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Keith Williams asked the following question-

>George says: "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."
>
>Can we have further explanation of this? My brain tells me that moving
>along an isobar (to make a simple analogy) can't move you to an area of
>higher pressure. Are equipotential shells not analogous to isobars?

======================

George replies-

I will try to explain a bit further.

In the case of isobars, they are surfaces of equal pressure, as Keith says,
and so moving along an isobar can not move you to an area of higher or
lower pressure. That's simple enough, and well understood.

Similarly, an equipotential shell is a surface of equal gravitational
energy, so in just the same way moving along that shell involves no change
at all in gravitational energy. That's EXACTLY the same as Keith's isobar
analogy.

With isobars, there's another quantity you can consider, however, the
CLOSENESS of their spacing, and anyone familiar with weather-maps knows
that when the isobars are closely-spaced then that means there's a high
pressure-gradient, and when there's a high pressure-gradient there's always
a lot of wind. There are some differences here with our potential-shell
analogy, in that we're used to seeing a pressure-contour map as a section
sliced through the atmosphere at or near surface level. Really, though, the
pressure distribution is three-dimensional, a series of isobar shells
(though I wouldn't like to guess just how the closeness of those shells in
the vertical direction affects the wind).

But I think the picture is clear: you can follow an isobar on a weather-map
and you can find it will lead you from a region of sparsely-spaced isobars
to other areas where they are closely spaced; from a weak wind to a storm,
all along that same isobar, at an identical pressure throughout.

Now the gravity analogy should be getting clear. The strength of the
gravity-field at any point; that is, the local value of g, is defined NOT
by which equipotential you are on at that point, but by how closely those
equipotentials are spaced there. If they are unusually close, then when you
put in a certain amount of work to move a certain mass from one
equipotential to another, you are doing it over a shorter distance. We know
that work = force x distance, and the work is the same, so that means that
the force must be greater. That force is the force of gravity on that mass,
or g. Where those same equipotentials are more widely spaced, the work in
moving between them is just the same, but the force, and therefore the
value of g is correspondingly less.

So g is not the same along an equipotential, it simply shows the closeness
of the spacings between them, and a map of the shells of equal-g would be
rather different from the map of equipotentials (and rather less useful).
There is some economic importance in this. The oil industry prospects for
deposits by looking for small changes in g from a survey vessel, which (of
course) travels along the sea-level equipotential. If g remained contant
over an equipotential, they would never observe any fluctuation.

However, we have to keep a sense of proportion here. On weather maps,
isobars can wander all over the place. Conversely, if the Earth was a
uniform sphere, the equipotential shells would be exactly spherical, and
so, as it happens, would be our equal-g contours. Even the ellipsoidal
effects caused by the Earth's spin give rise to changes which are less than
1%. And the hills and dips, that we've been discussing in such detail, are
molehills on that ellipsoid. So the gravity variations that we have been
discussing are small, and the surfaces of equal potential and the surfaces
of equal gravity, though not spherical shells, are pretty damn near to
that.

Has that helped, Keith?

George.

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