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Homework Help: Inverse fifth Law

  1. May 21, 2007 #1
    1. The problem statement, all variables and given/known data

    A charged particle of mass m moves non-relativistically around a circular orbit. The potential energy is U(r)= -A/r^4.

    What is the sign of A and why?
    Using Bohr's quantization for angular momentum to calculate the allowed values of the radius.


    2. Relevant equations

    F=-dV/dx
    F=mv^2/r
    mvr=nh/2*Pi


    3. The attempt at a solution

    Intuitively, I thought that A is positive, cause classically you consider that at infinity, the potential is zero. However, when I determined the radius of orbits, it turns out that r^2 is negative, and you cannot square root it!

    mv^2/r=-4*A/r^5 (Centripetal force to -dV/dx)

    Sub it v=nh/2*PI*m*r

    and r^2 = -4*m*A/n^2*H^2 where H=h/2*PI

    Could I ignore the negative sign here???

    Thanks

    Regards,
    The Keck
     
  2. jcsd
  3. May 22, 2007 #2
    Hello,

    This seems some standard homework or exercise.
    In your first equation:

    mv^2/r=-4*A/r^5 (Centripetal force to -dV/dx)

    You have a mistake in the sign of the right hand side when differentiating.


    Best regards
     
  4. May 22, 2007 #3
    No, I'm pretty sure I didn't make a mistake.

    dV/dx= (-4 x -A)/r^5 = (4xA)/(r^5)

    -dV/dx = F(x) = -4A/(r^5)

    Regards,
    The Keck
     
  5. May 22, 2007 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The signs are just fine. The problem is how to interpret them. The acceleration v^2/r is always pointed toward the center. When you are talking about the force of the potential, the positive r direction is usually outward toward larger r. So the negative sign on the force just means that it's directed inward also. The are both pointed in the same direction (for A positive). Just drop the superfluous sign.
     
  6. May 22, 2007 #5
    Thanks, I understand what you mean now, it's just I didn't really think about it using that reasoning

    Regards,
    The Keck
     
  7. May 22, 2007 #6
    Sorry, i did not expain very detail. I mean that:

    [tex]\vec{F}=-\hat{r}\frac{\partial}{\partial r}U(r)=\frac{4A}{r^5}(-\hat{r})[/tex]

    The acceleration in 2-D polar coordinate
    [tex]\vec{a}=(\ddot{r}-\dot{\theta}^2r)\hat{r}+(r\ddot{\theta}+2\dot{r}\omega)\hat{\theta}[/tex]
    The central acceleration in the case is [tex]\vec{a}=(-\dot{\theta}^2r)\hat{r}[/tex] only, where [tex]\dot{\theta} = \text{angular speed respect ro the origin} = \frac{v^2}{r}[/tex]

    Therefore,
    [tex]\frac{4A}{r^5}(-\hat{r})=m(-\frac{v^2}{r})\hat{r}\quad\Rightarrow\quad\frac{4A}{r^5}=m\frac{v^2}{r}[/tex]

    However, Dick expained very well already.



    Best regards
     
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