[extropy-chat] Juicy Classical Physics Problem Involving Gravitational Potential

[Lee Corbin]

First of all, a clock at the Earth's north pole runs *exactly* at the same
rate as does one on the equator. Did you know that?  Yet the polar radius
is about 13 miles less than the equatorial radius, so the first supposition
might be that the polar clock would run more slowly.  It is, after all,
immersed more deeply in the Earth's gravitational field (by general relativity).

But the speed of the clock on the equator, by special relativity, is less
because of its motion.  Amazingly, these two effects exactly cancel, and
the clocks progress at the same rate!  (I actually worked this out a year
or two ago, and found that the difference was less than 10^-16.)

Now this isn't really unexpected, I suppose, because of the shape of the
Earth.  Newton understood that ocean water at the equator was at the
same "gravitational potential" as water at a pole;  he had a memorable
diagram of a well that went down from the north pole to the center of 
the Earth, and another well dug  to the same point ninety degrees away
at a point on the equator. He argued that you wouldn't expect water
to flow "downhill" from the equator to the pole.  (A reason is, of course,
the centrifugal force that keeps the equatorial water 13 miles higher.)

Newton---but not I---was able to calculate the spherical oblateness of
the Earth by carefully regarding this notion of gravitational potential. (He'd
be wonderfully jazzed by the generalization to relativity theory spoken of
above.)

But my Newtonian calculation---problematical in some way---is off by a
factor of two! Perhaps someone here who really likes classical physics
can see what is going wrong.  Of course, I *could* find some alternate
derivations on the web, or find some in a book somewhere, but that
would take all the fun out of it.  Nothing like trying to work through
something yourself to really master the concepts.


Let  r  be the equatorial radius and R the polar radius.  The facts are
that R is 6357000 meters (or 6357 km), and r is 6378 km, a
twenty-one kilometer difference.

But my calculations below show that r "should" be only 6368 km,
that is, just 11 km more than R instead of the correct value of 21 more.

(One may point out that as a cc of water is raised by a hypothetical
elevator shaft to the equator, it gets kinetic energy from the walls of
the shaft.   This ends up being some momentum, (1/2)mv^2.)

Now v equals omega * r,   or, as I write it, wr.

The next equation, and what I'm trying to go by,  reads "the sum of
the K.E. of a particle at the equator  and  the remaining energy it
needs to get to infinity must equal the energy that the particle at the
pole would need to acquire to get to infinity".   Is there a problem
with that? It certainly raises questions in my mind, and one's that
I don't have clear answers for.

Anyway, in symbols

    (wr)^2  +   IntegralFromrToInfinity[ GM/r^2 ] =  IntegralFromRToInf[ GM/R^2 ]

where velocity = wr, and I suppress m (the mass of the particle) from
both sides of the equation.  (A note on GM:   we have F = GMm/r^2 from
Newton's formula, and so stripping m gives force per mass, or GM/r^2.)

The integral of 1/r^2 is  -1/r, so

   (wr)^2    -   [GM/r] eval from r to Inf   =   - [GM/R] eval from R to Inf

or

   (wr)^2  +  GM/r   =  GM/R

What is cool about this is that it allows one to solve for r, the equatorial
radius, in terms of R, the polar radius.  The only problem is omega, w,
the frequency of the Earth's rotation, which must be taken into account.
We have   w = 2pi/T  (see http://en.wikipedia.org/wiki/Angular_frequency)

Now T is the period of the Earth's rotation, which is not quite 86,400 
seconds because we must consider the sidereal day, not the solar day.
This turns out to be 86,160 seconds, because it's four minutes a day
different, or 4x60 = 240 seconds.

Then 2pi/86160   is, by my calculator, 7.2924x10^-5

What is GM?   G is 6.67 x 10^-11, and the Earth's mass is M = 6x10^24 kg
approximately.  When I do it carefully, I get GM = 3.9885 x 10^14, using
values I found online.

So then the equation   (wr)^2 + GM/r  =  GM/R  becomes

w^2Rr^3 - 2GMr + 2RGM  =  0

which is unfortunately a cubic.  Nonetheless, some calculators and some
on-line programs solve it.  The answer is, alas, r = 6368000 meters only,
not the hoped for 6378 km.

Lee

P.S.  Here is an intermediate step in the above calculation:

6.2742x10^7 = 3.9885x10^14/r + 5.318x10^-9xr^2 / 2