1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Orbital Motion Confusion

  1. Nov 25, 2013 #1
    Hi, I stumbled upon this while working on a problem on my physics homework. I still want to solve the problem myself if possible though so I won't post it here, instead, I'll post what is confusing me.

    Consider orbital motion with potential U(r), where U(r) is any arbitrary function of r.
    I was able to show that the quantity [itex]L=mvr[/itex] is conserved and I will call it L. Thus:
    [itex]v=\frac{L}{mr}[/itex]

    We know that the system has a total energy that is constant:
    [itex]E=T+U[/itex]
    [itex]E=\frac{mv^2}{2}+U(r)[/itex]
    [itex]U(r)=E-\frac{mv^2}{2}[/itex]
    [itex]U(r)=E-\frac{m}{2} \frac{L^2}{(mr)^2}[/itex]

    This shows that potential is only dependent on radius. Everything else is a constant. Furthermore, it shows that potential as a function of radius is ALWAYS equal to the same thing... This simply cannot be true... Where am I going wrong?

    edit-
    The problem that I'm working on gives me a function [itex]r(\theta)[/itex] and asks "What central force is responsible for this motion".

    Using the method above... I'm finding that F(r) is the same thing no matter what [itex]r(\theta)[/itex] is... (By taking the negative derivative of U(r) with respect to r.)
     
  2. jcsd
  3. Nov 25, 2013 #2
    The angular momentum is L=mvr only for special cases. Write it for the general case and you will get a more general result.
     
  4. Nov 25, 2013 #3
    Hmm, well this is why I thought it was for general U, maybe you can help me see what I did wrong:

    [itex]L_{Lagrangian}=\frac{1}{2}mv^2-U(r)[/itex]

    In polar, [itex]v^2=\dot{r}^2+r^2\dot{\theta}^2[/itex]

    [itex]L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-U(r)[/itex]

    And then in my text I found these equations:

    [itex]\frac{\partial L}{\dot{q}_i}=p_i[/itex]

    [itex]\frac{\partial L}{q_i}=\dot{p}_i[/itex]

    (Not quite sure why they're true yet)

    which implies:
    [itex]\frac{d}{dt}(\frac{\partial L}{\dot{q}_i})=\frac{\partial L}{q_i}[/itex]

    Applied to Lagrangian above:
    [itex]\frac{d}{dt}(\frac{\partial L}{\dot{\theta}})=\frac{\partial L}{\theta}[/itex]
    [itex]\frac{d}{dt}(mr^2\dot{\theta})=0[/itex]
    [itex]mr^2\dot{\theta}=const=mr(r\dot{\theta})=mvr=l[/itex]

    -edit
    I do see the problem now though! In this expression, this is only velocity in the theta direction. It doesn't account for velocity in the r direction, so I can't use it in the OP. Thanks :)
     
    Last edited: Nov 25, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Orbital Motion Confusion
  1. Orbital motion (Replies: 5)

  2. Orbital Motion (Replies: 17)

  3. Orbital motion (Replies: 10)

  4. Orbital motion (Replies: 1)

Loading...