- #1

SandraH

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## Homework Statement

A particle of mass m moves in a potential of the form

U(r) = (1/2)kr^2

k = const greater than zero

1) Determine the possible orbits r = r(theta) and show that they are closed

2) Solve the equations of motion (although it is sufficient to derive the time dependence

of the radial variable, r = r(t), in accordance with the general theory of the motion in

central potentials, there is a simpler way to address the problem !).

3) Analyze whether this potential respects Kepler’s first law: The orbits are ellipes with

the force centre at one focus.

4) Demonstrate that this potential does not respect Kepler’s third law.

## Homework Equations

The lagrange equations

(d/dt) (dL/dr) - (dL/dr_dot) = zero (sorry for the crappy formatting, you know what mean)

The Newton equation

m*acceleration = -(d/dr) U (r)

'Reduction to quadrature'

t-t0 = the integral of 1/ sqrt (E - U(r)) dr (again... sorry!)

This last one is helpful for finding r in terms of t but often yields tricky integrals

## The Attempt at a Solution

I think this is a doozy. WHat coordinates do I switch too? And what is the Lagrangian for a spherical harmonic oscillator?

I tried modelling my answer on a spherical pendulum problem (no oscillations)... There's a detailed worked pendulum problem in our course notes. Here the coordinate system was changed to

x = r sin (theta) cos (phi)

y = r cos (theta) cos (phi)

z = -rcos (theta)

why would you bother, when you can just work with r? Do you have to switch coordinate systems for the Lagrange equations to be valid?

Anyways I located the equation for a sperical pendulum Lagrangian in our text. It's:

L = (1/2)m (x˙^2 + y˙^2 + z˙^2) − mgz

(Note the dots... That rate in change of x squared plus rate of change in y squared plus...)

In the new coordinate system this is

ml2(˙θ2 + ˙φ2 sin2 θ) + mgl cos θ

For my spherical oscillator can I just replace the potential energy (mgz, the final term) with the given potential, (1/2)kr^2?

Also, is angular momentum conserved? That would mean I can set the change in in phi to a constant (Noether's theorem, for Lagrange aficionados). SOmehow I doubt I can... They do it in the pendulum example but... is angular momentum conserved in the spherical oscillator case? I'm finding it hard to picture.

I proceeded with my Lagrangian and assumed delta phi was constant and found

r = (+/-) 4nth root of (1/sin^2 theta) * (1 / (Veff + k theta/2m))

I have no idea if this is even close... that initial Lagrangian could be way off.

Also not sure how to ascertain whether this path is 'closed'. sin^2 theta moves between 0 and 1 and theta can go between a and pi so... I think the radius changes periodically, as an oscillator should.

On the other hand, that theta in the denominator on the right hand side is troubling... it suggests the radius goes to infinity when theta (the angle of the radial vector to z) is zero.

Help! Help, help, help. ANy advice about Lagrange or links to good resources would be enormously appreciated. Sorry for the scrappy summary- it's 2am here in Australia and I want to get this posted before all the AMericans start waking up :)