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Re: TIDES, was Asteroid Problem



At 05:38 PM 9/1/01 -0400, I wrote:

Write the asteroid's gravitational field g(X) as a Taylor series. Write
X = R + x
where R is the vector from the asteroid to the center of the earth, and x
is the vector from the center of the earth to the point where we want to
evaluate g(X).

Additional suggestion: It is also instructive to write out the asteroid's
gravitational _potential_ phi(X) as a Taylor series.

Since I suspect there are folks on the list who are not 1000% comfortable
with this, let me spell it out.

phi(X) = phi(R) (1a)
+ x_i del_i phi(R) (1b)
+ (1/2) x_i x_j del_i del_j phi(R) (1c)
+ (1/6) x_i x_j x_k del_i del_j del_k phi(R) (1d)
+ .... (1*)

where we are using the traditional albeit inelegant notation for evaluating
derivatives, namely:
del_i phi(R) means [(d/dx_i) phi(X)] evaluated at X=R (i.e. x=0)
because we are taking R to be a constant and it wouldn't make much sense to
set X=R before taking the derivative.

We are summing over repeated indices. For instance term (1b) is just (x
dot grad phi).

Homework: Show that equation (1) is the correct multi-dimensional Taylor
series. Grind it out component by component if you have to.

We know phi(X) is given by the law of gravitation, proportional to 1/|X|,
so you can in fact evaluate the RHS of equation (1) without any great brain
strain. Homework: Grind it out. This is very good exercise.

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

OK, enough math; let's do a little physics.

Why mention the potential? Why not just calculate the gravitational field
directly?
1) This is the clever way to address questions about the HEIGHT of the
tides. Ask yourself how the tidal potential perturbs the earth's built-in
potential.
2) Many people find a scalar field (such as phi(X)) easier to visualize
than a vector field (such as g(X)).

We can understand the terms in equation (1) individually:

Term (1a) is a constant contribution to the potential. It is not
interesting. We can make it go away by exercise of gauge invariance.

Term (1b) is a linear contribution to the potential. It represents a
plane, tangent to the exact asteroidal potential, tangent at the earth's
location. It represents a steady uniform gravitational pull. It is not
very interesting, because it is balanced by the earth's orbital
motion. Another way of saying almost the same thing: We can make this
term go away by exercise of Einstein's principle of relativity: it affects
our reference frame AND everything we would like to measure, affecting them
all equally.

Term (1c) is the lowest-order interesting term, the lowest-order term that
we can't get rid of. It is what drives the usual tides. It is a
saddle-shaped contribution to the potential. It curves downward in both
directions (+-) along the axis passing through earth and the asteroid,
representing tidal stretching. It curves upward in both directions (+-)
along the two perpendicular axes, representing tidal compression.

Term (1d) is there to address a question Ludwik has asked a couple of
times. When the asteroid is far away, this term and all higher-order terms
are small. When in doubt, calculate this term and see if it is big enough
to be interesting.

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

Once you have the potential you can calculate the field if you want:
g(X) = del phi(X)

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

At 07:38 PM 9/1/01 -0500, Kossom wrote:
Why should smaller bodies of water have smaller tides?

Think about term (1c). It is a second-order shape. Near the middle of the
saddle, things are pretty flat. As you go farther out, things get more
interesting.

In elementary physics, we like to use the lowest-order approximation to
everything. Well, folks, in this problem THERE IS NO LINEAR APPROXIMATION
that is of any use. The lowest-order interesting term is nonlinear.