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    Re: Global oceanic tides,
    From: Trevor Kenchington
    Date: 2003 Aug 24, 13:48 -0300

    Eppo Kooi wrote:
    > My simple understanding is that the centres of gravities of both earth
    > and moon, disregarding the sun, should remain at unchanged distances.
    > Therefore, when the moon pulls the water mass to one side of the earth,
    > it has to be compensated such that the centre of gravity remains
    > unchanged. That can be obtained by having an equal displacement of water
    > on the opposite side of the earth. I do recall that that opposite mass
    > is not precisely of the same shape, due to land masses disturbing the
    > equilibrium.
    > I am sure there is a lot more to it, but the prime reason suffices for
    > most people, I guess.
    If it was a matter of "has to be compensated", we would have to suppose
    that every water droplet in the ocean acted in concert with one another
    and with the solid parts of the Earth. That just doesn't happen.
    To put it in fairly simple principles:
    Any moving object maintains its course and speed unless acted on by some
    force. That is one of Newton's laws but it is also the law that takes
    charge when you round your boat up into the wind and she carries her way
    past your mooring buoy and right into the boat on the next mooring to
    Despite that basic principle, the Earth and Moon rotate around one
    another instead of flying off into space. (Actually, they both rotate
    around a common centre which lies 2430 nautical miles below the surface
    of the Earth.) They maintain this rotation because the force of gravity
    pulls them together. It acts as a "centripetal force".
    However, by another of Newton's laws, every action has an equal and
    opposite reaction. (Trying jumping from your small boat onto your yacht
    and you'll find that the little one moves away, leaving you to fall in
    the water between -- at least that is what I achieved one night.
    Newton's laws really do work.) In the case of a centripetal force, we
    call the opposite one the "centrifugal force". Try tying a knot in the
    end of a piece of rope and whirling it around your head. You will feel
    the centrifugal force pulling on your hand. The knot is pulled by
    the centripetal force which is equal and opposite to the centrifugal.
    So for the Earth/Moon system to remain in balance, gravity must pull the
    two bodies together, while a centrifugal force tries to whirl them
    apart. At the centres of mass of the two bodies, those forces are equal
    but opposite in direction.
    Now, the pull of gravity (as most of us learnt in high school) varies
    with the square of the distance between two objects. However, it turns
    out that the difference between the pull of gravity and the pull of the
    centrifugal force depends on the cube of the distance (as George has
    noted). So, although the side of the Earth towards the Moon is not a
    whole lot closer to it than is the side away (some 8,000 miles for the
    diameter of the Earth, versus some 500,000 to the Moon -- unless I have
    that number way off), the difference in the cubes of those distances is
    quite substantial. As a result, although the centripetal and centrifugal
    forces are exactly balanced at the centre of the Earth, the
    gravitational pull of the Moon exceeds the centrifugal force when the
    Moon is overhead (at your zenith), whereas the centrifugal exceeds
    the centripetal when the Moon is at your nadir. It turns out (through a
    bunch of math that I have never tried to understand) that those local,
    vertical forces are exactly equal, though opposite in direction.
    Those forces are, however, quite small. If you stood still on a hot,
    tropical night with the Moon directly overhead, the tidal effect would
    reduce your apparent weight by an amount equal to the weight of the bead
    of sweat that ran down your face and dripped off your nose. That much
    force wouldn't do anything much for the tides. However, when the Moon is
    45 degrees off your zenith or your nadir, the imbalance between
    centrifugal and centripetal forces has a significant horizontal
    component. It is that sideways pull, sliding water across the surface
    rather than trying to lift it, which primarily drives the ocean tides.
    Again, if the Moon is 45 degrees off the zenith, water is pulled towards
    the geographic position of the Moon by centripetal forces exceeding
    centrifugal. If it is 45 degrees off the nadir, the excess centrifugal
    force pulls water towards the antipode of the Moon's geographic
    position. The forces are, once more, equal and opposite.
    And so we seen semi-diurnal tides.
    Eppo also mentioned land masses disturbing "equilibrium". That is a
    common misunderstanding or perhaps half truth. My above explanation
    deals with tide generating forces and, if the water were free to
    respond, it would produce Newtonian tides that looked like two watery
    hills, rotating around the globe once per lunar day (ignoring for now
    the effect of the solar tides). It has been clear for over 200 years,
    however, that no such hills would appear, even in the absence of land
    and with a uniformly-deep ocean. (The hills would act as waves and yet
    would be so long that they would progress as shallow-water waves, like
    tsunamis or like wind waves approaching a beach. Shallow water waves run
    at a speed set by water depth and the ocean just isn't deep enough for
    any wave to get around the world in 24 hours.)
    What really happens is that the tide generating forces excite natural
    oscillations in the ocean basins, with basins of different shapes and
    sizes oscillating at different frequencies. Hence we see such oddities
    as the tides at the Magdalen islands being very small and strictly
    diurnal when a hundred miles due south the Bay of Fundy has almost
    perfectly semi-diurnal tides with the greatest amplitude of any on the
    planet. The Gulf of St.Lawrence oscillates with one pattern, with a null
    point near its centre, where the Magdalen Islands lie, whereas the
    waters in the Bay of Fundy follow a quite different oscillation, even
    though the pull of the Moon on each of them is almost indistinguishable.
    So ... semi-diurnal tide generating forces predominate but not all tides
    are semi-diurnal. The forces pull the water towards a point directly
    under the Moon and towards the antipode of that point but high water
    does not necessarily happen when the Moon is on your meridian. The
    complications are due less to land masses as to the shapes and depths of
    the sea.
    Yes, as Eppo noted, when it is high water where you are, it won't
    necessarily be either high or as high on the opposite side of the
    planet. But then again, it won't necessarily be either high or as high a
    dozen miles down the coast either.
    Trevor Kenchington
    Trevor J. Kenchington PhD                         Gadus@iStar.ca
    Gadus Associates,                                 Office(902) 889-9250
    R.R.#1, Musquodoboit Harbour,                     Fax   (902) 889-9251
    Nova Scotia  B0J 2L0, CANADA                      Home  (902) 889-3555
                         Science Serving the Fisheries

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