# [time-nuts] Q/noise of Earth as an oscillator

Tom Van Baak tvb at LeapSecond.com
Tue Jul 26 15:36:37 EDT 2016

```Hal Murray wrote:
> What is the Q of the Earth?  It might be on one of your web pages, but I
> don't remember seeing it.  Google found a few mentions, but I didn't find a
> number.

3.1415 * 86400 * 365 * 100 / 0.002 = 5e12, or 5 trillion.  Here's why:

(1)
Among other things, the quality-factor, or Q is a measure of how slowly a free-running oscillator runs down. There are lots of examples of periodic or damped oscillatory motion that have Q -- RC or LC circuit, tuning fork, pendulum, vibrating quartz; yes, even a rotating planet in space.

In the high vacuum of space you'd think the Earth would rotate forever, but the moon affects earth's rotation; it causes "tidal friction". Astronomers have measured the slowing as roughly "2 ms per day per century".

For example, the average length of a day in 1900 was about 86400.000 seconds and the average length of a day in 2000 was about 86400.002 seconds. That's a deceleration, or slight loss in rate, or slight gain in period, of 2 ms/day per century. Since the spinning earth oscillator is slowing down, you can calculate its Q factor.

(2)
The common definition of Q is 2pi times the number of periods it takes for the energy to fall to 37% (1/e). Although correct and simple, it is sometimes awkward to use this definition. I mean, it can take forever to lose 63% of your energy if Q is really high or the period is really long.

Another definition of Q is 2pi over dE/E, the latter term being called the "decrement", which is the relative energy loss per period. For example, if your wine glass oscillator loses 1% of its energy each period, then its Q = 2pi/1%, or 628.

This formula for Q is more convenient because you can do it with data from a single period. For earth the period is 1 day. So what's the decrement, the relative energy loss for one day?

(3)
Rotational energy goes as w^2, where w (omega) is the rotation rate. This means energy drops twice as fast as rotation rate drops.

And how fast does the earth rotation rate drop? That would be 2 ms per day per century, which is 2.3e-8 per century, or 2.3e-10 per year, or 6.3e-13 per day. So the relative loss of energy per period (day) is twice that: 1.3e-12. Plug that into the decrement definition of Q and your calculator gives 5e12.

So the Q of the earth is 5 trillion.

/tvb

p.s.

This Q is really high, and you might expect a similarly superb Allan deviation. But no, not at all:

http://leapsecond.com/museum/earth/

The reason the ADEV is so poor compared to the Q is that Earth, as a timekeeper, is an unstable mess inside and out. There are all sorts of massive perturbing factors from the liquid core to the swirling atmosphere to the annual redistribution of water and biomass; these all affect the short-term performance.

----- Original Message -----
From: "Hal Murray" <hmurray at megapathdsl.net>
To: "Tom Van Baak" <tvb at leapsecond.com>; "Discussion of precise time and frequency measurement" <time-nuts at febo.com>
Cc: <hmurray at megapathdsl.net>
Sent: Saturday, July 23, 2016 5:59 PM
Subject: Q/noise of Earth as an oscillator

>
> tvb at LeapSecond.com said:
>> Earth is a very noisy, wandering, drifting, incredibly-expensive-to-measure,
>> low-precision (though high-Q) clock.
>
> What is the Q of the Earth?  It might be on one of your web pages, but I
> don't remember seeing it.  Google found a few mentions, but I didn't find a
> number.
>
> I did find an interesting list of damping mechanisms in a geology book.
> Geology-nuts are as nutty as time-nuts.  Many were discussing damping of
> seismic waves rather than rotation.
>
> I've seen mention that the rotation rate of the Earth changed by a few
> microseconds per day as a result of the 2011 earthquake in Japan.  Does that
> show up in any data?  Your recent graph doesn't go back that far and it's got
> a full scale of 2000 microseconds so a few is going to be hard to see.
>

```