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Classical mechanics: orbits, force, potential

  1. Oct 18, 2013 #1
    1. The problem statement, all variables and given/known data

    a particle of mass m moves on the orbit [itex] r= a cos(θ), a>0[/itex].

    Find the force acting on the particle


    3. The attempt at a solution

    I had this formula in my notebook:


    [itex]U(r)= E-(L^2/2mr^2)(1+(1/r^2)(dr/dθ)^2)[/itex]

    Using it I got [itex] U(r)=E-L^2a^2/2mr^4[/itex]

    and [itex] F(r)=-dU/dr= (-5L^2a^2/2mr^5) \overline{r}[/itex]

    I would really appreciate if someone could confirm my result. I can't find other way to solve it but something smells fishy. Thanks!

    edit: I will detail it a bit more

    This is how I got the result

    [itex]dr/dθ=-a sen(θ)[/itex]
    [itex](dr/dθ)^2= a^2 sen(θ)^2 = a^2(1-cos(θ)^2)=(a^2-r^2)[/itex], then I just plugged this on the formula I'v written.
     
    Last edited: Oct 18, 2013
  2. jcsd
  3. Oct 18, 2013 #2

    mfb

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    The final result should not depend on the particle, so L and a should not be there. And you can remove the E, the offset of the potential does not matter.
    A 1/r^4-potential looks good. I'm not sure about the 4, but we had a similar problem here a while ago, and I think it was 4.

    Edit: This is the reverse problem
     
  4. Oct 18, 2013 #3
    How can I make them disappear?
     
  5. Oct 19, 2013 #4

    mfb

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    Find a in terms of L for your particle (or vice versa), put it there. That might lead to one parameter you cannot solve for, then let this stay unknown.
     
  6. Oct 19, 2013 #5
    I'm sorry but I'm not very good at this, can you provide me some equations that I should focus on to get such result? And I still don't understand what do you mean with "you can remove E", wouldn't that give a different result to the potential?
     
  7. Oct 19, 2013 #6

    mfb

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    Hmm, I guess this is not necessary. You probably get a circle for every constant prefactor of the potential. Just invent a new variable for L^2 a^2 /(2m).

    If two potentials differ by a constant, they lead to the same forces and therefore the same physics. Therefore, any added constant in the potential is arbitrary. You can simply remove it to make the formula shorter.
     
  8. Oct 19, 2013 #7
    So what you are saying is, for example:

    [itex]κ=L^2 a^2/2m[/itex]

    and so:

    [itex]U(r)=-κ/r^4[/itex]
    and
    [itex]F(r)=-κ/r^5[/itex]

    Also, help for finding the particle energy and angular momentum knowing that velocity= V when r=a, please.
     
  9. Oct 19, 2013 #8

    mfb

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    What is the kinetic energy of a particle with velocity V? The potential energy is just given by your potential. For angular momentum, you have to find the angle between particle velocity and r.
     
  10. Oct 20, 2013 #9
    Can i assume that when r=a r and v are perpendicular and thus L= amV ?
     
  11. Oct 20, 2013 #10

    mfb

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    r=a r does not make sense. v and r are perpendicular when the radius is maximal.
     
  12. Oct 20, 2013 #11
    I should have separated the sentence with a comma. I didn't meant r=a r, I meant r and v are perpendicular when r=a, which is the maximum radius like you said. Thanks !
     
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