[time-nuts] Q/noise of Earth as an oscillator
Peter Reilley
preilley_454 at comcast.net
Wed Jul 27 10:46:35 EDT 2016
If you consider viewing earth from above the equator at a long distance
and imagine a
spot on the surface of the earth. That spot will appear to have a
sinusoidal motion.
The frequency of the sinusoid exhibits a decay. That decay can be
considered as the Q
of the earths rotation.
Pete.
On 7/27/2016 9:00 AM, jimlux wrote:
> On 7/27/16 5:43 AM, Michael Wouters wrote:
>> On Wed, Jul 27, 2016 at 8:08 AM, Attila Kinali <attila at kinali.ch> wrote:
>>
>> "I am not sure you can apply this definition of Q onto earth."
>>
>> It doesn't make sense to me either.
>>
>> If you mark a point on the surface of a sphere then you can observe
>> that point as the sphere
>> rotates and count rotations to make a clock. If you think of just a
>> circle, then a point on it viewed in a rectilinear coordinate system
>> executes simple harmonic motion so the motion of that point looks like
>> an oscillator, so that much is OK.
>>
>> But unlike the LCR circuit, the pendulum and quartz crystal, the
>> sphere's rotational motion does not have a
>> resonant frequency. Another way of characterizing the Q of an
>> oscillator, the relative width of the resonance, makes
>> no sense in this context.
>>
>
> There's also the thing that "things that resonate" typically have
> energy transferring back and forth between modes or components: E
> field and H field for an antenna; kinetic vs potential energy for
> pendulums and weight/spring; charge and current (C & L, really E
> field/H field again).
>
> Spinning earth is more of an "rotational inertia and loss" thing, with
> zero frequency, just the exponential decay term.
>
> If you think of a single measurand in any of these scenarios you have
> at the core some sort of exp(-kt)*cos(omega*t+phi) and we're relating
> Q to the coefficient k.
>
>
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