djlewis Posted March 25, 2007 Share Posted March 25, 2007 >They are, indeed, due to the divergence of the moon's (and >sun's) gravitational field(s). >... >The tidal force is a "differential" force, meaning that it >acts on different parts of the earth with different forces. >In fact, if you look at the diagram, you can see that it has >a substantial component parallel to the earth's surface, >which can cause the oceans to flow. > >Once you have that, and the fact that water flows more >readily than land, it creates the tides. >... >What it is *not* is due to a centrifugal force. The earth >is in free-fall about the center of rotation of the >earth-moon system, and there is no force generated due to >this rotation. The best analogy is to think of people on a >spaceship orbiting the earth - they're weightless because >they're orbiting the earth, not because they're far away >from the earth. Well, I think you could say that the outward tidal bulge is due to differential centrifugal force, though not centrifugal force per se. That's as valid (or invalid) as using centrifugal force in any way. So, you get a neat symmetry -- near-side bulge due to differential centripetal force (that is gravity); far-side bulge due to differential centrifugal force. Centripetal and centrifugal force are two sides of the same coin; likewise the two bulges. Anyway, I still feel the model of earth falling toward the moon makes the point very neatly - removing the orbital motion from the picture as a thought experiment. That gets any notion of centrifugal force altogether out of the explanation, so it cannot confuse things. In this "just plain falling" scenario, the near-side ocean is pulled by the earth-moon gravity a tad more strongly than, say, the center of earth. So the near-side ocean falls a tad faster toward the moon, and bulges out relative to the solid earth. Likewise, the far-side ocean is pulled a bit less, falls a bit slower and bulges out the other way. In fact, seen this way, there aren't really two bulges, but all the oceans (if thee were just ocean covering the earth) form into a spheroid. (Hmmm... it's actually not a perfect spheroid, since the effects are not perfectly symmetric; maybe not even a spheroid modulo that -- you'd have to do the math -- but it is an ovoid shape of some kind, almost but not quite symmetric.) Of course, it all ends in a cosmic splat pretty soon, but you get the tidal bulge(s) in the meantime. When you put back orbital motion, it's the same, except that the "free fall" appears stable, that is, the earth and moon maintain about the same distance (apart from the elliptical orbit effect). That stability is what leads to puzzlement by many -- how can you call it "fall" when the earth and moon are not apparently falling anywhere; how can force of gravity pulling toward the moon cause the far-side bulge away from the moon? But the way to think of it is that the earth is trying to go straight, that is, fly way from the moon, but is restrained in orbit by gravity. That's the sense it is falling -- constantly being pulled toward the moon and away from where it would naturally go from linear momentum. The far side ocean, being restrained a tad less by earth-moon gravity, gets to go a bit straighter, that is it bulges away from the moon. The near side ocean is restrained a bit more and goes less straight, and bulges toward the moon. It's essentially the same as the really falling free fall thought experiment case. I haven't done the math (and not sure I know how), but I still have the gut feel that the equations would show that the far side ocean is essentially in a slightly wider orbit with the same period, and the near side ocean is in a tighter orbit. I'll do a diagram too, when I get some time. And thanks, John, for finally getting me to understand this! --David. Quote Link to comment Share on other sites More sharing options...
djlewis Posted March 25, 2007 Author Share Posted March 25, 2007 Is this a test? Gee, I didn't even pay tuition! ;-)) I vote for B, different radii. Gotta be. Just think of one super-massive object vs a light one (not unlike the sun and earth). The barycenter is very close to the super-massive one, practically at its center. The super-massive one can't be doing much rotating, that is, it has a very short, vanishingly short radius. In fact, in the extreme case of an infinite mass, it degenerates into the naive notion of the small object just plain rotating about the stationary big/infinite one. OK, so your point? Is it that the earth really isn't doing much rotating in the earth-moon system? Granted, but so what. The tides aren't very big either, in the larger scheme of things -- say 10 feet out of 25,000 miles, or 1 : 13,200,000. Besides, we already established that the orbital rotation is not the cause of the tides. Or are you pursuing the idea that the tidal bulge(s) are really describable as orbits of different radii? --David. Quote Link to comment Share on other sites More sharing options...
JohnHuth Posted March 26, 2007 Share Posted March 26, 2007 One of the perq's of being an NSPN member... Actually, both cases are possible in tides. In case 2, there has to be a significant torque on the object to cause it to rotate like that. Case 1 is closer to the earth's, I believe - there is a torque due to tidal forces and you can even quantify this to say that the earth's rotation slows down by a few milliseconds every number of years or so (I think it's a decade, but it may be more). Eventually, the earth will stop rotation of a 24 hour day, and will simply put the same face toward the moon all the time (as the moon currently does with us). So, the answer, I believe is "mostly case 1 with a tiny bit of case 2 mixed in". but, feel free to question this. Quote Link to comment Share on other sites More sharing options...
jeffcasey Posted March 26, 2007 Share Posted March 26, 2007 John is being particularly circumspect and noncommital, because I have challenged some of his statements, and he and I have been thrashing through this for awhile now (offline and politely, you pool-fools). (This is what we do for fun when not paddling). There is a certain amount of disagreement as to how to interpret cold hard results in layman's terms, plus there are some questions as to how to set the problem up in the first place, and hence which "cold hard results" are actually correct. As some of you are aware, there are a great many sources (websites, textbooks, experts, etc) out there, and much disagreement among them.....so they can't all be right -- that shouldn't be the case, but unfortunately is. We're trying to make sure we are solidly in agreement with each other before diving back into public proclamation mode. John's figure from message 2 of this thread is the most succinct demonstration of the different ways of looking at this problem that I've seen, and I think we're now converging on interpretation and less on what the correct setup is. If you wanna keep score, it looks like I'm going to owe John a few beers, but I'm not quite ready to buy yet. We'll keep you posted. Quote Link to comment Share on other sites More sharing options...
djlewis Posted March 26, 2007 Author Share Posted March 26, 2007 earth. And isn't the barycenter, by definition, stationary in this system? So isn't it like a drum major's baton, but with radically different knobs on the ends? The drum major is twirling the baton, but holding it much closer to the big knob to keep it in balance. And the two knobs trace concentric circles (ellipses for planets, of course, but forget that for now) of different sizes about the drum major's hand at the barycenter. Of course, the barycenter, and the whole earth-moon system are also revolving around the sun, but that's a different matter. And the sun is hurtling toward some other star at umpty-ump k/h, etc. So we don't worry that the drum major is also marching down the field and/or running around in other patterns, and/or moving his hand around while he twirls the baton in it. Anyway, that all looks exactly like your diagram 2. I can't see diagram 1 anywhere. So what'm I missin' -- could be lots; wouldn't be the first time my "intuition" has misled me in this discussion. And unlike Jeff, who actually kinda knows what he is talking about, I'm foolish enough to do this in public. ;-))) --David Quote Link to comment Share on other sites More sharing options...
JohnHuth Posted March 26, 2007 Share Posted March 26, 2007 Yup, Jeff and I are going through the details. It's actually kind of fun - I haven't thought hard about the tide issues, and he's forcing me to think about them. (that diagram was a result of us hashing through some of this). I'm not ready to be definitive until Jeff's on board, mind you. FYI - I'm not the only one to come up with that "case 1" - if you look at this website: http://www.vialattea.net/maree/eng/index.htm it has some good animations of that Case 1. (I'm not saying that "proof" exists in a URL - but just that there are people who have used that to help explain tides). Quote Link to comment Share on other sites More sharing options...
jason Posted March 27, 2007 Share Posted March 27, 2007 Jeff, I thinks that most of us owe John a few beers with all that he continues to teach us. :-) -Jason Quote Link to comment Share on other sites More sharing options...
JohnHuth Posted August 8, 2007 Share Posted August 8, 2007 Take a look at these two diagrams - tell me which one of the two most closely resembles your view of what you think happens with the rotation of the earth about the earth-moon center. Quote Link to comment Share on other sites More sharing options...
JohnHuth Posted August 8, 2007 Share Posted August 8, 2007 OK, having trudged through this, I think I have more or less an understanding of tides. They are, indeed, due to the divergence of the moon's (and sun's) gravitational field(s). You can represent the field as two pieces: 1.) a constant field that acts on the entire earth uniformly 2.) a tidal force that has two outward and two inward bulges (see figure attached) The tidal force is a "differential" force, meaning that it acts on different parts of the earth with different forces. In fact, if you look at the diagram, you can see that it has a substantial component parallel to the earth's surface, which can cause the oceans to flow. Once you have that, and the fact that water flows more readily than land, it creates the tides. Beyond that, there are the effects of the size of a tidal basin. The Bay of Fundy is a particularly interesting example because it has a natural period of about 13.3 hours, which is very close to the tidal period of 12.5 hours (give or take a few minutes). What it is *not* is due to a centrifugal force. The earth is in free-fall about the center of rotation of the earth-moon system, and there is no force generated due to this rotation. The best analogy is to think of people on a spaceship orbiting the earth - they're weightless because they're orbiting the earth, not because they're far away from the earth. So, in fact, Burch'es explanation is incorrect - at least my conclusion. Thanks to djlewis and eel for discussions! Quote Link to comment Share on other sites More sharing options...
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