What Did The Electron Say To The Proton?

This problem can be solved by integration. However we would like to present a different solution which does not require calculus. Since both, electrostatic and gravitational interactions obey an inverse square law, Kepler's Laws can also be applied to a system of two charges. The charges $+q$ and $-q$ could be set to orbit in ellipses around their center of mass. This seems to have little to do with the problem we are trying to solve. However, the motion of the charges in our problem can be treated as the degenerate case corresponding to $v_{A}\approx 0$ (see the figure below). The period corresponding to the circular motion is just $T=\frac{2 \pi r_{0}}{v_{0}}.$ Applying Newton's second Law we find $\frac{k q^{2}}{4 r_{0}^{2}}=m a_{c}= \frac{m v_{0}^{2}}{r_{0}}\rightarrow v_{0}=\sqrt{\frac{kq^{2}}{4 m r_{0}}}$ which gives us the period $T= 4 \pi \frac{r_{0}^{3/2}}{\sqrt{\frac{kq^{2}}{m}}}.$ Now, Kepler's third Law states that $\frac{T^{2}}{a^{3}}=\textrm{const}$ where $a$ is the semi-major axis of the orbit. Applying Kepler's Law for the circular orbit and the degenerate elliptic case we obtain $\frac{T^{2}}{(2 \tau)^2}=\frac{r_{0}^{3}}{(\frac{r_{0}}{2})^{3}} \rightarrow \tau=\frac{1}{4\sqrt{2}} T= \pi \frac{r_{0}^{3/2}}{\sqrt{\frac{2 kq^{2}}{m}}}=0.74~\mbox{s} .$

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