1 5 \left(\frac r R\right)^2
\left(\begin{matrix}
\sin\theta \cos\phi \\
\sin \theta \sin\phi \\
2\cos \theta
\end{matrix}\right)
\right]\right\},
\end{split}$$ with $M$ the mass of the object and $r$ its radius. In the particular case in which ${\boldsymbol{e}}$ is normal to the orbital plane, ${\boldsymbol{M}}_S$ precedes then around it with the same angular frequency $\om$ of the object motion.

In the general case, ${\boldsymbol{M}}_S(t)$ can be written as sum of a constant moment, which of course does not give rise to emission of electromagnetic radiation, and of a time-dependent moment of the form $$\sum_{i} {\boldsymbol{M}}_i \cos(\omega_i t -\psi_i),$$ with $\omega_i = \omega$, $2\omega$ and $\psi_i$ are suitable phases. It is then an exercise in classical electromagnetism to verify that the time-averaged (classical) electromagnetic energy radiated (on classical spacetime) per unit time by this variable magnetic moment is given, in natural units, by $$\label{eq:power}\begin{split}
\frac{d \cE}{dt} &= \frac{2}{9}\sum_{\omega_i = \omega_j}\omega_i^4 ( {\boldsymbol{M}}_i \cdot {\boldsymbol{M}}_j) \cos(\psi_i-\psi_j)\\
&= \frac2 9 e^2\lambda_P^4 M^2 R^2\omega^6\left(1+\sin^2 \theta +\frac1 2 \omega^2 R^2 \sin^2 \theta\right).
\end{split}$$ Therefore averaging over the unknown direction of ${\boldsymbol{e}}$ we get $$\label{eq:power2}
\frac{d \cE }{dt}   = \frac{8\pi}{27} e^2\lambda_P^4 M^2 R^2\omega^6 (5+ \omega^2 R^2) \simeq    e^2  \lambda_P^4 M^2 R^2 \omega^6  \simeq e^2  \left(\frac{\tau _P}{T} \right)^6 \left( 