R**2 is Fortran too, idnit?
(YourFortranKnowledge.gt.mine)
Actually C unfortunately doesn’t have any notation for exponentiation, other than the pow() function. The ^ is used for bitwise exclusive or, as I recall. C++ might have an exponentiation operator; I’m not familiar enough with it to say.
But since we have a superscript here, we might as well use it, since it’s prettier that way.
mea culpa. I started to use it, but it doesn’t show up on preview for me for some reason (I thought I might not have some button set), and I couldn’t find it in the online vB documentation. I see it worked anyway.
Revenons a nos moutons.
As far as I’m concerned, the issue is settled. RM Mentock is correct in stating that the centrifugal force has no contribution to tidal bulges - c.f. in an earth-centered rotating reference frame gives an equal force all the way around the equator of the moon, exactly equal to the c.f. you’d calculate just due to the once a month rotation of the moon by itself. The tidal bulges must be explained by differential gravity alone.
RM Mentock asks me:
Well, that’s another detail for which I lack the math.
dtilque points out that one of Saturn’s moons (Hyperion) is not tide-locked, but tumbles. I’d add that (as far as I know) nobody has looked closely enough at any of Jupiter’s small distant `irregular’ moons to know how they rotate, so it would be most accurate to say that we know of only one moon which is not tide-locked.
Hey, that doesn’t seem to be accurate at all. 
The next post after dtilque’s points out Saturn’s moon Phoebe as well.
I need to think about this some more. Changing reference frames always gives me a headache.
I have always said that you don’t need c.f. to explain tides; all you need is differential gravity. With no rotation or revolution, there would still be two tidal bulges; the way to imagine this is to have three rocks in a line pointed at the Earth and let them fall. The rocks will separate, due to the different forces on them. As judged from the center rock, the inner one will draw away toward the Earth, and the outer one will draw away away from the Earth. Since the Moon is in freefall around the Earth, you get the same thing. We’re done.
Having said that, has anyone seen the NOAA page about this?
They have c.f. all over the place. I think they’re wrong, and, worse, overly complicated. When I was writing my book, I started up the tides chapter trying to use the c.f. arguments, got about 5000 words into it, and scrapped the idea. I went back to my tried-and-true differential gravity.
This stuff can be fiercly complicated. That’s why Newton never solved it.
Crap. PREVIEW, PREVIEW, PREVIEW. Here’s the correct link to the NOAA page.
I fixed the first link for ya, Bad
Their diagram is a little more complicated, because they’re looking at tidal forces on the earth instead of the moon. One thing that complicates this is that the CG of the earth-moon system (the barycenter) is within the radius of the earth. However, the math is still the same. Each point will feel a centrifugal force (in the reference frame centered around the barycenter) of two vector components: the (constant) c.f. that the center of the earth feels, and one equal to the point’s position on the earth. The only variable is exactly equal to a once-a-month rotation of the earth. There is no force that’s not symmetric, which could stretch the earth along one axis, except for gravity. I’m pretty confident that the NOAA site got it wrong.
Excellent article, Chronos. 
But I want to add one small thing to what others said earlier in this thread disputing the claim that “All of the moons in the solar system are synchronized, or tidally locked, with their primary planet, and in the case of Pluto and its moon Charon, the planet is locked to its moon as well.”
Not only are many of the smaller moons not synchronized but even some large moons aren’t. The best example of this is undoubtedly Triton, the largest moon of Neptune. It orbits retrograde; several small moons of Jupiter and Saturn also do this, but Triton is unique in being such a large body orbiting retrograde. This means it orbits against the rotation of its parent, which puts tremendous strain on it – retrograde orbits are less stable than regular ones, and for the same reason that tidal forces tend to produce synchronous rotation.
Also, it is worth mentioning that the strains on Io are not just produced by the tidal lock between Io and Jupiter. Indeed, the tidal lock ought to reduce the strain somewhat. In addition to Jupiter-Io tides, however, Io is affected by the gravitation forces of its siblings. In addition to the fact that all four Galilean moons (Io, Europa, Ganymede, and Callisto) are tidally locked with Jupiter, there are also orbital resonances between some of them. Io orbits Jupiter exactly twice for every one Europa orbit, and Europa orbit Jupiter exactly twice for every one Ganymede orbit. Because of this resonance, Io wobbles as it goes around Jupiter, and this exacerbates the tidal strains on it. Io has tides as much as 100 meters!
This is why the Moon does not have a liquid interior, although Io, Europa, and Ganymede all do. There is no second large satellite of Earth to perturb the Moon’s synchronous rotation.
Triton is synchronized. Its orbital period is the same as its rotational period.
The moon’s synchronous rotation is “perturbed” by quite a bit, just because it is not a circular orbit. The rocking back and forth is known as libration.
And I admit, by the way, that I was wrong on the “all moons are synchronized” bit. I was going to put “almost” in there, but somewhere between my brain and the keyboard, it slipped out.
So wrong it’s ridiculous. That series of articles has been patched and reworked, until it doesn’t even make sense.
In the text of their chapter two–which also appears at the end of their chapter one, or introduction, they talk about the sublunar and subsolar points as if those are the only places on Earth where the tide is high–and that’s why there are two high tides! In the next paragraph, it makes a reference to the greater centrifugal force on the side farthest from the moon–contradicting their diagram and discussion in chapter three–the part linked before by the BA.
Still, chapter three begins the discussion of centrifugal force by saying “It is this little known aspect of the moon’s orbital motion which is responsible for one of the two force components creating the tides,” so it would appear to support arguments against what I’ve been saying. That’s probably why the BA posted the link.
But things get a little confused. It doesn’t have any computation, and it labels the centrifugal component F[sub]c[/sub] in the figure, but in the blue box it says it is labeled F[sub]g[/sub]. And it says “And, since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon,” without much explanation. Then it says “It is important to note that the centrifugal force produced by the daily rotation of the earth on it axis must be completely disregarded in tidal theory. This element plays no part in the establishment of the differential tide-producing forces.” So, that appears to support my position.
The next paragraph is key:
Besides the fact that the arrows actually do not appear to be the same length, the net result of that analysis is that the centrifugal force is shown to be constant across the entire Earth, except for that due to its rotation, which is ignored, pretty much what I’ve said all along (nevermind that this contradicts their chapter two!).
The bottom line is that an analysis in a different frame can produce centrifugal forces. And the results, in that frame, are equivalent to the results determined in any other frame–but that is not the same as saying that the comments that Chronos made are equivalent.
This NASA page shows a time-lapse sequence of the moon in which it appears to show libration.
I’m just finishing up Timothy Ferris’s book Seeing in the Dark. I highly recommend it–I discovered new and interesting things about subjects I’d already looked into a dozen other places.
In one of the appendices, he has a table of the known satellites of the planets. He lists less than half as synchronous, and in addition to Saturn’s Hyperion and Phoebe, lists four of Jupiter as definitely not synchronous: Himalia, Lysithea, Ananke, and Sinope. Nereid, a satellite of Neptune, is also listed as not synchronous. Except for Hyperion, they are all fairly far away from their planet–at least, the ones we know about.