x and y are constant, so dx/dt and dy/dt should be zero. The Coriolis force terms become important when one attempts to analyze the stability of the Lagrange points, something that's also done on the webpage I mentioned.
     
    Last edited: May 6, 2005
  7. May 7, 2005 #6

    rbj

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    i shoulda said "Lagrangian points".

    wikipedia seems to have a good article (sorta): http://en.wikipedia.org/wiki/Lagrangian_point

    the really cool ones are called the L4 and L5 points which are little sinks that space debris fall into and stay. i imagine there has to be some kind of blobs of space-crap in the L4 and L5 points in the Earth-Moon rotational system. i dunno if any telescope has seen this or not. (can anyone say?)

    r b-j
     
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