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A point particle $P$ of charge $Ze$ is fixed at the origin in 3-dimensions, while a point particle $E$ of mass $m$ and charge $-e$ moves in the electric field of $P$.

I have the Newtonian equation of motion as $$- \frac{Ze^2}{4 \pi \epsilon_{0}} \frac{\vec{r}}{r^3} = m \ddot{\vec{r}}$$ I derived this from Newton's law and Coulomb's law.

I then went on to show that the particle moves in a plane by showing it has a constant normal vector.

Can anyone help me find this if the orbit is circular about $P$ what it's orbital frequency is -in terms of the constants we have? I would be extremely grateful.

I have attempted this by parameterizing the orbit, differentiating with respect to time and substituting into the equation of motion, however all I get are two unsolvable differentials equations.

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Hint: $v=r\omega$ for circular orbits, and $a=v^2/r$. – Alex R. Oct 9 '12 at 19:11
up vote 2 down vote accepted

If you already know that the orbit you're interested in, you don't need to solve differential equations as such. It doesn't take much thought to guess that the speed must be the same everywhere in a cirular orbit, bu symmetry -- so just postulate that the orbit has the standard parameterization of uniform circular motion: $$ \vec r(t) = R\langle \cos(2\pi\nu t),\sin(2\pi\nu t)\rangle $$ for some $R$ and $\nu$.

Now compute $\ddot{\vec r}$ given this parameterization and plug into the equation of motion. What is left after some simplification is a system of two non-differential equations from which it should be easy to find the relation between $R$ and $\nu$ that make them true for all $t$.

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Thank you! I was using parameters as the angle as a function of $t$, if you don't mind, why can we simply state the parameter as time? – Freeman Oct 9 '12 at 20:06
I'm assuming that the circular motion is uniform, since it doesn't seem to make physical sense for symmetry reasons to have a non-uniform circular motion as a solution. Once we've decided to use uniform circular motion, the form of the parameterization is given up to a choice of radius, frequency (and initial angle which I'm here assuming for simplicity that we absorb into a choice of zero point for the time coordinate) – Henning Makholm Oct 9 '12 at 20:12
Ah I could kiss you! I've been trying for hours to solve non-linear differential equations.. – Freeman Oct 9 '12 at 20:16

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