[time-nuts] Q/noise of Earth as an oscillator
Chris Caudle
chris at chriscaudle.org
Wed Jul 27 11:33:58 EDT 2016
On Wed, July 27, 2016 8:58 am, Azelio Boriani wrote:
> Exciting the Earth with a new frequency (and an adeguate amount of
> energy) sets a new rotational speed: you cannot retune a (for example)
> quartz crystal in the same way...
Does that imply that this value is not constant:
> On Wed, Jul 27, 2016 at 3:42 PM, Tom Van Baak <tvb at leapsecond.com> wrote:
>> And if you take the classic definition Q = 2 pi * total energy / energy
>> lost per cycle then it would seem earth has a Q factor.
>>
>> In fact, if you use real energy numbers you get:
>>
>> - total rotational energy of earth is 2.14e29 J
>> - energy lost per cycle (day) is 2.7e17 J
>> - so Q = 2pi * 2.14e29 / 2.7e17 = 5e12, the same 5 trillion as my
>> earlier calculation.
My first intuition is that energy lost per cycle would not increase as the
square of angular velocity, but angular kinetic energy does increase as
the square of angular velocity. The energy losses are from a complicated
interaction of tidal forces and fluid friction between crust and
atmosphere/ocean and mantle and outer core, so I don't trust that my
intuition is correct at all. If however that turns out to be correct in a
rough sense then adding energy to speed up the rotation would change the
period and the Q, which isn't a property of most things with high Q that
we use as stable clocks.
Who has a globe on magnetic bearings in a vacuum chamber and will run the
experiment for us?
--
Chris Caudle
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