# Date: 23 Feb 94 17:10:16 To: All Subject: Kent Hovind Arguments In article 18FEB1994142309

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Date: 23 Feb 94  17:10:16
From: Karl Hahn
To: All
Subject: Kent Hovind Arguments

From: hahn@newshost.lds.loral.com  (Karl Hahn)
Organization: Loral Data Systems

In article <18FEB199414230985@skyblu.ccit.arizona.edu> James J. Lippard,
lippard@skyblu.ccit.arizona.edu writes(quoting Hovind):

> 5). The Moon is receding a few inches each year.  A few million years
> ago the moon would have been so close that the tides would have
> destroyed the earth twice a day.  #3 P.25, #6 P.43.

Here are some real calculations for the recession of the moon:
(Note that products and quotients are done only on quantities
that are in the kg-meter-second units system).

If an object changes from an orbit of period P_0 and radius R_0
to one of period P, and radius R, Kepler's 3rd law gives:

P/P_0  =  (R/R_0)^1.5                             eq 1

or

P = P_0 * (R/R_0)^1.5                             eq 1a

Where "^" indicates exponentiation.

The angular momentum, rho, of the object is given by:

rho = M * (2pi/P) * R^2                           eq 2

or by substituting eq 1a into eq 2:

rho = (2pi * M) * R_0^1.5 * R^0.5                 eq 3

and solving for R we have

R = rho^2 * P_0^2 / (4pi^2 * R_0^3 * M^2)         eq 4

The angular momentum of the earth due to rotation is given as

rho_e = (2/5) * M_e * R_e^2 * omega               eq 5

where mass of the earth, M_e = 5.97e24 kg, radius of the earth,
R_e = 6.37e6 meters, angular rate of the earth's rotation,
are dimensionless, omega = 7.27e-5 /sec.

Multiplying out eq 5 we get angular momentum of the earth,
rho_e = 7.04e33 kg-meters^2/sec.

62nd CRC Handbook of Chemistry and Physics gives, on page F-159, the rate
at which the earth's rotation is slowing as a fraction of its current rate.

eta = 2.8e-8 /century   or   9.0e-18 /sec

Multiplying eta times rho_e we get the rate at which the earth's angular
momentum is decreasing:

drho/dt = 6.4e16 kg-meter^2/sec^2

The moon's angular momentum must be increasing at an identical rate, as
angular momentum must be conserved.  Taking the derivative of eq 4:

dR/dt = 2 * rho * P_0^2 * drho/dt / (4pi^2 * R_0^3 * M^2)   eq 6

We have current distance to the moon, R_0 = 3.84e8 meters, current
orbital period of the moon, P_0 = 2.36e6 sec, mass of the moon,
M = 7.35e22 kg, and angular momentum of the moon given by eq 2,
rho = 2.89e34 kg-meter^2 /sec.  Multiplying all that out gives:

dR/dt = 1.7e-9 meters/sec  or 5.3e-2 meters / year

A few million years of lunar recession at that rate does not
appreciably effect the tides.  Let's try 10 million years.  That
puts delta-R at 5.3e5 meters, which is less than 2/10 or 1% of
the distance the moon is today.  This would have affected the tides
by less than 1/2 of 1%.

Let's try a bigger number of years.

If the moon has been receding at a constant rate, 4 billion years
ago it would have been at 45% of its distance today.  The magnitude
of tides, which is inversely proportional to the cube of distance,
would have been about 11 times what it is today.  Mean tide height
is about 0.6 meters today, so at that time it would have been
6.6 meters -- a big tide, but not earth-destroying.

The sidereal lunar period would have been 30% of what it is today,
in accordance with eq 1.  Applying the angular momemtum equations
above, we get that the rotational rate of the earth would have been
about 3.7 times its rate today, or in other words, the day would have
been 6.4 hours long.

Of course we have made the assumption several paragraphs back that the
recession rate has been constant, in order to make the demonstration
simpler.  That assumption is probably a good one over the short
term (that is the 10 million year calculation), but not where we
see a 55% change in R.

There are a number of competing factors that make the rate nonconstant.
The first is a mathematical one.  Eq 6 shows that the recession rate,
dR/dt, is proportional to the moon's angular momentum, rho.  Eq 3 shows
that when R is smaller, rho is smaller as well.  So this factor works
toward a slower recession rate in times past.  It would be a simple
matter to solve the differential equation suggested by eq 6 for R as
a function of time, but there are other effects to contend with.

The remaining effects all have to do with the term drho/dt, which
is the rate at which angular momentum is tranfered from the earth to
the moon.  There are two factors that would tend to increase this
term (and hence increase the recession rate) in times past.  The first
is that the earth was rotating faster then.  Faster rotation would
mean more energy loss due to friction, which is the cause of the
angular momentum transfer to begin with.  The second is the greater
tides.  You must understand that bodies of water are not the only
things subject to tides.  The entire body of the earth is deformed
by lunar tidal forces.  This deformation is the source of the friction
that slows the earth and causes the moon to recede.  Greater tides
cause more friction.  Both of these effects can be modelled
mathematically, and a numerical solution, at least, exists for the
resulting differential equation.  But there is more.

The last factor works in the opposite direction.  This is that a younger
earth was undoubtably much hotter.  This is because a) less of the
earth's primordial heat would have had a chance to escape, and b) there
would be a greater concentration of radio nuclides in the earth since
they would have had less time to decay.  A hotter earth means that it had
less solid mantle and more liquid core.  Tidal friction is due largely
to inelasticity in the mantle.  Less mantle means less friction, and
hence a lower recession rate.  The profile of the earth's interior in
geologic times past is, of course, the domain of geologists, and I
claim no ability to figure the details of this one out.

Hence, I am in no position to determine whether the recession rate has
been increasing or decreasing.  I will offer this editorial opinion
though: anybody unable to follow the calculations (for which one needs
only 1 semester of calc and 2 semesters of basic phisics) or the
arguments for nonconstancy I've made above is in no position to speculate
on whether the moon's recession rate proves or disproves anything.

--
|         (V)              |  "Tiger gotta hunt.  Bird gotta fly.
|   (^    (`>              |   Man gotta sit and wonder why, why, why.
|  ((\\__/ )               |   Tiger gotta sleep.  Bird gotta land.
|  (\\<   )   der Nethahn  |   Man gotta tell himself he understand."
|    \<  )                 |
|     ( /                  |                Kurt Vonnegut Jr.
|      |                   |
|      ^                   |

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Number: 309 (Read 0 times)              Date: 24 Feb 94  04:06:42
From: glenn r morton
To: All
Subject: Kent Hovind Arguments

From: xdegrm@oryx.com (glenn r morton)
Organization: Oryx Energy

An old but very thorough discussion of the Earth-Moon-Sun system over the
past several billion years is Gordon J. F. MacDonald, "Tidal Friction"
Reviews of Geophysics 2:3, Aug 1964 p. 467-541 As we go back in time, the
moon gets closer to the earth in at an ever increasing rate. But even as
far back as 3.5 billion years the moon is at 20 earth radii.  Prior to
that the earth moon system goes bananas.  Some have suggested that the
moon was captured at that time because of the strange behavior of the moon
prior to that time.  One solution to the problem would be to have the average
lag between the tidal bulge and the moon's position to be less than 2.16
degrees for the 5 billion years.  An average lag less than the present
value would solve the difficulty.  There are I am sure much more recent
work on this problem than MacDonald's but I haven't chased them down. Has
anyone else?

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