firstname.lastname@example.org (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 nonexpertsI 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?
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