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Fred Hebard said-

>I might note,
>however, that work would be done moving across a gradient in the
>gravitational field even if there were no change in sea-surface
>elevation.

We have to be careful here. The Earth is immersed in its own gravitational
field. This consists of a series of shells called equipotentials, a long
word which means nothing more than that there is NO work done at all
against gravity in moving from any point to any other on the same
equipotential. These shells fit inside each other like Russian dolls, each
one having a different potential, so work is done (or regained) if you move
from one shell to another, as you do when climbing (or descending) a
mountain.

Each of these shells is roughly ellipsoidal, the combination of the effect
of gravitational attraction ang the centifugal effects of a rotating Earth;
but because of the uneven mass distribution within the Earth, there are
some superimposed dips and valleys, the objects we have been discussing. At
any point on an equipotential shell, its surface is exactly at right angles
to the local direction of gravitational attraction,"g" (including those
centrifugal effects into the gravity). This simply has to be so: if it
wasn't, then as you moved along that surface, there would be some component
of g causing a force along your direction of travel, and you would be
giving-up or gaining energy, which would defeat the definition of an
equipotential.

The most important of these equipotentials for our purpose, is that at
sea-level. In the absence of disturbing factors, such as driving winds or
barometric differences or local waves or tides, sea-level adjusts itself to
conform to the sea-level equipotential. If not, one patch would find itself
with more gravitational energy than another, which would cause a water-flow
to adjust the levels until there was no longer a difference. When they say
"water finds its own level", that's the level it finds. Because that
equipotential  surface has dips and bumps, so does the sea-level.

However, those dips and bumps are just differences in the distance to the
earth's centre. They are not like the hills and valleys on the Earth's
solid surface. Because they are exactly on an equipotential, there is no
work involved in travelling from any one point on the Earth's ocean surface
to any other. They are NOT dips and bumps in the gravitional sense: there's
no uphill and no downhill. And that's just as true for the ship floating in
the sea as it is for the sea surface itself. Except for those minor
disturbing factors listed above, there are NO changes of gravitational
energy AT ALL when moving around ANYWHERE on the ocean surface.

So how does all this relate to "g", which describes the STRENGTH of the
Earth's gravitational pull? Well, g is related to the spacing between the
equipotentials, rather in the same way as on a land-map, in which the
ground-slope corresponds to the closeness of the contours. The closer the
equipotentials are (which happens near the poles) the greater is the value
of g. Near the equator, the potential contours are more widely spaced, and
so g is less. You could if you wished draw out a set of shells of equal
values of g but it would not be very helpful to do so, because such a shell
of equal g is NOT itself an equipotential, and the sea surface does NOT
follow it.

It's possible to go from one value of g to another, always following an
equipotential path, and there will be changes in elevation above the
Earth's centre along that path, but being an equipotential path, there will
be no work done against gravity. None at all. In fact, this is exactly what
happens when a ship travels from one latitude to another and from one value
of g to another, over the oceans.

So when Fred Heberd states-

>I might note,
>however, that work would be done moving across a gradient in the
>gravitational field even if there were no change in sea-surface
>elevation.

It's not entirely clear to me what he means here, but whatever it is, I
think it must be wrong.

Then he adds-

>It would seem that accounting for a 400-meter-equivalent change in
>gravity might be a worthwhile consideration in ship routing.

I ask Fred what exactly he means by "a 400-meter-equivalent change in gravity".

I would like to point out that the matters I have been describing are not
those of any scientific controversy. They have been accepted and understood
rather well for over 100 years, and are not a matter of current bickering
or dispute. However, like all scientific theory, it's always open to
challenge by anyone who can show measurements or logical arguments that
contradict it.

George Huxtable.

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01865 820222 (from outside UK, +44 1865 820222), or by mail at 1 Sandy
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