Relative orientation in a 1:1 resonance of a planet and a satellite

Question

To my understanding, in a two-body problem of a planet and a satellite, a 1:1 resonance means that the orbital period of the satellite is the same as its angular frequency (maybe not, so please correct me).

In section 2.1 of Makarov and Efroimsky, 2013 No pseudosynchronous rotation for terrestrial planets and moons they say:

For a nonzero eccentricity $e$, and in a sufficient proximity of the 1:1 resonance, the relative orientation of the perturber and the bulge changes twice over an orbital period*

I don't understand why the relative orientation of the tidal bulge changes twice, the way I see it the relative orientation should stay the same. Can someone please clarify that statement?

Answer 1

The mean motion $n$ and the time-derivative of the true anomaly ${\nu}$ are related by $$ n\,=\,\frac{(1-e^2)^{3/2}}{(1+e\cos \nu)^{2}}~\dot{\nu}~~. $$ Therefrom it is clear that in the pericentre (i.e., for $\nu=0$) we have $\dot\nu>n$. By distinction, in the apocentre (for $\nu=\pi$) we obtain $\dot\nu<n$. This is why an observer standing on the primary will see that sometimes the orbiter leads and sometimes trails the bulge it generates. Mathematically, in eqn (1b) of our paper, the torque changes its sign twice over an orbital period.

PS.
You said "the orbital period of the satellite is the same as its angular frequency". In fact, the sidereal prbital period $P_{orb}$ is related to the mean motion $n$ via $P_{orb}=2\pi/n$.

PPS.
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