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Angular momentum & Energy using Yukawa's potential

  1. May 19, 2015 #1
    Hello there!
    I was doing my Gravitation problems and I found this problem that I'm unable to solve.

    Yukawa's theory for nuclear forces states that the potential energy corresponding to the attraction force produced by a proton and a neutron is:
    [tex]U(r) = \frac{k}{r}e^{-\alpha r},\ k<0,\ \alpha > 0[/tex]
    From the expression of it's effective potential, find the module of it's angular momentum and it's energy, for which it's possible a circular movement with a radius r0

    I've tried several things, none of them leading to something meaningful. In fact, I know that expression for effective potential is:
    [tex]U_{ef}(r)=U(r)+\frac{L}{2r^2}[/tex]
    So I imagine I would need to find L fist in order to get the expression for Uef, but I'm not able to remember nor find any kind of formula linking U and L. Would you please help me out?

    PS: Once I know how to find L I know how to end it, since:
    [tex]\frac{dU_{ef}}{dr} = 0 \Leftrightarrow r = r_0 [/tex]
    is the expression of the energy of a circular movement with a radius r0

    Thanks in advance.
     
  2. jcsd
  3. May 19, 2015 #2

    Orodruin

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    I think you are misunderstanding the problem task. You need to use ##r_0## to find the angular momentum and energy. Your work so far is fine and you should continue by differentiating the effective potential.
     
  4. May 19, 2015 #3
    Okay, I have it differentiated. How do I find angular momentum? As far as I know, I only find the energy with what I've done.
     
  5. May 20, 2015 #4

    Orodruin

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    How does the energy look like for a circular orbit?
     
  6. May 20, 2015 #5
    Do you mean that the Potential is twice the kinetic?
     
  7. May 20, 2015 #6

    Orodruin

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    No. I mean: How do you express the total energy for a circular orbit? There is a very simple expression.
     
  8. May 20, 2015 #7
    I don't really know that formula. The only thing that I can think of is [tex]E=\frac{U}{2}[/tex]
     
  9. May 20, 2015 #8

    Orodruin

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    No, it is much simpler than you are thinking. What contributions are there to the total energy?
     
  10. May 20, 2015 #9
    I really don't know. Can you give me a hint?
     
  11. May 20, 2015 #10

    Orodruin

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    What types of energy do you know of?
     
  12. May 20, 2015 #11
    Kinetic and potential.
     
  13. May 20, 2015 #12

    Orodruin

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    Right, so what are their values for a circular orbit?
     
  14. May 20, 2015 #13
    The potential is twice the kinetic, so [tex] \left| E\right| = \left| \frac{U}{2} \right| = \left| K \right| [/tex]
     
  15. May 20, 2015 #14

    Orodruin

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    No. This is simply wrong. You have to be aware when certain theorems hold and when they do not. The answer is much simpler and does not require anything else than very basic mechanics.
     
  16. May 20, 2015 #15
    Could you please give me another hint? I'm really struggling to get anything clear.
     
  17. May 20, 2015 #16
    In theory [tex]
    L=(-mkr_0(1+\alpha r_0)e^{-\alpha r_0})^{1/2}
    [/tex]
     
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