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Hamiltons equations for a satellite

  1. Nov 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Find Hamiltons canonical equations for a satellite of mass m moving about a star of mass M at the origin. Consider motion in a plane, using polar coordinates r,[tex]\theta[/tex].

    Repeat the process using cartesian coordinates x,y from the start.


    2. Relevant equations
    H=T+V


    3. The attempt at a solution
    I did both questions but i'm not exactly sure if I defined the kinetic and potential energies correctly for each case. For polar coordinates I found T=m/2 * (r[tex]\dot{}[/tex]2 +r2 [tex]\theta[/tex][tex]\dot{}[/tex]2) and potential energy V=V(r)

    For cartessian coordinates I found the Kinetic T=m/2(x[tex]\dot{}[/tex]2)
    and potential energy V=[tex]\frac{-GMm}{\sqrt{x^2+y^2}}[/tex].

    So if anyone can confirm if I set up the kinetic and potential energy equations correctly that would be great.
     
    Last edited: Nov 21, 2009
  2. jcsd
  3. Nov 21, 2009 #2

    Pengwuino

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    Gold Member

    That is the correct kinetic energy in the Lagrangian formulation, what did you have for your kinetic energy and how did you form the Hamiltonian?
     
  4. Nov 21, 2009 #3
    Taking care to rearrange the equation I found H=[tex]\frac{1}{2m}[/tex] [pr2+[tex]\frac{p\theta^2}{r^2}[/tex]]. Using this I got the hamilton equations, which there 4 of in this case. I'm sure i did that part correctly i just wasn't 100% sure about setting up the kinetic and potential energy equations correctly.

    Oh and when doing the same problem in cartessian coordinates was I correct in defining the Kinetic energy T=m/2(x[tex]\dot{}[/tex]2)
    and potential energy V=[tex]\frac{-GMm}{\sqrt{x^2+y^2}}[/tex].
     
  5. Nov 21, 2009 #4

    ideasrule

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    Homework Helper

    Not quite. H=T+V, and you only included the kinetic energy. The potential energy, V=-GMm/r, must also be included.

    The potential energy is correct, but don't forget the y component of velocity counts towards the kinetic energy!
     
  6. Nov 22, 2009 #5
    Ah, that makes more sense. Thanks for the help, its much appreciated.
     
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