## The Wobbling of the Earth and Other Curiosities

#### Jim Heckman's Reply

#### 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... :-)
To continue click here.

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