Re: slowing down the earth's spin

[David Bowman <dbowman@TIGER.GEORGETOWNCOLLEGE.EDU>]

Regarding Carl Mungan's question about the spindown of the Earth:

> Barger and Olsson page 295 claim that the earth's rotation is slowing
> down at 44 ns per day due to tidal friction. I have checked the math
> and come up with a similar answer to within a factor of 2. (Most of
> the discrepancy is probably due to what value to use for the moment
> of inertia I of the earth. I used the naive value of 0.4MR^2,

The actual moment of inertia of the Earth about its spin axis is much
closer to I = 0.33078*M*R^2 where R is the equatorial radius.

> undoubtedly an overestimate because of earth's dense core.

True.

> BTW, is
> measurement of the rate at which the earth-moon distance is
> increasing, namely 0.5 cm/month, an accurate way to determine I?)

Of course, depending on how sophisticated the calculation is, it most
probably ignores the tidal effects of the Sun on the Earth's spindown
as well as the inclination of the moon's orbit w.r.t. the equatorial
plane, etc., etc.  So it probably is not all that accurate.

Also, I think I recall that the average recession rate of the Moon
from the Earth is more like 3.8 cm/yr (~= 0.32 cm/month), but my
memory could be wrong about this figure.

> Anyhow, they then say this amounts to 28 s/century. Unless I'm being
> really dumb (always possible, if not probable), this is about 4
> orders of magnitude too large.

It seems ok to me as long as we interpret the 28 s/century as the net
*accumulated* time mismatch over time due to the ever increasing
period of rotation.

Let r = rate of spindown per day.
Let N = the number of total days for which the net spindown mismatch
        is to be observed.

Then the net accumulated time mismatch M is:

M = 1*r + 2*r + 3*r + ... + N*r = r*N*(N+1)/2)

Now we are told that r = 44 ns and N = 100*365.2425 = 36524.25, so:

Thus M = (44 ns)*(36524.25)*(36525.25)/2 = 29.35 s

which is close enough to 28 s for me.

> But, they claim that this is confirmed by the fact that astronomical
> events such as eclipses have been found to run systematically ahead
> of calculations based on observations over preceding centuries. Is
> this claim correct?

I don't know it the claim is correct, but any mismatch in the timing
of eclipses need to be in terms of the net accumulated shift in time
based on the two methods of timekeeping (i.e. time by earth days
and time by atomic clocks).

> If so, what is the correct explanation of it,
> since the 44 ns/day is not enough?

It is if you realize that the 28 s/century is *not* the lenthening
of a day after 100 years, but is, instead, the total sccumulated time
mismatch between earth rotation time and atomic time.

David Bowman

This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.