Classical Mechanics: Central Force Problem

In summary, the conversation discusses the motion of a particle under an attractive central force of Kr^4 with an angular momentum L. The energy for circular motion is found by setting the potential energy to zero, resulting in a circular orbit at r= \sqrt[7]{\frac{2L^2}{μK}}. The frequency of radial oscillations can be found by considering the effective potential and expanding it in a power series, with the second order term being equivalent to a spring-mass system with a spring constant of \left.\partial^2 U_s/\partial r^2\right|_{r=r_{\text{min}}}. This results in a frequency of ω_0 = \sqrt{\frac{\left
  • #1
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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 Us(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 Us>0

Secondly, I want to know the extrema of Us:

[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 Us

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 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 [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!
 
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  • #2
The second term in your original energy expression is missing a factor of 2, since it is coming from [itex]\mu r^2\dot{\phi}^2/2=L^2/(2\mu r^2)[/itex], where [itex]L=\mu r^2 \dot{\phi}[/itex]. I think the radius can also be calculated easier by just taking [itex]\dot{r}=0[/itex] in the radial equation of motion:
[tex]
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
[/tex]
which gives you [itex]r=\left[L^2/(\mu K)\right]^{1/7}[/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 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 [itex]C[/itex] 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 [itex]r_{\text{min}}[/itex], the second order term looks just like the potential of a spring-mass system, with spring constant [itex]\left.\partial^2 U_s/\partial r^2\right|_{r=r_{\text{min}}}[/itex]. How is this "effective spring constant" related to the oscillation frequency it would have?
 
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  • #3
Ah yes, I see I now: so [itex](ω_0)^2 = \frac{k}{m}[/itex] with k being the second derivative evaluated at r_(min). Thank you jk86!
 

1. What is the central force problem in classical mechanics?

The central force problem in classical mechanics is a type of two-body problem where one body (called the "central body") exerts a force on the other body (called the "orbiting body") that is always directed towards the central body. This means that the orbiting body moves in a single plane around the central body, with the force acting along the line connecting the two bodies.

2. What are some examples of central forces?

Some examples of central forces are the gravitational force between two objects, the electrostatic force between two charged particles, and the magnetic force between two charged particles moving at constant velocities.

3. How is the central force problem solved?

The central force problem is typically solved using the laws of motion and energy conservation. By writing out the equations of motion and applying energy conservation, the trajectory of the orbiting body can be determined.

4. What is Kepler's First Law and how does it relate to the central force problem?

Kepler's First Law states that the orbit of a planet is an ellipse with the Sun at one of the two foci. This law is related to the central force problem because it describes the shape of the orbit of an orbiting body under the influence of a central force (such as the gravitational force).

5. How does the mass of the orbiting body affect its motion in the central force problem?

The mass of the orbiting body does not affect its motion in the central force problem. This is because the mass of an object does not affect how it responds to a given force. However, the mass of the central body does affect the orbital period of the orbiting body, with larger central masses resulting in shorter orbital periods.

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