# Particle moving in an inverse-cube force field?

1. Sep 29, 2004

### Farina

Before I get yelled at for cross-posting, it was advised that I post this query here, as opposed to the celestial mechanics forum.

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:

$$r=r_0 \cos \theta$$

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 (exponential form; sin, cos form, etc.
-- whatever the basic alternative forms are to solving 2nd order ODEs) 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
$$f(u^{-1}) \text { = the central force}$$

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

Any ideas?

Last edited by a moderator: Oct 2, 2004
2. Oct 2, 2004

### Mk

how's inverse cube different from inverse square?

3. Oct 2, 2004

### ehild

The equation for the function u=1/r takes the form of

$$\frac{d^2u}{d\theta^2} = B u$$

where B is a constant, and includes the angular momentum, the mass and the parameter of the force.

If B>0

$$u(\theta) = c_1\cos(\sqrt{B}\theta) + c_2\sin(\sqrt{B}\theta)$$

If B=0

$$u(\theta) = c_1\theta + c_2$$.

If B<0

$$u(\theta) = c_1\exp(\sqrt{-B}\theta)+c_2\exp(-\sqrt{-B}\theta)$$.

ehild

4. Oct 2, 2004

### Integral

Staff Emeritus
Looks to me like your DE is not dimensionally correct. Where did you get it? Perhaps you could give us a reference.

It is not clear to me just exactly what you are trying to do. Could you please define the force which is acting?

Last edited: Oct 2, 2004
5. Oct 2, 2004

### Odyssey

Greetings,
say a point mass experiences a net force F = -Amr^-3, where r is the distance, A is some constant, and m is the mass of the point mass, how would you go about in describing the possible orbits of the point mass with non-zero angular momentum and zero energy? What if E>0 and E<0? I know this involves some differential equaotions and integration with trig sub.

6. Oct 2, 2004

### ehild

Well, I think I understand what Farina's problem was.

As for the dimension, l should not be the angular momentum but the angular momentum divided by mass. Otherwise the DE is OK.
The starting point was the equation of motion of a point mass in a central field, using circular polar coordinates. One of the two equations results in conservation of angular momentum. The other differential equation contains the second time derivative of r. We change the variable t to theta, and replace the time derivative of theta by l/r^2. At the end we replace r by 1/u.

I've just noticed that my solutions are not quite correct as they do not exclude negative values for r. Anyway, it would be really interesting to live in a world with inverse-cube gravitation but it would be a bit difficult to survive.

ehild

7. Oct 2, 2004

### pervect

Staff Emeritus
As I recall, the orbits should not be closed figures for an inverse cube force.

Goldstein gives the applicable diffeq as

$$\frac{d^2u}{d \theta^2} + u = -\frac{m}{l^2}\frac{d}{du}V(\frac{1}{u})$$

here
$$\frac{dV(r)}{dr} = a/r^3$$
so
$$V(r) = -\frac{a}{2*r^2}$$

Hence V(u) = (-a/2)u^2, dV/du = -a*u

So the diffeq as written wasn't correct (or my textbook is in error - could happen, I haven't double-checked the answer).

The previous solution offered looks correct, though, as the equation is indeed of the form

$$\frac{d^2u}{d \theta^2} = cu$$

c = a*m/l^2 - 1

so you get the previously mentioned solutions of the form
$$cos(\sqrt{c}\theta), sin(\sqrt{c}\theta), e^{\sqrt{c}\theta}, e^{-\sqrt{c}\theta}$$

depending on whether c is positive or negative. I'm not sure about the "r being negative" issue. Actually I'm not positive I haven't dropped a bit here or there, but it's encouraging that my answer agrees with someone else's.

8. Oct 3, 2004

### Farina

I erroneously (sorry) said that one solution was:

$$r=r_0cos\theta$$

I should have said one solution is:

$$r=r_0e^{k\theta}$$

The following comes from Fowles & Cassiday "Analytical Mechanics" and is basically Newton's DE of motion using two-dimensional polar coordinates instead of three:

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

where:

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

And, yes, l = angular momentum per unit mass (thanks ehild!).

There is a fair amount of derivation provided in the text. The DE is labled as "the differential equation of the orbit of a particle moving under a central force."

I erroneously (sorry) said that one solution was:

$$r=r_0cos\theta$$

I should have said one solution is:

$$r=r_0e^{k\theta}$$

Setting u = 1/r you thus have:

$$u={r_0}^{-1}e^{-k\theta}$$

Referring back to the DE and using

$$\text{u, } \frac {d^2u}{d\theta^2}$$

then simply re-arranging and solving algebraically
for $$f(u^{-1} )$$ you get

$$f(r)=-(k^2+1)ml^2\cdot \frac {1}{r^3}$$

which shows that the central field corresponds to an inverse-cube force.

I'm stuck at the proposition of having to find two other types of possible orbits (and their equations). I was thinking it was a simple matter of expressing the solution to the DE in basic alternative forms -- but maybe not??

9. Oct 3, 2004

### pervect

Staff Emeritus
OK, I should have looked a little longer before I posted. Goldstein has two formula's for the equation of orbit, the one I posted and

$$\frac{l^2 u^2}{m} (\frac {d^2u}{d\theta^2} + u}) = -f(1/u)$$
where f is the force. This is basically the same as your text, except that in Goldstein's eq. the angular momentum is not normalized. So all the formulas posted are equivalent, except for the issue about whether or not to normalize angular momentum.

In any event, the answer $$u = c_1 cos(k \theta)$$ should still work.

more generally, $$u = c_1 cos( k \theta) + c_2$$ should work
come to think of it, you could add a c_3 \theta term in there, too

$$r = \frac{r_0}{cos(k \theta)+ c_1 \theta + c_2}$$
[end edit]

So that's two solutions. (You can replace cos(theta) with sin(theta)).

*idea*. We've noted that

$$\frac {d^2 u}{d \theta^2} = B \theta$$

and we've come up with solutions for B positive, and B negative. The third solution occurs when B = 0

This is just $$r = \frac{c_1}{1 + c_2 \theta}$$

Note that an attractive cubic force will be given by

f(r) = -a/r^3

So that B = a * m / l^2 - 1 (my version of l), or B = a/ (m l^2) -1 (your version of l). This demonstrates that B can indeed by positive, zero, or negative, as a, m, and l^2 are all positive quantites. A large l gives a negative B and the inverse trig solution, a critical value of l gives the linear solution (this class of solution includes the not-quite stable circular orbits of constant r), and a low value of l gives the exponential solution.

Last edited: Oct 3, 2004
10. Oct 3, 2004

### pervect

Staff Emeritus
grrr - can't edit this now
Anyway, my 'cos(theta)' solution is wrong, it should be

1/r = c_1 cos(theta) + c_2 sin(theta) or just read ehild's response :-)