= Solution
Write $C=e^2/(4\pi\epsilon_0)=mc^2r_0$. For a slowly evolving circular <Coulomb orbit>, force balance and the mechanical energy give
$$
v^2=\frac{C}{mr}=\frac{c^2r_0}{r},\qquad |\boldsymbol\alpha|=\frac{C}{mr^2},\qquad E=-\frac{C}{2r}.
$$
The supplied <Larmor formula> therefore yields
$$
P=\frac{e^2|\boldsymbol\alpha|^2}{6\pi\epsilon_0c^3}=\frac{2mc^3r_0^3}{3r^4}.
$$
Set $dE/dt=-P$. Since $dE/dt=C\dot r/(2r^2)$,
$$
\dot r=-\frac{4cr_0^2}{3r^2},\qquad\frac{d(r^3)}{dt}=-4cr_0^2.
$$
Thus the <classical radiation collapse of a Coulomb orbit> obeys
$$
\boxed{r(t)^3=a_0^3-4r_0^2ct.}
$$
The decreasing mechanical energy means a more tightly bound orbit, not a smaller speed. The formal collapse time is $a_0^3/(4r_0^2c)$. The nonrelativistic, small-loss-per-orbit approximation eventually fails; the result demonstrates the classical instability rather than describing an actual <hydrogen atom>.
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