 Check out the bookstore at IRBS.com |
|
|
|
|||||
|
||||||
From: George Huxtable (no email)
Date: Fri Aug 15 2003 - 09:50:31 EDT
Sharon Casey (Blackwood <>) wrote-
>I remember hearing in an astrophysics class that the gravitational pull
>causing peaks and dips on the ocean surface was caused by the
>interaction of the earth's gravity and the gravitational pull from space
>and the moon. Without the tidal swells, the surface of the ocean is
>as varied as the surface of land.
>
>Sharon Casey
Response from George.
Sharon's contribution is part-right.
The STATIC variation of the ocean surface from a spheroid (that is, its
knobbliness, the hills and valleys that we are discussing) is caused by
local variations in gravity due to non-uniformities in the crust. It's
mainly the effect of continental masses which swim slowly on the magma
below. Above an underwater sea-mount made of lead, there would be an
increase in gravity, and the sea-level would be a bit lower there than it
would be if the lump of lead was absent. The sea-level would everywhere
follow the gravitational equipotential, a word which means that there's no
energy-change due to gravity as you travel anywhere around that surface,
which applies the same to a ship and to a water-drop.
The DYNAMIC effect of other bodies in the universe (effectively just the
Moon and Sun combined, other planets contributing rather negligibly,
"space" itself having no effect) is to add a local gravity-gradient which
tries to pull the ocean surface, as described above, by an ellipsoid,
pointing in the general direction of that combined force. That ellipsoid
would shift the water-surface by no more than a foot or so: a very small
effect compared with the hills and valleys we have been discussing.
However, that ellipsoid points in a direction in space which changes only
slowly, shifting only as the Moon and Sun move against the stars. On the
other hand, the Earth, with its oceans, is spinning with respect to that
ellipsoid once a day, so with respect to the Earth these tidal forces are
trying to move an wave-ellipsoid of water, 1 foot or so high and low, right
around the Earth each day..
Because of the inertia of the immense water-masses involved, together with
the obstruction caused by the continents (which restrict free flow from one
ocean basin to another) the ocean surface can't, anywhere near, respond to
these forces in the necessary time. Instead, the water swirls and surges
around the ocean basins, in a pattern that's so complex that it defies
calculation. In midocean, the resulting tides are usually only inches high,
but near the coasts the shapes of basins can amplify these alternating
water-flows to produce the 40-foot ranges we see in certain places.
Because energy is being put in to the oceans in this way, the surface
doesn't, over a short term, quite follow the equipotential, but on average,
it does. The energy comes from the gradual slowing-down of the earths
rotation, mainly caused by these tides; interestingly, this relates to the
leap-second clock changes we have recently been discussing.
These effects also allow small-boat sailors in tidal waters to quicken
their passages if they play their tides right, and trap them in rough
waters off a tidal headland when they get things wrong (as many of us well
know).
There are also local effects caused by short-term barometric changes which
can shift the surface, sometimes by a foot or so.
Geophysics isn't really my subject, so I hope I've got things about right.
If not, someone please tell me.
==========================
Now for David Hoyte <>, who said-
>Do we have any serving or retired merchant-marine or naval officers
>on this List who could comment on this question from their experience?
>I find it is usually best to get out of the classroom and look at what
>happens in >real-life when simplifying assumptions can produce any answer
>you please.
Well, I ask David what "simplifying assumptions" are being made, that he
objects to, and that "can produce any answer you please"? The basic
physical law is that the free surface of the water is a surface of equal
gravitational potential, for every water-drop and for any ship floating on
it, and there's NO gravitational energy change between any point on that
surface and any other point, not for a water-drop and not for a ship.
Centrifugal forces due to Earth rotation will affect the position of the
surface, but will affect the water, and a ship floating in it, in exactly
the same way. Tidal and atmospheric effects can make short-term
fluctuations, as noted before (but again, affect ship and water just the
same).
David may prefer to "get out of the school-room", but an attempt to change
the basic laws of science by asking for the opinions of mariners will be
futile. To support his case he needs either contrary numerical observation
or contrary logical argument, but presents neither.
He added-
>The hoary school-book question about the time taken by a man to swim a
>mile against a current , then turning round and swimming with it,
>compared to the time taken by the man swimming both directions in still
>water,
Perhaps he can explain the relevance of this analogy, a man swimming in a
current, because I have not understood it. If he is still thinking about
peaks and valleys, and making up on the downhill bits what has been lost on
the uphill bits, it's not like that, not at all. Gravitationally speaking,
the oceans are everywhere FLAT, no favoured patches or disadvantaged ones,
hills or valleys, anywhere. It's just that some bits of ocean are nearer to
the centre of the Earth than others are: a fact that no mariner or ship is
(or needs to be) aware of.
Then he goes on to say-
>... compared to travel on a surface of uniform 'g' . . . depending on what
>assumptions >you care to make.
"Uniform 'g'" is a completely different matter. The acceleration due to
gravity, for which the symbol g is used, varies with latitude (because of
the centrifugal force and the ellipsoidal shape of the Earth) and altitude.
If we take it to be 981 cm per square second at Paris latitude at
sea-level, it's about 983 at the poles at sea-level, and about 978 at the
equator, at sea-level, though it has local lumpinesses too. 'g' actually
measures the rate-of-change of the gravity potential with height. A surface
of equal 'g' is NOT an equipotential, and the water level does NOT follow
it. So it would only be possible to "travel on a surface of uniform 'g' "
(as David suggests) across changing latitudes, by devising a submarine that
could also fly. 'g' is irrelevant to our argument.
George Huxtable
================================================================
contact George Huxtable by email at , by phone at
01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
Lane, Southmoor, Abingdon, Oxon OX13 5HX, UK.
================================================================
|