<p>This is not a hypothetical question since the Earth's rotation rate is slowing due to tidal interactions with the Moon. Currently, the Earth rotates once every 86164 seconds (sidereal time) for about 366 days in a year.  If there were 360 rotations a year, then a day would last <span class="math-container">$86164*366/360\approx 87600$</span> seconds which is 1436 seconds slower than now. The Earth's rotation rate is slowing by about 2 seconds every hundred thousand years according to <a href="https://iopscience.iop.org/article/10.3847/0004-6256/151/4/103" rel="nofollow noreferrer">Deines and Williams</a>.  This means there will be 360 days in a year in about 71.8 million years.</p>
<p>The <a href="https://www.vanderbilt.edu/AnS/physics/astrocourses/ast201/angular_momentum.html" rel="nofollow noreferrer">Earth's current rotational angular momentum</a> is about <span class="math-container">$7.1 \times 10^{33} \text{kg} \text{m}^2 \text{s}^{-1}$</span>.  The momentum 71.8m years from now will be about 360/366 of this, or <span class="math-container">$6.98 \times 10^{33} \text{kg} \text{m}^2 \text{s}^{-1}$</span>.  This is a loss of <span class="math-container">$1.2 \times 10^{32} \text{kg} \text{m}^2 \text{s}^{-1}$</span>.  From conservation of angular momentum, the momentum lost by the slowing of Earth is gained in the Moon's orbit around the Earth.</p>
<p>The angular momentum of the Moon in orbit is defined as</p>
<p><span class="math-container">$L=I\omega = ma^22\pi/p$</span></p>
<p>where <span class="math-container">$I=ma^2$</span> is the moment of inertia, and <span class="math-container">$\omega = 2\pi/p$</span> is the angular orbital speed and <span class="math-container">$m = 7.3459 \times 10^{22} \text{kg}$</span>, <span class="math-container">$a=3.85 \times 10^8 \text{m}$</span>, and <span class="math-container">$p = 2.36 \times 10^6 \text{s}$</span> are the mass, semi-major axis, and period of the Moon respectively.</p>
<p>Plugging these into the above formula for angular momentum gives us the current angular momentum of the Moon as:
<span class="math-container">$L=2.9 \times 10^{34} \text{kg} \text{m}^2 \text{s}^{-1}$</span>.</p>
<p>If we add Earth's angular momentum loss to the Moon's current angular momentum, in 71.8m years, the Moon's angular momentum will be <span class="math-container">$L=2.912 \times 10^{34} \text{kg} \text{m}^2 \text{s}^{-1}$</span>.  We know from Kepler's 3rd law that <span class="math-container">$p^2 = ka^3$</span> for some constant <span class="math-container">$k=9.7598202 \times 10^{-14}$</span>.  Which means <span class="math-container">$p=\sqrt{k}a^{3/2}$</span>.</p>
<p>Substituting <span class="math-container">$p$</span> back into our equation for angular momentum and simplifying gives us:</p>
<p><span class="math-container">$L= m\sqrt{a}2\pi/\sqrt{k}$</span></p>
<p>Solving for <span class="math-container">$a$</span>:</p>
<p><span class="math-container">$a=(\frac{L\sqrt{k}}{2m\pi})^2\approx 3.885 \times 10^8 m$</span></p>
<p>solving for a new <span class="math-container">$p$</span> gives us <span class="math-container">$p=2.39 \times 10^6 \text{s}$</span></p>
<p><strong>So, when the Earth's rotation rate slows to 360 rotations per orbital revolution, the Moon's average distance will increase by almost a percentage from 385,000 km to 388,500 km (a total change of 3,500km).  The Moon's orbital period will increase by more than a percentage by about 30,000 seconds or a bit over 8 hours.</strong></p>

CC BY-SA 4.0
2022-04-11T23:48:35.237Z
The journal *Astronomy and Astrophysics* has a good open-access article which contains polynomial expressions for the lunar orbit parameters, [A new determination of lunar orbital parameters, precession constant and tidal acceleration from LLR measurements](https://www.aanda.org/articles/aa/abs/2002/20/aa2201/aa2201.html), J. Chapront, M. Chapront-Touzé and G. Franco (2002). doi: 
10.1051/0004-6361:20020420 LLR measurements have improved even further in the last two decades.
PM 2Ring
CC BY-SA 4.0
2022-04-12T00:29:56.360Z
Here's a Sage script that calculates the synodic month length using the polynomial in that article by Chapront et al. See https://en.wikipedia.org/wiki/Lunar_month#Derivation for details.
PM 2Ring
CC BY-SA 4.0
2022-04-12T00:30:03.457Z
[Synodic month](https://sagecell.sagemath.org/?z=eJxtUcFu2zAMvesrHryD7TZ1FcUxkqAZhq4DhmIFhjbADsNQuA5tC7WlQJaz5u9Hqy3Qw3QgSL4n8pGMoggPJ2P3ukJvjW_RkWl8K3457T0ZPJ3w8w7qXpsGSiqVyTyTCxFFkRCf8N06Qw4H250y_NCGSgdPrkdXDj4Te6rRl8_0WI-mSipLdZ1uBPg12EIGr7YOI7RBgF_Rd0Zz9oJzjCHnyI_OoJna3pDTx9LrI8HW8C0FBaHUzSzE1FnTMMOad8adZb92tg_Rw2hE-LNFwuXnhVTrgu2ch5urGS6KbJWvVziDmjEuMymL5XrJ8YLBKZRMXOWcyEX6e7O5mP-ZlH215kjOoyLDcjUNOPB-SlcNVMFb7MvTa8rRUVxz90WxVKFsIdm-GSGmhTF6jcsPC5wEp0IMrf2btL7vkjiOr9r88_8OeHXJgPhm-BrYTeu9HTtdmg_CBm0qwq3iSTLJldJUfNETv6x8uNxjstvKt3sdHENJUPGy3aUzxNMkcfoPoBylBQ==&lang=sage)
PM 2Ring
CC BY-SA 4.0
2022-04-12T02:04:04.030Z
@PM2Ring thanks for the links.  My answer above could be done much better, this is just a "back of the napkin".   Though, in some ways, the best answer to the question is that a synchronization with rotation period and lunar cycle doesn't matter.
Connor Garcia
CC BY-SA 4.0
2022-04-11T20:24:11.677Z
2022-04-11T20:24:11.677Z
Connor Garcia
<p>In principle, nothing.  There is no direct link between the length of a day, the length of a month, and the length of a year. If you have a wizard on hand to do the magic, you can change the rotation speed without changing the orbit of the moon, or the orbit of the Earth.</p>
<p>If you don't have a wizard, then the outrageous amounts of energy required to change any of these three values may have a secondary effect on the others. How much of an effect might depend on how you slow down the rotation period of the Earth.</p>

CC BY-SA 4.0
2022-04-11T12:22:32.257Z
There's kind of a tidal link between the length of the day and the length of the month.
PM 2Ring
CC BY-SA 4.0
2022-04-11T19:18:16.353Z
Thanks for the answers!  They make good sense.  Does someone also have simple reference(s), so that I may repeat the answers elsewhere and back myself up?
Peter Heffner
CC BY-SA 4.0
2022-04-11T20:27:26.930Z
@PM2Ring would you care to elaborate.  The tides are currently transferring momentum from the Earth to the moon, causing the Earth to slow down (and the moon to increase its orbit, which slows it down too) but the length of a day is not a rational fraction of the length of a month (however you define month)
James K
CC BY-SA 4.0
2022-04-11T23:33:47.100Z
Yes, I was just referring to the slowing down of the Earth's rotation and the increase in the Earth-Moon mean orbital radius due to the tidal transfer of angular momentum. I certainly wasn't implying that there are any simple rational relationships between the day and month lengths. FWIW, here's a recent question on approximating the ratios of the various lunar months, which can be helpful in the study of eclipse cycles: https://math.stackexchange.com/q/4412720/207316
PM 2Ring
CC BY-SA 4.0
2022-04-10T18:59:23.543Z
2022-04-10T18:59:23.543Z
James K
<p>Hypothetical Question</p>
<p>If the earth’s rotation slowed down to align to the lunar cycle so that the orbital period began and ended at the same lunar phase, what other things would result?</p>
<p>For example, take the orbital period as exactly 360 earth rotations.  The rotation would have decelerated. Would the speed of the lunar cycle also change, and by how much?</p>
<p>How might a change in rotation speed effect the orbital period?</p>
<p>How would all that be estimated?</p>
<p>Thank you!</p>

CC BY-SA 4.0
2022-04-10T00:50:55.290Z
I think you're using the wrong term in the title where you say "rotation", it sounds more like you mean the "orbital period".  The answer is yes, any of the Sun-Earth Lagrange points would produce this result (though you wouldn't be able to see it at L3).
Greg Miller
CC BY-SA 4.0
2022-04-10T03:31:30.783Z
Thanks.  I’ll look into that.  But actually I should have said that if the orbital period remains constant, but the rotation slowed to 360 days in a complete orbit, what happens to the speed of the moon’s cycles?
Peter Heffner
<the-sun><the-moon><rotation><time><hypothetical>
Earth’s rotation speed vs rate of lunar cycle
Peter Heffner
# Earth’s rotation speed vs rate of lunar cycle

(Asked by: username_0 on 2022-04-09)

Hypothetical Question

If the earth’s rotation slowed down to align to the lunar cycle so that the orbital period began and ended at the same lunar phase, what other things would result?

For example, take the orbital period as exactly 360 earth rotations. The rotation would have decelerated. Would the speed of the lunar cycle also change, and by how much?

How might a change in rotation speed effect the orbital period?

How would all that be estimated?

Thank you!

**Tags:** the-sun, the-moon, rotation, time, hypothetical
> **Comments:**
> **username_1**: I think you're using the wrong term in the title where you say "rotation", it sounds more like you mean the "orbital period".  The answer is yes, any of the Sun-Earth Lagrange points would produce this result (though you wouldn't be able to see it at L3).  
> **username_0**: Thanks.  I’ll look into that.  But actually I should have said that if the orbital period remains constant, but the rotation slowed to 360 days in a complete orbit, what happens to the speed of the moon’s cycles?  

---

## 2 answers

### Answer by username_2 on 2022-04-11. Score: 3 | Accepted answer

This is not a hypothetical question since the Earth's rotation rate is slowing due to tidal interactions with the Moon. Currently, the Earth rotates once every 86164 seconds (sidereal time) for about 366 days in a year. If there were 360 rotations a year, then a day would last $86164\*366/360\approx 87600$ seconds which is 1436 seconds slower than now. The Earth's rotation rate is slowing by about 2 seconds every hundred thousand years according to [Deines and Williams](https://iopscience.iop.org/article/10.3847/0004-6256/151/4/103). This means there will be 360 days in a year in about 71.8 million years.

The [Earth's current rotational angular momentum](https://www.vanderbilt.edu/AnS/physics/astrocourses/ast201/angular_momentum.html) is about $7.1 \times 10^{33} \text{kg} \text{m}^2 \text{s}^{-1}$. The momentum 71.8m years from now will be about 360/366 of this, or $6.98 \times 10^{33} \text{kg} \text{m}^2 \text{s}^{-1}$. This is a loss of $1.2 \times 10^{32} \text{kg} \text{m}^2 \text{s}^{-1}$. From conservation of angular momentum, the momentum lost by the slowing of Earth is gained in the Moon's orbit around the Earth.

The angular momentum of the Moon in orbit is defined as

$L=I\omega = ma^22\pi/p$

where $I=ma^2$ is the moment of inertia, and $\omega = 2\pi/p$ is the angular orbital speed and $m = 7.3459 \times 10^{22} \text{kg}$, $a=3.85 \times 10^8 \text{m}$, and $p = 2.36 \times 10^6 \text{s}$ are the mass, semi-major axis, and period of the Moon respectively.

Plugging these into the above formula for angular momentum gives us the current angular momentum of the Moon as:
$L=2.9 \times 10^{34} \text{kg} \text{m}^2 \text{s}^{-1}$.

If we add Earth's angular momentum loss to the Moon's current angular momentum, in 71.8m years, the Moon's angular momentum will be $L=2.912 \times 10^{34} \text{kg} \text{m}^2 \text{s}^{-1}$. We know from Kepler's 3rd law that $p^2 = ka^3$ for some constant $k=9.7598202 \times 10^{-14}$. Which means $p=\sqrt{k}a^{3/2}$.

Substituting $p$ back into our equation for angular momentum and simplifying gives us:

$L= m\sqrt{a}2\pi/\sqrt{k}$

Solving for $a$:

$a=(\frac{L\sqrt{k}}{2m\pi})^2\approx 3.885 \times 10^8 m$

solving for a new $p$ gives us $p=2.39 \times 10^6 \text{s}$

**So, when the Earth's rotation rate slows to 360 rotations per orbital revolution, the Moon's average distance will increase by almost a percentage from 385,000 km to 388,500 km (a total change of 3,500km). The Moon's orbital period will increase by more than a percentage by about 30,000 seconds or a bit over 8 hours.**

> **Comments:**
> **username_3**: The journal *Astronomy and Astrophysics* has a good open-access article which contains polynomial expressions for the lunar orbit parameters, [A new determination of lunar orbital parameters, precession constant and tidal acceleration from LLR measurements](https://www.aanda.org/articles/aa/abs/2002/20/aa2201/aa2201.html), J. Chapront, M. Chapront-Touzé and G. Franco (2002). doi:  10.1051/0004-6361:20020420 LLR measurements have improved even further in the last two decades.  
> **username_3**: Here's a Sage script that calculates the synodic month length using the polynomial in that article by Chapront et al. See https://en.wikipedia.org/wiki/Lunar_month#Derivation for details.  
> **username_3**: [Synodic month](https://sagecell.sagemath.org/?z=eJxtUcFu2zAMvesrHryD7TZ1FcUxkqAZhq4DhmIFhjbADsNQuA5tC7WlQJaz5u9Hqy3Qw3QgSL4n8pGMoggPJ2P3ukJvjW_RkWl8K3457T0ZPJ3w8w7qXpsGSiqVyTyTCxFFkRCf8N06Qw4H250y_NCGSgdPrkdXDj4Te6rRl8_0WI-mSipLdZ1uBPg12EIGr7YOI7RBgF_Rd0Zz9oJzjCHnyI_OoJna3pDTx9LrI8HW8C0FBaHUzSzE1FnTMMOad8adZb92tg_Rw2hE-LNFwuXnhVTrgu2ch5urGS6KbJWvVziDmjEuMymL5XrJ8YLBKZRMXOWcyEX6e7O5mP-ZlH215kjOoyLDcjUNOPB-SlcNVMFb7MvTa8rRUVxz90WxVKFsIdm-GSGmhTF6jcsPC5wEp0IMrf2btL7vkjiOr9r88_8OeHXJgPhm-BrYTeu9HTtdmg_CBm0qwq3iSTLJldJUfNETv6x8uNxjstvKt3sdHENJUPGy3aUzxNMkcfoPoBylBQ==&lang=sage)  
> **username_2**: @PM2Ring thanks for the links.  My answer above could be done much better, this is just a "back of the napkin".   Though, in some ways, the best answer to the question is that a synchronization with rotation period and lunar cycle doesn't matter.  

---

### Answer by username_4 on 2022-04-10. Score: 3

In principle, nothing. There is no direct link between the length of a day, the length of a month, and the length of a year. If you have a wizard on hand to do the magic, you can change the rotation speed without changing the orbit of the moon, or the orbit of the Earth.

If you don't have a wizard, then the outrageous amounts of energy required to change any of these three values may have a secondary effect on the others. How much of an effect might depend on how you slow down the rotation period of the Earth.

> **Comments:**
> **username_3**: There's kind of a tidal link between the length of the day and the length of the month.  
> **username_0**: Thanks for the answers!  They make good sense.  Does someone also have simple reference(s), so that I may repeat the answers elsewhere and back myself up?  
> **username_4**: @PM2Ring would you care to elaborate.  The tides are currently transferring momentum from the Earth to the moon, causing the Earth to slow down (and the moon to increase its orbit, which slows it down too) but the length of a day is not a rational fraction of the length of a month (however you define month)  
> **username_3**: Yes, I was just referring to the slowing down of the Earth's rotation and the increase in the Earth-Moon mean orbital radius due to the tidal transfer of angular momentum. I certainly wasn't implying that there are any simple rational relationships between the day and month lengths. FWIW, here's a recent question on approximating the ratios of the various lunar months, which can be helpful in the study of eclipse cycles: https://math.stackexchange.com/q/4412720/207316  

---

