Like virtually any physical system, the governing effects can be virtually limitless. The reason why we have tides, is due to gravity. The specific effects that tidal forces have on a physical system depend on many more factors. You could calculate all the forces exerted on every molecule of water to model tides even more accurately, but you would have long exceeded a point of dimished returns.

Per Alfnapper's reference above: http://www.cepo.interacesso.pt/Artigos/Astrol/GCalcE.htm

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The average gravitational acceleration the Moon creates on something on Earth is

gm = G ´ Mm / Dem = 3.36E-05 N/Kg

To get the gravitational acceleration for something on Earth closest to the Moon and furthest from the Moon, we just need to replace the distance value of Dem by, respectively, this value minus the Earth’s radius and this value plus the Earth’s radius. The result is:

gm(closest) = 3.48E-05 N/Kg
gm(furthest) = 3.25E-05 N/Kg

So the Moon’s gravitational pull varies by 2.24E –6 N/Kg between when the Moon is directly over our heads and when it is on the other side of the Earth.

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This is the part that puzzles me, that the delta "gm" between the closest and furthest surface on Earth pointing towards the Moon is not a positive-negative relationship, but merely a differential between two negative forces. By this logic, the bulge on the face towards the moon should bulge by a value of 3.48e-05 N/Kg, while the far side of the planet should be 'depressed' (not bulge) by 3.25e-05 N/Kg. This is not how the 3 point analogy (P1, P2, P3), mentioned by both Shambolic and Wintermute11, is understood to mean, where rather (per them) the far side bulging out, it is merely pulled less... not the same thing. Though this is currently accepted theory, it does not hold water. I understand how the gravitational pull causes particles on the facing surface to gravitate outwards, and how this pull is lessened at the Earth's core, but I am not convinced the pull on the far side is so minimized as to cause water to bulge outwards again; it should merely "un-bulge" less.

Likewise per Shambolic's:

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The moon will pull on P1 more than on P2, so they will separate. This is equivalent to the bulge facing the moon. But it will also pull on P2 more than on P3, causing those two particles to separate as well. If you consider P2 to be at the center of the earth, this explains the tidal bulge facing away from the moon.

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... the separations from P1 to P2 to P3 would not be enough to cause the far side to bulge outwards; rather, it should only translate into it bulging 'inwards' less. However, this is not what actually happens, since it bulges 'outwards', so this explanation, though mainstream, does not satisfy enough to be acceptable prima facie. Yet, though I have not read upon relativistic explanations for the tidal bulge on the far side of the planet, I believe this is the hinge on which this gravitational theory pivots, that the lessened gravitational pull for P3 causes a bulge. I find this hard to swallow, however.

Of course, I truly hate going up against accepted theory (since I expect to get a lot of flack), but reason forces me to question it. This is why I suggested an alternative idea, where the electrostatic like-charge of the hemispheric ocean waters repel. Mind you, this is not a proven scientific theory, but merely a suggestion of how to look at it from another perspective. Coastal topography, land mass, ocean currents, ocean depths, all have their modifying affects, but the principle should be good a priori. Shambolic rightly points out that the idea of a magnetic field causing charge imbalance in the ocean mass is not currently accepted theory. However, this could be easily tested in a lab with any spherical vessel of saline solution spinning in a bipolar magnetic field: the upper hemisphere should see its fluid separate from any fluid displacement on the opposite side of the sphere. So if you bulge out fluid on one side (the fluid being magnetically charged), the fluid on the opposite side should bulge away in like manner, which would cause it to bulge proportionally on both sides. Has this experiment been tried by anyone to disprove this proposal?

Like I said, this is only a suggestion. For the present, the "relaxed" bulge on the opposite face of the globe is what is generally acceptable. Not to knock your textbooks, just a questioning of the reasoning behind this currently accepted theory. The Coriolis effect, mentioned by tmorten, would influence oceanic currents in how the water mass is redistributed around the planet, but I don't think it can be used to explain the counter-face tidal bulge. I had lived both in Maine and the South China seas, and can personally vouch for tides on the New England coast being very high, while those near the Equator are comparatively negligible... but I could be wrong.

So in my opinion, the question as to why there is a tidal bulge opposite from the Lunar gravitational pull, or the Sun, remains unresolved. If so, then it boils down to: Is there a magnetic charge effect on a hemisphere's water mass, and is this water mass dual charge measurable? Has anyone bothered to look for it? I'd like to know that it is not so, otherwise the far sided bulge of the planet is better explained via electrostatic charge than the 3 point gravity effect.
__________________
I have formally 'resigned' (tactical withdraw) from the Space-Talk boards; mine were many questions, ideas, but no real answers. Thanks. 04/10/04.

Disclaimer: Please note the ideas expressed here by me are cutting edge theory, very speculative in nature, and not physics as it is being currently taught. Caveat lector.

The magnetic and electric dipole moments of water molecules are known very well. They could not lead to 'bulges'. Any fluid, without external forces, forms a sphere, because it is most energy efficient.

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This is the part that puzzles me, that the delta "gm" between the closest and furthest surface on Earth pointing towards the Moon is not a positive-negative relationship, but merely a differential between two negative forces. By this logic, the bulge on the face towards the moon should bulge by a value of 3.48e-05 N/Kg, while the far side of the planet should be 'depressed' (not bulge) by 3.25e-05 N/Kg. This is not how the 3 point analogy (P1, P2, P3), mentioned by both Shambolic and Wintermute11, is understood to mean, where rather (per them) the far side bulging out, it is merely pulled less... not the same thing

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... because the water on the far side from the moon is 'pulled less', it shows up as an inertial force. So, the water is pulled away from the earth: it bulges.

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Originally posted by Coppernicus2
Why should the water on the far side of the Earth be 'pulled less'? Is it not responsive to the same gravitational force that pulls the planet's mass towards the moon as is all of the planet's mass? What makes that particular mass different, so that it responds instead as an 'inertial force'?

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What you're missing, I think, is that the moon does not simply orbit the Earth. They orbit each other around a common point. The Earth's motion around this point would cause the oceans to want to slosh to the side of the Earth furthest from the moon due to "centrifugal force". Meanwhile, the moon's gravity is tugging the water in the opposite direction. On the side of the Earth closest to the moon, the force of gravity is stronger...strong enough to overcome the "centrifugal force". On the far side of the Earth, "centrifugal force" wins out over gravity. Hence, a bulge on opposite sides.

I believe that explanation is correct...erm, with the exception of mentioning "centrifugal force" which I think makes physicists antsy.

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Originally posted by Coppernicus2
questioning whether the current explanation really addresses it, or is it merely a "brush off theory", since we don't really know?

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There is no such thing as a "brush off theory" in the context you describe.

As always, all the data is available to you. You're free to run the numbers yourself and see if they work.

Sorry, JoeC, but the bulge on the farside is not due to centrifugal force. And yes, "centrifugal force" does make me grind my teeth in disgust "Centrifugal force" would account for a much smaller force than whats causing that, and you would not get a distinct force gradient (change) between longitude.

Coppernicus2, the reason for the two bulges has already been explained. Just because you don't understand or don't like the answer doesn't mean its wrong. I don't know how better to explain it then "thats just the way it works. Get over it."

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Why should the water on the far side of the Earth be 'pulled less'? Is it not responsive to the same gravitational force that pulls the planet's mass towards the moon as is all of the planet's mass? What makes that particular mass different, so that it responds instead as an 'inertial force'?

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Thats quite a homework assignment Wintermute11 gave you Cop2,don't forget to check out the Roche limit,lagrangian points,etc,etc,and if you do any of the math,tidal forces falloff as the cube of the distance between the masses.
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alien life force

http://www.seamanship.co.uk/deck/navigator/TideWiz/history_of_tidal_predictions.htm

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Originally posted by wintermute11
Sorry, JoeC, but the bulge on the farside is not due to centrifugal force.

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My bad. I'm not sure how that bit of incorrect info got inside my head...but in searching the net after you corrected me, I saw lots of places that claim "centrifugal force" is the reason for the farside bulge. So I guess I have company in being wrong.

It did finally dawn on me though that the tidal bulges are, I guess, a similar effect to the "spaghettification" often discussed with regards to black holes and falling in one.

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And yes, "centrifugal force" does make me grind my teeth in disgust

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The price you pay for dealing with laymen.

Thanks tmorten for this very fine reference. I found most useful "Tides and the Moon" at:
http://www.seamanship.co.uk/deck/navigator/TideWiz/tides_and_the_moon.htm

This page explains in simple terms why there is a counterbulge of tide opposite the moon's declination orbital position, so that the bulge on the far side is greatest opposite the greatest Lunar attraction. This does not exactly fit my suggestion, however, since the larger counterbulge should be in the same hemisphere rather than the opposite hemisphere. Facts are facts, so no argument there.

The explanation that caught my attention particularly regarding the equatorial tides is:

"Although you may think that the points where the force is vertical have the greatest effect on the tides, here the tide producing force is acting against the earth's own gravitational force, which is about 9 million times stronger. Instead, the greatest force is produced where the direction of the tide producing force is horizontal with respect to the earth, as although the force is smaller, it is unopposed by any other force. This results in the water moving to create two bulges, at the points where the earth is closest to and furthest from the moon. The bulges would reach equilibrium when the tendency to flow away from the bulge under gravity is balanced by the tide raising force. The range of this tide on the equator is less than 1 metre."

So the tides at the equator are substantially lower than the northern and southern hemisphere tides, as I had noted, but the reasoning for this, which in essence incorporates the 'centripetal-centrifugal effects', is because the direction of a declined angle force from the moon is not directed at the equator. If it were, then the tidal bulges would have been greatest there instead.

Well, it looks like gravity is a better explanation than the one I presented, but it was fun thinking alternatively... blame it on the desert sun. Thanks for explanations guys!


__________________
I have formally 'resigned' (tactical withdraw) from the Space-Talk boards; mine were many questions, ideas, but no real answers. Thanks. 04/10/04.

Disclaimer: Please note the ideas expressed here by me are cutting edge theory, very speculative in nature, and not physics as it is being currently taught. Caveat lector.