<p>The moon's rotation is very slowly decreasing. The page you link to is correct but confusing, and you need to understand one of the weird things about orbits to understand it:</p>

<p><em>If you speed up in an orbit you go slower</em></p>

<p>Here is how this apparent paradox works.  If you are orbiting the Earth and you fire your rockets to speed up, you will lift yourself to a higher orbit, where you will be moving slower. </p>

<p>The tidal bulge is a large mass and this large mass is in front of the moon. Any large mass has a gravitational effect and there is a gravitational force on the moon by the bulge, and an equal and opposite force on the bulge by the moon. The bulge is in front of the moon, so the gravitational action of the bulge pulls the moon forward.</p>

<p>So the tidal bulge acts to pull the moon forward in its orbit, which makes it slow down, and since the moon is tidally locked it also slows its rotation speed. However the moon is already spinning quite slowly, only one complete rotation per month, so the amount of slow down is very small.</p>

<p>The article writes</p>

<blockquote>
  <p>The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth. [...] The increase in speed of revolution of moon around the Earth also results in the increase in radius of orbit of moon.</p>
</blockquote>

<p>You should understand that when it says "increase in speed of the moon" it means that the moon is actually slowing down due to the increase in radius.</p>

CC BY-SA 4.0
2018-05-11T15:59:54.883Z
So overall, the moon's revolution speed is slowing down?
StellarExile
CC BY-SA 4.0
2018-05-11T16:04:15.837Z
the tidal bulge acts to pull the moon forward in its orbit, which makes it slow down, and since the moon is tidally locked it also slows its rotation speed.
James K
CC BY-SA 4.0
2018-05-11T16:17:22.423Z
What do you mean exactly by acts to make it move forward? Do you mean move away from the Earth?
StellarExile
CC BY-SA 4.0
2018-05-11T16:21:12.417Z
edited. The "act" is gravitational.
James K
CC BY-SA 4.0
2018-05-11T17:44:51.547Z
I'm actually talking about the moon's orbital speed.
StellarExile
CC BY-SA 4.0
2018-05-11T17:48:32.397Z
Yes, I'm talking about the same thing. I'll only repeat myself.  The bulge pulls the moon forward, causing it to slow down. This seems like a paradox, but is not, as explained in the answer.
James K
CC BY-SA 4.0
2018-05-11T18:11:50.313Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/77373/discussion-between-calvin-face-and-james-k).
StellarExile
CC BY-SA 4.0
2018-05-11T19:42:22.857Z
I would say that "The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth." is flat out wrong. Whether "speed of rotation" means angular velocity or orbital velocity, both are decreasing.
David Hammen
CC BY-SA 4.0
2018-05-11T15:48:47.920Z
2018-05-11T16:20:46.947Z
2018-05-11T16:20:46.947Z
James K
<p>From the cited article,</p>

<blockquote>
  <p>The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth.</p>
</blockquote>

<p>I suspect that this is the statement that is causing the OP to be confused, and rightly so. This statement is incorrect. Whether "speed of rotation" means orbital velocity or angular velocity, both are decreasing as the Moon slowly retreats from the Earth.</p>

<p>The Moon is retreating very slowly from the Earth, currently about 3.78 cm per year. Dividing by 385000 km (the Earth-Moon mean orbital radius) yields about 10<sup>-10</sup> parts per year. That qualifies as "very slowly". For simplicity, I'll assume the Moon's orbit is essentially circular, plus this very slow spiraling out. For a circular orbit, the relation between orbital radius and angular velocity is $$r^3\omega^2 = G(M+m)$$
This essentially is Kepler's third law. Assuming $G$, $M$, and $m$ are constant, differentiating both sides with respect to time yields
$$r^2\omega (3\dot r \omega + 2 r \dot \omega) = 0$$
meaning
$$r \dot \omega = -\frac32 \dot r \omega$$
Suppose some quantity $q$ is of the form to $q=r^n\omega^m$. Differentiating with respect to time yields $\dot q = r^{n-1}\omega^{m-1}\left(n\dot r\omega + m r \dot\omega\right)$. Since $r\dot\omega = -\frac32 \dot r \omega$, this can be rewritten as $$\dot q = \left(n - \frac 3 2 m\right) \dot r r^{n-1}\omega^m$$
Quantities of this form include angular velocity ($q=\omega$, so $n=0, m=1$ in this case), orbital velocity $(q=r\omega$, so $n=m=1$), and angular momentum (q$=r^2\omega$, so $n=2, m=1$).  Note that $(n-\frac 32m)$ is negative for angular and orbital velocity but positive for orbit angular momentum. Orbit angular momentum thus has the same sign as does $\dot r$ while angular velocity and orbital velocity have the opposite sign as does $\dot r$.</p>

<p>As a sanity check, look to centripetal acceleration. This is given by $a = r\omega^2$. In terms of the above, $n=1$ and $m=2$. The term $n-\frac32m$ is clearly negative in this case, so centripetal acceleration decreases as the Moon retreats from the Earth. Centripetal acceleration is also give by Newton's law of gravitation: It is proportional to the inverse of the square of the distance between the Earth and the Moon, so this does indeed decrease as the Moon retreats from the Earth.</p>

CC BY-SA 4.0
2018-05-11T22:08:15.670Z
I think I'm understanding it better, So, the moon's orbital speed is decreasing because the gravitational force is weaker as distance between the two bodies increases? I'm not very good with the mathematics behind the concept, sorry.
StellarExile
CC BY-SA 4.0
2018-05-11T20:15:53.647Z
2018-05-11T22:46:38.097Z
2018-05-11T22:46:38.097Z
David Hammen
<p>Some confusion may have arisen in parts of the literature due to the facts that (a) the moon's mean rate of angular motion (and revolution speed) does appear to be speeding up if it is measured by mean solar time or by sidereal time, but also (b) the mean rate appears to be slowing down if it is measured by an atomic or dynamical time-scale.  </p>

<p>That is because there are (at least) three real effects in operation: </p>

<p>(1) an <em>apparent</em> increase in the moon's rate of motion caused by the long-term slowing-down of earth rotation (when the day is longer the moon would appear to go further even if its real rate is unchanged),</p>

<p>(2) a <em>real</em> increase in the moon's rate due to the long-term reduction in the eccentricity of the earth-sun orbit, and </p>

<p>(3) a <em>real</em> decrease in rate due to tidal retardation of the moon's motion itself.  </p>

<p>The sum of all three effects is an apparent net positive acceleration. (When effect (1) is included, that of course implies the use of a mean solar or sidereal clock and time-scale.) But if apparent effect (1) is taken out, by using an atomic timescale instead, the remaining real effect (3) more than offsets real effect (2), leaving a net negative acceleration (retardation). </p>

<p>Historically these were discovered in the order, first the aggregate of (1)+(2)+(3) from the 1690s onwards, then effect (2) from the late 1700s, and finally effects (1) and (3) were suspected in the later 1800s and after much work separately verified during the 20th century. </p>

<p>Edmond Halley in the 1690s suspected a net positive acceleration; but he could not evaluate it quantitatively, for lack of good geographical longitudes of the places where the ancient moon observations had been taken. He appealed for such measurements in the last two pages (174-5) of a paper of 1693 (<a href="http://rstl.royalsocietypublishing.org/content/19/218/160" rel="nofollow noreferrer">http://rstl.royalsocietypublishing.org/content/19/218/160</a>). </p>

<p>In the 1740s Richard Dunthorne had better data, and by evaluating ancient moon observations in the near East he estimated the mean acceleration as +10 arc"/cy/cy (arc-seconds, centuries) (sources cited in <a href="https://en.wikipedia.org/wiki/Richard_Dunthorne" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Richard_Dunthorne</a>). A closer estimate is about +12.  </p>

<p>Then Laplace in 1788 published results of theoretical calculations of celestial mechanics to identify the real accelerative effect on the moon caused by the (by-then established) slow and slight reduction in eccentricity of the earth-sun orbit ("Sur l'equation seculaire de la Lune", (<a href="http://gallica.bnf.fr/ark:/12148/bpt6k77599c/f248" rel="nofollow noreferrer">http://gallica.bnf.fr/ark:/12148/bpt6k77599c/f248</a>)). Unfortunately Laplace treated as negligible too many of the smaller components of the infinite series that he had to deal with. His result overestimated the real effect by about double. By chance this seemed to match the whole of the observed apparent acceleration. The apparent agreement between theory and observation quietened further enquiry for about half a century, until in 1853 J C Adams established that a correct calculation of this accelerative effect accounted only for about half of the apparent acceleration observed ("On the secular variation of the moon's mean motion", Phil. Trans. R. Soc. Lond. 1853 vol.143 397-406) (<a href="http://rstl.royalsocietypublishing.org/content/143/397" rel="nofollow noreferrer">http://rstl.royalsocietypublishing.org/content/143/397</a>). Adams' finding caused a controversy amongst astronomers that lasted some years, but it was finally accepted that Adams (and Delaunay, who confirmed his calculations) were in the right. But that left about half of the observed apparent lunar acceleration still to be accounted for. The size of effect (2) (real positive acceleration due to changing solar eccentricity) was calculated at about +6 arc"/cy/cy from the later 19th-c. onwards.</p>

<p>Irregular slowing-down of earth rotation was established more recently, in the 1920s-1930s: previous suspicions were converted to practical certainty by works of E W Brown (1926), W de Sitter (1929) and a more complete quantitative study by H Spencer Jones (1939) ("The rotation of the earth, and the secular accelerations of the sun, moon and planets") (<a href="http://articles.adsabs.harvard.edu/full/1939MNRAS..99..541S" rel="nofollow noreferrer">http://articles.adsabs.harvard.edu/full/1939MNRAS..99..541S</a>). This last paper set the basis for a practical time-scale freed from irregular earth-rotation effects, and led via Ephemeris Time (<a href="https://en.wikipedia.org/wiki/Ephemeris_time" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Ephemeris_time</a>) to the later varieties of atomic time. </p>

<p>The tidal deceleration was first well-measured at about -26 arc"/cy/cy as recently as 1975 by Morrison and Ward (<a href="http://adsabs.harvard.edu/abs/1975MNRAS.173..183M" rel="nofollow noreferrer">http://adsabs.harvard.edu/abs/1975MNRAS.173..183M</a>) by an analysis of traditional optical results (perhaps surprisingly, by a long series of Mercury solar transit timings). As the period of lunar laser-ranging measurements has grown longer and the technique more precise, the 1975 value has been refined more recently to about -25.858 (Chapront et al., 2002 (<a href="http://adsabs.harvard.edu/abs/2002A%26A...387..700C" rel="nofollow noreferrer">http://adsabs.harvard.edu/abs/2002A%26A...387..700C</a>)).  (The 3.8cm or so of annual lunar recess from the earth established by lunar laser-ranging is believed to be a result of the tidal retardation, with relatively small effect on velocity.)</p>

<p>The net deceleration in modern estimates of lunar motion on atomic timescales amounts in aggregate to about -11.5 "/cy/cy (relative to equinox of date).  </p>

CC BY-SA 4.0
2018-05-12T02:15:28.597Z
2018-05-12T10:28:54.307Z
2018-05-12T10:28:54.307Z
terry-s
<p>I've come across multiple answers to this question. So, the moon is moving away from the Earth a couple of centimeters per year and the Earth's rotation is slowing down. <a href="http://astro-interest.com/2014/10/the-speed-of-rotation-of-earth-is-decreasing-due-to-revolution-of-moon/" rel="nofollow noreferrer">Here</a> it says the moon's revolution is speeding up, another place it says it is slowing down. So, due to tidal effects, is the moon's revolution slowing down or speeding up?</p>

CC BY-SA 4.0
2018-05-11T22:30:45.670Z
Ah, the "beauty" of orbital mechanics, where you need to speed up if you want to slow down. Awfully counter-intuitive. I'd recommend Kerbal Space Program, if you want to *really* understand WTF is wrong with how this whole thing works and why and how (the heck) both statements are simultaneously correct.
SF.
<the-moon><tidal-locking>
Is the moon's revolution speed increasing or decreasing
StellarExile
# Is the moon's revolution speed increasing or decreasing

(Asked by: username_0 on 2018-05-11)

I've come across multiple answers to this question. So, the moon is moving away from the Earth a couple of centimeters per year and the Earth's rotation is slowing down. [Here](http://astro-interest.com/2014/10/the-speed-of-rotation-of-earth-is-decreasing-due-to-revolution-of-moon/) it says the moon's revolution is speeding up, another place it says it is slowing down. So, due to tidal effects, is the moon's revolution slowing down or speeding up?

**Tags:** the-moon, tidal-locking

> **Comments:**
> **username_1**: Ah, the "beauty" of orbital mechanics, where you need to speed up if you want to slow down. Awfully counter-intuitive. I'd recommend Kerbal Space Program, if you want to *really* understand WTF is wrong with how this whole thing works and why and how (the heck) both statements are simultaneously correct.  

---

## 3 answers

### Answer by username_2 on 2018-05-11. Score: 2 | Accepted answer

The moon's rotation is very slowly decreasing. The page you link to is correct but confusing, and you need to understand one of the weird things about orbits to understand it:

*If you speed up in an orbit you go slower*

Here is how this apparent paradox works. If you are orbiting the Earth and you fire your rockets to speed up, you will lift yourself to a higher orbit, where you will be moving slower.

The tidal bulge is a large mass and this large mass is in front of the moon. Any large mass has a gravitational effect and there is a gravitational force on the moon by the bulge, and an equal and opposite force on the bulge by the moon. The bulge is in front of the moon, so the gravitational action of the bulge pulls the moon forward.

So the tidal bulge acts to pull the moon forward in its orbit, which makes it slow down, and since the moon is tidally locked it also slows its rotation speed. However the moon is already spinning quite slowly, only one complete rotation per month, so the amount of slow down is very small.

The article writes

> The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth. [...] The increase in speed of revolution of moon around the Earth also results in the increase in radius of orbit of moon.

You should understand that when it says "increase in speed of the moon" it means that the moon is actually slowing down due to the increase in radius.

> **Comments:**
> **username_0**: So overall, the moon's revolution speed is slowing down?  
> **username_2**: the tidal bulge acts to pull the moon forward in its orbit, which makes it slow down, and since the moon is tidally locked it also slows its rotation speed.  
> **username_0**: What do you mean exactly by acts to make it move forward? Do you mean move away from the Earth?  
> **username_2**: edited. The "act" is gravitational.  
> **username_0**: I'm actually talking about the moon's orbital speed.  
> **username_2**: Yes, I'm talking about the same thing. I'll only repeat myself.  The bulge pulls the moon forward, causing it to slow down. This seems like a paradox, but is not, as explained in the answer.  
> **username_0**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/77373/discussion-between-calvin-face-and-james-k).  
> **username_3**: I would say that "The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth." is flat out wrong. Whether "speed of rotation" means angular velocity or orbital velocity, both are decreasing.  

---

### Answer by username_3 on 2018-05-11. Score: 5

From the cited article,

> The slowing of the speed of rotation of Earth also results in the increase in speed of revolution of moon around the Earth.

I suspect that this is the statement that is causing the OP to be confused, and rightly so. This statement is incorrect. Whether "speed of rotation" means orbital velocity or angular velocity, both are decreasing as the Moon slowly retreats from the Earth.

The Moon is retreating very slowly from the Earth, currently about 3.78 cm per year. Dividing by 385000 km (the Earth-Moon mean orbital radius) yields about 10-10 parts per year. That qualifies as "very slowly". For simplicity, I'll assume the Moon's orbit is essentially circular, plus this very slow spiraling out. For a circular orbit, the relation between orbital radius and angular velocity is $$r^3\omega^2 = G(M+m)$$
This essentially is Kepler's third law. Assuming $G$, $M$, and $m$ are constant, differentiating both sides with respect to time yields
$$r^2\omega (3\dot r \omega + 2 r \dot \omega) = 0$$
meaning
$$r \dot \omega = -\frac32 \dot r \omega$$
Suppose some quantity $q$ is of the form to $q=r^n\omega^m$. Differentiating with respect to time yields $\dot q = r^{n-1}\omega^{m-1}\left(n\dot r\omega + m r \dot\omega\right)$. Since $r\dot\omega = -\frac32 \dot r \omega$, this can be rewritten as $$\dot q = \left(n - \frac 3 2 m\right) \dot r r^{n-1}\omega^m$$
Quantities of this form include angular velocity ($q=\omega$, so $n=0, m=1$ in this case), orbital velocity $(q=r\omega$, so $n=m=1$), and angular momentum (q$=r^2\omega$, so $n=2, m=1$). Note that $(n-\frac 32m)$ is negative for angular and orbital velocity but positive for orbit angular momentum. Orbit angular momentum thus has the same sign as does $\dot r$ while angular velocity and orbital velocity have the opposite sign as does $\dot r$.

As a sanity check, look to centripetal acceleration. This is given by $a = r\omega^2$. In terms of the above, $n=1$ and $m=2$. The term $n-\frac32m$ is clearly negative in this case, so centripetal acceleration decreases as the Moon retreats from the Earth. Centripetal acceleration is also give by Newton's law of gravitation: It is proportional to the inverse of the square of the distance between the Earth and the Moon, so this does indeed decrease as the Moon retreats from the Earth.

> **Comments:**
> **username_0**: I think I'm understanding it better, So, the moon's orbital speed is decreasing because the gravitational force is weaker as distance between the two bodies increases? I'm not very good with the mathematics behind the concept, sorry.  

---

### Answer by username_4 on 2018-05-12. Score: 4

Some confusion may have arisen in parts of the literature due to the facts that (a) the moon's mean rate of angular motion (and revolution speed) does appear to be speeding up if it is measured by mean solar time or by sidereal time, but also (b) the mean rate appears to be slowing down if it is measured by an atomic or dynamical time-scale.

That is because there are (at least) three real effects in operation:

(1) an *apparent* increase in the moon's rate of motion caused by the long-term slowing-down of earth rotation (when the day is longer the moon would appear to go further even if its real rate is unchanged),

(2) a *real* increase in the moon's rate due to the long-term reduction in the eccentricity of the earth-sun orbit, and

(3) a *real* decrease in rate due to tidal retardation of the moon's motion itself.

The sum of all three effects is an apparent net positive acceleration. (When effect (1) is included, that of course implies the use of a mean solar or sidereal clock and time-scale.) But if apparent effect (1) is taken out, by using an atomic timescale instead, the remaining real effect (3) more than offsets real effect (2), leaving a net negative acceleration (retardation).

Historically these were discovered in the order, first the aggregate of (1)+(2)+(3) from the 1690s onwards, then effect (2) from the late 1700s, and finally effects (1) and (3) were suspected in the later 1800s and after much work separately verified during the 20th century.

Edmond Halley in the 1690s suspected a net positive acceleration; but he could not evaluate it quantitatively, for lack of good geographical longitudes of the places where the ancient moon observations had been taken. He appealed for such measurements in the last two pages (174-5) of a paper of 1693 (<http://rstl.royalsocietypublishing.org/content/19/218/160>).

In the 1740s Richard Dunthorne had better data, and by evaluating ancient moon observations in the near East he estimated the mean acceleration as +10 arc"/cy/cy (arc-seconds, centuries) (sources cited in <https://en.wikipedia.org/wiki/Richard_Dunthorne>). A closer estimate is about +12.

Then Laplace in 1788 published results of theoretical calculations of celestial mechanics to identify the real accelerative effect on the moon caused by the (by-then established) slow and slight reduction in eccentricity of the earth-sun orbit ("Sur l'equation seculaire de la Lune", (<http://gallica.bnf.fr/ark:/12148/bpt6k77599c/f248>)). Unfortunately Laplace treated as negligible too many of the smaller components of the infinite series that he had to deal with. His result overestimated the real effect by about double. By chance this seemed to match the whole of the observed apparent acceleration. The apparent agreement between theory and observation quietened further enquiry for about half a century, until in 1853 J C Adams established that a correct calculation of this accelerative effect accounted only for about half of the apparent acceleration observed ("On the secular variation of the moon's mean motion", Phil. Trans. R. Soc. Lond. 1853 vol.143 397-406) (<http://rstl.royalsocietypublishing.org/content/143/397>). Adams' finding caused a controversy amongst astronomers that lasted some years, but it was finally accepted that Adams (and Delaunay, who confirmed his calculations) were in the right. But that left about half of the observed apparent lunar acceleration still to be accounted for. The size of effect (2) (real positive acceleration due to changing solar eccentricity) was calculated at about +6 arc"/cy/cy from the later 19th-c. onwards.

Irregular slowing-down of earth rotation was established more recently, in the 1920s-1930s: previous suspicions were converted to practical certainty by works of E W Brown (1926), W de Sitter (1929) and a more complete quantitative study by H Spencer Jones (1939) ("The rotation of the earth, and the secular accelerations of the sun, moon and planets") (<http://articles.adsabs.harvard.edu/full/1939MNRAS..99..541S>). This last paper set the basis for a practical time-scale freed from irregular earth-rotation effects, and led via Ephemeris Time (<https://en.wikipedia.org/wiki/Ephemeris_time>) to the later varieties of atomic time.

The tidal deceleration was first well-measured at about -26 arc"/cy/cy as recently as 1975 by Morrison and Ward (<http://adsabs.harvard.edu/abs/1975MNRAS.173..183M>) by an analysis of traditional optical results (perhaps surprisingly, by a long series of Mercury solar transit timings). As the period of lunar laser-ranging measurements has grown longer and the technique more precise, the 1975 value has been refined more recently to about -25.858 (Chapront et al., 2002 (<http://adsabs.harvard.edu/abs/2002A%26A...387..700C>)). (The 3.8cm or so of annual lunar recess from the earth established by lunar laser-ranging is believed to be a result of the tidal retardation, with relatively small effect on velocity.)

The net deceleration in modern estimates of lunar motion on atomic timescales amounts in aggregate to about -11.5 "/cy/cy (relative to equinox of date).


---

