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 <[email protected]> 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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