# General Relativity – Is There a Gravitational Analogue of a Classical Rutherford Atom?

In a Rutherford-atom, the electron classically emits EM radiation on an average rate of,
$$-\frac{dE}{dt}=\frac {\omega^4 e^2 R_0^2}{3c^3(4π\epsilon_0)}$$
Where $$\omega$$ is the angular frequency, $$R_0$$ is the initial orbit-radius and the electron will ultimately spiral into the nucleus. Now, as far as I know, the entire phenomenon can be explained via Maxwell equations. Building on the same, and assuming a set of gravitoelectromagnetic equations to be valid (which is a weak field approximation) for masses, can't we say that the Earth itself should one day spiral into the sun, radiating continuously at a rate:
$$-\frac {dE}{dt} = \frac {\omega^4 G M_e M_s R_e^2}{3c^3}$$
Where symbols have their usual meanings? The mean decay time comes to be $$4.75 × 10^{11}$$ years.

Good question! Indeed, there is an analogous phenomenon in GR to the emission by a Rutherford atom, but your formula does not describe it: as written, it would apply to dipolar emission of gravitational radiation, but gravitational emission only starts at the quadrupole order of the perturbative expansion, which is suppressed by a factor $$(v/c)^2$$ compared to the dipole.
The formula describing the energy loss of a binary system due to quadrupolar emission can be found in this Wikipedia article; note the $$c^5$$ term in the denominator as opposed to your $$c^3$$. There is also a hidden factor of $$\omega^6$$ (as opposed to $$\omega^4$$) in the Wikipedia formula: an alternative way to express the radiated power (averaged over a period) is $$- \left\langle \frac{\mathrm{d} E}{\mathrm{d} t} \right\rangle = \frac{32}{5} \frac{G \mu^2}{c^5} R^4 \omega^6\,,$$ where $$\mu = (1/M_s + 1/M_e)^{-1}$$ is the reduced mass of the Earth-Sun system. This reflects the higher power of $$v$$.
To give a sense of the difference, I've plugged some numbers in and your expression would predict a rate of energy loss of about $$3 \times 10^{14} \,\mathrm{W}$$ in the current Sun-Earth configuration, while the correct GR expression predicts $$2 \times 10^{2} \,\mathrm{W}$$.