<p>This is an excellent question.</p>
<p>Presently, the mean distance between the Moon and the Earth is about
<span class="math-container">$$
R_{EM} \simeq 3.84 \times 10^8 \mbox{m}\;.
$$</span>
[ For a crude estimate, I make no distinction between the mean distance and the semimajor axis. ]</p>
<p>As explained <a href="https://astronomy.stackexchange.com/a/50836/30705">here</a>, at the moment of synchronisation of the Earth's rotation by the Moon (assuming the Sun does not intervene, and the synchronisation happens) the distance between the two will be
<span class="math-container">$$
R_{EM} \simeq 5.72 \times 10^8 \mbox{m}\;.
$$</span>
Assuming that the mean distance between the Earth and the Sun stays about the same as today,
<span class="math-container">$$
R_{ES} \simeq 1.50 \times 10^{11} \mbox{m}\;,
$$</span>
and knowing the masses of the Moon and the Sun, compare the tidal torques wherewith they are acting on the Earth's spin.</p>
<p>To this end, we recall that the tidal torque scales as the distance to power -6.</p>
<p>[ While the disturbing potential scales as the inverse cube of distance, the resulting tidal potential and tidal torque scale as the inverse sixth power. see e.g. eqn (116) in <a href="https://link.springer.com/article/10.1007/s10569-011-9397-4" rel="nofollow noreferrer">this paper</a>. ]</p>
<p>So, by the time of the Earth's synchronisation by the Moon, the value of the lunar torque acting on the Earth will be equal to its present value multiplied by  <span class="math-container">$(3.84/5.72)^6\approx 0.09$</span>, assuming that the Earth's quality function (the quadrupole Love number divided by the quadrupole quality factor) is frequency-independent.</p>
<p>Again, assuming the quality function constant, we conclude that, while today the lunar tides are about <a href="https://physics.stackexchange.com/a/149499/63683">2.3 times stronger</a> than the solar tides, the solar tides will take over before the synchronisation is achieved. Thence, the Sun will overpower the Moon and will force the Earth to keep slowing down its rotation rate also after the synchronisation. So the Moon will then start its tidal descent towards the Earth.</p>
<p>In reality, this will never happen, because the quality function is not a constant but a function -- a function of frequency.  So the lunar torque will still be stronger, and the synchronism will be achieved. For bodily tides, this is explained in our recent <a href="https://www.aanda.org/component/article?access=doi&amp;doi=10.1051/0004-6361/202245533" rel="nofollow noreferrer">paper</a></p>
<p>Generalisation of this explanation to ocean tides may require some effort, due to a different shape of the frequency-dependence. Numerical integration shows, however, that the synchronisation takes some 30 Bln years (Valeri V. Makarov, private communication). The Sun will have evolved into a red giant in 5 Bln years only, and no ocean will be left after that. So yes, the cited explanation based on bodily tides is applicable.</p>
<p>PS.
As demonstrated <a href="https://astronomy.stackexchange.com/a/50836/30705">there</a>, synchronism will be achieved <em>before</em> the Moon leaves the reduced Hill sphare.</p>
<p>PPS.
As mentioned in the cited <a href="https://www.aanda.org/component/article?access=doi&amp;doi=10.1051/0004-6361/202245533" rel="nofollow noreferrer">paper</a>, the achieved synchronism will actually be <em>quasi</em>-synchronism, due to the nonzero solar torque. But the deviation from synchronism will be minuscule, and so will be the resulting orbital evolution of the Moon.</p>

CC BY-SA 4.0
2023-07-23T12:04:08.950Z
2023-07-24T02:07:50.503Z
2023-07-24T02:07:50.503Z
Michael_1812
<p>At some point, won't the tidal forces from the Moon be less than the ones from the Sun? Would that mean that the Moon stops moving away, or would the process still continue. Would the Earth start trying to lock to the Sun instead?</p>

<the-sun><the-moon><earth><tidal-forces><tidal-locking>
Will the Earth tidally lock to the Moon (ignoring Sun's expansion)
user138890
# Will the Earth tidally lock to the Moon (ignoring Sun's expansion)

(Asked by: username_0 on 2023-07-21)

At some point, won't the tidal forces from the Moon be less than the ones from the Sun? Would that mean that the Moon stops moving away, or would the process still continue. Would the Earth start trying to lock to the Sun instead?

**Tags:** the-sun, the-moon, earth, tidal-forces, tidal-locking


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## 1 answers

### Answer by username_1 on 2023-07-23. Score: 1 | Accepted answer

This is an excellent question.

Presently, the mean distance between the Moon and the Earth is about
$$
R\_{EM} \simeq 3.84 \times 10^8 \mbox{m}\;.
$$
[ For a crude estimate, I make no distinction between the mean distance and the semimajor axis. ]

As explained [here](https://astronomy.stackexchange.com/a/50836/30705), at the moment of synchronisation of the Earth's rotation by the Moon (assuming the Sun does not intervene, and the synchronisation happens) the distance between the two will be
$$
R\_{EM} \simeq 5.72 \times 10^8 \mbox{m}\;.
$$
Assuming that the mean distance between the Earth and the Sun stays about the same as today,
$$
R\_{ES} \simeq 1.50 \times 10^{11} \mbox{m}\;,
$$
and knowing the masses of the Moon and the Sun, compare the tidal torques wherewith they are acting on the Earth's spin.

To this end, we recall that the tidal torque scales as the distance to power -6.

[ While the disturbing potential scales as the inverse cube of distance, the resulting tidal potential and tidal torque scale as the inverse sixth power. see e.g. eqn (116) in [this paper](https://link.springer.com/article/10.1007/s10569-011-9397-4). ]

So, by the time of the Earth's synchronisation by the Moon, the value of the lunar torque acting on the Earth will be equal to its present value multiplied by $(3.84/5.72)^6\approx 0.09$, assuming that the Earth's quality function (the quadrupole Love number divided by the quadrupole quality factor) is frequency-independent.

Again, assuming the quality function constant, we conclude that, while today the lunar tides are about [2.3 times stronger](https://physics.stackexchange.com/a/149499/63683) than the solar tides, the solar tides will take over before the synchronisation is achieved. Thence, the Sun will overpower the Moon and will force the Earth to keep slowing down its rotation rate also after the synchronisation. So the Moon will then start its tidal descent towards the Earth.

In reality, this will never happen, because the quality function is not a constant but a function -- a function of frequency. So the lunar torque will still be stronger, and the synchronism will be achieved. For bodily tides, this is explained in our recent [paper](https://www.aanda.org/component/article?access=doi&doi=10.1051/0004-6361/202245533)

Generalisation of this explanation to ocean tides may require some effort, due to a different shape of the frequency-dependence. Numerical integration shows, however, that the synchronisation takes some 30 Bln years (Valeri V. Makarov, private communication). The Sun will have evolved into a red giant in 5 Bln years only, and no ocean will be left after that. So yes, the cited explanation based on bodily tides is applicable.

PS.
As demonstrated [there](https://astronomy.stackexchange.com/a/50836/30705), synchronism will be achieved *before* the Moon leaves the reduced Hill sphare.

PPS.
As mentioned in the cited [paper](https://www.aanda.org/component/article?access=doi&doi=10.1051/0004-6361/202245533), the achieved synchronism will actually be *quasi*-synchronism, due to the nonzero solar torque. But the deviation from synchronism will be minuscule, and so will be the resulting orbital evolution of the Moon.


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