The Wobbling of the Earth and Other Curiosities

John Baez

December 26, 1999

Okay, now we've seen that the period of the earth's wobble depends only on the moments of inertia I1 and I2 together with the period of rotation of the earth - known to nonphysicists as "a day".

Jim Heckman instantly guessed an almost correct answer:

>Well... On dimensional grounds alone, it seems like it would almost
>*have* to be ~( I1 / (I1-I2) ) times the period of the earth's rotation...
>(I'm *guessing* that the wobble period -> infinity rather than -> 0 as
>earth -> spherical.)
Actually the answer is I2/(I1 - I2) times the period of the earth's rotation, but this doesn't matter much, since I1 is so close to I2.

So the only part of this problem where you need to know *facts about the earth* is in guessing I2/(I1 - I2). This is related to the oblateness of the earth.

When I got this part of the problem, I guessed that the earth is round to about 1 part in 1000. So I guessed that the wobble period was about 1000 days.

Jim Heckman knew more numbers:

>Well... Again dimensionally, inertia is (mass * distance^2). I'm pretty 
>sure earth's radius is ~6700 kilometers and I *think* the polar vs. 
>equatorial flattening is ~50 kilometers, so
>( I1 / (I1-I2) ) ~ ( 6700^2 / (6700^2 - (6700-50)^2) ) ~ 70
So he guessed a wobble period of about 70 days!

According to Goldstein's _Classical Mechanics_, the actual number is

(I1 - I2)/I1 = 0.0033

so Goldstein calculates a wobble period of 300 days.

But what's the real answer?

Let me quote Goldstein:

"An observer on the earth should therefore find that the axis of rotation traces out a circle about the North Pole once every 10 months. Something vaguely resembling this phenomenon has actually been observed. The amplitude of the precession is quite small, the axis of rotation never wandering more than 15 feet from the North Pole. But ther orbit is quite irregular, and the fundamental period seems to be about 427 days rather than the 300 days predicted. The flucutuals are ascribed to small shifts in the mass distribution such as are caused by atmospheric motion, while the difference in the period arises from the fact that the earth is not completely rigid but has the elastic properties of a material like steel."

Atmospheric motion??? This sounds suspicious. Ain't the atmosphere really light compared to the rest of the earth?

Anyway, both Jim and I were right to within an order of magnitude.

Now... when Jim posted his answer, I gave him some flak about guessing the formula I1/(I1 - I2). It's not the only dimensionally correct formula. How do we guess the right one?

I don't know any super-slick method. But if you gaze at Euler's equations for the motion of a rotating body, you'll eventually see that the right answer is I2/(I1-I2). The equations look like this:

I2 ----  = (I3 - I1) W3 W1
and cyclic permutations thereof, where (W1,W2,W3) is the angular momentum as measured in the rotating coordinate system of the body itself. Remember that I2 = I3 in this problem, so W1 is constant, while the vector (W2,W3) keeps rotating around.

Finally, I asked a bunch of questions about precession and nutation... more on those later.

To continue click here.

© 1999 John Baez