<p>Short answer: YES. I will assume you mean the eccentricity in the orbit of the Moon around the Earth.</p>

<p>In general, tidal forces in binary systems (like the Earth-Moon system, or a binary star, etc...) effect the binary in three main ways: in order of longest timescale to shortest timescale</p>

<p>1) circularization of the orbit (eccentricity goes to zero, binary separation goes to minimum).</p>

<p>2) alignment of the binary components' spin angular momenta with the orbital angular momentum (the directions of <span class="math-container">$S$</span> and <span class="math-container">$L$</span> are the same).</p>

<p>3) synchronization of the binary components' rotational frequency with the orbital frequency.</p>

<blockquote>
  <p>but why?</p>
</blockquote>

<p>There's different ways of answering "why," and <a href="https://arxiv.org/abs/0807.4870" rel="nofollow noreferrer">here</a> is a great conceptual answer given by the God Father of astrophysical tides himself, Z. P. Zahn:</p>

<blockquote>
  <p>A fundamental property of closed mechanical systems is that they conserve their
  total momentum. This is true in particular for binary stars, star-planet(s) systems,
  whether they possess or not a circumstellar disc, if one can ignore the angular
  momentum that is carried away by winds and by gravitational waves. Through
  tidal interaction, kinetic energy and angular momentum are exchanged between
  the rotation of the components their orbital motion and the disc. In the absence
  of such a disc, which is the case that we shall consider here, they evolve due
  to viscous and radiative dissipation to the state of minimum kinetic energy, in
  which the orbit is circular, the rotation of both stars is synchronized with the
  orbital motion, and their spin axis are perpendicular to the orbital plane. How
  rapidly the system tends to that state is determined by the strength of the tidal
  interaction, and thus by the separation of the two components...</p>
</blockquote>

<p>So, basically, the tidal torque drives dissipation, and this dissipation brings the binary to a minimum kinetic energy state, i.e. circular orbit, synchronized spins with the orbit, aligned spins with the orbit.</p>

<p>Here's a contour plot of the timescales provided by equations #<span class="math-container">$9 - 13$</span> from Hut's 1981 seminal <a href="http://adsabs.harvard.edu/abs/1981A&amp;A....99..126H" rel="nofollow noreferrer">paper</a>, assuming the separation does not change much relative to the other quantities, for a binary composed of a black hole and a Wolf-Rayet star, which is a similar system to a planet-satellite system, where the timescales are parameterized in terms of the mass of the WR star and the separation of the binary:</p>

<p><a href="https://i.sstatic.net/0kZol.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/0kZol.png" alt="enter image description here"></a></p>

<p>The dotted black line is the merger timescale for the binary due to gravitational waves, meaning below that line you are binary that mergers within the lifetime of the universe. The synchronization timescale is independent of the initial spin of the WR star, which is why there is only one synchronization line in the plot, but the alignemnt timescale does depend on the initial spin of the component feeling the tides. Points below the contours achieve that process (below the red dashed line are synchronized). The quantity <span class="math-container">$f_{B}$</span> is the break-up fraction parameter, is between <span class="math-container">$0$</span> and <span class="math-container">$1$</span>, and the choice of which determines the initial WR star spin as a fraction of its break-up spin. (the tidal timescale contours were made by equating the WR lifetime with the tidal timescale). I did not include the circularization timescale because it is essentially always less than the sync. timescale (seen simply from the fact that generally, <span class="math-container">$S &lt;&lt; L$</span>).</p>

<p>Lastly, it must be said that a WR star is quite different from a moon/satellite, however the effects of tides on each is similar enough to make a conceptual comparison (since a WR star is much less massive than the BH, as is the moon than the Earth, the hierarchy of tidal timescales is the same, but you must keep in mind that the WR's apsidal response to the tidal gradient is very different than a rocky moon's). </p>

<p>As you pointed out, the Moon's orbit is slightly eccentric, but the moon is also tidally synchronized with the Earth's rotation (which are both sync'ed to the orbital rotation). This should make sense since it takes longer for tides to circularize than to synchronize.</p>

<p>I'm sure somebody can post an example of using tides in an actual planet-moon system, rather than my black hole-star system. :)</p>

CC BY-SA 4.0
2019-01-01T23:32:55.063Z
I would like to add to this, that if not for the Sun, an Earth-Moon system, then the Moon would likely have a very near circular orbit.   The Sun's pull on the Moon is what causes the Moon's eccentricity as well as what pulls the moon 6 degrees off an equatorial orbit. The tidal forces do circularize the orbit, but the Sun is the bigger factor, so as the Moon moved away from the Earth it's orbit has likely become less circular over time for much of it's history.
userLTK
CC BY-SA 4.0
2019-01-02T00:38:05.350Z
Re "the magnitudes of $S$ and $L$ are the same": shouldn't it be the frequencies that are the same? For example, the Moon is already tidally locked to the Earth, but its orbital angular momentum $L$ is *much* larger than its spin angular momentum $S$ because the relevant moments of inertia are quite different.
rob
CC BY-SA 4.0
2019-01-02T02:30:26.903Z
@userLTK that is a very good point, thank you for making it. According to observation, the moon's orbit has increasing eccentricity due to the Sun which is well understood, and there is an anomalous increase that is currently unexplainable ( https://arxiv.org/abs/1102.0212 ). And so clearly, the Sun's effect is dominating over the tides, since the eccentricity is increasing. I was merely trying to explain how tides act on the moon's orbit to cause circularization, but in reality our Moon's eccentricity is increasing.
Daddy Kropotkin
CC BY-SA 4.0
2019-01-02T02:33:23.747Z
@rob thank you for pointing this out rob, I agree I should be talking about the rotational frequency of the binary component and the orbital frequency in terms of synchronization.
Daddy Kropotkin
CC BY-SA 4.0
2019-01-02T02:41:59.160Z
@rob it's perhaps worth noting that one can work with the dimensionless spin of the star, rather than the frequencies (this is what I do in my own work since the star eventually forms a BH).
Daddy Kropotkin
CC BY-SA 4.0
2019-01-02T15:58:33.863Z
@userLTK The Moon's orbital plane is slightly more than 5° off an *ecliptic* orbit. Its orbital plane precesses quite quickly, and its inclination to the celestial equator varies between 18° & 28°. If you look at the Moon's *solar* orbit on a computer screen, it looks quite circular, since the eccentricity of the Earth-Moon system is fairly small, and the perturbations on the Moon's solar orbit due to the Earth are hard to see, amounting to no more than a pixel or 2 on a typical monitor.
PM 2Ring
CC BY-SA 4.0
2019-01-01T17:41:54.403Z
2019-01-03T15:21:59.520Z
2019-01-03T15:21:59.520Z
Daddy Kropotkin
<p>Stumbled across this a year later, but thought I'd post a simple, non-math answer.</p>

<p>The Moon's gravitational influence on the Earth is like that of any orbiting body on its primary, to wit that it creates a bulge in the Earth's surface, and to a lesser extent vice versa.  As the Earth rotates that bulge moves around its circumference so it always points <em>almost</em> toward the Moon.</p>

<p>Because the Earth's crust is not perfectly elastic this moving deformation is never exactly beneath the moon, but leads it slightly.  (This would happen even if the crust <em>were</em> perfectly elastic, due to the inertia of the material being displaced.)  This leads to three phenomena.</p>

<p>The first is that some of the Earth's rotational energy is dissipated as heat.  It's not much, but it's important in the last stage of tidal locking.</p>

<p>The second is that the slightly increased gravity of that bulge, leading the Moon as it does, pulls the Moon forward in its orbit.  As it adds kinetic energy to the Moon, it tends to lift the Moon higher in its orbit.  This same force pulls backward against Earth's rotation, slowing it down.</p>

<p>The third is that when the Moon is closest to Earth in its not-quite-circular orbit this force is greater, lifting it more when it is close to Earth.  This tends to iron out any eccentricity in the Moon's orbit.</p>

<p>The sum total of all this is that the Earth's rotation will eventually slow down to the rate that the Moon revolves around it, and it reduces the eccentricity of the Moon's orbit.  This force decreases as the two become synchronized, with the last remnant of Earth's "excess" rotational energy being lost through tidal heating so that one side always faces the Moon.</p>

<p>Mind you, as noted by other posters, the Sun's gravity tends to make the Moon's orbit slightly eccentric, which would eventually cause Earth / Moon to be tidally locked to the Sun, but a bit of Fermi arithmetic suggests that the Solar System itself won't last quite that long.</p>

CC BY-SA 4.0
2020-02-19T15:05:38.727Z
About your last sentences: What would be the position of the Moon in this tidal lock, maybe in a Lagrange-point?
peterh
CC BY-SA 4.0
2020-02-19T19:08:16.650Z
@peterh, I've forgotten too much math to turn this general description into a good quantitative answer, and just don't know some factors, e.g. energy loss through tidal heating.  Still, doing sliderule-style adding and subtracting orders of magnitude, it looks like 100BN years for Earth's rotation to lock to the Moon, and the Moon to wind up in maybe a 700K km orbit.  But the more I think of it, Earth becoming tidally locked to the Sun will likely shrink the Moon's orbit, so that last sentence may not be quite right.  I'll do some reading and get back.
RLWatkins
CC BY-SA 4.0
2020-02-19T19:45:31.500Z
Ok. Somewhere I've read that 3- or more-body systems, except some rare case (0-measure metastable points, Lagrange-points and so) always evaporate away ultimately.
peterh
CC BY-SA 4.0
2020-02-20T21:24:03.350Z
@peterh, yes, ultimately that will happen, a lot later on.  Here is a lecture which addresses just that question, among others: http://www.aleph.se/Trans/Global/Omega/dyson.txt
RLWatkins
CC BY-SA 4.0
2020-02-18T23:54:24.683Z
2020-02-18T23:54:24.683Z
RLWatkins
<p>The moon has an elliptical orbit (of course), but as it inches away from the Earth each year, does its orbit become more circular and how come? Io's orbit is becoming more circular due to tidal heating effects, but why?</p>

<the-moon>
Is the moon's orbit circularizing? Why does tidal heating circularize orbits?
StellarExile
# Is the moon's orbit circularizing? Why does tidal heating circularize orbits?

(Asked by: username_0 on 2019-01-01)

The moon has an elliptical orbit (of course), but as it inches away from the Earth each year, does its orbit become more circular and how come? Io's orbit is becoming more circular due to tidal heating effects, but why?

**Tags:** the-moon


---

## 2 answers

### Answer by username_5 on 2020-02-18. Score: 5 | Accepted answer

Stumbled across this a year later, but thought I'd post a simple, non-math answer.

The Moon's gravitational influence on the Earth is like that of any orbiting body on its primary, to wit that it creates a bulge in the Earth's surface, and to a lesser extent vice versa. As the Earth rotates that bulge moves around its circumference so it always points *almost* toward the Moon.

Because the Earth's crust is not perfectly elastic this moving deformation is never exactly beneath the moon, but leads it slightly. (This would happen even if the crust *were* perfectly elastic, due to the inertia of the material being displaced.) This leads to three phenomena.

The first is that some of the Earth's rotational energy is dissipated as heat. It's not much, but it's important in the last stage of tidal locking.

The second is that the slightly increased gravity of that bulge, leading the Moon as it does, pulls the Moon forward in its orbit. As it adds kinetic energy to the Moon, it tends to lift the Moon higher in its orbit. This same force pulls backward against Earth's rotation, slowing it down.

The third is that when the Moon is closest to Earth in its not-quite-circular orbit this force is greater, lifting it more when it is close to Earth. This tends to iron out any eccentricity in the Moon's orbit.

The sum total of all this is that the Earth's rotation will eventually slow down to the rate that the Moon revolves around it, and it reduces the eccentricity of the Moon's orbit. This force decreases as the two become synchronized, with the last remnant of Earth's "excess" rotational energy being lost through tidal heating so that one side always faces the Moon.

Mind you, as noted by other posters, the Sun's gravity tends to make the Moon's orbit slightly eccentric, which would eventually cause Earth / Moon to be tidally locked to the Sun, but a bit of Fermi arithmetic suggests that the Solar System itself won't last quite that long.

> **Comments:**
> **username_6**: About your last sentences: What would be the position of the Moon in this tidal lock, maybe in a Lagrange-point?  
> **username_5**: @username_6, I've forgotten too much math to turn this general description into a good quantitative answer, and just don't know some factors, e.g. energy loss through tidal heating.  Still, doing sliderule-style adding and subtracting orders of magnitude, it looks like 100BN years for Earth's rotation to lock to the Moon, and the Moon to wind up in maybe a 700K km orbit.  But the more I think of it, Earth becoming tidally locked to the Sun will likely shrink the Moon's orbit, so that last sentence may not be quite right.  I'll do some reading and get back.  
> **username_6**: Ok. Somewhere I've read that 3- or more-body systems, except some rare case (0-measure metastable points, Lagrange-points and so) always evaporate away ultimately.  
> **username_5**: @username_6, yes, ultimately that will happen, a lot later on.  Here is a lecture which addresses just that question, among others: http://www.aleph.se/Trans/Global/Omega/dyson.txt  

---

### Answer by username_1 on 2019-01-01. Score: 9

Short answer: YES. I will assume you mean the eccentricity in the orbit of the Moon around the Earth.

In general, tidal forces in binary systems (like the Earth-Moon system, or a binary star, etc...) effect the binary in three main ways: in order of longest timescale to shortest timescale

1) circularization of the orbit (eccentricity goes to zero, binary separation goes to minimum).

2) alignment of the binary components' spin angular momenta with the orbital angular momentum (the directions of $S$ and $L$ are the same).

3) synchronization of the binary components' rotational frequency with the orbital frequency.

> but why?

There's different ways of answering "why," and [here](https://arxiv.org/abs/0807.4870) is a great conceptual answer given by the God Father of astrophysical tides himself, Z. P. Zahn:

> A fundamental property of closed mechanical systems is that they conserve their
> total momentum. This is true in particular for binary stars, star-planet(s) systems,
> whether they possess or not a circumstellar disc, if one can ignore the angular
> momentum that is carried away by winds and by gravitational waves. Through
> tidal interaction, kinetic energy and angular momentum are exchanged between
> the rotation of the components their orbital motion and the disc. In the absence
> of such a disc, which is the case that we shall consider here, they evolve due
> to viscous and radiative dissipation to the state of minimum kinetic energy, in
> which the orbit is circular, the rotation of both stars is synchronized with the
> orbital motion, and their spin axis are perpendicular to the orbital plane. How
> rapidly the system tends to that state is determined by the strength of the tidal
> interaction, and thus by the separation of the two components...

So, basically, the tidal torque drives dissipation, and this dissipation brings the binary to a minimum kinetic energy state, i.e. circular orbit, synchronized spins with the orbit, aligned spins with the orbit.

Here's a contour plot of the timescales provided by equations #$9 - 13$ from Hut's 1981 seminal [paper](http://adsabs.harvard.edu/abs/1981A&A....99..126H), assuming the separation does not change much relative to the other quantities, for a binary composed of a black hole and a Wolf-Rayet star, which is a similar system to a planet-satellite system, where the timescales are parameterized in terms of the mass of the WR star and the separation of the binary:

[![enter image description here](https://i.sstatic.net/0kZol.png)](https://i.sstatic.net/0kZol.png)

The dotted black line is the merger timescale for the binary due to gravitational waves, meaning below that line you are binary that mergers within the lifetime of the universe. The synchronization timescale is independent of the initial spin of the WR star, which is why there is only one synchronization line in the plot, but the alignemnt timescale does depend on the initial spin of the component feeling the tides. Points below the contours achieve that process (below the red dashed line are synchronized). The quantity $f\_{B}$ is the break-up fraction parameter, is between $0$ and $1$, and the choice of which determines the initial WR star spin as a fraction of its break-up spin. (the tidal timescale contours were made by equating the WR lifetime with the tidal timescale). I did not include the circularization timescale because it is essentially always less than the sync. timescale (seen simply from the fact that generally, $S << L$).

Lastly, it must be said that a WR star is quite different from a moon/satellite, however the effects of tides on each is similar enough to make a conceptual comparison (since a WR star is much less massive than the BH, as is the moon than the Earth, the hierarchy of tidal timescales is the same, but you must keep in mind that the WR's apsidal response to the tidal gradient is very different than a rocky moon's).

As you pointed out, the Moon's orbit is slightly eccentric, but the moon is also tidally synchronized with the Earth's rotation (which are both sync'ed to the orbital rotation). This should make sense since it takes longer for tides to circularize than to synchronize.

I'm sure somebody can post an example of using tides in an actual planet-moon system, rather than my black hole-star system. :)

> **Comments:**
> **username_2**: I would like to add to this, that if not for the Sun, an Earth-Moon system, then the Moon would likely have a very near circular orbit.   The Sun's pull on the Moon is what causes the Moon's eccentricity as well as what pulls the moon 6 degrees off an equatorial orbit. The tidal forces do circularize the orbit, but the Sun is the bigger factor, so as the Moon moved away from the Earth it's orbit has likely become less circular over time for much of it's history.  
> **username_3**: Re "the magnitudes of $S$ and $L$ are the same": shouldn't it be the frequencies that are the same? For example, the Moon is already tidally locked to the Earth, but its orbital angular momentum $L$ is *much* larger than its spin angular momentum $S$ because the relevant moments of inertia are quite different.  
> **username_1**: @username_2 that is a very good point, thank you for making it. According to observation, the moon's orbit has increasing eccentricity due to the Sun which is well understood, and there is an anomalous increase that is currently unexplainable ( https://arxiv.org/abs/1102.0212 ). And so clearly, the Sun's effect is dominating over the tides, since the eccentricity is increasing. I was merely trying to explain how tides act on the moon's orbit to cause circularization, but in reality our Moon's eccentricity is increasing.  
> **username_1**: @username_3 thank you for pointing this out rob, I agree I should be talking about the rotational frequency of the binary component and the orbital frequency in terms of synchronization.  
> **username_1**: @username_3 it's perhaps worth noting that one can work with the dimensionless spin of the star, rather than the frequencies (this is what I do in my own work since the star eventually forms a BH).  
> **username_4**: @username_2 The Moon's orbital plane is slightly more than 5° off an *ecliptic* orbit. Its orbital plane precesses quite quickly, and its inclination to the celestial equator varies between 18° & 28°. If you look at the Moon's *solar* orbit on a computer screen, it looks quite circular, since the eccentricity of the Earth-Moon system is fairly small, and the perturbations on the Moon's solar orbit due to the Earth are hard to see, amounting to no more than a pixel or 2 on a typical monitor.  

---

