Classical Mechanics: Central Force Problem

[Wavefunction]

Homework Statement

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.

Homework 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}

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 U_s(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 U_s>0

Secondly, I want to know the extrema of U_s:

\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 U_s

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?

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

\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!

 

[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:
<br /> 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<br />
which gives you r=\left[L^2/(\mu K)\right]^{1/7}.

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.

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

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?

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

As for calculating the radial frequency I'm not sure how to go about this either

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?

 

[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!