<p>The mean motion <span class="math-container">$n$</span> and the time-derivative of the true anomaly <span class="math-container">${\nu}$</span> are related by
<span class="math-container">$$
n\,=\,\frac{(1-e^2)^{3/2}}{(1+e\cos \nu)^{2}}~\dot{\nu}~~.
$$</span>
Therefrom it is clear that in the pericentre (i.e., for <span class="math-container">$\nu=0$</span>) we have <span class="math-container">$\dot\nu&gt;n$</span>. By distinction, in the apocentre (for <span class="math-container">$\nu=\pi$</span>) we obtain <span class="math-container">$\dot\nu&lt;n$</span>. 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.</p>
<p>PS.<br />
You said &quot;the orbital period of the satellite is the same as its angular frequency&quot;. In fact, the <em>sidereal</em> prbital period <span class="math-container">$P_{orb}$</span> is related to the mean motion <span class="math-container">$n$</span> via <span class="math-container">$P_{orb}=2\pi/n$</span>.</p>
<p>PPS.<br />
If you have questions about our papers, please do not hesitate to write to us directly. Our current addresses are given <a href="https://www.mdpi.com/2218-1997/8/4/211" rel="nofollow noreferrer">here</a>.</p>

CC BY-SA 4.0
2022-06-20T19:14:31.310Z
different question entirely, but potentially of interest: [Is Venus in some way tidally locked to... Earth?](https://astronomy.stackexchange.com/q/36488/7982)
uhoh
CC BY-SA 4.0
2023-04-02T01:12:03.940Z
@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.
Michael_1812
CC BY-SA 4.0
2022-06-20T02:02:33.397Z
2023-01-26T12:26:28.687Z
2023-01-26T12:26:28.687Z
Michael_1812
<p>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).</p>
<p>In section 2.1 of Makarov and Efroimsky, 2013 <a href="https://arxiv.org/abs/1209.1616" rel="nofollow noreferrer">No pseudosynchronous rotation for terrestrial planets and moons</a> they say:</p>
<blockquote>
<p>For a nonzero eccentricity <span class="math-container">$e$</span>, 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*</p>
</blockquote>
<p>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?</p>

<orbital-mechanics><mathematics><tidal-forces><orbital-resonance>
Relative orientation in a 1:1 resonance of a planet and a satellite
ValientProcess
# Relative orientation in a 1:1 resonance of a planet and a satellite

(Asked by: username_0 on 2021-12-08)

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](https://arxiv.org/abs/1209.1616) 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?

**Tags:** orbital-mechanics, mathematics, tidal-forces, orbital-resonance


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## 1 answers

### Answer by username_1 on 2022-06-20. Score: 4

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. 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](https://www.mdpi.com/2218-1997/8/4/211).

> **Comments:**
> **username_2**: different question entirely, but potentially of interest: [Is Venus in some way tidally locked to... Earth?](https://astronomy.stackexchange.com/q/36488/7982)  
> **username_1**: @username_2 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.  

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