## The Wobbling of the Earth and Other Curiosities

#### December 20, 1999

```John Baez wrote:

> Jim Heckman wrote:

> >> Okay, here comes the puzzle. The earth is not spinning exactly
> >> around the first principal axis. It's a bit off, so it wobbles. Estimate
> >> the period of this wobble!

> >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.)

> Are you also guessing that the wobble period doesn't depend on
> the wobble tilt, or something like that?  And is there some reason
> you're guessing this?
```
Hmm... Wobble tilt would indeed seem to be an initial condition that I overlooked, wouldn't it? :-) Even worse, I didn't even think through clearly what you meant by "wobble" -- are you talking about *precession*, or *nutation* (assuming that the latter is possible in this situation)?

As to the reason for my original guess, I was shooting from the hip, but after thinking about it more, I see my intuition was that since L ~ I/T and L is conserved, dI/I = dT/T -- which of course says nothing at all obvious about the *period* of the wobble, but would instead seem to give an *amplitude* for the relative change in the length of the day in terms of the amplitude of the nutation -- assuming again that nutation is even possible. :-)

To get *period*s out of all this, my first reaction is that you need to do some kind of Fourier analysis, which is a little beyond my abilities and/or motivation right now -- at least not in your 3 minutes, you stud you!

It occurs to me that I could probably get a handle on this by digging out Goldstein or some other buried text on classical kinematics/dynamics -- but that would be cheating, wouldn't it? (Anyway, I'm not sure where G. is hidden in all my unpacked boxes of books. :)

```> >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... ~70 days?

> I'd prefer not to comment on this until I get a few more takers - I don't
> want to spoil the fun.
```
Well, in light of my comments above, you can probably throw my guess out for now, but I'll try to find time to think about it some more... :-)