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Re: [Phys-l] Sharing a problem for students



On Dec 21, 2007, at 5:03 PM, John Mallinckrodt wrote:

Ludwik wrote:
In the first sentence of your reply (see above) you are saying
that my electrical "system is highly unstable." The same is true
electrical "system is highly unstable." The same is true for our
solar system. Is this what you have in mind? If so then I agree.

Yes and no. To first order, the solar system could hardly be more
different from your system. It consists of a central body that is a
thousand times more massive than all of the rest of the bodies in the
system put together. That body is surrounded by negligibly massive
objects in enormous circular orbits with widely varying radii (again
to first order) so that they hardly interact with each other at all.
As far as I know, the stability of the solar system may be an open
question, but it is most certainly not "highly unstable."

On the other hand, the solar system certainly is unstable in the
sense that, for instance, it includes comets which do occasionally
interact significantly with other low mass objects in the system and
that may be flung out of the system (or crash into the Sun (or
Jupiter (or even Earth!))) as a result. . . .

This forced me to look critically into what I wrote.

1) First I found an error (see point 5 and Figure 2) in what was posted at

http://pages.csam.montclair.edu/~kowalski/cf/339students.html

Yes, U' increases (becomes less negative) when the central charge is virtually displaced up or down. And the same is true when one of the positive charges is vertically displaced up or down. But this is certainly not true for non-vertical displacements. A horizontal virtual displacement of the central charge, for example, would make U' lower (more negative). Was I drank when I wrote that "the sum of negative terms becomes smaller?" No, but I probably made an arithmetic error is checking this numerically, most likely confusing + with - somewhere. In any case, it is clear that a horizontal virtual displacement of the negative charge will at once create a net force toward the nearest positive charge. The same is true for a horizontal displacement of a positive charge, and for any displacement that has a horizontal component.

2) My first reaction was to think that my electrical system is just as durable (stable) as a pencil standing on its sharp point. But then I realized the same is true for the similar solar system with two identical, symmetrically located, planets (M>>m, as emphasized by John). There is something wrong with my attempt to discuss durability (three-dimensional stability) in terms of the U' The solar system with two symmetrically revolving planets (any circular orbit and M>>m) would certainly be durable. I have no doubt that the electrical analog I invented would also be durable; it would not behave like a pencil standing on its sharp point. But I need a convincing logical argument for stability in the third dimention. Please help. I hope this can be done at the elementary level of my discussion.

3) By the way, I just simulated the sun (M=2e30 kg) with two identical planets (m=6e24 kg) revolving on a circular orbit (R=1.5e11 m). The system potential energy is -1e55 J while its kinetic energy is only +5e33 J. For obvious reason, the magnitude of potential energy is much higher than kinetic energy. For my electrical system, on the other hand, potential and kinetic energies differ only by the factor of two (-13.5e7 J versus 6.75 J.)

4) Also note that the mass of the negative particle, in the electrical orbiting system, was not used in my formulas. That mass is practically irrelevant because gravitational forces are about 19 orders of magnitude smaller than electrical forces. Assign one billion kilograms to the central particle, instead of 1 kg, and the posted numerical results will be the same.
_______________________________________________________
Ludwik Kowalski, a retired physicist
5 Horizon Road, apt. 2702, Fort Lee, NJ, 07024, USA
Also an amateur journalist at http://csam.montclair.edu/~kowalski/cf/