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Equations of orbit

  1. Dec 4, 2008 #1
    I am trying out the Lagrange equations for the first time, in this case for a particle in orbit around a massive body using classical mechanics. I come up with

    [tex]
    2 \dot r \dot \theta + r \ddot \theta = 0[/tex]

    [tex]
    \ddot r - r \dot \theta ^2 + \frac {GM}{r^2}=0

    [/tex]

    These seem credible to me but are unfamiliar. Are they correct?
     
  2. jcsd
  3. Dec 5, 2008 #2
    OK, I just gave them a second look and now think they are correct. The first one represents conservation of angular momentum (just take the first derivative of [tex]mr^2 \dot \theta[/tex] with respect to time, set it equal to zero then divide both sides by mr), and multiplying the second one by m represents the net radial force on the particle as the difference between the centrifugal force acting on it and the gravitational force.

    Just in case anyone was interested...
     
    Last edited: Dec 5, 2008
  4. Dec 8, 2008 #3
    By the way, how does one go about solving equations like these? I (vaguely) remember learning how to solve 2nd-order differential equations back in college but those were of only one variable. Do I just treat the other one as if it's a constant?
     
  5. Dec 9, 2008 #4
    unfortunately, to solve these you have to express r in terms of theta, probably integrate at least once, and do another substitution. Sometimes reformulating the problem in terms of the hamiltonian helps. It definitely is an analytically solvable system though. Off the top of my head the first equation looks right, second one is at least close. If i have time i'll do this tomorrow but I need to find someone to help me with some questions on fluid mechanics first!

    also, i should be more clear. there are many more powerful ways to solve 2nd order PDE's, but i think the method I described will work.
     
  6. Jan 12, 2009 #5
    sorry,i don't know how to type an equation here,generally AlexGreen was right,in details,as you said,r^2*(dA/dt)=Constant(A is theta),you get dA/dt=C/r^2,you can substitute this to the second equation first,but the essential step is to express dotdot r as:r''=dr'/dA*dA/dt=C/r^2*dr'/dA,similarly, r'=dr/dA*dA/dt=dr/dA*C/r^2=-C*d(1/r)/dA,so r''=-C^2/r^2*(d^2)(1/r)/dA^2,and let u=1/r,put everything in the second equation,you get something like u''+ku=H (k,H are constants,and u'' is the second order derivative with respect to A),then you can solve it by standard method.
     
  7. Jan 12, 2009 #6
    Thanks, kof9595995. I get

    [tex]

    \frac {d^2 u}{d \theta ^2} + u = \frac {GM}{C^2} [/tex]

    where

    [tex]

    C=r^2 \dot \theta [/tex].

    which yields - I think -

    [tex]

    r = \frac {1}{k sin \theta + \frac {GM}{C^2}}

    [/tex]

    - a conic section, as one would expect. Now, to get [tex] \theta [/tex] in terms of t..

    If you want to know how to type an equation, just left-click on it and the details will appear! :smile:
     
    Last edited: Jan 12, 2009
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