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Strange behavior for orbits of inverse cube forces and higher?

  1. Oct 19, 2014 #1
    After working a homework assignment which required sketching effective potential energy for the gravitational/coloumb forces, I went and looked at a few effective potentials for inverse cube and inverse quartic (not sure if this is the right word; 1/r^4 force) forces, with inverse square and inverse cube potentials, respectively.

    [itex]F(r) = \frac{k}{r^3}[/itex]
    [itex]U(r) = \frac{k}{2r^2}[/itex]
    ##U_{eff}(r) = \frac{L^2}{2mr^2} + \frac{k}{2r^2} ##

    And redefining the first terms constants into one constant:

    ##U_{eff}(r) = \frac{a}{r^2} + \frac{k}{2r^2} ##

    And similarly for the r^4 force:

    ##U_{eff}(r) = \frac{a}{r^2} + \frac{k}{3r^3} ##

    Having graphed a few effective potentials messing around with the k and a values, I've come to a few conclusions that I'd like to see if I was right on:

    In both cases, there are potentials where the curve always lies above the origin, indicating that E is necessarily non-negative for any orbit.

    For the inverse cube force, there don't appear to be any bound orbits. This I anticipated since the centrifugal term was also of order r^2 (since the centrifugal term was derived from the Lagrangian and is independent of U(r), I suspect this is unique to inverse cube forces). However, for any value where k < -1, and |k| > a, result is a term that looks like:

    ## U_{eff}(r) = \frac{-1}{r^2} ##

    Which would seem to indicate that with E < 0, an object with L and m chosen to meet the condition above ( |k| > [l^2]/[2m]) would get a maximum distance away from the second object (the one generating the force), and then move in and collide with the object (since there's no other point on the curve where dr/dt would become 0 again).

    The inverse quartic force has an even stranger aspect to it. Where k > 0, it produces the unbound, positive energy orbits expected. When k < 0, the curve produced looks like the potential curves for gravity flipped; there are two points where dr/dt = 0, but both have positive energy. This seems to suggest that for an object subject to an inverse quartic force, bound orbits may have positive energy, depending on the sign of the force (which I'm guessing that, as with gravity/electrostatics, indicates whether the force is repulsive or attractive). Further, if E < 0 for these same curves, the implication seems to be that of a collision again.

    Are any of these conclusions valid, or have I made some grave mistake in trying to determine the effective potentials?

    For the "bound orbits with positive energy" possibility, I've thought of another possible alternative, which seems to make more sense. For those curves, might there some "critical" values of r, say ## r_{c1}, r_{c2} ##, where a particle closer than the smaller critical value would reach ## r_{c1} ## and then turn back to collide with the other object, and a particle farther than the larger critical value would reach ## r_{c2} ## and escape back to infinite separation?
    Last edited: Oct 19, 2014
  2. jcsd
  3. Oct 19, 2014 #2
    I am not completely sure I understand what you really wanted to say, but I think you are mistaken. For central attractive forces, bound orbits definitely exist. For example, in the cases you consider, there is always a circular orbit - can you find it?
  4. Oct 20, 2014 #3
    I attempted to go through the effective potentials of inverse cube and inverse quartic forces to see if I could find the types of orbit involved, as the textbook (Classical Mechanics, Taylor) did the same thing for an attractive inverse square force, and I had to do a repulsive inverse square force as a homework problem. I was hoping it'd be readable; I'll have to go back to the drawing board to see if I can rephrase it in a more easily comprehensible way.

    Hmm.... If:

    ## k = -2a ## would give: ## U_{eff} = \frac{a}{r^2} - \frac{a}{r^2} = 0 ##

    For the attractive inverse cube force, this, and the case where |k| > a are the only ones that don't require that E > 0 (E = 0 in the first case, E < 0 in the second). Since a circular orbit corresponds to the lowest possible orbital energy of a given orbit, I'm guessing that one of the two should be the circular orbit of the inverse cube force, but it doesn't seem quite right. Am I at least moving in the right direction?
  5. Oct 20, 2014 #4
    It seems to me you are making things very complicated and the complexity gets out of control. Let's look at the elementary picture. In a circular orbit of radius ##r## and constant velocity ##v##, acceleration is radial and equal to ## v^2 \over r ##. On the other hand, a radial attractive inverse cubic force has magnitude ## f \over r^3 ##, where ##f ## is some constant, which corresponds to acceleration ## {f \over mr^3} = {a^2 \over r^3} ##. Equating the accelerations, we obtain ## v = {a \over r} ##.
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