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The gravitational acceleration on Earth is approximately $ 10 \mathrm{m}/\mathrm{s}^2 $. Compared to this, the tidal effect of the Moon's gravity gives a local variation in the acceleration of approximately $ 9 \cdot 10^{-7} \mathrm{m}/\mathrm{s}^2 $, that is, seven orders of magnitudes less. The level of the water can rise $ 1 \mathrm{m} $ during high tide, which is only three and a half orders of magnitude smaller than the depth of the ocean.

How can this small variation in gravity move so much water and cause such a strong effect on the seas?

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A particularly nice explanation of the tides is given by Feynman in his Character of Physical Law lectures at Cornell, Part 1, The Law of Gravitation, at about 25 minutes in. –  Emilio Pisanty Oct 2 '13 at 17:55

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The relevant "100%" from which you should calculate the percentage isn't the depth of the ocean but the radius of the Earth $$ R\sim 6,378,000\,{\rm m} $$ Multiply this $R$ by $10^{-7}$ and you will get $0.6$ meters, a reasonable estimate for average tides.

You must understand that the surface of the ocean always tries to create an "equipotential surface" – connect all points that have the same gravitational potential. The Earth's gravity (plus the centrifugal potential) adds the major contribution to the potential and, as you said, the Moon modifies this function by corrections that are 7 orders of magnitude smaller and that are anisotropic (different in different directions). That's why the ellipse we get because of the Moon will differ from the previous one by corrections of order $10^{-7}$, too.

For example, if you imagine the Moon-less Earth to be a sphere, its ocean is spherical, i.e. ellipsoid with semi-axes $a=b=c$. A correction to the original potential that is 7 orders of magnitude smaller will create $|a-b|/a$ of order $10^{-7}$. All these calculations may be done much more accurately although the precise shape of continents etc. is needed for learning the precise shape of tides at various points of the real globe.

Whether there is 100 meters of water, 11 km of water, or (unrealistically) 3,000 km of water beneath a point on the ocean surface plays no role in the fact that the ocean will be elevated or suppressed by a few meters or so.

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Oh! This also explains why the difference between the polar and equatorial radius of Earth is about 3•10^(-3) times the radius, as the rotation of Earth contributes about 3•10^(-3) times the gravitational acceleration on the Equator surface. –  b_jonas Jan 11 '13 at 10:46
Although I think this is alluded to, it isn't really explained why lakes have such small tides then. Or, take an extreme case - my coffee does not show tide variances of ~0.6m. Thankfully! :-) –  Johan Jan 11 '13 at 10:53
Lakes cannot have strong tides because they do not span enough of the globe to "see" the large-scale variations in the gravitational potential. As far as a lake is concerned all the moon does is add a small linear perturbation, so the lake's surface will incline in response, shifting water from one side to the other. This will be damped from the full effect by a factor of approximately (lake radius)/(Earth radius). –  Emilio Pisanty Jan 11 '13 at 14:03
Note, also, that tidal effect from the sun is of the same order of magnitude as that from the moon. So now we can easily explain plus-or-minus nearly a meter of tidal action. Then add some dynamics and see what happens. –  dmckee Jan 11 '13 at 14:30
@b_jonas - yup, absolutely. Concerning the fat Earth, you may also want to see another SE question physics.stackexchange.com/questions/8074/… motls.blogspot.cz/2011/07/why-is-earth-so-fat.html?m=1 –  Luboš Motl Jan 11 '13 at 14:56

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