Re: Circular orbit


Ben Rudiak-Gould wrote:
> PD wrote:
> > n=5 is not a case that permits a closed orbit solution.
> > http://scienceworld.wolfram.com/physics/CentralOrbit.html
>
> There will always be circular orbits for any 1/r^n central force, since you
> can always find m,v,r that satisfy mv^2/r = k/r^n. The orbits may not be
> stable, but they'll exist.
>
> -- Ben

Apparently this is an old Goldstein problem I had forgotten, with
Goldstein's remarks as follows (taken from another post):

L^2*u^2/m*(d^2u/dtheta^2 + u) = -f(1/u), (1)

where u = 1/r, as the differential equation that describes orbits for
central forces.

r = d*cos(theta) is an equation of a diameter d circle that passes
through the orgin (centre of force). Take f = -k/r^n, with k > 0.

Then

d^2u/dtheta^2 = (1 + sin^2(theta))/(d*cos^3(theta)).

Substituting this, u, and f into (1), and simplifying gives

2*L^2/(m*d^3*cos^5(theta) = k/(d^n*cos^n(theta)).

L constant => n = 5.
=================================

However, the OPs point is well-taken. Regardless of the *magnitude* of
the force at r=0 (and it's infinite, not zero), an orbit that is
tangent to the force center is asymptotically radial as it approaches
that force center. Since the force is also radial, where does the
particle get any azimuthal velocity after passing through the force
center?

Perhaps what I'm missing is that it is a *one-time-only* circular
orbit, and once it passes through the force center, it proceeds along a
straight line, like a rounded "P"?

PD

.

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