lunar rotation

Cecil's Columns/Staff Reports
41

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.

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

42

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.

43

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.

44

That’s why I agree with Cecil. :slight_smile:

45

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.)

46

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.

47

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.

48

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… :slight_smile:

Well, I too think we shouldn’t be talking about the centrifugal bulge. That’s been my point all along.

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.

49

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.

50

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.

51

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.

52

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

53
  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.

54

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”.
  1. 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.
55

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?

56

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.

57

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.

58

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

w[sup]2[/sup]sqrt(R[sup]2[/sup]+r[sup]2[/sup])

You can indicate superscripts with the ‘sup’ and ‘/sup’ tags.

59

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

60

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.