- #1

Wavefunction

- 99

- 4

## Homework Statement

A particle of mass m moves under an attractive

__central__force of [itex]Kr^4[/itex] with an angular momentum [itex]L[/itex]. For what energy will the motion be circular? Find the frequency of the radial oscillations if the particle is given a small radial impulse.

## Homework Equations

[itex]E=\frac{1}{2}μ\dot{θ}^2+\frac{L^2}{μr^2}+U(r)~and~\vec{F(r)}=-\vec{\nabla} U(r)=-Kr^4\hat{e_r}[/itex]

## The Attempt at a Solution

Part a)

First I need to find the potential from the central force, also because its a central force (acting only antiparallel to the radial vector) [itex]\frac{\partial L}{\partial t} = 0~\Rightarrow~L=C [/itex]

When I integral the force I get that [itex]U(r)=\frac{Kr^5}{5}+C[/itex] I set C=0 then [itex] U(r) = \frac{Kr^5}{5}[/itex]

Now, [itex] U_s (r)= \frac{L^2}{μr^2}+\frac{Kr^5}{5}[/itex]

Next I seek the zero(s) of U

_{s}(r):

[itex]0=\frac{L^2}{μr^2}+\frac{Kr^5}{5}~→~r=-\sqrt[7]{\frac{5L^2}{μK}}[/itex]

because, a radial distance can't be negative this implies for C=0 that U

_{s}>0

Secondly, I want to know the extrema of U

_{s}:

[itex] \frac{\partial U_s}{\partial r} = \frac{-2L^2}{μr^3}+Kr^4 = 0~→~r=\sqrt[7]{\frac{2L^2}{μK}}[/itex]

Finally, I need to verify whether this is a minimum or maximum:

[itex] \frac{\partial}{\partial r} (\frac{\partial U_s}{\partial r}) = \frac{6L^2}{μr^4}+4Kr^3~→~∂^2_rU_s(r= \sqrt[7]{\frac{2L^2}{μK}})>0~\Rightarrow~r=\sqrt[7]{\frac{2L^2}{μK}}[/itex] is a min of U

_{s}

So since the minimum of energy occurs at [itex]r= \sqrt[7]{\frac{2L^2}{μK}}[/itex] the energy of the particle in a circular orbit will be (for [itex]\dot{r}=0[/itex]) [itex] E = \frac{L^2}{μ[\sqrt[7]{\frac{2L^2}{μK}}]}+\frac{K[\sqrt[7]{\frac{2L^2}{μK}}]}{5}[/itex]

That was part A) I'm not quite sure if I'm justified in setting C=0 in this case; however I am dealing with a central force that increases with r

^{4}so I don't immediately see anything wrong with my answer. Now for part B) I'm not quite sure on how to do this one, but this is what I have so far:

A radial impulse implies that [itex]\dot{r}≠0[/itex] which means that its radial position will change with time so its orbit will no longer be just circular.

*Find the frequency of the radial oscillations*When I read this part of the question it had me thinking that particle will oscillate along the radial direction like a mass-sping system so that the motion of this particle is like a sinusoidal wave wrapped around a circle (think of sin(x) except that the x-axis is bent into a circle and the oscillations occur perpendicular to the circumference of the circle in the plane of the circle. Am I correct in my description and way of thinking?

As for calculating the radial frequency I'm not sure how to go about this either, but here is what I do know:

[itex]\vec{L}=\vec{r}X\vec{p}~→~\frac{\partial \vec{L}}{\partial t} = \dot{\vec{r}}X\vec{p}+\vec{r}X\dot{\vec{p}}[/itex] I'm also going to take a slight leap of faith here and say that [itex] \dot {\vec{L}} = 0 [/itex] so that [itex]\dot{\vec{r}}X\vec{p} = -\vec{r}X\dot{\vec{p}}[/itex] beyond this however, I am at a loss. Thank you for any guidance you can provide in advance!