Radiative collapse of a classical atom

Question

I'm trying to determine the time it takes for a classical electron to spiral towards a proton, but I can't seem to get an answer on the right timescale. Using the energy of the electron $E = -\dfrac{ke^2}{r}+\dfrac{1}{2}mv^2$ and assuming the force on the electron is $\dfrac{mv^2}{r} = \dfrac{ke^2}{r^2}$ gives $$E = -\dfrac{1}{2}\dfrac{ke^2}{r},$$ then differentiating $\dfrac{dE}{dt} = \dfrac{dE}{dr} \dfrac{dr}{dt} = \dfrac{1}{2}\dfrac{ke^2}{r^2}\dot{r}$ and setting equal to the Larmor formula gives $\dfrac{1}{2}\dfrac{ke^2}{r^2}\dot{r} = -\dfrac{2}{3}\dfrac{e^2a^2}{c^3} = -\dfrac{2}{3}\dfrac{e^2k^2e^4}{c^3m^2r^4}$ or $$r^2dr = -\dfrac{4}{3}\dfrac{ke^4}{c^3m^2}dt$$ Integrating from some initial radius $r_i$ to some final radius $r_f$ gives $$t=-\dfrac{1}{4}\dfrac{c^3m^2}{ke^4}\big(r_f^3-r_i^3\big)$$ Plugging in $r_i = 1\rm{e}-10\rm{m}$ and $r_f = 1\rm{e}-15\rm{m}$, gives a spiral time of $0.944$ seconds. However this doesn't seem to agree with timescales I've seen online, giving something around $1.1\rm{e}-10$ seconds. It seems like I might be off by a factor of $1/k$ but I can't see where I might have messed up the algebra. This has been bugging me for a few days, so any help is greatly appreciated!

Answer 1

If $k = 1/4\pi \epsilon_0$, then I think Larmor's formula is $$\frac{dE}{dt} = \frac{2e^2a^2}{3c^3}k$$

i.e. You have mixed SI and cgs units.