# Is there a gravitational analogue of a classical Rutherford-atom?

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 reason for this, in Newtonian terms, is that when differentiating the gravitational dipole you find the total momentum of the emitting system, which is conserved, so the second derivative of the gravitational dipole vanishes (see here) --- this reflects the fact that while in electromagnetism we only have one conservation law for the charge, in GR the full four-momentum is conserved.

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 $$2 \times 10^{2} \,\mathrm{W}$$.

Building on the same, and assuming a set of gravitoelectromagnetic equations to be valid …

Gravitoelectromagnetic equations cannot be used to write the equations of gravitational radiation because this approximation is concerned with equations for metric perturbations component $$h_{00}$$, which is an analogue of scalar potential $$\Phi$$ and $$h_{0i}$$, which is an analogue of vector potential $$\mathbf{A}$$. But for radiation one must also consider equations for components $$h_{ij}$$ (that does not have an electromagnetic analogue), which is a higher order of approximation. Instead, one has to use the full set of linearized gravity equations.

Gravitational radiation has quadrupole nature unlike electromagnetic radiation which has dipole character. As a result, radiation of gravitational waves is an effect of fifth order in powers of $$1/c$$, whereas EM radiation is of third order.

The requisite equation for the power of gravitational radiation from a planet in a circular orbit around the star: $$\frac{dE}{dt}=\frac{32 G \mu^2 \omega^6 r^4}{5 c^5},$$ where $$\mu=m_1 m_2 /(m_1+m_2)$$ is the reduced mass of the system “star–planet” (if $$m_1\ll m_2$$, then $$\mu\approx m_1$$).

… can't we say that the Earth itself should one day spiral into the sun …

For Earth–Sun system this formula gives approximately $$200\,\text{W}$$ of gravitational radiation. This gives a “lifetime” due to gravitational radiation of about $$10^{23}\,$$years. Of course, this figure has little to do with the actual fate of the Earth since it will likely be swallowed by the Sun when it enters the red giant phase or ejected from Solar system by some passing star.