# Particle moving in an inverse-cube force field?

1. Sep 29, 2004

### Farina

PRE-QUESTION:

This questions pertains to celestial mechanics and
classical mechanics. Given that cross-posting is
frowned on -- what's a person to do to make sure
the both (appropriate) audiences are exposed to this post?

--------
Ok -

I have a particle moving in an inverse-cube force field.

One type of orbit for this particle is given by:

$$r=r_0cos\theta$$

Apparently there are thus two additional types of
orbit possible.

What are their equations?

The only thing I can think of is that there are multiple
solution expressions to the governing differential equation
for a particle moving in a central force field:

$$\frac {d^2u}{d\theta^2} + u = -\frac {1}{ml^2u^2}f(u^{-1})$$

where:

u = 1/r
m = mass
l = angular momentum
$$f(u^{-1}) \text {is the central force}$$

Other than remembering this stuff -- I'm at a loss
to see how to cough-up the other two orbit-type
equations.

Any ideas?

2. Sep 29, 2004

### pervect

Staff Emeritus
I'd recommend clasical mechanics, the problem of motion with a generalized central force law is discussed in, for instance, "Goldstein, Classical Mechanics". The problem does not appear much in astronomy :-)

OK, I'll present a brief summary, if you want more detail you can try and find a copy of Goldstein's text and read about it for yourself, or ask some more questions.

Note that angular momentum will be conserved for any central force.

The motion of a particle under a general central force can be reduced to the equivalent one-dimensional problem by making use of the conservation of angular momentum.

First off, we should probably introduce the potential energy V(r). We demand that the radial force fr be the gradient of the potential energy V(r)

$$f_r = -\frac{\partial V}{\partial r}$$, i.e. that the force be the gradient of the potential energy.

The "effective potential" will then be

Veff = V(r) + L^2/(2*m*r^2)

where L is the constant angular momentum, and V(r) is the potential energy as defined above.

The motion of the body will be the same as that of a one-dimensional body in the effective potential. The intial position of the body, it's intial energy, and it's invariant angular momentum will suffice to specify the motion of the body. The radial motion will be given by solving the equation

$$m\frac{d^2r}{dt^2} = -\frac{\partial V_{eff}}{\partial r}$$

The total energy will be a constant of motion, thus

$$\frac{1}{2} m (\frac{dr}{dt})^2 = E$$ where E is a constant.

This equations of motion can also be written in terms of the original force law fr

$$m\frac{d^2r}{dt^2} = f_r + \frac{L^2}{mr^3}$$

For the inverse cubic force, f= -k/r^3, the potential V(r) is -k/(2*r^2), so we have

Veff = -k/(2*r^2)+ L^2/(2*m*r^2) = (-k/2 + L^2/(2*m))/r^2

and

m r'' = a/r^3

For a circlar orbit, Veff is constant, and we see by the above equation that it must be zero. This is a marginal condition for stability.

The orbits will in general not be closed, they are closed for all initial conditions only for force laws f=-k/r and f=-k/r^2

Probably the easiest way to solve for the radial motion is to make use of the fact that the energy is constant

You'll wind up with an equation for dr/dt = g(r), which you can re-write as dr/g(r) = dt. By integrating both sides, you can find r as a function of t.

The result will be messy though, the general result is

$$dt = \frac {dr}{\sqrt{\frac{2}{m}(E-V_eff)}}}$$ where the equation for Veff has been previously given.

3. Oct 2, 2004

### Odyssey

Greetings,

say a point mass experiences a net force F = -Amr^-3 (r = distance, A = some constant, m = mass of point mass), how should I go about in describing the particle's possible orbits with non-zero angular momentum? I know I have to consider the energy if it's <0 or >0 too in the process, and it requires some integation with differential equations?