# Classical Mechanics: Central Force Problem

1. Feb 23, 2014

### Wavefunction

1. The problem statement, all variables and given/known data

A particle of mass m moves under an attractive central force of $Kr^4$ with an angular momentum $L$. For what energy will the motion be circular? Find the frequency of the radial oscillations if the particle is given a small radial impulse.

2. Relevant equations

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

3. 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) $\frac{\partial L}{\partial t} = 0~\Rightarrow~L=C$

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

Now, $U_s (r)= \frac{L^2}{μr^2}+\frac{Kr^5}{5}$

Next I seek the zero(s) of Us(r):

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

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

Secondly, I want to know the extrema of Us:

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

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

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

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

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 r4 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 $\dot{r}≠0$ 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?

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

2. Feb 23, 2014

### jk86

The second term in your original energy expression is missing a factor of 2, since it is coming from $\mu r^2\dot{\phi}^2/2=L^2/(2\mu r^2)$, where $L=\mu r^2 \dot{\phi}$. I think the radius can also be calculated easier by just taking $\dot{r}=0$ in the radial equation of motion:
$$0=\frac{d}{dt}\left(\mu \dot{r}\right) = \mu r \dot{\phi}^2 - \frac{\partial U}{\partial r} =\mu r \left(\frac{L}{\mu r^2}\right)^2 - Kr^4$$
which gives you $r=\left[L^2/(\mu K)\right]^{1/7}$.

The potential energy is defined up to an additive constant, so the choice of $C$ is not an issue.

That is a good way of thinking of what is happening.

Think about what you said regarding thinking of the radial oscillations like a mass-spring system. If you take the effective potential and expand it in a power series about its minimum $r_{\text{min}}$, the second order term looks just like the potential of a spring-mass system, with spring constant $\left.\partial^2 U_s/\partial r^2\right|_{r=r_{\text{min}}}$. How is this "effective spring constant" related to the oscillation frequency it would have?

3. Feb 23, 2014

### Wavefunction

Ah yes, I see I now: so $(ω_0)^2 = \frac{k}{m}$ with k being the second derivative evaluated at r_(min). Thank you jk86!