<p>Derivation of the right-hand side of formula (17.1) from the cited <a href="https://web.gps.caltech.edu/classes/ge131/notes2016/Ch17.pdf" rel="nofollow noreferrer">lecture notes</a> requires some work, because this equation of equilibrium is written in a noninertial frame associated with one of the bodies, not with their barycentre.  In <a href="https://link.springer.com/article/10.1007/s10569-009-9204-7" rel="nofollow noreferrer">our paper with Jim Williams</a>, also <a href="https://arxiv.org/abs/0803.3299" rel="nofollow noreferrer">available in ArXiV</a>, we wrote this derivation in detail. From eqn (11), you will see that the leading term (the one proportional to <span class="math-container">$(R/a)^2$</span>, also called ``quardupole'') reads, in simplified notation, as
<span class="math-container">$$
-~\frac{GM}{r}~\left(\frac{R}{r}\right)^2~~,
$$</span>
where <span class="math-container">$r$</span> is the distance to the perturber. (I omit the sub- and superscripts used in our paper.)</p>
<p>For a perturber describing an elliptic orbit, a variation of the distance <span class="math-container">$r$</span> is <span class="math-container">$er$</span>, so the tidal variation of the above expression will be of order
<span class="math-container">$$
\sim~e~\frac{GM}{r}~\left(\frac{R}{r}\right)^2~~.
$$</span></p>
<p>Also be vigilant about some details in the <a href="https://web.gps.caltech.edu/classes/ge131/notes2016/Ch17.pdf" rel="nofollow noreferrer">chapter</a> you cited.</p>
<ul>
<li><p>On page 2, the author says that ``A non-synchronously rotating satellite is despun to synchroneity''. This rule is not universal, Mercury being an obvious counterexample.</p>
</li>
<li><p>The subsequent story of tidal dissipation is very confusing and at some point plainly misleading. Look at the author's eqn (17.5) and ask yourself if the dissipation rate can be expressed without the viscosity entering the formula.</p>
</li>
<li><p>In the paragraph above eqn (17.5), the author refers to the Maxwell model, without realising that within that model the dissipation rate is inversely proportional to the viscosity and does not depend upon the rigidity at all (which, of course, is indicative of the said model being symplistic).</p>
</li>
<li><p>On page 8, the author complains that eqn (17.5) underestimates the heat production in Enceladus by two orders of magnitude. It sure does!<br />
...while the correct expression would have rendered him the measured heat production rate with an amazing precision, had the ice viscosity value near the melting point being used in it.</p>
</li>
</ul>

CC BY-SA 4.0
2024-03-23T01:02:34.867Z
2024-03-23T20:16:12.083Z
2024-03-23T20:16:12.083Z
Michael_1812
<p>I'm reading <a href="http://web.gps.caltech.edu/classes/ge131/notes2016/Ch17.pdf" rel="nofollow noreferrer">these excellent lecture notes</a> about tidal dissipation, and they mention at some point that the time-dependent part of the tidal potential is approximated by (Eq. 17.1 on the third page):</p>
<p><span class="math-container">$$ ~ e\frac{GMR^2}{a^3} $$</span></p>
<p>Where <span class="math-container">$M$</span> is the host planet mass, <span class="math-container">$R$</span> is the satellite radius and <span class="math-container">$a$</span> is the orbital radius (which I guess it's the semi-major axis since the orbit is eccentric?), and <span class="math-container">$e$</span> is the eccentricity as far as I understand.</p>
<p>Can someone shed some light on how they got to this approximation for the tidal potential?</p>

CC BY-SA 4.0
2022-09-21T14:35:42.363Z
possibly helpful is the article cited in the first paragraph: [Origin and Evolution of the Natural Satellites](https://doi.org/10.1146/annurev.astro.37.1.533) also in [researchgate](https://www.researchgate.net/publication/228687644_Origin_and_Evolution_of_the_Natural_Satellites)
uhoh
CC BY-SA 4.0
2022-09-21T14:53:25.283Z
Sadly this article doesn't provice insight about this specific approximation
ValientProcess
<tidal-forces>
Tidal potential expression using eccentricity
ValientProcess
# Tidal potential expression using eccentricity

(Asked by: username_0 on 2022-09-21)

I'm reading [these excellent lecture notes](http://web.gps.caltech.edu/classes/ge131/notes2016/Ch17.pdf) about tidal dissipation, and they mention at some point that the time-dependent part of the tidal potential is approximated by (Eq. 17.1 on the third page):

$$ ~ e\frac{GMR^2}{a^3} $$

Where $M$ is the host planet mass, $R$ is the satellite radius and $a$ is the orbital radius (which I guess it's the semi-major axis since the orbit is eccentric?), and $e$ is the eccentricity as far as I understand.

Can someone shed some light on how they got to this approximation for the tidal potential?

**Tags:** tidal-forces

> **Comments:**
> **username_1**: possibly helpful is the article cited in the first paragraph: [Origin and Evolution of the Natural Satellites](https://doi.org/10.1146/annurev.astro.37.1.533) also in [researchgate](https://www.researchgate.net/publication/228687644_Origin_and_Evolution_of_the_Natural_Satellites)  
> **username_0**: Sadly this article doesn't provice insight about this specific approximation  

---

## 1 answers

### Answer by username_2 on 2024-03-23. Score: 4

Derivation of the right-hand side of formula (17.1) from the cited [lecture notes](https://web.gps.caltech.edu/classes/ge131/notes2016/Ch17.pdf) requires some work, because this equation of equilibrium is written in a noninertial frame associated with one of the bodies, not with their barycentre. In [our paper with Jim Williams](https://link.springer.com/article/10.1007/s10569-009-9204-7), also [available in ArXiV](https://arxiv.org/abs/0803.3299), we wrote this derivation in detail. From eqn (11), you will see that the leading term (the one proportional to $(R/a)^2$, also called ``quardupole'') reads, in simplified notation, as
$$
-~\frac{GM}{r}~\left(\frac{R}{r}\right)^2~~,
$$
where $r$ is the distance to the perturber. (I omit the sub- and superscripts used in our paper.)

For a perturber describing an elliptic orbit, a variation of the distance $r$ is $er$, so the tidal variation of the above expression will be of order
$$
\sim~e~\frac{GM}{r}~\left(\frac{R}{r}\right)^2~~.
$$

Also be vigilant about some details in the [chapter](https://web.gps.caltech.edu/classes/ge131/notes2016/Ch17.pdf) you cited.

* On page 2, the author says that ``A non-synchronously rotating satellite is despun to synchroneity''. This rule is not universal, Mercury being an obvious counterexample.
* The subsequent story of tidal dissipation is very confusing and at some point plainly misleading. Look at the author's eqn (17.5) and ask yourself if the dissipation rate can be expressed without the viscosity entering the formula.
* In the paragraph above eqn (17.5), the author refers to the Maxwell model, without realising that within that model the dissipation rate is inversely proportional to the viscosity and does not depend upon the rigidity at all (which, of course, is indicative of the said model being symplistic).
* On page 8, the author complains that eqn (17.5) underestimates the heat production in Enceladus by two orders of magnitude. It sure does!  
  ...while the correct expression would have rendered him the measured heat production rate with an amazing precision, had the ice viscosity value near the melting point being used in it.


---

