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Converting effective potential to dimensionless function

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Convert the effective potential to a dimensionless function by scaling to the radius of a circular orbit.

    2. Relevant equations

    [itex]U_{eff}(r) = - \frac{k}{r}+ \frac{L^2}{2mr^2}[/itex]
    [itex]\frac{U_{eff}}{U_0} = -\frac{2}{r / r_0} + \frac {1}{(r/r_0)^2}[/itex]
    [itex]r_0= \frac{L^2}{mk}[/itex], [itex]U_0= \frac{k}{2r_0}[/itex]

    3. The attempt at a solution

    I would love to try solving this, but unfortunately I can't decipher what the question actually is. What does "scaling to the radius of a circular orbit" mean, in this case?
     
  2. jcsd
  3. Oct 16, 2013 #2
    In the equation for the potential, one of the terms represents repulsion and the other term represents attraction. Scaling this problem like this reduces the number of parameters involved from 5 to 2. This boils the equation down to its bare essence. Solving problems in terms of dimensionless variables often simplifies the compilation of the results, and often also reduces the algebraic complexity and the potential for making typos.
     
  4. Oct 16, 2013 #3
    OK, so I'm supposed to do what exactly? Again, I can't figure out what the prof wants me to actually DO to it.

    I'm sorry. Pretend I am a retarded marmoset taking physics. :-)

    EDIT: I'm not asking for anyone to solve this for me-- I just want to understand what action I am supposed to take.
     
  5. Oct 16, 2013 #4
    You already did what you are supposed to do.

    Chet
     
  6. Oct 16, 2013 #5
    Dude, not meaning to be a jerk or anything, but the Zen koan doesn't help me see what's happening here. It's a bit frustrating. He wants me to make it dimensionless, and maybe my being up all night is clouding my reasoning. Or maybe the whole thing is a lot simpler than it looks.
     
  7. Oct 16, 2013 #6
    I think he might have wanted you to substitute your expression for r0 into your equation for U0:
    [tex]U_0=\frac{m}{2}\left(\frac{k}{L}\right)^2[/tex]

    This eliminates U0 so that
    [tex]\frac{u}{U_0}=\frac{2U}{m}\left(\frac{L}{k}\right)^2[/tex]
     
    Last edited: Oct 16, 2013
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