2
$\begingroup$

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?

$\endgroup$

1 Answer 1

4
$\begingroup$

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.
If you have questions about our papers, please do not hesitate to write to us directly. Our current addresses are given here.

$\endgroup$
2
  • 1
    $\begingroup$ different question entirely, but potentially of interest: Is Venus in some way tidally locked to... Earth? $\endgroup$
    – uhoh
    Jun 20, 2022 at 19:14
  • 1
    $\begingroup$ @uhoh I never studied the rotation of Venus. Used to think that it is defined, mainly, by the interplay of two torques -- the one created by the bodily tides generated by the Sun, and the one created by the atmospheric tides caused by the Sun. I would presume, from the top of my head, that the tides generated by the Earth should be weaker. $\endgroup$ Apr 2, 2023 at 1:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .