[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.


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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