## The Wobbling of the Earth and Other Curiosities

#### 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?