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Quick mechanics question

  1. Sep 17, 2004 #1
    I have a central force motion with vector form r''=(k^2)*r, where k^2>0. It's trivial to solve for the vector r(t), but I'm having a little trouble solving for r(theta). I get a second order differential equation of the form r''=A/r^3+Br. Any tips on solving this?
     
  2. jcsd
  3. Sep 18, 2004 #2

    Gokul43201

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    [tex]r'' = \frac {A}{r^3} + Br = - \frac {d}{dr}(\frac{A}{2r^2} - \frac{Br^2}{2})[/tex]

    Multiply both sides with r' = dr/dt

    [tex]LHS = r'r" = \frac{1}{2} \frac {d}{dt} r'^2 [/tex]
    [tex]RHS = -\frac{dX}{dr}.\frac{dr}{dt} = -\frac{dX}{dt} [/tex]
    This gives :
    [tex]r'^2 = -2(X) + const. [/tex]
     
  4. Sep 18, 2004 #3
    think i got it, thanks gokul
     
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