# Same side of the Moon always faces the Earth: The Moon's recession rate. **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [November 18, 2004, 4:57pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486 "2004-11-18T16:57:39Z")\ **Posts on this page:** 16\ **Page:** 2
**Author:** ![John\_W.Kennedy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_w.kennedy/32/1031_2.png) [@John\_W.Kennedy](https://boards.straightdope.com/u/John_W.Kennedy)\ **Post date:** [November 21, 2004, 1:17am UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/21 "2004-11-21T01:17:18Z")
> [@Podkayne](#): > > Weird! I cannot imagine what mechanism they could be invoking to bring the Moon back in. Over a _huge_ amount of time, of course, there’s atmospheric drag, such as it is. Or (I don’t have the necessary maths to hand) would Solar tides on the Earth-Moon system steal angular momentum? ---
**Author:** ![Podkayne](https://avatars.discourse-cdn.com/v4/letter/p/edb3f5/32.png) [@Podkayne](https://boards.straightdope.com/u/Podkayne)\ **Post date:** [November 21, 2004, 1:45am UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/22 "2004-11-21T01:45:37Z")
> [@John W. Kennedy](#): > > Over a _huge_ amount of time, of course, there’s atmospheric drag, such as it is. Even on long timescales, even at the Moon’s current distance, it’d be insignificant. > [@](#): > > Or (I don’t have the necessary maths to hand) would Solar tides on the Earth-Moon system steal angular momentum? You may be right. The major effect of solar tides is slowing the Earth’s rotation, so that could take some energy out of the Moon’s orbit, but that could only slow the Moon’s recession. In order to turn the Moon around, it would have to tidally lock the Earth to the Moon (so the Earth always keeps one face toward the Moon), but the Moon would escape before that point . . . but maybe the parameters they were using at the time showed the Earth locking up before the Moon escaped, and then solar tides brought it back in. ---
**Author:** ![John\_W.Kennedy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_w.kennedy/32/1031_2.png) [@John\_W.Kennedy](https://boards.straightdope.com/u/John_W.Kennedy)\ **Post date:** [November 21, 2004, 2:19am UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/23 "2004-11-21T02:19:21Z")
> [@Podkayne](#): > > In order to turn the Moon around, it would have to tidally lock the Earth to the Moon I’m pretty sure that that was, in fact, stated. ---
**Author:** ![Jinx](https://avatars.discourse-cdn.com/v4/letter/j/c6cbf5/32.png) [@Jinx](https://boards.straightdope.com/u/Jinx)\ **Post date:** [November 21, 2004, 1:41pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/24 "2004-11-21T13:41:20Z")
OK, maybe I’ve missed something here, but isn’t the logic argument as follows? First, the earth’s rotation is slowing down. Due to conservation of angular momentum, the moon is moving away (i.e.: longer orbit, slower speed), so the moon’s revolution is slowing down, and for the same reason (as some form of compensation) the moon’s rotation has slowed down to the point where a rotation is ABOUT equal to one revolution (that’s why we see 59% of the surface). Anyway, if everything is slowing down in this scenario, then how is angular momentum being conserved? I WAG you might argue the moon keeps the same period as its orbit receeds (i.e.: grows larger) but, if this is true, where is the extra energy to account for the increase in speed required to maintain the same period? I would then WAG that the extra energy comes from the moon’s rotation slowing down. But again, this presumes the period must stay the same, but why should it? - Jinx 😕 ---
**Author:** ![Podkayne](https://avatars.discourse-cdn.com/v4/letter/p/edb3f5/32.png) [@Podkayne](https://boards.straightdope.com/u/Podkayne)\ **Post date:** [November 21, 2004, 4:09pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/25 "2004-11-21T16:09:29Z")
> [@Jinx](#): > > OK, maybe I’ve missed something here, but isn’t the logic argument as follows? > First, the earth’s rotation is slowing down. Due to conservation of angular momentum, the moon is moving away (i.e.: longer orbit, slower speed), so the moon’s revolution is slowing down, and for the same reason (as some form of compensation) the moon’s rotation has slowed down to the point where a rotation is ABOUT equal to one revolution (that’s why we see 59% of the surface). Actually, its rotation period exactly equals its revolution period. The reason that we can see 59% of the surface (libration) is that the Moon’s speed in its orbit is not constant because its orbit isn’t a perfect circle. Now, the Moon’s rotation period equaling its period of revolution is a separate process ocurring at the same time. The Earth raises tides on the Moon, so the Moon always wants to keep its long axis lined up with the Earth. Imagine that the Moon hops into a slightly higher orbit. Its rotation rate is a bit too fast for the new orbit, so its tidal bulge gets a little bit ahead of where it should be, no longer pointed exactly at Earth. Earth’s gravity drags back on the Moon’s tidal bulge. This decreases the Moon’s rotational rate, and also makes the Moon’s orbit larger to conserve angular momentum (see below). However, since the Moon is already synchronized, this effect is small compared to the effect of the Moon’s tides on Earth (which is really just the Moon trying to slow the Earth down so that _Earth’s_ tidal bulge stays line up with the Moon.) > [@](#): > > Anyway, if everything is slowing down in this scenario, then how is angular momentum being conserved? The Moon’s orbital speed decreases as the size of its orbit increases, and its period increases. Kepler’s Third Law, the Harmonic Law, p^2=a^3, is still in effect. The angular momentum of an object in orbit is mvr: the mass of the object times its orbital speed times the radius of its orbit. The (average) orbital speed an object, v, is proportional to 1/sqrt(r) (due to Kepler’s 3rd, above). This means that the angular momentum of the orbit is proportional to sqrt(r). Thus, as the Moon’s orbit gets bigger, it’s angular momentum increases. ---
**Author:** ![chorpler](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chorpler/32/2888_2.png) [@chorpler](https://boards.straightdope.com/u/chorpler)\ **Post date:** [November 21, 2004, 6:23pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/26 "2004-11-21T18:23:57Z")
Yeah, it turns out to be solar tides. On pages 159-160 of the paperback edition of _A Choice of Catastrophes_, after completing his discussion of the receding moon and concluding with the moon and earth circling each other, “dumbbell-fashion,” every 47 days, Asimov says: > [@](#): > > **The Approaching Moon** > > If there were no tidal effects on the Earth or the moon, the dumbbell revolution would last forever. However, the sun’s tidal effects would still exist. These effects would work in a rather complicated fashion to speed up the rotations of the Earth and the moon and to draw the two bodies closer together, at a rate slower than that at which they are now separating. Apparently, this increasing closeness would continue indefinitely so that the moon would, it might be supposed, eventually fall to the Earth after all (though I began by saying it couldn’t happen) since its angular momentum of revolution will finally be shifted all the way to angular momentum of rotation. It will not fall in the usual sense of the word, however, but will inch its way toward us in an excruciatingly slow and gradual decreasing spiral. Yet it will not truly fall even in this way, for no contact will be made. > > As the two bodies approach closer and closer, the tidal effects will increase as the cube of the decreasing distance. By the time the Earth and moon are separated, center to center, by a distance of only some 15,500 kilometers (9,600 miles), so that the two surfaces are separated by only 7,400 kilometers (4,600 miles), the tidal effect of the moon on the Earth will be 15,000 times as strong as it is now. The tidal effect of the Earth on the moon will be 32.5 times stronger still, or nearly 500,000 times the moon’s tidal effect on Earth today. > > By that time, then, the tidal pull on the moon will be so great that the moon will simply be pulled apart and will break up into small fragments. The lunar fragments will, as a result of collisions (and further fragmenting), spread out through the moon’s orbit and the Earth will end up with a ring, like that of Saturn, but far more brilliant and dense. ---
**Author:** ![John\_W.Kennedy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_w.kennedy/32/1031_2.png) [@John\_W.Kennedy](https://boards.straightdope.com/u/John_W.Kennedy)\ **Post date:** [November 21, 2004, 9:19pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/27 "2004-11-21T21:19:10Z")
> [@Podkayne](#): > > The reason that we can see 59% of the surface (libration) is that the Moon’s speed in its orbit is not constant because its orbit isn’t a perfect circle. Entirely? No precession, nutation, _etc.?_ ---
**Author:** ![Trunk](https://avatars.discourse-cdn.com/v4/letter/t/87869e/32.png) [@Trunk](https://boards.straightdope.com/u/Trunk)\ **Post date:** [November 22, 2004, 6:03pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/28 "2004-11-22T18:03:38Z")
Could the earth fall into “synchronous” rotation with the sun? Has that already happened with Mercury? Or is it settling into such motion? Doesn’t Mercury have days that are longer than its years? ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [November 22, 2004, 8:20pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/29 "2004-11-22T20:20:30Z")
> [@](#): > > Could the earth fall into “synchronous” rotation with the sun? > > Has that already happened with Mercury? Mercury is a very interesting case. A few decades ago, it was believed that it was locked, but in actuality, it’s in a resonant lock. Mercury’s rotation period is exactly two thirds its orbital period. This works because Mercury’s orbit is particularly eccentric: It has a permanent elongation, and every orbit when it’s closest to the Sun (and therefore, the tidal forces are strongest), the long axis lines up pointing towards or away from the Sun. If anything, though, Venus is even more interesting. It’s almost locked to the Sun (its rotation period is close to the same as its orbital period), but it’s also got a resonance with the Earth. Whenever Venus passes closest to the Earth in our orbits, it always presents the same face towards us. ---
**Author:** ![Podkayne](https://avatars.discourse-cdn.com/v4/letter/p/edb3f5/32.png) [@Podkayne](https://boards.straightdope.com/u/Podkayne)\ **Post date:** [November 22, 2004, 9:15pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/30 "2004-11-22T21:15:07Z")
> [@John W. Kennedy](#): > > Entirely? No precession, nutation, _etc.?_ Yes, contribute to libration also. My major point was that libration is not caused by a difference between the Moon’s rotational period and revolution period. ---
**Author:** ![Trunk](https://avatars.discourse-cdn.com/v4/letter/t/87869e/32.png) [@Trunk](https://boards.straightdope.com/u/Trunk)\ **Post date:** [November 23, 2004, 3:30pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/31 "2004-11-23T15:30:50Z")
> [@Chronos](#): > > Mercury is a very interesting case. A few decades ago, it was believed that it was locked, but in actuality, it’s in a resonant lock. Mercury’s rotation period is exactly two thirds its orbital period. This works because Mercury’s orbit is particularly eccentric: It has a permanent elongation, and every orbit when it’s closest to the Sun (and therefore, the tidal forces are strongest), the long axis lines up pointing towards or away from the Sun. Wha? If it’s rotation period is exactly 2/3 it’s orbital period, then when it makes one orbit, it has rotated through 240 degrees. Are you saying that because it has such a stretched out orbit, that the 2/3 and 1/3 of an orbit fall exactly at the points of the elongated orbit where its closest. [Here](http://www.fellspointchess.com/images/mercury.html) is what I’m visualizing you’re saying in a crude drawing. I highlighted one side of mercury just to indicate it’s orientation. “T2” in the first diagram corresponds to “T2” in the second diagram. > [@Chronos](#): > > If anything, though, Venus is even more interesting. It’s almost locked to the Sun (its rotation period is close to the same as its orbital period), Like the moon to the earth? > [@](#): > > but it’s also got a resonance with the Earth. Whenever Venus passes closest to the Earth in our orbits, it always presents the same face towards us. That sounds incredible. That point at which we’re closest must “move” every year, say in relation to where the Earth is during our equinoxes. ---
**Author:** ![Trunk](https://avatars.discourse-cdn.com/v4/letter/t/87869e/32.png) [@Trunk](https://boards.straightdope.com/u/Trunk)\ **Post date:** [November 23, 2004, 3:32pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/32 "2004-11-23T15:32:28Z")
> [@Trunk](#): > > Wha? > > If it’s rotation period is exactly 2/3 it’s orbital period, then when it makes one orbit, it has rotated through 240 degrees. I mean 540 degrees, or 3 \* 180 degrees, so that it is exactly flipped every time it returns to the starting point. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [November 23, 2004, 7:12pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/33 "2004-11-23T19:12:37Z")
> [@](#): > > That sounds incredible. That point at which we’re closest must “move” every year, say in relation to where the Earth is during our equinoxes. It does move every year. But if we neglect eccentricity (which is relatively small for Venus and Earth), it’s always the same amount of time between closest approaches (this period is called the synodic period of Venus relative to Earth). So the planet can accomplish this by rotating at a constant rate (provided it’s just the right constant rate, which it is). Venus isn’t completely locked to the Sun, though, like the Moon is to the Earth. For the Moon, the two periods aren’t just close to equal, they’re exactly equal, and the Earth doesn’t rise or set (except for a small region at the edges, due to libration). For Venus, this is not quite the case, so the Sun does rise and set over any given point on Venus. Presumably, the Sun was trying to lock Venus, but as Venus’ rotation slowed, it reached the point where it had that resonance with Earth, and decided it liked that better. And Mercury is, in fact, exactly flipped whenever it reaches the starting point. Don’t think of one side of the planet being “special”, think of it as shaped like a football (it actually is, to a very small extent). When the pointy end of a football is pointing away from the Sun, the other pointy end is pointing towards the Sun, so it doesn’t matter that it’s flipped. Your diagram is a bit misleading, since it seems to show two separate points which are closest to the Sun. If you take care to draw a true ellipse, and put the Sun at one focus, then there will be one unique closest point, at one of the pointy ends of the ellipse. ---
**Author:** ![Trunk](https://avatars.discourse-cdn.com/v4/letter/t/87869e/32.png) [@Trunk](https://boards.straightdope.com/u/Trunk)\ **Post date:** [November 23, 2004, 7:57pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/34 "2004-11-23T19:57:27Z")
Oh, OK. I out-thought myself. I was thinking that if the rotation period was 2/3 it’s orbital period, then it was pointing in a different direction at the same point around the eclipse it traces. What I overlooked (even though I drew it) was that it points in the exact opposite direction for every one of it’s orbital periods, i.e., it’s always exactly 1.5 days later. ---
**Author:** ![rwjefferson](https://avatars.discourse-cdn.com/v4/letter/r/a6a055/32.png) [@rwjefferson](https://boards.straightdope.com/u/rwjefferson)\ **Post date:** [November 30, 2004, 11:00pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/35 "2004-11-30T23:00:00Z")
> [@Podkayne](#): > > The reason that we can see 59% of the surface (libration) is that the Moon’s speed in its orbit is not constant because its orbit isn’t a perfect circle. One [Moonth](http://antwrp.gsfc.nasa.gov/apod/ap010218.html) rwj ---
**Author:** ![user\_hostile](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/user_hostile/32/474_2.png) [@user\_hostile](https://boards.straightdope.com/u/user_hostile)\ **Post date:** [December 1, 2004, 1:58pm UTC](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486/36 "2004-12-01T13:58:59Z")
**rwjefferson** Now that was cool! The Man in the Moon is winking at you! 🆒 [Previous page](https://boards.straightdope.com/t/same-side-of-the-moon-always-faces-the-earth-the-moons-recession-rate/275486.md?page=1)