Misconceptions about tides

[JohnHuth]

I had to fly to NYC yesterday and grabbed David Burch's book on kayak navigation and had a "grrr....." moment in reading his chapter on tides. So, in a bit of a rant, with nowhere to go - I thought I'd mention this.

Tides are complicated. The explanation I saw in Burch's book is not consistent with my understanding of tides. They're not due to a bulge because of the moon on one side and "centrifugal force" on the opposite side of the earth. I even had the experience where a few high-end kayak instructors repeated the "Burch" explanation as gospel (although it's not really Burch's - and I think he has a doctorate in physics even!).

Tidal forces arise because the gravitational field lines from the moon are spreading out. To be a bit technical - the sum of the earth's plus the moon's gravitational fields creates a kind of "egg-shape" that causes the tides. Water has different elastic properties than land, and this difference, coupled with the "egg-shape" bulges is responsible for tides.

An easy way to convince yourself that this is true is the following: the effect of the moon on tides is about 5 times that of the sun, yet the gravitational pull of the sun is far, far larger than the moon. So, you would naively expect the sun to dominate the tide based on the erroneous explanations.

The proximity of the moon and the fact that its gravitational field lines are spreading out far more rapidly than the sun's (which are nearly parallel at the distance of the orbit of the earth) is responsible for the fact that the moon's effect on tides is so much larger than the sun's.

Here's a great URL that you can check out that will explain this.

http://www.lhup.edu/~dsimanek/scenario/tides.htm

So, no more use of the "centrifugal force" explanation, OK? (now that I know I'm safely among nerds!)

[EEL]

>The proximity of the moon and the fact that its

>gravitational field lines are spreading out far more rapidly

>than the sun's

If I understand this correctly, the field lines radiate outward in all directions from the object so that the angle between a moon field line which hits the top of the earth and the line which hits the bottom of the earth is far greater at the moon and far less at the earth than similar lines from the sun and thus the force from the moon's lines has a great impact in bulging the earth?

Is it also true that the tide in the Gulf of Maine is not true ocean tide or perhaps I should say not part of the Atlantic Ocean tide, but rather it is similar to the tidal flow into a bay? Not that I understand the distinction.

I read somewhere that the tide generally floods north and east when north of Cape Ann and south and west when south of Cape Ann. Why does that happen?

>

>(now that I know I'm safely among nerds!)

I think your post demonstrates there are nerds and then there are Nerds.

Ed Lawson

Wannabe nerd

[JohnHuth]

>If I understand this correctly, the field lines radiate

>outward in all directions from the object so that the angle

>between a moon field line which hits the top of the earth

>and the line which hits the bottom of the earth is far

>greater at the moon and far less at the earth than similar

>lines from the sun and thus the force from the moon's lines

>has a great impact in bulging the earth?

Yup, that's precisely correct!!

>

>Is it also true that the tide in the Gulf of Maine is not

>true ocean tide or perhaps I should say not part of the

>Atlantic Ocean tide, but rather it is similar to the tidal

>flow into a bay? Not that I understand the distinction.

I don't think there's really a distinction - it's that different bodies of water have different natural periods. For example, harbors may have periods of something like a few minutes. If you "tweak" a body of water, it will oscillate. In the case of harbors, these are sometimes called seiches. As the body of water gets larger, the period gets longer.

If you take a system and periodically tweak it, it will respond. If you happen to tweak it at its natural frequency, it will show a huge response. The best analogy to this is when you pump a swing. The swing, being a pendulum, has a natural period to it, and if you pump faster or slower than the natural period, it won't go so high, but if you pump it just right, it will respond dramatically.

The size of the Gulf of Maine is such that it has a natural response which is close to the periodicity of the tides that tweak it. Nantucket Sound, on the other hand, is much smaller, so it has a shorter natural period and doesn't see as large an extreme. The middle of the ocean sees very little tidal variation.

>

>I read somewhere that the tide generally floods north and

>east when north of Cape Ann and south and west when south of

>Cape Ann. Why does that happen?

>>

>

I don't know so much about Cape Ann, but I do know that there are some odd places - for example, at the tip of Monomoy, the flood current goes east and ebb goes west. That's because the predominant flood somes into both Long Island Sound and Nantucket Sound with flows that sort of originate around Newport RI. A lot of this has to do with the nature of the semi-trapped body of water that's being flooded, its natural period and how most efficient flow of water develops.

I'd have to think about the topography of Cape Ann a bit, in order to give you a thoughtful answer.

>I think your post demonstrates there are nerds and then

>there are Nerds.

>

I hesitate to ask if I'm a candidate for the capital "N".

[djlewis]

OK, you're a world-class physicist... I take your word for it.

To check my understanding of the article, I've tried to put it in fairly simple, direct terms. (The article really has a lot of verbiage that gets in the way, IMHO)

* centrifugal force is a fictional "force" used to explain the radial stability of an object in a rotating system

* fictional or not, the centrifugal force explanation of opposite side tidal bulge is simply wrong; there would be an opposite side tidal bulge even if the earth and moon were not rotating. (Is this because the centrifugal "force" acts equally on all points on and within the earth, so that it cannot cause one part to move relative to another?)

* the real cause of tidal bulge one the side of earth toward the moon is that the gravitational pull of the moon upward on the top of the ocean on that side of earth is (ever so slightly) stronger than the upward pull at the bottom of the ocean. That's because the bottom of the ocean there is a tad farther from the moon than the top of the ocean, and gravitation falls off over distance (by a cubic factor). All together, that creates a net upward force acting on the water, pulling it toward the moon, hence a bulge toward the moon (though ever so slight).

* likewise on the the side of earth opposite the moon, the moon's gravitational pull down toward the moon on the ocean is (ever so slightly) stronger at the ocean floor (because it's nearer the moon) than on the surface, creating a slight net force away from the moon, or upwards on the ocean and hence a bulge away from the moon.

* The above effects create tidal bulges, but not, per se, tidal flow, because the forces on the ocean are straight up and down

* in between the points on earth nearest and farthest from the moon, there are sideways components of this net force which essentially pull the water around toward the nearest bulge, creating tidal flow (do I have this right?)

* These "ever so slight" effects actually have an equally slight effect on the total shape of the earth (including water), generally just a few (1-50) feet up and down, which is pretty small in terms of the 25,000 mile diameter of the earth. But that still generates the tides we observe and, in human terms, move a lot of water around.

The presentational problem for me is that centrifugal force, fictional and irrelevant though it may be, corresponds to a vivid everyday experience that we know from childhood -- hang onto the rim of a rotating platform and the spin seems to draw your body outward, flinging you off the platform. I can't come up with an equally vivid and convincing image for the differential gravitational effects, especially the far side one. That may be because they are indeed so slight. But until we can, it's gonna be hard to convince people of this analysis.

--David

[JohnHuth]

I've thought about a good presentation for this - I think I can do with with some simple visual diagrams.

First, centripetal acceleration does have effect on the planet. The earth is slightly oblate - bulges out at the equator because of it's rotation.

Second - the earth-moon system also has a centripetal effect associated with it. The earth actually orbits a common center-of-mass in the earth moon system.

What distinguises the tides from these effect is that the tides are created by the divergence of the moon's gravitational field.

You can look at it this way: the "first order" effect is having a constant field from the moon, acting on the earth. This, combined with the centripetal effects of the mutual orbits, acts to position the center of the earth, due to an "average" gravitational field, which could be represented by a set of parallel lines all pointing toward the moon.

The tides are caused by any residual variation of the moon's gravitational field over the earth - because the field lines are both diverging from the moon, and the field is getting weaker as you go farther away from the moon.

The residual variations must add up to zero, if you add them up over the surface of the earth. If they didn't, then there would be some left-over constant field, that would recenter the earth's center.

What's left will sum to zero.

The easiest way to see the bulge both nearest and farthest from the moon at the equator, and compare it to a "reference" or average field at the center of the earth.

At the side of the earth closest to the moon - the gravitational field is pointing toward the moon, and is stronger than the value at the center of the earth, so it tends to make the earth bulge toward the moon. This is the easiest to buy and is intuitive.

If you look at the opposite side of the earth from the moon, the force of the moon is still pointing toward the moon (down toward the center of the earth). BUT, the value of the field is smaller because it's farther from the moon than the center of the earth. As a result the *difference* between the average field and the field at the center of the earth points *away* from the center of the earth, and will also produce a bulge that sticks away from the center of the earth.

At ninety degrees from a line connecting the moon and the earth's center, you have field lines that are pointing toward the moon. If you subtract out the average component (parallel lines, if you want to picture that), then the difference between the average and the actual fields tends to point inward toward the earth's center.

This is a relatively straightforward way to see that you get this "egg-shape" force that causes bulges on the side pointing toward and away from the moon, and indentations on the sized at 90 degrees with respect to the bulge.

Centripetal forces only come into play when you talk about the displacement of the earth's center from the center of rotation of the earth-moon system and also the oblateness of the earth.

Another "centrifugal force" effect that does come into play is so-called "tidal drag" - the fact that water flows better than the earth's mantle, causes effects that slow down the earth's rotation, ever so slightly - I think it's a millisecond per year.

Because the earth's field on the moon is so much stronger than the moon's on the earth, the tidal drag forces are much larger and tidal drag has basically frozen the moons rotation so that it always shows the same face toward the earth.

When I get back to the States, if anyone is interested, I can draw up some diagrams that I think will make it more clear.

[EEL]

John:

Based on this explanation, I believe it is now quite safe for you to use the appellation of Nerd

Ed Lawson

[jonsprag1]

Tides hmmm Lemme see---they go in and they go out---twice daily in most areas of the world although a very few have diurnal (once a day tides) They are an hour of so later each day and the amount of flow can vary greatly depending on local topography---yep that's about it--anymore questions? lol

[djlewis]

Ok, looking for a good, "intuitive" presentation. The paragraph below seems to me to be the crucial part of the explanation for the opposite side bulge, the one that's hard to understand...

> If you look at the opposite side of the earth from the moon, the force

> of the moon is still pointing toward the moon (down toward the center

> of the earth). BUT, the value of the field is smaller because it's

> farther from the moon than the center of the earth. As a result the

> *difference* between the average field and the field at the center of

> the earth points *away* from the center of the earth, and will also

> produce a bulge that sticks away from the center of the earth

I have to say that doesn't do it for me. How can a force (the moon's gravity) which pulls in one direction -- toward the moon; down on the water on the far side of the earth -- produce a bulge in the opposite direction -- away from the moon; upward on the water? It makes no sense, visually and intuitively speaking.

But, if you grant that the slightly larger force at the ocean floor actually pulls the ocean floor down, or maybe the whole earth, then the water bulge would appear at the surface of the ocean by being kind of left behind. In other words, the ocean surface does not so much bulge out as the ocean floor and the water nearer the ocean floor (that is, the whole earth) pulls away from the ocean surface and that's what generates a bulge shape on the surface.

Is that right? Does the earth actually move like that to make the bulge? Or is it just resolution of forces?

If that's true, then is the apparent upward bulge evidence of the fictional centrifugal force? of course, the statement in the article that this would all work even of there were no rotation of the moon about the earth (or both about the barycenter) is nonsense, because none of this would work at all if the two bodies were not so rotating -- they'd just crash into each other, as in a 1960's catastrophe movie. In that sense, it really IS centrifugal force, to the extent that centrifugal force is an accurate and convenient model of what happens in rotating systems, even though it's not a true force.

Aha, just got it (I think). The sense in which centrifugal force acts like a force in a rotational system is the same sense in which the earth moves toward the moon and creates the opposite side bulge by "pulling away" from the far side ocean surface.

Or maybe, trying to get back to basics, the opposite side bulge happens because the opposite side ocean is indeed trying to go in a straight line, despite being held in place in an elliptical trajectory by the moon's gravity. If it were not held in place, it would go in a perfectly straight line, that is, it would fly away on a tangent, creating the ultimate tide, also worthy of a catastrophe movie.

Well, that's silly -- if the moon let go, the whole earth and its oceans would fly off in a straight line (neglecting the sun's effect, of course). But let's say if magically the moon's gravity stopped pulling on the ocean but kept pulling on the solid earth, then the water would fly off in a straight line, which would look to us on earth like it went straight up into the sky. That would be the ultimate tidal "bulge".

But the ocean surface IS held in place, though with slightly less force than the water at the ocean floor, so the surface does indeed "go straight" a bit more than the deeper water, which makes it bulge. I think that's it, right? But it IS "centrifugal" force then, and the textbooks are not altogether wrong... just a bit simplified, the same way centrifugal force itself is a conceptual simplification. Oh wait, it's centrifugal force interacting with the gravity differential. That's the part many oceanography textbooks leave out.

And that explains why I couldn't come up with a good visual model. It really does not work without the rotation of the earth about the moon/barycenter. But it also explains why a ball twirling on a string model does not work... because there is no force differential on the opposite side -- the kid pulling on the string exerts force at just one point on the ball where the string is attached. So there is a force differential, from some specific "gravitational" force at the attachment point to none on the rest of the ball, and voila, you get a bulge on the near side of the ball (say a tennis ball, not a billiard ball, to make the effect visible), but not on the far side.

--David.

[JohnHuth]

>OK, you're a world-class physicist... I take your word for

>it.

Actually - PLEASE DON'T. The great physicist Richard Feynman was very adamant that people should only be content when they believe in their heart the explanations. Physics does provide good tools for thought. I'm thinking through this myself, so I don't claim to have the definitive answer.

>* centrifugal force is a fictional "force" used to explain

>the radial stability of an object in a rotating system

Yes - although the rotational motion of the moon and earth around a common center-of-mass is the thing that keeps the moon and earth from crashing into each other.

>

>* fictional or not, the centrifugal force explanation of

>opposite side tidal bulge is simply wrong; there would be an

>opposite side tidal bulge even if the earth and moon were

>not rotating. (Is this because the centrifugal "force" acts

>equally on all points on and within the earth, so that it

>cannot cause one part to move relative to another?)

Yes - in essence - the effect of the orbit of the moon and the force of gravity can be said to be acting as if the earth were concentrated at a point - its center-of-mass. If you add up every force on every bit of the earth, you can say that is is as if the force, added over every little chunk of earth were happening at the center of the earth. What's left over are the forces that differ from one piece of the earth to the next.

Now, "centrifugal forces" could create some differences between the near-side and the far side of the earth. I'm trying to figure out its magnitude compared to the effect of the divergence of the moon's gravitational field. There is some effect associated with the rotation of the earth, which makes it oblate - so, it's plausible that some residual effect from the rotation of the earth-moon system could create some effect, once you take out the overall average effect on the earth. Two things - I suspect that the effect is small, and can be neglected (this is something that one can calculate easily), and secondly, it should be relatively symmetric - where only the "differences" would create any kind of bulges. In any case, I think it's misleading to say that any bulge on the far side of the earth (from the moon) is "caused by" centrifugal forces.

>

>* the real cause of tidal bulge one the side of earth toward

>the moon is that the gravitational pull of the moon upward

>on the top of the ocean on that side of earth is (ever so

>slightly) stronger than the upward pull at the bottom of the

>ocean. That's because the bottom of the ocean there is a tad

>farther from the moon than the top of the ocean, and

>gravitation falls off over distance (by a cubic factor).

>All together, that creates a net upward force acting on the

>water, pulling it toward the moon, hence a bulge toward the

>moon (though ever so slight).

Yes, what's left over is the force of deformation on the earth. The forces of deformaion has to add up to zero, but there will be different forces acting on different bits of the earth, which will "deform them" - as if the earth were a slightly elastic rubber ball.

>

>* likewise on the the side of earth opposite the moon, the

>moon's gravitational pull down toward the moon on the ocean

>is (ever so slightly) stronger at the ocean floor (because

>it's nearer the moon) than on the surface, creating a slight

>net force away from the moon, or upwards on the ocean and

>hence a bulge away from the moon.

Yup - although, the force acts on the entire earth. If it were a rubber ball, the entire earth would be uniformly deformed. This is where the difference between the properties of water and the mantle of the earth come into play. Water is more readily deformed by the tidal forces than the mantle, which is why the water seems to move relative to the shores.

>

>* The above effects create tidal bulges, but not, per se,

>tidal flow, because the forces on the ocean are straight up

>and down

Yup.

>

>* in between the points on earth nearest and farthest from

>the moon, there are sideways components of this net force

>which essentially pull the water around toward the nearest

>bulge, creating tidal flow (do I have this right?)

Yes.

>

>* These "ever so slight" effects actually have an equally

>slight effect on the total shape of the earth (including

>water), generally just a few (1-50) feet up and down, which

>is pretty small in terms of the 25,000 mile diameter of the

>earth. But that still generates the tides we observe and,

>in human terms, move a lot of water around.

Yes.

>

>The presentational problem for me is that centrifugal force,

>fictional and irrelevant though it may be, corresponds to a

>vivid everyday experience that we know from childhood --

>hang onto the rim of a rotating platform and the spin seems

>to draw your body outward, flinging you off the platform. I

>can't come up with an equally vivid and convincing image for

>the differential gravitational effects, especially the far

>side one. That may be because they are indeed so slight. But

>until we can, it's gonna be hard to convince people of this

>analysis.

>

That's a helpful point - one to keep in mind if I (or you or an enlightened instructor) make a presentation on the tides.

The idea of centrifugal force has been invoked many times in the description of tides, so it's difficult to abandon.

If you want to think about "centrifugal force" - in what us physicists call a "fictitious frame of reference" - it seems to be a real force. In the "frame of reference" of a rotating system like the earth-moon system, it might appear that a "centrifugal force" is pointing in the direction opposite to the moon's gravity. This "fictional force" keeps the earth from falling toward the moon. It "balances" the force of gravity.

This isn't what happens if you adopt a non-accelerating frame of reference - as if you were floating above the earth-moon system and independent of the earth-moon orbit,then, you see the moon and earth describing a circular orbit about their mutual centers-of-gravity.

In any case, the effect of any "centrifugal" force would be to balance the earth-moon system, and wouldn't cause any significant deformation of the earth, in any case. The only place where you get a modest deformation is from the rotation of the earth about its axis, and this causes a slight oblateness of the earth - but not relevant to the tides. Again, this deformation has to do with the difference in forces that different chunks of the earth experience.

[djlewis]

>>OK, you're a world-class physicist... I take your word for it.

>Actually - PLEASE DON'T.

Don't worry -- just kidding. All that your status as a professional physicist did for me was to make me pay attention, even though you were contradicting a lot of published texts and other material. I still gotta figure this out for myself.

Please read my other response -- yes, it's overly long, but it represents my current understanding and attempt to come up with a way of explaining all this that will appeal to people.

And just to prove that I'm not just taking your word for it, I'll even dispute something you've said, and in the process summarize how I might teach this.

My dispute is that, in a sense, it actually IS centrifugal force that causes the far-side bulge, or at least that's one legitimate way to think of it. But it's ~varying~ centrifugal force that causes the tides, not centrifugal force per se.

Let me explain... On the ocean surface on the far side of earth from the moon, the centrifugal force from the rotating earth/moon system is a tad smaller than it is at the far side ocean floor. That's because the centriPETAL force -- which is the moon's gravitational pull on earth -- is a tad smaller at the far-side surface, because gravitation decreases with distance. So the balancing centriFUGAL force there is a tad smaller. That gives the net force vector upwards and causes the bulge.

But force vectors aren't very convincing to most people. In fact, you get the apparent fallacy of the moon pulling in one direction and the far-side tide bulging in the other direction, which does not compute, intuitively speaking.

The way out of the paradox is that, contrary to the article you referenced, you DO need to look at the rotation of earth and moon about each other. But it actually might work best to discard the centrifugal force argument and go back to what centrifugal force really means. That's simply the tendency of a body held in a rotational system by centripetal force (the force needed on a string to keep a tennis ball from flying off) to try to head in a straight line -- normal inertia of a body in motion wanting to remain in motion on a straight line.

So the "normal thing" for earth and its oceans is to head in straight line which, from the point of the view of the moon, is heading away, despite the moon's gravity holding it in an elliptical orbit. If the whole earth and ocean were equally affected by the moon's gravity, they would just be held in orbit and there would be no bulge, no tide. But since the moon's pull at the surface of the far-side ocean is a tad smaller than at the far-side ocean floor, the surface of the ocean is a bit freer to move in its "normal" straight line, which is away from the moon, hence it bulges away from the moon. But it's not rally bulging UP so much as it just being allowed to relax a bit into where it would "normally" go without the moon holding it in orbit, that is, in a straight line in space, which is away from the moon.

So, it isn't that the moon's pull produces a paradoxical bulge in the opposite direction. It's that the surface ocean does get to go a teeny bit more in that straight line, escaping ever so slightly more from the moon's pull, than the water at the bottom of the ocean (and that varies continuously from top to bottom). Hence, a bulge. And at other points in between the far and near sides of earth, that same principal produces sideways forces, which actually make tidal currents to accompany the bulges.

One might try to illustrate this with the proverbial tennis ball, but with some extra props. First, attach a string to the surface of a tennis ball and swing it around your head. There's a force differential there -- a lot of force at the attachment point and zero force applied elsewhere on the ball -- well, less force anyway, because the tennis ball shell deforms and does not convey 100% of the force to the rest of the ball. Result -- the attachment point tends to bulge toward the twirling string, or you could equally say that the rest of the ball pulls away form the attachment point, leaving a bulge behind.

It's a little trickier on the other side. Imagine a billiard ball inside a spherical ball made of balloon-type rubber (a balloon ball, as it were), dead center, with some stable but squishy, conformable stuff filling the space between the two balls. When at rest, the billiard ball is centered inside the balloon ball.

Now, run a string through a small hole in the balloon ball, and attach it firmly to the relatively rigid billiard ball inside. Swing that contraption around your head. What should happen? It's not hard to believe that the outer ball and squishy stuff in between will deform and the far side of the balloon ball will tend to bulge away from the string. That's again because there is a difference in the centriPETAL force, a certain amount on the billiard ball and a lot less on the balloon ball and squishy stuff, because it doesn't convey the force 100%. If the whole system of the two balls were rigid and equally pulled by the string, there would be no bulging.

Actually, you could think of some kind of jelly between the two balls, and then you really have modeled the relatively rigid earth (billiard ball) and conformable oceans (jelly and balloon ball surface). The only thing you haven't modeled very well is the ~continuous~ variation in force between the the balloon ball and the billiard ball. But there is a variation, and without it, no bulge.

Well, maybe something like that.

--David.

[JohnHuth]

OK, so I broke down and did the calculation. I was surpised to find that the differential of the "centrifugal force" was about the same size as the difference in gravity from one side of the earth to the other. I thought the term was negligible.

Give me some time to play with the calculation, and I'll try to respond in a bit.

John

[djlewis]

OK, while you're doing that, here's another thought. Is it valid to say it this way, physics-ly speaking...

The radius of a planetary orbit is partly a function (what order?) of the mutual gravitation between the orbiting bodies (again, treating them as both rotating about their barycenter). This is a combination of Kepler and Newton. See, for example...

http://www.glenbrook.k12.il.us/gbssci/Phys...cles/u6l4a.html

(Well, those equations also involve the orbit's period, and I'm not sure what to do with that.)

But plunging ahead... since the water at the earth's ~ocean surface~ on the side opposite the moon is a tad farther from the moon than the water at the far side ~ocean floor~, the gravitation is slightly weaker, and surface is really in a different, slightly wider orbit than the water at the ocean floor. And, of course, there's a continuous variation from the surface to the ocean. Hence the bulge.

Put simply, the far side ocean is in a slightly wider orbit than the rest of the earth, and that's what the bulge is.

But that raises the question of why solid land on the opposite side doesn't bulge out into a wider orbit. Is this why... the solid earth "fights back" against the gravity differential, that is, it exerts a downward force on the far side surface to keep it from going a little straighter (that is, bulging), thus holding it in place. Put another way, this extra force supplements the moon's slightly weaker gravity at the solid surface, and the orbit of the solid land remains the same.

But water flows rather than pulling down on the surface to counteract it. So the oceans relieve the force imbalance by flowing into a slightly wider orbit, that is, bulging.

Make any sense??

Hmmm... as I look at the equations, it seems that perhaps the decrease in gravitation at the far-side ocean surface vs the ocean floor might also/instead decrease the period of the orbit, that is, make the water bulge out "behind" the earth as it orbits the earth/moon barycenter. Yike! Might that effect simply make the tide asymmetric, with the leading edge of the tide slightly flatter and/or delayed versus the trailing edge (or do I have it backwards?). Trouble with that is, for the period really to decrease, the water would have to fall farther and farther behind, and fly into space??? So maybe earth's gravity prevents that, and forces the radius change alone to do the rebalancing of forces.

--David.

[JohnHuth]

There are two pieces to the puzzle. The first is the fact that some kind of "differential" forces cause the earth to deform. The question is - how much of this is the "centrifugal" contribution and how much comes directly from the moon's gravity.

Once you get the differential forces that tend to distort the earth, then you can talk about the relative influence on water versus land. Since land is less elastic, it doesn't deform the same way under stress and strain as water.

I think that with any stress/strain - water would deform more readily than land.

At least that's how I'm trying to figure it.

Besides the raw physics, I'm trying to think through the "clues" we know about tides -

1.) They're very symmetric on the near-side and far-side of the moon/sun (high and low tides are pretty much the same heights on the near and far sides).

2.) The sun has an influence of about 1/5th of the moon.

I'm pretty sure that for the sun, the centrifugal contribution has to be small - but there's probably some contribution from the diverging gravity field lines.

The symmetry makes it difficult for me to "buy" the centrifugal explanation. I'd expect the tides to be asymmetric (highest on the far side of the moon) if this were the whole story.

These are all hunches, mind you.

I tried to calculate the center of rotation of the earth/moon system, and got that it was about 4000 km from the center of the earth, so, it's inside the earth. Maybe you can recheck my math on this?

[JohnHuth]

Try this website

http://www.vialattea.net/maree/eng/index.htm

[JohnHuth]

OK, I think I got it - that website I posted in the previous helped me out a lot. After all the talk about centrifugal forces, I began to wonder if I'd missed something altogether, and actually did the "wrong" calculation that is shown in the website listed.

The issue is that the earth is always in "free-fall" about the center of the earth-moon system, and, as a result there is no force felt from the rotation. The only force available is gravity.

As the guy on the website points out, if you assume that there is come centrifugal force, by using the wrong frame of reference, you end up with a force that's about 70 times the force due to the moon. It would also have some strange characteristics - it would be largest at the equator and also largest at the side opposite the moon, it would also diminish with latitude. None of these are characteristics of the tides.

The main thing is that the centrifugal force is zero - not even small - it is the divergence of the moon (and sun's) gravitational field lines.

So, I'm now convinced.

[djlewis]

>OK, I think I got it

>...

>The issue is that the earth is always in "free-fall" about

>the center of the earth-moon system, and, as a result there

>is no force felt from the rotation. The only force

>available is gravity.

>...

>So, I'm now convinced.

(Another good reference -- http://www.jal.cc.il.us/~mikolajsawicki/ -- click on the tides/gravity link in the lower left)

Anyway, I think I got it too. So you and I may be convinced, but can we convince anyone else? ;-)) For example, how can the force of the moon's gravity cause the ocean on the far side of the earth to bulge the other direction?! Seems very unintuitive. The answer is what you said, that both bulges are due to gravitational decrease over distance, not to gravity per se. But how you gonna teach that? Gravitational force differentials and force vectors are awfully abstract and hard to visualize.

Lemme try...

Let's do drop centrifugal force from the discussion. Yes, the earth is in free fall -- meaning it's acted on by no other force than gravity. But there's free fall and there's free fall. One kind of free fall is like on a parachutist (or any dropped object) falling toward earth before she opens the parachute (and neglecting air friction). There's no (or insufficient) rotational momentum to keep the parachutist up in the air, so she falls to earth at an increasing speed, and pretty soon -- end of story.

That's obviously not what is happening between earth and moon (or earth and sun). But let's pretend. During that brief period of free fall before the spectacular collision, there would still be tides on earth! Why? Because the ocean on the side nearest the moon gets a little stronger tug from the moon's gravity, so it falls toward the moon a tad faster. Likewise, earth's far-side ocean gets a slightly weaker gravitational pull from the moon and it falls a tad slower toward the moon. And because the oceans are water rather than relatively solid land, they do bulge away from the relatively rigid earth in the middle by flowing upwards.

The net effect in this non-rotational thought experiment is a (symmetric) lengthening of the earth/ocean's shape into an ovoid, with bulges toward and away from the moon. I think they are also ever increasing bulges, because if they really are going different speeds at all times, they will be separating ever more, until, of course, the world ends with a bang.

This is still a tad abstract, but much easier to see because the far-side bulge is just due to that water lagging behind, and the near-side bulge is due to that water getting a bit ahead during the brief fall toward the moon. That's obviously tides with an extra twist -- ever increasing -- even though the whole system is doomed.

So, someone may have thought they understood the near side bulge as due to gravity and found the far-side bulge more complex. But in fact, they didn't even have the right mental model for the near-side bulge -- that's also due entirely to the drop-off in gravity with distance, not to the gravity per se.

OK so far?

Now consider the rotational case, the one that really happens between earth and moon (and earth and sun). There's some rotational momentum -- actually just straight-line momentum constrained into an ellipse by the moon's gravity and it's just enough to make things stable, with the earth staying the same distance from the moon (neglecting the elliptical shape of the orbit). Right?

[blockquote] Footnote -- where did that rotational momentum come from? I assume it was put there when the sun and planets formed, by some primeval process. But that's a problem for cosmologists. All we need to know is that it's there, and in the (near) absence of friction or opposing forces, it continues forever, or at least a darn long time.[/blockquote]

So, the moon's gravity is perfectly counterbalanced by the sideways momentum of the earth, and the two stay about the same distance apart (neglecting the ellipse). I think that the "force" needed to explain that stability is what we call centrifugal force. It's really nothing more than the tendency of the earth to go in a straight line even while held in place my the moon's gravity.

As another thought experiment, what if the moon's gravity suddenly let go but the rotational momentum of the earth continued, as inertia dictates? Then it would appear from a vantage on the moon that the earth was flying off into space on a tangent from its former orbit, that is, both straight outward and to the side. Right?

Anyway, back at reality. That situation between earth and moon is also free fall, even though there is no actual falling going on, because the only force on the earth is the moon's gravity (neglecting the sun, again). And now the far- and near-side tidal bulges still happen, for the same reason (differential gravity). In this case, however, the bulges don't steadily increase because the rotational momentum and force of the moon's gravity are in balance.

But what does happen, is that the far-side ocean gets to go a little more where its inertia wants it to go, that is, straighter and farther away from the moon. Likewise, the near side ocean is pulled a little more by the moon which makes it go less straight and closer to the moon. It's like the case of the moon altogether letting go and the earth flying off into space, but a whole lot less dramatic. The result is the two tidal bulges.

And the role played by "centrifugal force", its only role, is to keep the whole system in balance, not to create variations within that balance, that is, not to make tidal bulges, or sideways tidal currents between the near- and far-sides.

So both tidal bulges are due strictly to the differential force of gravity. Both are due to the oceans "falling" differentially as well, just like in the non-rotational case, even though the "falling" appears stable. The near side ocean "falls" a little more and the far side ocean "falls" a little less than the rest of the rigid earth. Hence they bulge.

I still wonder if the tidal bulges in the rotational case could be understood by saying that the far-side oceans actually have a slightly wider orbit than the solid earth, and the near-side oceans a slightly tighter orbit. I don't have the chops to work through the equations -- just a gut feel. It's probably complicated by material dynamics -- water flows, etc -- but there still may well be a simple model that makes the point.

Whew.

--David

[JohnHuth]

Progress.

I guess there are two pieces - one is to make the convincing argument that the centrifugal term doesn't exist (you had me going for a few days there) and that it's due to the divergence of the moon's (and sun's) gravitational field. The way I think of the near-far side argument is that any net gravitational force will cause the center of the earth-moon rotational system to shift, so any residual force must be symmetric and a bulge on the near side must be the same as the bulge on the far side - an effect which is born out from experience.

If you buy the "egg-shape" potential, then the only thing left is the elastic properties of water and land. I think of it this way - the elastic properties of water and land are such that they will stretch under tidal forces and start to exert a "spring-like" force that holds the water and land from flying apart from the tidal forces. This "stretching" is necessary to hold things together. If you put a book on a table, the reason the book doesn't continue to accelerate downwards is because the table flexes ever so slightly, providing a counter-force to gravity. Now, water, being less elastic than land, will distort more than land, and hence will cause tides.

We still have a bit of clean-up to do, here, because tides imply flow of water and we have to describe clearly why a flow is created.

I think we're converging on the answer, however. I started out not-believing David Burch,then "almost" believing it, and now, I think that explanation he gives doesn't cut it. Now, it's only a paragraph, and you can readily argue that it doesn't matter. "Just hand me the friggin tables and a copy of Eldridge" - but it's been bugging me for some time.