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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.

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  • $\begingroup$ What is the numerical value predicted by your formula and how long will it take for us to fall into the sun? $\endgroup$ – my2cts Feb 18 at 9:05
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    $\begingroup$ The earth-sun system does radiate see e.g. en.wikipedia.org/wiki/Gravitational_wave $\endgroup$ – AlmostClueless Feb 18 at 9:06
  • $\begingroup$ Related: Does the Earth emit gravitational waves? $\endgroup$ – Michael Seifert Feb 18 at 16:32
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    $\begingroup$ Planets are not point-masses and therefore tidal/rotational influences usually dominate the evolution of two-body systems. Beyond a certain distance (which may be synchronous orbit, not sure) the bodies tend to move farther apart (such as the moon and earth). However, if they are closer that a certain point, they will tend to spiral inward. Multi-body systems, like the solar system are even more complex. $\endgroup$ – RBarryYoung Feb 18 at 18:00
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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 $2 \times 10^{2} \,\mathrm{W}$.

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    $\begingroup$ It might also be worth explicitly pointing out that the two extra powers of $v$ translate into a $\omega^6$ dependence for the power loss, rather than the familiar $\omega^4$ dependence for dipole radiation given in the OP. $\endgroup$ – Michael Seifert Feb 18 at 16:35
  • $\begingroup$ @MichaelSeifert Good idea! I have added an alternate expression for the power to address this point. $\endgroup$ – Jacopo Tissino Feb 18 at 18:00
  • $\begingroup$ I often hear it repeated, but I don't believe the quadrupole nature of the radiation has anything to do with the signs of the charges available. In the E&M case you can certaintly get dipole radiation by only moving negative charges around: that's what antennas do after all. $\endgroup$ – wnoise Feb 18 at 22:15
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    $\begingroup$ Neat. I can harness the power of an entire solar system to power two lightbulbs. Or ten lightbulbs, if I get the newfangled LED kind! Isn't physics marvelous? $\endgroup$ – user253751 Feb 19 at 9:26
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    $\begingroup$ @wnoise You're right, that argument doesn't really work. I've fixed my answer. $\endgroup$ – Jacopo Tissino Feb 19 at 10:28
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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.

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