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…?

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.

And the Staff Report being referenced is Why does the same side of the moon always face the earth?

Unless of course you’re talking about Cecil’s column Why does the same side of the moon always face the earth?

Glad to see that Cecil got it right.

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?

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.

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?

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.

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.

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.

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 ?

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.

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.

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.

As to **jezzaOZ**’s question about the symmetry of the principal semidiurnal tidal components 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.

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?

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

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.

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

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.

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.