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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
windward!

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
                      http://home.istar.ca/~gadus

   

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