<p>The last part of the answer you linked to actually says (right at the end) that tidal locking will never be achieved, with reasoning similar to what I gave in <a href="https://astronomy.stackexchange.com/questions/36915/how-can-the-earth-and-the-moon-be-in-synchronous-rotation-if-the-moon-wont-be-i/36917#36917">this answer</a>.</p>
<p>That said, even though the Moon and the Earth will never actually achieve tidal synchronization, we can still do the thought experiment and ask, &quot;If there were enough time for the current Earth-Moon system to achieve tidal synchronization, what would the length of the day be at which the rotation of the Earth and the orbit of the Moon would be synchronized?&quot;</p>
<p>To do this, we can assume that the Moon spirals outward due to an exchange of <a href="https://en.wikipedia.org/wiki/Angular_momentum" rel="noreferrer">angular momentum</a> between the Earth's rotation and the Moon's orbit.  The Earth's spin slows down as it loses angular momentum, and the Moon moves into a larger (and thus higher angular momentum) orbit as it gains that same angular momentum.  The rotation of the Moon would presumably stay locked to the Moon's orbital period, so it would slow down as well.</p>
<p>So, using <span class="math-container">$L$</span> to represent angular momentum, the key equation is</p>
<p><span class="math-container">$$ L_{\rm now} = L_{\rm then} $$</span></p>
<p>where &quot;then&quot; is some time in the future when lock is achieved.  The total angular momentum in the system is constant.</p>
<p>The angular momentum of any object is <span class="math-container">$L = I\omega$</span>, where <span class="math-container">$I$</span> is the moment of inertia, and <span class="math-container">$\omega = \frac{2\pi}{P}$</span> is the orbital frequency, related to the orbital period <span class="math-container">$P$</span>.   For a constant-density sphere of mass <span class="math-container">$M$</span> and radius <span class="math-container">$R$</span> rotating on its axis, <span class="math-container">$I = 0.4 M R^2$</span>.  The Earth and the Moon are somewhat more centrally condensed, so their moments of (rotational) inertia are a little smaller than the 0.4 for a uniform sphere.  The leading coefficient is <a href="https://en.wikipedia.org/wiki/Moment_of_inertia_factor" rel="noreferrer">0.33 for the Earth, and 0.39 for the Moon</a>.</p>
<p>For the Moon orbiting the Earth, it's a good approximation to just treat it as a point mass (since its size is small compared to its distance from Earth), so it has <span class="math-container">$I = M_{\rm Moon}R_{\rm Earth-Moon}^2$</span>.</p>
<p>Putting all three of these motions (Earth rotation, Moon orbit, Moon rotation) together, we get</p>
<p><span class="math-container">$$L_{\rm now} = 2\pi \left( \frac{0.33 M_\oplus R_\oplus^2}{P_\oplus} +  \frac{0.39 M_{\rm Moon} R_{\rm Moon}^2}{P_{\rm Moon}} +  \frac{M_{\rm Moon} R_{\rm Earth-Moon}^2}{P_{\rm Moon}} \right) $$</span></p>
<p>All of the values there represent current, known values, i.e. <span class="math-container">$P_\oplus = 1$</span> day, and <span class="math-container">$P_{\rm Moon} = 1$</span> month = 27.3 days.  Similarly, at some point in the (hypothetical) future, we would have</p>
<p><span class="math-container">$$
L_{\rm then} = 2\pi \left( \frac{0.33 M_\oplus R_\oplus^2}{P_{\rm then}} +  \frac{0.39 M_{\rm Moon} R_{\rm Moon}^2}{P_{\rm then}} +  \frac{M_{\rm Moon} R_{\rm Earth-Moon,\ then}^2}{P_{\rm then}} \right) $$</span></p>
<p>or</p>
<p><span class="math-container">$$ 
L_{\rm then}  = \frac{2\pi}{P_{\rm then}} \left(0.33 M_\oplus R_\oplus^2 + 0.39 M_{\rm Moon} R_{\rm Moon}^2 + M_{\rm Moon} R_{\rm Earth-Moon,\ then}^2 \right)
$$</span></p>
<p>Notice that there's just a single period here, since everything is now assumed to be synchronized.  So we could set this equal to <span class="math-container">$L_{\rm now}$</span> and solve for <span class="math-container">$P_{\rm then}$</span> – except that we have a second unknown in the equation, <span class="math-container">$R_{\rm Earth-Moon,\ then}$</span>, the new orbital distance of the Moon from the Earth.   Fortunately, we can use <a href="https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law_of_Kepler" rel="noreferrer">Kepler's third law</a> to relate this distance to the orbital period:</p>
<p><span class="math-container">$$ P_{\rm then}^2 (M_\oplus + M_{\rm Moon} ) = \frac{4 \pi^2 }{G} R_{\rm Earth-Moon,\ then}^3  $$</span></p>
<p>To make life a little easier when we substitute into the equation, we could write this as a proportion with the current values, which makes some of the constants cancel:</p>
<p><span class="math-container">$$ \frac{P_{\rm then}^2}{P_{\rm Moon}^2} = \frac{R_{\rm Earth-Moon,\ then}^3}{R_{\rm Earth-Moon}^3} $$</span></p>
<p>which means that</p>
<p><span class="math-container">$$ R_{\rm Earth-Moon,\ then}^2 = R_{\rm Earth-Moon}^2 \left(\frac{P_{\rm then}}{P_{\rm Moon}}\right)^{4/3} $$</span></p>
<p>Substituting that into our expression for <span class="math-container">$I_{\rm then}$</span>, we finally end up with</p>
<p><span class="math-container">$$L_{\rm then} = \frac{2\pi}{P_{\rm then}} \left(0.33 M_\oplus R_\oplus^2 + 0.39 M_{\rm Moon} R_{\rm Moon}^2 + M_{\rm Moon} R_{\rm Earth-Moon}^2 \left(\frac{P_{\rm then}}{P_{\rm Moon}}\right)^{4/3} \right) $$</span></p>
<p>So in principle, we're done - we set this equal to <span class="math-container">$L_{\rm now}$</span> and solve for <span class="math-container">$P_{\rm then}$</span>.  It's not a simple equation to solve analytically, but not hard to solve numerically.</p>
<p>Symbolically there's a lot going on, but we know most of these values, so if we plug in numbers for everything we know and simplify, this becomes</p>
<p><span class="math-container">$$
P_{\rm then} = 0.16809413\ {\mathrm d} + 0.27626727\ {\mathrm d}^{-1/3} P_{\rm then}^{4/3}
$$</span></p>
<p>where the &quot;d&quot; represents units of days.  Solving this equation gives a period of 46.9 days, so that is how long the day, the month (i.e. the orbital period of the Moon), and the rotation period of the Moon would be if all were to become tidally locked to each other.</p>
<p>If you want to see the calculation done in Python, I've posted the code in a gist <a href="https://gist.github.com/elnjensen/234e6045dd02c06dac1a227334f5a6a0" rel="noreferrer">here</a>. It's a nice example of the usefulness of Python's quantities and astropy's constants.</p>

CC BY-SA 4.0
2020-07-18T03:48:06.917Z
Very nice! I didn't believe that the Earth would nearly stop and the Moon would move out so little, so I went and calculated the rotational angular momentum of the Earth and orbital angular momentum of the Moon and was surprised to find out that the latter is five times bigger than the former. Wow that factor of 50 in $r$ really helps!
uhoh
CC BY-SA 4.0
2020-07-18T06:10:15.210Z
Thank you very much! One question: what is "Python"?
Ioannes
CC BY-SA 4.0
2020-07-18T12:54:25.973Z
@uhoh  Yes, the long lever arm of the Moon’s orbit makes a big difference.  In the solar system overall, most of the mass is in the Sun, but most of the angular momentum is in the orbits of the planets, especially the giant planets.
Eric Jensen
CC BY-SA 4.0
2020-07-18T12:56:19.593Z
yep yep, for example despite Jupiter being so much larger, Neptune does quite a job pushing the Sun around the solar system's barycenter.
uhoh
CC BY-SA 4.0
2020-07-18T13:00:30.573Z
@Greenhorn  Python is a computer programming language.  It is used a lot in astronomy research these days (though of course for a lot of other things, too), and [astropy](http://learn.astropy.org) is a set of specialized astronomy routines written in Python.
Eric Jensen
CC BY-SA 4.0
2020-07-18T15:40:23.010Z
@Greenhorn you can see that it is the only computer language that has its own tag here in Astronomy SE [`python`](https://astronomy.stackexchange.com/questions/tagged/python) as well as in Space Exploration SE [`python`](https://space.stackexchange.com/questions/tagged/python)
uhoh
CC BY-SA 4.0
2020-10-07T17:49:50.370Z
@ELNJ I've noticed your gist is not licensed. Would you be willing to license it so that it remains available for future readers of this answer?
called2voyage
CC BY-SA 4.0
2020-10-07T17:50:36.487Z
@ELNJ If interested in licensing, see the comments here for a couple ideas of how to do that for a gist: https://softwareengineering.stackexchange.com/questions/302555/what-is-the-license-of-a-github-gist?noredirect=1&lq=1
called2voyage
CC BY-SA 4.0
2022-10-23T07:56:29.563Z
Can you run through where your expression for the Earth-Moon orbital angular momentum comes from? Is it a good approximation to treat the earth as fixed (at the level of 3 sig figs for the answer)? Does the orbital eccentricity come into this?
ProfRob
CC BY-SA 4.0
2022-10-23T08:14:09.807Z
I think the correction is small and you don't have to make these assumptions. $L_{\rm orbit} = M_{\rm moon} \sqrt{GM_{\rm Earth}a(1-e^2)}$.
ProfRob
CC BY-SA 4.0
2020-07-18T01:31:21.997Z
2020-07-18T01:31:21.997Z
Eric Jensen
<p>Most of the answer is wonderfully written above by Eric Jensen. I shall only add one final touch. As demonstrated by Eric, synchronisation is attained when the Moon's orbital period reaches <span class="math-container">$P_{then} = 46.9$</span> days. Implying that <span class="math-container">$R$</span> stands for the semimajor axis, we take, for an estimate, <span class="math-container">$R_{Moon} = 3.84\times 10^8$</span> m. Also, <span class="math-container">$P_{Moon} = 27.3$</span> days. Insertion of these numbers into the Kepler law renders
<span class="math-container">$$
R_{then} = R_{Moon}\,\left(\frac{P_{then}}{P_{Moon}} \right)^{2/3} =  5.72\times 10^8\;\mbox{m}\;\,.
$$</span>
This value must be residing within the Earth's Hill sphere, i.e. within the domain where the Moon is attracted by the Earth stronger than by the Sun, so it does not fly away.</p>
<p>Naively, the radius of the Hill sphere of the Earth is
<span class="math-container">$$
r_H = a_E \,(1-e_E) \,\left(\frac{M_E}{3M_\ast} \right)^{1/3} = 1.47\times 10^9\;\mbox{m}\;\,, 
$$</span>
where <span class="math-container">$a_E$</span> and <span class="math-container">$e_E$</span> are the Earth's semimajor axis and eccentricity, while <span class="math-container">$M_E$</span> and <span class="math-container">$M_\ast$</span> are the Earth's and Sun's masses.</p>
<p>However, numerical study by <a href="https://www.nature.com/articles/nature01622" rel="nofollow noreferrer">Astakhov et al. (2003)</a> and <a href="https://academic.oup.com/mnras/article/373/3/1227/1063626" rel="nofollow noreferrer">Domingos et al.
(2006)</a> demonstrated that the orbits too closely approaching <span class="math-container">$r_H$</span> are unstable in the long term, while those confined to a smaller region (which we may call ''the reduced Hill sphere'') stay stable. The radius of the reduced Hill sphere (call it ''the reduced Hill radius''), for a circular (e = 0) lunar orbit, is
<span class="math-container">$$
r^{\,\prime}_H = B\, a_E\, (1 - b\, e_E)\,\left(\frac{M_E}{3M_\ast} \right)^{1/3} \;\,.
$$</span>
For prograde moons, <span class="math-container">$B=0.49$</span> and <span class="math-container">$b = 1.03$</span>, wherefrom<br />
<span class="math-container">$$
R^{\,\prime}_H = 7.09\times 10^8\;\mbox{m}\;\,.
$$</span>
We see that the distance <span class="math-container">$R_{then}=5.72\times 10^8$</span> m will <em>barely</em> fit into the reduced Hill sphere. Anyhow, by sheer luck, the synchronous configuration will be stable.</p>

CC BY-SA 4.0
2022-10-23T07:43:34.567Z
You don't have to neglect eccentricity to easily write down the angular momentum of the orbit.
ProfRob
CC BY-SA 4.0
2022-10-23T08:07:41.943Z
@ProfRob Yes. You are right. I shall remove those words. Thank you.
Michael_1812
CC BY-SA 4.0
2022-10-23T07:21:54.260Z
2022-10-25T10:06:45.943Z
2022-10-25T10:06:45.943Z
Michael_1812
<p><a href="https://astronomy.stackexchange.com/a/18795/7982">This answer</a> to <em>Will the Earth ever be tidally locked to the Moon?</em> supports the widely held thinking that during the Sun's red giant phase or later the Earth and the Moon should be tidally locked to each other. The Earth is said to have a slower rotation then.</p>
<p>Do we have any idea how long a sidereal day and the mean solar day on Earth would be in several billion years?</p>

CC BY-SA 4.0
2020-07-14T06:54:43.787Z
This is an interesting question, but can you mention where you read that they should be tidally locked when the Sun is in a red giant phase? *Thanks!*
uhoh
CC BY-SA 4.0
2020-07-14T07:13:59.107Z
@uhoh On one occasion it was in an answer to one of the questions in the following link. But it's the general assumption among many scientists that the Earth and the Moon are to be tidally locked once in case they'd survive the red giant phase. https://www.astronews.com/frag/index.shtml
Ioannes
CC BY-SA 4.0
2020-07-14T07:21:07.120Z
In Stack Exchange it is a good idea to support each question's premise with facts or supporting links. In this case there's some information here already! :-)
uhoh
CC BY-SA 4.0
2020-07-18T02:12:59.623Z
It would surely be well after the Sun becomes a white dwarf.  There's not enough time in six or seven billion years for the Earth-Moon system to become tidally locked.
Brian Tung
<earth><rotation><tidal-locking><prediction>
As the Moon and the Earth are predicted to get into tidal lock, how slow would the Earth rotate?
Ioannes
# As the Moon and the Earth are predicted to get into tidal lock, how slow would the Earth rotate?

(Asked by: username_0 on 2020-07-13)

[This answer](https://astronomy.stackexchange.com/a/18795/7982) to *Will the Earth ever be tidally locked to the Moon?* supports the widely held thinking that during the Sun's red giant phase or later the Earth and the Moon should be tidally locked to each other. The Earth is said to have a slower rotation then.

Do we have any idea how long a sidereal day and the mean solar day on Earth would be in several billion years?

**Tags:** earth, rotation, tidal-locking, prediction
> **Comments:**
> **username_1**: This is an interesting question, but can you mention where you read that they should be tidally locked when the Sun is in a red giant phase? *Thanks!*  
> **username_0**: @uhoh On one occasion it was in an answer to one of the questions in the following link. But it's the general assumption among many scientists that the Earth and the Moon are to be tidally locked once in case they'd survive the red giant phase. https://www.astronews.com/frag/index.shtml  
> **username_1**: In Stack Exchange it is a good idea to support each question's premise with facts or supporting links. In this case there's some information here already! :-)  
> **username_2**: It would surely be well after the Sun becomes a white dwarf.  There's not enough time in six or seven billion years for the Earth-Moon system to become tidally locked.  

---

## 2 answers

### Answer by username_3 on 2020-07-18. Score: 8

The last part of the answer you linked to actually says (right at the end) that tidal locking will never be achieved, with reasoning similar to what I gave in [this answer](https://astronomy.stackexchange.com/questions/36915/how-can-the-earth-and-the-moon-be-in-synchronous-rotation-if-the-moon-wont-be-i/36917#36917).

That said, even though the Moon and the Earth will never actually achieve tidal synchronization, we can still do the thought experiment and ask, "If there were enough time for the current Earth-Moon system to achieve tidal synchronization, what would the length of the day be at which the rotation of the Earth and the orbit of the Moon would be synchronized?"

To do this, we can assume that the Moon spirals outward due to an exchange of [angular momentum](https://en.wikipedia.org/wiki/Angular_momentum) between the Earth's rotation and the Moon's orbit. The Earth's spin slows down as it loses angular momentum, and the Moon moves into a larger (and thus higher angular momentum) orbit as it gains that same angular momentum. The rotation of the Moon would presumably stay locked to the Moon's orbital period, so it would slow down as well.

So, using $L$ to represent angular momentum, the key equation is

$$ L\_{\rm now} = L\_{\rm then} $$

where "then" is some time in the future when lock is achieved. The total angular momentum in the system is constant.

The angular momentum of any object is $L = I\omega$, where $I$ is the moment of inertia, and $\omega = \frac{2\pi}{P}$ is the orbital frequency, related to the orbital period $P$. For a constant-density sphere of mass $M$ and radius $R$ rotating on its axis, $I = 0.4 M R^2$. The Earth and the Moon are somewhat more centrally condensed, so their moments of (rotational) inertia are a little smaller than the 0.4 for a uniform sphere. The leading coefficient is [0.33 for the Earth, and 0.39 for the Moon](https://en.wikipedia.org/wiki/Moment_of_inertia_factor).

For the Moon orbiting the Earth, it's a good approximation to just treat it as a point mass (since its size is small compared to its distance from Earth), so it has $I = M\_{\rm Moon}R\_{\rm Earth-Moon}^2$.

Putting all three of these motions (Earth rotation, Moon orbit, Moon rotation) together, we get

$$L\_{\rm now} = 2\pi \left( \frac{0.33 M\_\oplus R\_\oplus^2}{P\_\oplus} + \frac{0.39 M\_{\rm Moon} R\_{\rm Moon}^2}{P\_{\rm Moon}} + \frac{M\_{\rm Moon} R\_{\rm Earth-Moon}^2}{P\_{\rm Moon}} \right) $$

All of the values there represent current, known values, i.e. $P\_\oplus = 1$ day, and $P\_{\rm Moon} = 1$ month = 27.3 days. Similarly, at some point in the (hypothetical) future, we would have

$$
L\_{\rm then} = 2\pi \left( \frac{0.33 M\_\oplus R\_\oplus^2}{P\_{\rm then}} + \frac{0.39 M\_{\rm Moon} R\_{\rm Moon}^2}{P\_{\rm then}} + \frac{M\_{\rm Moon} R\_{\rm Earth-Moon,\ then}^2}{P\_{\rm then}} \right) $$

or

$$
L\_{\rm then} = \frac{2\pi}{P\_{\rm then}} \left(0.33 M\_\oplus R\_\oplus^2 + 0.39 M\_{\rm Moon} R\_{\rm Moon}^2 + M\_{\rm Moon} R\_{\rm Earth-Moon,\ then}^2 \right)
$$

Notice that there's just a single period here, since everything is now assumed to be synchronized. So we could set this equal to $L\_{\rm now}$ and solve for $P\_{\rm then}$ – except that we have a second unknown in the equation, $R\_{\rm Earth-Moon,\ then}$, the new orbital distance of the Moon from the Earth. Fortunately, we can use [Kepler's third law](https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law_of_Kepler) to relate this distance to the orbital period:

$$ P\_{\rm then}^2 (M\_\oplus + M\_{\rm Moon} ) = \frac{4 \pi^2 }{G} R\_{\rm Earth-Moon,\ then}^3 $$

To make life a little easier when we substitute into the equation, we could write this as a proportion with the current values, which makes some of the constants cancel:

$$ \frac{P\_{\rm then}^2}{P\_{\rm Moon}^2} = \frac{R\_{\rm Earth-Moon,\ then}^3}{R\_{\rm Earth-Moon}^3} $$

which means that

$$ R\_{\rm Earth-Moon,\ then}^2 = R\_{\rm Earth-Moon}^2 \left(\frac{P\_{\rm then}}{P\_{\rm Moon}}\right)^{4/3} $$

Substituting that into our expression for $I\_{\rm then}$, we finally end up with

$$L\_{\rm then} = \frac{2\pi}{P\_{\rm then}} \left(0.33 M\_\oplus R\_\oplus^2 + 0.39 M\_{\rm Moon} R\_{\rm Moon}^2 + M\_{\rm Moon} R\_{\rm Earth-Moon}^2 \left(\frac{P\_{\rm then}}{P\_{\rm Moon}}\right)^{4/3} \right) $$

So in principle, we're done - we set this equal to $L\_{\rm now}$ and solve for $P\_{\rm then}$. It's not a simple equation to solve analytically, but not hard to solve numerically.

Symbolically there's a lot going on, but we know most of these values, so if we plug in numbers for everything we know and simplify, this becomes

$$
P\_{\rm then} = 0.16809413\ {\mathrm d} + 0.27626727\ {\mathrm d}^{-1/3} P\_{\rm then}^{4/3}
$$

where the "d" represents units of days. Solving this equation gives a period of 46.9 days, so that is how long the day, the month (i.e. the orbital period of the Moon), and the rotation period of the Moon would be if all were to become tidally locked to each other.

If you want to see the calculation done in Python, I've posted the code in a gist [here](https://gist.github.com/elnjensen/234e6045dd02c06dac1a227334f5a6a0). It's a nice example of the usefulness of Python's quantities and astropy's constants.

> **Comments:**
> **username_1**: Very nice! I didn't believe that the Earth would nearly stop and the Moon would move out so little, so I went and calculated the rotational angular momentum of the Earth and orbital angular momentum of the Moon and was surprised to find out that the latter is five times bigger than the former. Wow that factor of 50 in $r$ really helps!  
> **username_0**: Thank you very much! One question: what is "Python"?  
> **username_3**: @uhoh  Yes, the long lever arm of the Moon’s orbit makes a big difference.  In the solar system overall, most of the mass is in the Sun, but most of the angular momentum is in the orbits of the planets, especially the giant planets.  
> **username_1**: yep yep, for example despite Jupiter being so much larger, Neptune does quite a job pushing the Sun around the solar system's barycenter.  
> **username_3**: @Greenhorn  Python is a computer programming language.  It is used a lot in astronomy research these days (though of course for a lot of other things, too), and [astropy](http://learn.astropy.org) is a set of specialized astronomy routines written in Python.  
> **username_1**: @Greenhorn you can see that it is the only computer language that has its own tag here in Astronomy SE [`python`](https://astronomy.stackexchange.com/questions/tagged/python) as well as in Space Exploration SE [`python`](https://space.stackexchange.com/questions/tagged/python)  
> **username_4**: @ELNJ I've noticed your gist is not licensed. Would you be willing to license it so that it remains available for future readers of this answer?  
> **username_4**: @ELNJ If interested in licensing, see the comments here for a couple ideas of how to do that for a gist: https://softwareengineering.stackexchange.com/questions/302555/what-is-the-license-of-a-github-gist?noredirect=1&lq=1  
> **username_5**: Can you run through where your expression for the Earth-Moon orbital angular momentum comes from? Is it a good approximation to treat the earth as fixed (at the level of 3 sig figs for the answer)? Does the orbital eccentricity come into this?  
> **username_5**: I think the correction is small and you don't have to make these assumptions. $L_{\rm orbit} = M_{\rm moon} \sqrt{GM_{\rm Earth}a(1-e^2)}$.  

---

### Answer by username_6 on 2022-10-23. Score: 1

Most of the answer is wonderfully written above by Eric Jensen. I shall only add one final touch. As demonstrated by Eric, synchronisation is attained when the Moon's orbital period reaches $P\_{then} = 46.9$ days. Implying that $R$ stands for the semimajor axis, we take, for an estimate, $R\_{Moon} = 3.84\times 10^8$ m. Also, $P\_{Moon} = 27.3$ days. Insertion of these numbers into the Kepler law renders
$$
R\_{then} = R\_{Moon}\,\left(\frac{P\_{then}}{P\_{Moon}} \right)^{2/3} = 5.72\times 10^8\;\mbox{m}\;\,.
$$
This value must be residing within the Earth's Hill sphere, i.e. within the domain where the Moon is attracted by the Earth stronger than by the Sun, so it does not fly away.

Naively, the radius of the Hill sphere of the Earth is
$$
r\_H = a\_E \,(1-e\_E) \,\left(\frac{M\_E}{3M\_\ast} \right)^{1/3} = 1.47\times 10^9\;\mbox{m}\;\,,
$$
where $a\_E$ and $e\_E$ are the Earth's semimajor axis and eccentricity, while $M\_E$ and $M\_\ast$ are the Earth's and Sun's masses.

However, numerical study by [Astakhov et al. (2003)](https://www.nature.com/articles/nature01622) and [Domingos et al.
(2006)](https://academic.oup.com/mnras/article/373/3/1227/1063626) demonstrated that the orbits too closely approaching $r\_H$ are unstable in the long term, while those confined to a smaller region (which we may call ''the reduced Hill sphere'') stay stable. The radius of the reduced Hill sphere (call it ''the reduced Hill radius''), for a circular (e = 0) lunar orbit, is
$$
r^{\,\prime}\_H = B\, a\_E\, (1 - b\, e\_E)\,\left(\frac{M\_E}{3M\_\ast} \right)^{1/3} \;\,.
$$
For prograde moons, $B=0.49$ and $b = 1.03$, wherefrom  
$$
R^{\,\prime}\_H = 7.09\times 10^8\;\mbox{m}\;\,.
$$
We see that the distance $R\_{then}=5.72\times 10^8$ m will *barely* fit into the reduced Hill sphere. Anyhow, by sheer luck, the synchronous configuration will be stable.

> **Comments:**
> **username_5**: You don't have to neglect eccentricity to easily write down the angular momentum of the orbit.  
> **username_6**: @ProfRob Yes. You are right. I shall remove those words. Thank you.  

---

