The Wobbling of the Earth and Other Curiosities

John Baez

December 26, 1999

Okay, I guess it's time for me to come clean and give the answer to my puzzles about the wobbling of the earth. I'm really disappointed with how little discussion this puzzle raised. The problem was that Jim Heckman instantly nailed the original puzzle, so to stir up discussion I was forced to raise some minor objections to his reasoning.

Remember the setup:

We're ignoring external forces and treating the earth as a rigid rotating body. The earth is roughly an oblate spheroid, and it's rotating almost around the principal axis with the biggest moment of inertia, I1. But actually the axis of rotation is slightly tilted away from the principal axis, so the earth wobbles. The other two moments of inertia, I2 and I3, are almost equal. Let's pretend they are equal and guess the period of the earth's wobble.

First of all, the wobble period doesn't depend much on the angle of tilt, as long as this angle is small. This follows from general principles of classical mechanics. As Ted Bunn put it:

>[...] I'd guess that one possible answer is based on the wonderfully
>useful fact that everything in nature is linear, to first order.  When
>you work out the equation of motion for this system, you'll be able to
>do perturbation theory in some small quantity.  The first-order
>equation will be the harmonic oscillator equation, since it always is
>for small departures from equilibrium, and we all know that the period
>of a harmonic oscillator doesn't depend on the amplitude.
>
>Of course, this doesn't *prove* anything: I'm just guessing that the
>first-order equation of motion will be the harmonic oscillator
>equation, because that's what it always seems to be in these
>situations.
One can actually make this guess rigorous! In any classical mechanics problem where a system has 1 degree of freedom and oscillates slightly about a generic stable equilibrium, to first order its equations of motion will be those of a simple harmonic oscillator. This is a theorem.

Note the word "generic" - a generic stable equilibrium is one that doesn't go away or become unstable under a small perturbation. One needs to check this, but it's not hard to do in the problem at hand. Perhaps it's just obvious that a slight perturbation ain't gonna destabilize the earth's spin.

"But wait," you cry, "this problem doesn't have just one degree of freedom!"

Ah, but the trick is to use conserved quantities to eliminate degrees of freedom! The rigid rotating body has 3 degrees of freedom, but we also have 3 commuting conserved quantities: energy, total angular momentum, and angular momentum about the z axis. Ultimately this enables us to completely solve the problem: it's "completely integrable". But if we hold off and only use 2 of our conserved quantities, we can treat the wobbling of the earth's axis as a problem with 1 degree of freedom.

"Hmm. Do we really need to do this? After all, even with *more* degrees of freedom, when a system oscillates slightly about a generic stable equilibrium, to first order its equations of motion will be those of a harmonic oscillator. It'll just be an oscillator with more degrees of freedom!"

True, but these can have different frequencies depending on which *way* they wiggle. Different vibrational modes! So it's good to know that in our problem this complication doesn't arise.

So.... with a little dimensional analysis we see that the answer depends only on I1, I2, and the period of rotation of the earth.

Now what?

To continue click here.


© 1999 John Baez
baez@math.removethis.ucr.andthis.edu

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