0
$\begingroup$

I was doing my Gravitation problems and I found this one 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: $$U(r) = \frac{k}{r}e^{-\alpha r},\ k<0,\ \alpha > 0$$ 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 $r_0​$.

I've tried several things, none of them leading to something meaningful. In fact, I know that expression for effective potential is: $$U_{ef}(r)=U(r)+\frac{L}{2r^2}$$

So I imagine I would need to find $L$ fist in order to get the expression for $U_ef$, 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: $$\frac{dU_{ef}}{dr} = 0 \Leftrightarrow r = r_0$$ is the expression of the energy of a circular movement with a radius $r_0$

$\endgroup$

closed as off-topic by Kyle Kanos, John Rennie, Jim, ACuriousMind, JamalS May 24 '15 at 20:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Kyle Kanos, John Rennie, Jim, ACuriousMind, JamalS
If this question can be reworded to fit the rules in the help center, please edit the question.

1
$\begingroup$

Since this is a homework problem, I won't provide a full solution, but here's a nudge in the right direction. Take a look at these two plots of the effective potential:

k = -1, $\alpha$ = 1, L = 0.25

k = -1, $\alpha$ = 1, L = 1

What's different about these two effective potentials? We only changed $L$ between the two graphs; what does that imply about the allowed values of $L$ if we want a circular orbit? (If you can't answer this question immediately, try plotting the effective potential for a few more values of $L$. If you have access to Mathematica and know how to use it, the Manipulate function is your friend.)

$\endgroup$
  • $\begingroup$ I think I get it. The value of L should be $r_0$. Meaning that L = d(Uef)/dr at $r_0$ $\endgroup$ – Ignasi Sánchez May 20 '15 at 14:45
  • $\begingroup$ Not quite. It's more that for certain values of $L$, no $r_0$ will exist. This is different than the usual Newtonian potential, for which the effective potential has a minimum for every value of $L$. $\endgroup$ – Michael Seifert May 20 '15 at 17:34
  • $\begingroup$ Okay, so, how should I get the angular momentum with the potential energy? I really don't know how to solve it. $\endgroup$ – Ignasi Sánchez May 20 '15 at 17:54
  • $\begingroup$ Given the above information, you should be able to find a relationship between $L$ and $r_0$. Remember, the question you're being asked is "for a given $r_0$, what should $L$ be to have a circular orbit at that radius?" (At least, I think that's what the question is asking. It's no model of clarity.) $\endgroup$ – Michael Seifert May 20 '15 at 18:29
  • $\begingroup$ Hmm I don't think you're getting it right. In theory, $$L=(-mkr_0(1+\alpha r_0)e^{-\alpha r_0})^{1/2}$$ $\endgroup$ – Ignasi Sánchez May 20 '15 at 21:38

Not the answer you're looking for? Browse other questions tagged or ask your own question.