## The Wobbling of the Earth and Other Curiosities

#### Jim Heckman's Reply

#### December 17, 1999

baez@galaxy.ucr.edu (John Baez) wrote:
> Here's a nice puzzle about the wobbling of the earth's axis
> of rotation. I'm sure some real experts on this subject are
> lurking out there, so I'd appreciate it if they kept quiet
> until some of the nonexperts

I hope I qualify as a "nonexpert" -- especially if I'm wrong!
> [...]
>
> Let's call the biggest moment of inertia I1, the middle one
> I2, and the smallest one I3.
>
> [...]
>
> Now, the earth is roughly an oblate spheroid, so it's rotating
> almost about the first axis, the one with the biggest moment of
> inertia. It's a bit like a spinning frisbee, but not so
> dramatically flattened. I2 and I3 are almost equal - not
> quite, but for starters let's pretend they are.
>
> 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.)
> The nice thing about this puzzle is that you don't need
> to look up a bunch of numbers to solve it! You need to
> understand physics, but there's just one non-obvious number
> that you need to know -

I'm guessing ( I1 / (I1-I2) )...
> and if you're reasonably lucky, you
> can guess it to within an order of magnitude.
>
> Takers, anyone?

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?

To continue click here.

© 1999 John Baez

baez@math.removethis.ucr.andthis.edu