lunar rotation

[ace22]

Just a quick question. I understand why lunar rotations sync up with their revolutions. Why don't planetary rotations do the same with regards to the sun? Why doesn't the same side of the earth face the sun all the time? Is the orbital pattern more complicated, or are the gravitational forces not strong enough...?

:confused:

[Chronos]

The gravitational force is stronger for the planets, but the tidal forces, i.e., the difference in gravitational forces on the near and far side of a planet, is actually weaker. That's because tidal forces fall off more quickly with distance than do gravitational forces, and the planets are all much further from the Sun than the moons are from the planets. For most of the planets, there just hasn't been enough time for a lock to develop. Mercury, as mentioned in the report, is sort of locked to the Sun, and Venus has a very slow rotation and is on its way to becomming locked, but it's not there yet. As for the Earth, we can't lock to the Sun while we still have the Moon, which has greater tidal effects on us: Even if we were to become locked, it'd be to the Moon, not to the Sun. All of the rest of the planets are too far out for the effects to be noticeable at all.

[Duck Duck Goose]

And the Staff Report being referenced is Why does the same side of the moon always face the earth? (http://www.straightdope.com/mailbag/mlunarrotate.html)

Unless of course you're talking about Cecil's column Why does the same side of the moon always face the earth?
(http://www.straightdope.com/classics/a2_112.html)

Glad to see that Cecil got it right. :D

[zimaane]

A quick question:

In the article it stated that:

"This constantly-changing distortion heats up the rock and causes energy to be lost from the rotation."

Does this mean that the heating of the rock causes energy to be lost from the rotation? If so, how?

[John W. Kennedy]

Heat is energy. Energy has to come from somewhere.

It's exactly the same way that brakes work. When you step on the brakes, your kinetic energy is converted into heat.

[stochastic]

I recently had an argument with one of my fellow high school science teachers about this subject. He was of the mind that lunar rotation synchrony was one of those grand coincidences of nature, and taught his students as such. I tried to convince him, unsucessfully, that it was due to the tidal effects as beautifully spelled out in the articles here. What's really distressing is he's been teaching Physical and Earth science for 30 years longer than I have and he should know this stuff better than I do.

However, one thing he claimed got me wondering: he says that as the moon recedes from the Earth the rotational period of the moon will remain unchanged, causing the synchrony to fade away, and the rotation of the moon will once again become apparent to those on Earth. (assuming, of course, that the solar system is still around by that time) Cecil seems to imply that the tidal lock will become even more stable as the moon drifts away. Is my friend even remotely correct or do I do the right thing and believe Cecil?

[RM Mentock]

I'm glad to see Chronos using the term centrifugal force in public. Nothing to be ashamed of.

I do have a small nit, though. In his article, he mentions that "centrifugal force gets stronger as you get further from the center, while gravity gets weaker". It seems that the center is the Earth, but in fact, in that reference frame, the centrifugal force doesn't increase.

[John W. Kennedy]

stochastic -> Your friend is an ignoramus and shouldn't be teaching science.

RM Mentock -> I'm not quite sure what you're saying. All other things being equal, a rotating system with a large radius exhibits a greater centrifugal force than one with a small radius. That's why astronaut-training centrifuges have long arms.

[RM Mentock]

I mean in the context of the moon. The moon rotates, of course, but slowly. This causes a certain amount of centrifugal force, but should it be considered in the Earth-moon system?

If the moon in a circular orbit were not rotating, then each particle of the moon follows a circular path in its orbit which has a different center, but the same radius and period. In that sense, the centrifugal force, even as a vector, is the same at all points of the moon.

I think zimaane's question had to do with the loss of rotational energy to heat--if rotational energy is lost, how the moon regains it.

[jezzaOZ]

Originally posted by Chronos
The gravitational force is stronger for the planets, but the tidal forces, i.e., the difference in gravitational forces on the near and far side of a planet, is actually weaker. On a tangent, why are tides equal height? If the force of gravity falls off with distance, you would expect the bulge on the moon side would be bigger than on the anti-moon side. You would also expect the sun induced tide to be much more symmetrical because the sun is further away and so the field is more uniform.

In fact, the sun and moon tides (M2. S2) are totally symetrical and perfect sine waves.

Why is this so ?

[CurtC]

zimaane wrote:
Does this mean that the heating of the rock causes energy to be lost from the rotation? If so, how?This is one of those little things that I don't like. People who work with physics have become comfortable enough with the idea of conservation of energy, that they use it as a shortcut to really understanding something. I postulate (the "CurtC Postulate") that conservation of energy will explain nothing to the layman. The real world works with forces and masses, and it so happens that energy is conserved, but our brains don't really understand it that way. If a ball reaches a certain speed after falling a certain height, saying that its potential energy was converted to kinetic energy, and using those measurements to calculate its speed, gives the correct answer and is quick, but doesn't really explain how the ball achieved that speed.

I think a good intuitive explanation of the tidal-forces-causing-slowed-rotation effect is this. At one instant in time, there will be a bulge toward earth. As the object rotates, that bulge will be pointed not exactly toward the earth anymore. But because of those tidal effects, the earth's gravity will be pulling slightly harder on the side with the bulge, which causes the rotation of the moon to slow down, and then the bulge will shift back towards the earth by moving the rock around. But while the moon was still rotating, the bulge location was always a little behind, so there was a slight force on the bulge that slowed the rotation.

[John W. Kennedy]

It's not the rotation of the Moon that's being talked about, but its orbit around the Earth. The point nearest the Earth is going slower (in miles per hour) than the center, and the point furthest from the Earth is going faster. Therefore, while the center has exactly enough centrifugal force to counteract the Earth's gravity, the point nearest the Earth doesn't have enough, while the point furthest from the Earth has too much.

On top of that, the point nearest the Earth receives more gravity from the Earth, and the point furthest from the Earth receives less, because gravity falls off with distance.

Therefore, for both reasons, the near side bulges in and the far side bulges out.

As to the symmetry, you have to work with both the gravity and the centrifugal force effects. I suspect the asymmetry cancels out -- if it didn't, the whole system would be unstable.

[RM Mentock]

Originally posted by John W. Kennedy
It's not the rotation of the Moon that's being talked about, but its orbit around the Earth.
Yes, that's what I mean. I think the rotation of the moon should be considered separately, but it's not. The rotation of the moon induces a bulge on all sides--the moon then is slightly oblate. Without that bulge, the reasoning--that the centrifugal force increases with distance from the Earth--fails, as the centrifugal force is the same at all points of the moon then.

That's why I don't really like the centrifugal force arguments for tides. What if there were no centrifugal force? What if the moon were free-falling to the Earth? The Earth's gravity would induce a tide then too--of a magnitude the same as it does now.

Well, the Earth is free-falling towards the Earth. And, as I point out, the centrifugal force is the same across the moon, so it doesn't contribute at all to the differential. Plus, we don't have to use the c word, as the BA doesn't in his explanation (http://www.badastronomy.com/bad/misc/tides.html).

As to jezzaOZ's question about the symmetry of the principal semidiurnal tidal components M2 (http://co-ops.nos.noaa.gov/glossary/gloss_m.html#M2) and S2, the intuition is correct. There is a lack of symmetry, but it gets complicated by the fact that the Earth/moon/Sun system is tilted--so that there are not only components like O1 (the lunar diurnal component), K1 (the lunisolar diurnal component), and P1 (the solar diurnal component) which together express the effect of the moon's and Sun's declination, but there are also M1 and S1. M1 and S1 are just a lot smaller, as you might expect, since the gravity fall-off at those distances can be approximated linearly (that is, the increase on one side of the Earth is approx. the same as the decrease on the other side).

Why are M2 and S2 symmetrical? Well, that's the definition. They're sorta like the coefficients in the second degree component of a power spectrum--it's the second degree component function that is symmetrical, not the tide itself.

Of course, that's not the complete story.

[bronto]

It was pointed out to me a year or two ago (in news:rec.arts.sf.science, i think) that Mercury's 2:3 lock is not arbitrary.

Mercury's orbit is the most elliptical of the major planets, and the solar tide across Mercury is therefore significantly stronger at perihelion, when Mercury is also moving most rapidly. So the sun as seen from Mercury sits roughly still around perihelion. (More precisely, it makes a little loop.)

Do the math (Cartesian geometry, algebra and perhaps a wee bit of first-semester calculus) and you'll find that 2:3 rotation will make the sun stop at perihelion of an orbit whose eccentricity is 1/5. Mercury's eccentricity is 0.2056; the difference is presumably due to tidal effects on the rest of the orbit, which though smaller are not zero.

What I'd like to know, and lack the math skilz to find out, is: what are the attractors? If you run your Solar System Simulator with a million different initial settings for Mercury's rotation and eccentricity, what does it look like after a billion years? For appropriate eccentricity, is 3:5 as stable as 2:3? How about 4:7? How about 401:701?

[RM Mentock]

How are you measuring the amount of stability? And what is the stability of 2:3?

[John W. Kennedy]

Without that bulge, the reasoning--that the centrifugal force increases with distance from the Earth--fails, as the centrifugal force is the same at all points of the moon then.

No, because the Moon is not of zero thickness. The far side has to move faster than the near side, because it's moving in a larger ellipse. On the simplifying assumption that the Moon's orbit is a circle, every month, the near side travels about 2,404,300 km, and the far side about 2,426,200 km.

[RM Mentock]

Originally posted by John W. Kennedy
No, because the Moon is not of zero thickness. The far side has to move faster than the near side, because it's moving in a larger ellipse.
Only if it's rotating. That's my point.

[dtilque]

from the Staff Report
In fact, this is the situation of moons in general: All of the moons in the solar system are synchronized, or tidally locked, with their primary planet
Not quite true. One of Saturn's moons (and I can't remember which one) rotates chaotically.

But otherwise, I liked your article, Chronos. I thought it better than Cecil's on the same subject.

[RM Mentock]

Originally posted by dtilque
Not quite true. One of Saturn's moons (and I can't remember which one) rotates chaotically.
Hyperion (http://www.planetscapes.com/solar/eng/hyperion.htm), but there's also Phoebe (http://www.planetscapes.com/solar/eng/phoebe.htm).

[John W. Kennedy]

Only if it's rotating. That's my point.

No, it's in orbit. Not to mention that it is rotating. I don't know of any body in the Solar System that isn't.

[marcos alvarez]

I would like to point out a conceptual mistake in Chronos' description of orbits. He says that "An orbit can be considered as a balance between gravity and centrifugal force." That is incorrect. If that were so the resulting force would be zero and therefore the acceleration of the orbiting mass would be zero as well (Newton's second law). Uniform circular motion is, however, accelerated motion: the velocity vector changes in time. Contradiction.

The correct description of a circular orbit is thus: there is only one force, namely the gravitational attraction F. We have an expression for this force F, the Newtonian law of gravitation. Once we know that the orbit is circular we can use geometry to prove that the acceleration of the orbiting mass is a=v^2/R where v is the tangential velocity and R the orbit radius. Equating F=ma gives the equation of the orbit. Only one force, hence the accelerated (circular) motion. This is the correct description based on Newtonian mechanics in the inertial frame where the source of the gravitational force is at rest. Non-circular orbits are not much harder, but let's keep it simple.

Chronos' mistake, which appears in many textbooks, originates in that m v^2/R is also the expression for the centrifugal force felt by a rotating observer in the absence of gravity. Did he mean to
describe the orbit in the frame of reference of the orbiting mass, the only frame where the centrifugal force exists?. Maybe, but that is a non-inertial frame, where Newtonian Mechanics doesn't apply. Things are a lot harder then. For example one has to argue that free gravitating masses move in precisely the right way for the centrifugal force that exists in their rest frame to cancel the gravitational force. That is, one has to invoke a form of Einstein's equivalence principle, even though we are only working out a simple nonrelativistic problem.

I am under the strong impression that Chronos really believes that the orbits of a free gravitating mass seen from an inertial frame are a balance between two opposing forces, even though one of them only exists in non-inertial frames. Perhaps he can clarify this point in his own words.

[RM Mentock]

Originally posted by John W. Kennedy
No, it's in orbit. Not to mention that it is rotating. I don't know of any body in the Solar System that isn't.
I know it's in orbit. :)

But the difference in centrifugal force from the near side to the far side can be completely attributed to its rotation. If you only look at the contribution from its orbit, the difference is zero.

[John W. Kennedy]

No, the rotation of the Moon is immaterial. At any given instant, the point on the Moon facing the Earth is moving more slowly (linearly) than its opposite point on the other side of the Moon.

[RM Mentock]

Originally posted by John W. Kennedy
No, the rotation of the Moon is immaterial. At any given instant, the point on the Moon facing the Earth is moving more slowly (linearly) than its opposite point on the other side of the Moon.
Why is it immaterial?

If the moon were not rotating, then each point of the moon has identical instananeous "linear" velocity. If you say that's not true, then we have something to discuss, to clear up the misunderstanding. But you have to consider a case where the moon is not rotating, to do that.

[Chronos]

RM Mentock: I was considering the center to be at the Earth. Given that, the "top" point of the Moon is experiencing a larger centrifugal force than the "bottom" point. While it's true that the Moon itself must be rotating for this distinction to be meaningful, I'm not sure that that's relevant.

marcos alvarez, it depends on what reference frame you're using. If one is working in an inertial reference frame, then you're correct, one should not be using centrifugal force. In an inertial reference frame, the only force is gravity, which is the centripetal force, and since there is a net force, there is an acceleration.

However, I was not working in the inertial frame; I was working in the (non-inertial) co-rotating frame. In that frame, the Moon is not moving, and remains at rest due to the balance of two forces, the gravitational force and the centrifugal force. Since we're in a non-inertial frame, we need that centrifugal force.

With regards to the energy versus force methods for explaning the effect, it may just be a reflection of my training, but I find the conservation of energy explanation to be more intuitive. If you prefer to use the forces instead, go ahead.

And as for tidal bulges, for small bulges, the asymmetries cancel out, but for large bulges, they don't, even in simple two-body systems. In the exreme case, an object will be distorted into a sort of teardrop shape called a Roche lobe, with a pointy cusp pointing towards the other object. This is often the case in binary star systems where one has become a red giant.

[John W. Kennedy]

I really don't know how to explain this any better without a picture. http://pws.prserv.net/jwkennedy/moonorbit.png

The red circle is larger than the green circle.

[RM Mentock]

Originally posted by Chronos
RM Mentock: I was considering the center to be at the Earth. Given that, the "top" point of the Moon is experiencing a larger centrifugal force than the "bottom" point. While it's true that the Moon itself must be rotating for this distinction to be meaningful, I'm not sure that that's relevant.
What would it take before it were relevant?

John W. Kennedy, thanks for the graphic! I didn't intend for you to go to that work, I'd produced a graphic of my own a while back and it seems I've lost track of it. I'll see what I can dig up.

[RM Mentock]

Originally posted by Chronos
And as for tidal bulges, for small bulges, the asymmetries cancel out
I missed that comment. They don't completely cancel out, see my post above answering jezzaOZ.

I reread the thread, and I realized I hadn't been as specific as I thought. If you calculate the centrifugal force on the points of the moon using John W. Kennedy's diagram, you notice that the centrifugal force is proportional to distance. However, part of the centrifugal force on the surface of the moon is due to the rotation of the moon--not its orbit around the Earth. If you calculate that, and subtract it, you find that the result is that the centrifugal force left is constant, no variation with distance from the Earth!

If you draw another diagram where the moon orbits the Earth without rotation, it's apparent that each point of the moon follows a similar shaped orbit, but around an offset center. That offset is enough to cause the centrifugal force not to vary by distance from the Earth.

[Chronos]

OK, I think I see what you're getting at, RM, and yes, you can analyze it that way, but it seems to be more compicated conceptually for giving the same result. As for the symmetry of the bulges, what I should properly have said is that the low-order terms are symmetrical, but the higher order terms are not. Technically, those higher order terms are always there, but for small bulges, the symmetrical portion is much more significant than the asymmetric part.

[RM Mentock]

Originally posted by Chronos
OK, I think I see what you're getting at, RM, and yes, you can analyze it that way, but it seems to be more compicated conceptually for giving the same result.
Saying they give the same result is kinda like saying that the asymmetries cancel out. They don't. :)

What if the moon rotated seven times faster than it does now? In your analysis, would you back out seven revolutions, or six? If you back out the rotation of the moon, then your claim is not true. Your analysis may have some heuristic value, but it is certainly wrong. You should have consulted Cecil on this (http://www.straightdope.com/classics/a5_199.html).

[MonkeyMensch]

Since it looks like this thread is active I thought I'd post a related question that has been on my mind from time to time.

Is the abundance of mare on the near side of the Moon and the absence of mare on the far side realted to the uneven distribution of mass in the Lunar body? It's hard to believe it could be a coincidence.

If the Lunar highlands (non-mare) are similar to continental formations on Earth and the mare similar to oceanic basalts then this would indicate the center of mass is on the sub-Earth side of the Moon.

Just curious 'cause I've never read any sort of reference to this sort of thing.

[RM Mentock]

Originally posted by MonkeyMensch
If the Lunar highlands (non-mare) are similar to continental formations on Earth and the mare similar to oceanic basalts then this would indicate the center of mass is on the sub-Earth side of the Moon.
More or less. There are a lot of webpages out there that explain the tidal locking of the moon by saying that this displacement of the moon's center of mass is towards the Earth, and that is why it is "locked." The center of mass is displaced a large amount (about 2 kilometers) but this website (http://history.nasa.gov/EP-95/orbital.htm) says it's towards a point midway between Mare Serenitatis and Mare Crisium--probably not where you expected? (Moon map (http://www.oarval.org/MoonMapen.htm))

[Squink]

Here's a bit of the back side of the moon revealed to earth dwellers by a favorable libration: Mare Orientale (http://skyandtelescope.com/observing/objects/moon/article_723_1.asp)

[John W. Kennedy]

However, part of the centrifugal force on the surface of the moon is due to the rotation of the moon--not its orbit around the Earth. If you calculate that, and subtract it, you find that the result is that the centrifugal force left is constant, no variation with distance from the Earth!First, this cannot even be calculated without specifying how fast the Moon is rotating. Second, the centrifugal force due to rotation at the innermost point (assuming that everything's in the same plane, which is close enough for the present) is toward the Earth, reducing (even negating, if the Moon is rotating a great deal faster than it in fact is) the already weaker centrifugal force due to the orbit at that point. Third, the centrifugal force due to rotation on the outermost point is away from the Earth, increasing the already stronger centrifugal force due to the orbit.

Innermost point: Orbital centrifugal force is away from the Earth, but is less than at the center. Rotational centrifugal force is toward the Earth. Gravity toward the Earth is more than at the center.

Center: Orbital centrifugal force exactly balances gravity. Rotational centrifugal force is zero. Gravity toward the Earth exactly balances orbital centrifugal force.

Outermost point: Orbital centrifugal force is away from the Earth, but is more than at the center. Rotational centrifugal force is away from the Earth. Gravity toward the Earth is less than at the center.

All three factors, orbital centrifugal force, rotational centrifugal force, and gravity, draw the innermost point in and drive the outermost point out.

[RM Mentock]

Originally posted by John W. Kennedy
First, this cannot even be calculated without specifying how fast the Moon is rotating.[b]
We do know how fast the moon is rotating.
[b]Second, the centrifugal force due to rotation at the innermost point (assuming that everything's in the same plane, which is close enough for the present) is toward the Earth, reducing (even negating, if the Moon is rotating a great deal faster than it in fact is) the already weaker centrifugal force due to the orbit at that point.
It is not "already weaker". That's my point. Without the effect of the moon's rotation, there is no variation in centrifugal force across the body of the moon.

Trace the path of the points on the moon as it orbits. If the moon does not rotate, but translates, each point will describe curves that are identical but displaced from each other by a traslation. The reaultant vectors are identical. So, if you back out the rotation, there is no variation of the centrifugal force across the body of the moon.

[John W. Kennedy]

We do know how fast the moon is rotating.We know how fast it's rotating now, but the original point of the question was how it got that way.

Trace the path of the points on the moon as it orbits. If the moon does not rotate, but translates, each point will describe curves that are identical but displaced from each other by a traslation. The reaultant vectors are identical. So, if you back out the rotation, there is no variation of the centrifugal force across the body of the moon.Yes, but now you are describing a situation that never existed, will never arise under natural conditions (because the locked configuration is stable), and, even if it were created, would be unstable (because, unless the Moon were infinitely rigid, purely gravitational tidal effects would drag the Moon around until the locked configuration was reached).

[RM Mentock]

Originally posted by John W. Kennedy
We know how fast it's rotating now, but the original point of the question was how it got that way.
All I'm talking about is the analysis, not whether the moon lags, leads, or is locked. The point is, it is rotating, and that rotation has nothing to do with its orbit. It could be rotating ten times as fast and the orbit would not change significantly.

You have to back out the effects of that rotation whether it is rotating ten times as fast, just as fast, or not at all. If it were not at all, it would be easy.

That rotation also produces a bulge on the left and right sides of the moon--the sides from the point of view from the Earth. And that bulge is the same as the bulge produced by the rotation on the front and back sides.

[John W. Kennedy]

...that bulge is the same as the bulge produced by the rotation on the front and back sides....which only suggests that the rotational centrifugal force has no net 1st-order effect on tidal braking at all.

[RM Mentock]

Originally posted by John W. Kennedy
....which only suggests that the rotational centrifugal force has no net 1st-order effect on tidal braking at all.
Not only. It also suggests that the orbital centrifugal force has no net 1st-order effect on tidal braking, or on the generation of tides.

[CurtC]

RM, could you do me a favor and explain what it is you're saying? Your points up to now have been a little too fragmented for me to understand, but I have a great deal of respect for your points in general. Chronos's tides explanation using a co-rotating reference frame and centrifugal force seemed satisfactory to me, and I can't figure out what you're point is about the lunar rotation about its axis.

[RM Mentock]

Thanks for the faith CurtC!

My point, if I have one, is that the rotation of the moon really doesn't have anything to do with the shape of the tide on the moon or the moons orbit around the Earth. So we should back out, or ignore, any contribution of that rotation.

If you draw a diagram of the moon orbiting the Earth, without rotation of the moon, you'll see that a point on the farside of the moon ends up on the nearside after one half orbit. Assuming a circular orbit of the moon, if you trace the path of any point, its orbit is a circle of size identical to every other point. Therefore, they experience identical centrifugal force in magnitude, and it turns out that the direction is the same as well.

So, the only reason that there is a difference in centrifugal force from the farside to the nearside is because the moon is rotating! But that rotation causes a bulge on all sides of the moon, around its equator. We shouldn't consider it part of the tide anymore than the 20 kilometer bulge of the Earth is part of the tide. They certainly shouldn't be considered in any explanation of why the moon doesn't fall to the Earth. And that's what Cecil said (http://www.straightdope.com/classics/a5_199.html).

It's not just a matter of two equivalent explanations.

[CurtC]

Cecil said that the centrifugal force explanation is wrong. Chronos said that it's satisfactory. I, too, find that CF is an OK explanation.

If you look from the co-rotating frame with earth at its center, then the only way to do so easily is in the case when the moon always has the same side facing earth. Fortunately that's the case we have nowadays. In that case, the earth "grabs" the moon at its center. On the far side, the CF is stronger than gravity, so it tends to make a bulge on the far side, and the gravity in the moon keeps that material from being flung off. On the near side, the earth's gravity is stronger than the CF, so it bulges out too.

In neither of these did I feel it was helpful to consider the equatorial bulge around the moon due to its once a month rotation. Are you saying that this would have a significant effect?

I have to say that I disagree with Cecil that the CF explanation is wrong. It seems that the two are equivalent, just different reference frames.

[RM Mentock]

Originally posted by CurtC
In neither of these did I feel it was helpful to consider the equatorial bulge around the moon due to its once a month rotation. Are you saying that this would have a significant effect?
Significant? It's the only effect. That is, the equatorial bulge is what you guys are talking about!

That's my point, made another way.

I too do not have a problem with the term centrifugal force. That's not my objection at all. But the centrifugal force does not contribute to any front-to-back difference in force that is not also experienced side-to-side.

[RM Mentock]

That's why I agree with Cecil. :)

[John W. Kennedy]

But it's not the rotational centrifugal force that _causes_ tidal braking, it's the combination of gravity and orbital centrifugal force. And the orbital centrifugal force is there in all cases except your hypothetical and unstable non-rotating model.

(Of course, it is possible to analyze the entire system without employing the fictional "centrifugal force" at all. But we're not doing it that way.)

[RM Mentock]

Originally posted by John W. Kennedy
But it's not the rotational centrifugal force that _causes_ tidal braking, it's the combination of gravity and orbital centrifugal force. And the orbital centrifugal force is there in all cases except your hypothetical and unstable non-rotating model.
Tidal braking is caused by the tidal bulge being rotated forward, causing an asymmetry. It is balanced, but not quite because of the added distance, by the tidal bulge on the backside. In order to have tidal braking, the asymmetry is necessary. However, the centrifugal bulge that you guys are talking about is completely symmetrical about the moon's equator. You just haven't computed it for the sides, yet, to realize that.

The centrifugal bulge that you guys are talking about can be computed two ways. Your way (except that ignores the sides, so far) and my way. My way shows that it is a sum of a non-rotating moon (completely constant centrifugal force across the body of the moon) and an equatorial bulge due to the moon's rotation that is completely symmetrical. Your way will show the same thing, go ahead and try it--similar triangles will give you two components at each side, one component equal to the force at the center of the moon, and the other equal to the differential component at the near and far side, all directed away from the center of the moon.

That asymmetrical tidal bulge is not caused by centrifugal force--your way or my way. The tidal bulge is only due to gravitational stretching.

[CurtC]

RM, I think the only one who's even mentioned the equatorial bulge, due to the moon's rotation, is you. The bulges we're talking about (I am, anyway) are the ones that make the moon have a football-like shape (an American football), or an ellipsoid, with one end pointing toward the earth and the other away from it. I'm sure that the moon's z-axis is compressed slightly due to rotation, but I've been assuming that I could neglect that.

I completely agree with what you said at the first of that most recent post, that tidal braking is caused by the asymmetry of the gravity on the front and back (slightly rotated) tidal bulges. I don't see why it's necessary to bring the centrifugal bulge into the discussion, because it's tiny, and its effects will be cancelled by symmetry.

Oh, and I'm not sure whether the front and back tidal bulges would be the same size. It seems to me that the front one might be larger. But in either case, even with symmetric bulges, there would be braking, because the earth's tug on the front bulge is stronger than the tug on the back side one. This differential tug slows the moon's rotation. I still don't see a problem with the centrifugal force explanation.

[RM Mentock]

Originally posted by CurtC
RM, I think the only one who's even mentioned the equatorial bulge, due to the moon's rotation, is you. The bulges we're talking about (I am, anyway) are the ones that make the moon have a football-like shape (an American football), or an ellipsoid, with one end pointing toward the earth and the other away from it. I'm sure that the moon's z-axis is compressed slightly due to rotation, but I've been assuming that I could neglect that.
z-axis? OK, let's concentrate on a disk through the moon in the plane of the orbit of the moon. Is that z=0? The centrifugal force is w^2R, right, where R is the distance to the point and w is the rotational speed. The force is directed out along the line connecting the center of the orbit with the point.

For the point on the near-side, the force is w^2(R-r) where r is the radius of the moon. That's just the value of force at the moon's center minus w^2r. For the point on the far-side, it's fairly clear that it would be just the value of c.f. at the moon's center plus w^2r. In both cases, you can consider the result to be w^2R plus a force of w^2r directed away from the center of the moon.

Now do your analysis for a point at the side. The force directed away from the Earth and is w^2sqrt(R^2+r^2), or slightly larger than at the moon's center. It can be resolved into two components--one which is identical to the centrifugal force at the center of the moon, and one which is equal to w^2r, directed away from the center of the moon. Same as for the far and near side points.

So, every point of the moon, by your way of analysis, experiences a centrifugal force equal to that at the center of the moon, plus a component equal to w^2r, where r is its distance from the rotational axis, directed away from the rotational axis.

In other words, the bulges that you guys are talking about are the same bulges that I'm talking about! I'm the only one who's mentioned the "equatorial bulge" by name, but it is the same thing.

If I only had a whiteboard... :)

I completely agree with what you said at the first of that most recent post, that tidal braking is caused by the asymmetry of the gravity on the front and back (slightly rotated) tidal bulges. I don't see why it's necessary to bring the centrifugal bulge into the discussion, because it's tiny, and its effects will be cancelled by symmetry.
Well, I too think we shouldn't be talking about the centrifugal bulge. That's been my point all along.

Oh, and I'm not sure whether the front and back tidal bulges would be the same size. It seems to me that the front one might be larger. But in either case, even with symmetric bulges, there would be braking, because the earth's tug on the front bulge is stronger than the tug on the back side one. This differential tug slows the moon's rotation. I still don't see a problem with the centrifugal force explanation.
See, the centrifugal bulge that you guys are talking about is symmetric around the equator of the moon. There's no asymmetry that can cause tidal slowing.

[pHabala]

To RM and John W:
Guys I think you can't agree because you are missing a point. The bulge is irrelevant and the rotation as well. John cannot disregard non-rotating moons, because when the Solar system started, pieces of dirt flew around in any way they pleased. The question is exactly how did it happen that they locked, so we have to start from a general situation, not a stable one.

RM makes one mistake when he says that in a non-rotating Moon, all points have the same path, ergo the same "force". This is a confusion of notions. Force is a thing that exists only in a specific time moment. It does not add over time (when you add force, you are talking energy, and energy has much less information in it then force, which makes it unsuitable for deeper analysis). RM in fact says that all particles of a non-rotating orbiting Moon have the same total energy output/input after one revolution of Moon. This is true. However, their energy balance changes during the rotation, and not in the same way for all of them! At the time when some particles have "energy downs", others have "energy ups", which means we are talking certain time points, which means we are talking forces, and that is exactly where things start happening.

Look again at the picture of your non-rotating Moon, at a certain specific time. Consider pieces of the Moon on the extreme ends from the point of view of Earth: the furthermost and closest. Then consider a "cut" of Moon that connects them, sort of a tube. This tube moves around, so as whole it is stable (at least the point at its center). Sit for a moment at its center. The end that is further from the Earth is experiencing pull away from you, in the direction away from the Earth, and the end which is closest to the Earth experiences pull towards the Earth. This therefore tends to position the stick exactly so that it is pointing away (or towards) the Earth. Now imagine how this stick behaves as it moves around in a non-rotating Moon (or a Moon which rotates too slowly/too fast). It does not stay in the "pointing away" position, it rotates with respect to the Earth. For instance, in your non-rotating Moon, after a quarter of orbit, this stick is actually level with the Earth's surface. This trend goes against the pull of forces, therefore we see that the stick has the tendency to stay in the radial position, in other words, your non-rotating Moon will have a tendency to start rotating. In the same way, a Moon that rotates too fast will have a tendency to slow down. This analysis works for the whole moon, because you can imagine it, at any given time, asa collection of radial (with respect to Earth) sticks and all fo them will have tendency to say locked.
I am aware that this explanation is lacking in some respects (momentum of a rotating mass etc), but I have a feeling that it at the same time illuminates the most crucial point. At least I think it clear up some misunderstanding and you ahve a chance to clear up on some other points as well. Or I am dead wrong, wouldn't be the first time :-).

Have a nice non-rotating day, pH.

[RM Mentock]

Originally posted by pHabala
This trend goes against the pull of forces
Can you calculate those forces? I think if you do, you'll find that their genesis is from the stretching of gravity. That's all I'm trying to say.

[CurtC]

RM, I a whiteboard isn't necessary now - I drew a diagram as I was reading your post, and I'm pretty sure that I follow what you were saying. Let me see if I can state it another way.

Hypothetically, if the moon were tied to the earth on a string, connected at its center, instead of gravitationally, then the centrifugal forces around the perimeter of the moon at each point would be just a summation of the c.f. of the moon's center back to the earth, plus a c.f. due to the moon's once-a-month rotation. Therefore the c.f. would not tend to make the moon have either a bulge on its far side, nor its near side. Ergo the c.f. is not responsible for the tidal bulges, and they must be completely explained by the earth's differential gravity.

OK, you've made a convert - it's OK to talk about c.f. in a rotating reference frame, but in the case of tides the c.f. does not explain anything because it's symmetric around the full perimeter of the moon. I now think that Chronos's explanation was wrong in using c.f. to explain the bulge, and that Cecil was right.

Each time I've been in a discussion with you, I either already agree with you, or I've learned something.

[RM Mentock]

I know you're not talking about the scissors and the speed of light.

[John W. Kennedy]

1. sqrt(x+y) is not sqrt(x)+sqrt(y) and x+y is not meaningfully characterized as "just the same as x-y except for the sign of y".

[quote] Guys I think you can't agree because you are missing a point. The bulge is irrelevant and the rotation as well. John cannot disregard non-rotating moons, because when the Solar system started, pieces of dirt flew around in any way they pleased.2. That does not describe the situation by the time the Earth and Moon can be described as distinct objects. The Moon was moving, and rotating, and faster than it is today.

[John W. Kennedy]

Ugh! Repost.

1. sqrt(x+y) is not sqrt(x)+sqrt(y) and x+y is not meaningfully characterized as "just the same as x-y except for the sign of y".

Guys I think you can't agree because you are missing a point. The bulge is irrelevant and the rotation as well. John cannot disregard non-rotating moons, because when the Solar system started, pieces of dirt flew around in any way they pleased.2. That does not describe the situation by the time the Earth and Moon can be described as distinct objects. The Moon was moving, and rotating, and faster than it is today.

[RM Mentock]

Originally posted by John W. Kennedy
1. sqrt(x+y) is not sqrt(x)+sqrt(y) and x+y is not meaningfully characterized as "just the same as x-y except for the sign of y".
But the vectors can meaningfully be represented as the sum of a large constant component plus a small radial component.

I'm not sure what your comment about sqrt(x+y) is in objection to. Do you think I made a math error?

[John W. Kennedy]

Yes. I though that your posting suggested that sqrt(x+y) is equal to sqrt(x)+sqrt(y), but on reexamining it, I see that it actually says that sqrt(x+y) is equal to sqrt(x)+y.

[RM Mentock]

Doing math in public is a horrible thing.

As near as I can tell, the only time I used sqrt, I used it to represent the resultant w^2sqrt(R^2+r^2), which is the sum of two vectors at right angles to each other, one of magnitude w^2R and the other of w^2r. Still seems right to me, though.

[CurtC]

Yep, you got the math right as far as I can read. It might be helpful if we used a superscript:

w2sqrt(R2+r2)

You can indicate superscripts with the 'sup' and '/sup' tags.

[RM Mentock]

Isn't R^2 understood to be the same as R2? Or is that just some computer languages? Argh, I must have some button turned off.

[CurtC]

I think that R^2 is universally understood, it's just harder for my brain to parse. In "w^2sqrt(", it looked like the "2" went with the "sqrt" when I first read it.

Although "R^2" is how exponentiation is done in most computer languages, the vast majority of programming is done in C nowadays, and C uses the notation "R**2". But I sure wouldn't recommend that for a message board.

[RM Mentock]

R**2 is Fortran too, idnit?

[CurtC]

(YourFortranKnowledge.gt.mine)

[Chronos]

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.

[RM Mentock]

Originally posted by Chronos
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.

[John W. Kennedy]

Revenons a nos moutons. Now do your analysis for a point at the side. The force directed away from the Earth and is w^2sqrt(R^2+r^2), or slightly larger than at the moon's center. It can be resolved into two components--one which is identical to the centrifugal force at the center of the moon, and one which is equal to w^2r, directed away from the center of the moon. Same as for the far and near side points.

So, every point of the moon, by your way of analysis, experiences a centrifugal force equal to that at the center of the moon, plus a component equal to w^2r, where r is its distance from the rotational axis, directed away from the rotational axis.

[CurtC]

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.

[bronto]

RM Mentock asks me: How are you measuring the amount of stability? And what is the stability of 2:3? Well, that's another detail for which I lack the math.

[bronto]

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.

[RM Mentock]

Hey, that doesn't seem to be accurate at all. :)

The next post after dtilque's points out Saturn's moon Phoebe as well.

[The Bad Astronomer]

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 (http://co-ops.nos.noaa.gov/restles3.html)?
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.

[The Bad Astronomer]

Crap. PREVIEW, PREVIEW, PREVIEW. Here's the correct link to the NOAA page (http://co-ops.nos.noaa.gov/restles3.html).

[C K Dexter Haven]

I fixed the first link for ya, Bad

[CurtC]

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.

[calliarcale]

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.

[RM Mentock]

Originally posted by calliarcale
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.

Triton (http://www.planetscapes.com/solar/eng/triton.htm) is synchronized. Its orbital period is the same as its rotational period.
There is no second large satellite of Earth to perturb the Moon's synchronous rotation.
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.

[Chronos]

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.

[RM Mentock]

Originally posted by CurtC
I'm pretty confident that the NOAA site got it wrong. 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 (http://www.co-ops.nos.noaa.gov/restles2.html)--which also appears at the end of their chapter one, or introduction (http://www.co-ops.nos.noaa.gov/restles1.html), 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 (http://www.co-ops.nos.noaa.gov/restles3.html)--the part linked before by the BA.

Still, chapter three (http://www.co-ops.nos.noaa.gov/restles3.html) 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 Fc in the figure, but in the blue box it says it is labeled Fg. 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:
While space does not permit here, it may be graphically demonstrated that, for such a case of revolution without rotation as above enumerated, any point on the earth will describe a circle which will have the same radius as the radius of revolution of the center-of-mass of the earth around the barycenter. Thus, in Fig. 1, the magnitude of the centrifugal force produced by the revolution of the earth and moon around their common center of mass (G) is the same at point A or B or any other point on or beneath the earth's surface. Any of these values is also equal to the centrifugal force produced at the center-of-mass (C) by its revolution around the barycenter. This fact is indicated in Fig. 2 by the equal lengths of the thin arrows (representing the centrifugal force Fc) at points A, C, and B, respectively.
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.

[ehymel]

This NASA page shows a time-lapse sequence of the moon in which it appears to show libration.

http://antwrp.gsfc.nasa.gov/apod/ap991108.html

[RM Mentock]

Originally posted by Chronos
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.
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.