Re: Moon's synchronism (very long)

[David Bowman <dbowman@tiger.gtc.georgetown.ky.us>]

Concerning the excerpt of Dave Dockstader's post (subject: astronomy) of
19 MAR 98:
> ...
> #14. If the Moon had oceans like the Earth's, what would the tidal effect be
> like there?   ....   How would
> the variations in height compare to those on the Earth?  ....

I finally got a chance to look more deeply into this subject.  In general
this problem is very difficult to solve, but with a number of simplifying
assumptions (some more drastic than others) we might get some decent
estimates.  The assumptions are as follows:

1.  We consider only a bound 2-body system of a body of mass M
gravitationally bound to another body of mass m such that the separation
between these bodies' centers of mass is R.  We use M for the mass of the
body whose tidal distortion is to be calculated, and m for the body whose
tidal external gravitational field gradient is responsible for the
induced tidal effect in the body of mass M.  (This assumption means we
neglect the effect of the Sun's gravity on the problem.)
2.  Related to assumption # 1 above we assume that R is enormous compared
to the radii of both bodies M & m.  This means that we need to consider
no multipolar interactions higher than the quadrupole interaction, that
the induced tidal distortions in each body are tiny perturbations on the,
otherwise, spherical shape of each body, and that the field gradient of
body m at the location of body M is essentially that of a point mass m
at a distance R away.  IOW, the bodies are separated so far that the
quadrupole moment of each body only interacts with the field gradient of
the (point mass-like) monopole moment of the other body.
3. Somewhat similar to assumption 2, we assume that any spin-induced
centrifugal distortions of body M are tiny perturbations on its,
otherwise, spherical shape.  IOW, we will not consider a body spinning
so fast that it is so spin-flattened that it is about to fly apart.
4.  We assume the M - m orbital plane is the equatorial plane of body M
so that the spin axis of body M is always perpendicular to the line of
sight connecting bodies M & m.  This assumption is somewhat drastic for
calculating lunar-induced tidal effects on the Earth since the Moon's
orbital plane is inclined between 18.3 and 28.6 deg (depending on
precessional effects) wrt the Earth's equatorial plane.  This assumption
is less drastic for the Earth-induced tidal effects on the Moon whose
equatorial plane is inclined only 6.68 deg wrt its orbit about the Earth.
5.  We assume that body M is a perfectly relaxed plastic fluid-like body
whose instantaneous shape at any given time reflects an equi-potential
surface as calculated in a frame in which the surface of body M is at
rest.  As we shall see, this assumption is the most drastic of all of our
assumptions for the real Earth - Moon system, esp. for the Moon since it
is throughly frozen and is too stiff to respond very well to the nuances
of the Earth's tidal influences.  Even the Earth doesn't obey this
assumption very well (As Jim Green is wont to remind us), but at least
the Earth does have a mostly fluid (oceanic) surface which can relatively
rapidly respond to the forces acting on it.  This response, not being
instantaneous in practice, allows the actual induced tidal bulge of the
Earth to be somewhat rotated out in front of the Moon by the Earth's
spin.  Our analysis here ignores this complication when calcuating the 
tidal height.  So even though this instantaneous equilibrium assumption
is not a very good one in practice for the Moon, it *does* seem to be
just the case for which Dave Dockstader was asking about above.
6. Since our answers will depend, to some extent, on the detailed
internal density profile of body M we will need to make a model for the
internal mass distribution for the planet.  This is done as follows:
Since the center of mass body M is supposedly a point of inversion
symmetry, and since the density ought to be an analytic function of
position there, we assume that the density [rho] approaches its maximal
central value [rho]_0 quadratically as a function of radial position.
Let [rho]_s be the density at the surface of body M and let t
(0 <= t <=1) be a radial parameter such that t = 0 at the center of the
body and t = 1 at the surface.  The simplest analytic function which is
quadratic at t = 0 and fits the proper densities at the center and at the
surface is the function: [rho](t) = [rho]_0 - ([rho]_0 - [rho]_s)*t^2 .
We assume that body M is sufficiently well approximated by such a density
profile.

Since body M is going to be tidally distorted (and spin distorted) its
surface shape will not be spherical, so the parameter t is a
directionally dependent measure of radial distance. Each value of t
(between 0 & 1) labels a surface of constant density for the interior of
body M.  We take the equation for the surface of body M to be the
tri-axial ellipsoid: 1 = (x/a)^2 + (y/b)^2 + (z/c)^2 where the x-axis 
points along the direction connecting the centers of bodies M and m, the
z-axis points along the spin axis of body M, and the y-axis is the
equatorial direction in M that is perpendicular to the M - m line.
Since the distortions of body M are assumed to be small we know that
a, b, and c (a > b > c) are all close to each other in value and that the 
parameters e == (a - b)/a and f == (a - c)/a  are *tiny* compared to 1.
Here f represents the polar flattening of the body M relative to the
equatorial direction of the tidal bulge, and e represents the equatorial 
elongation of the tidal bulge.  (In the absence of spin but in the
presence of tidal distortion then e would equal f and that would
represent the prolateness of the tidal bulge.  In the presence of spin
but the absence of tidal effects e would be 0, and f would represent the
spin-induced oblateness of the body.)  Using the smallness of e and f we
can rewrite (to leading order in e & f) the equation of the surface of M
in spherical-polar coordinates as:
r = a*(1 - e*sin^2([theta])*sin^2([phi]) - f*cos^2([theta])).  Since the
pressure inside the body increases with depth and since most materials
become stiffer with increasing pressure we assume that the body M becomes
more spherically symmetric with depth, and has it maximal anisotropy at
the surface. We can model this by simply relating the equation for the
radius to the radial depth parameter t according to this simple
generalization of the equation for the surface:
r = a*t(1 - (t^n)*(e*sin^2([theta])*sin^2([phi]) + f*cos^2([theta]))).
This effectively makes each surface of constant density have the
effective shape parameters e*t^n and f*t^n (instead of e & f like at the
surface has).  So as long as the parameter n > 0 the surfaces of
constant density become more and more isotropic with depth.  Notice that
if we set [rho]_s = [rho]_0 then the profile boils down to a simple
uniform (slightly eccentric) ellipsiod.  Also, if we keep [rho]_0 greater
than [rho]_s but set n = 0 then the anisotropy of the profile is
constant with depth.  If we take the limit n --> [infinity] then the
body's interior remains spherically symmetric, with only its surface
layer participating in the induced distortion.  (This would be a decent
approximation for the Earth's tidal distortion which would seem to be
confined to mostly ocean depths or less.)

Using a combination of the above two parametric equations relating (via
t) [rho] to the spherical coordinates r, [theta], and [phi] we can use
this model to evaluate the quadrupole moment and the inertia tensor (and
its component differences) for body M.  By our judicious choice of
coordinate system, our assumption of instantaneous equilibrium (#5), and
our assumption of zero orbital-equatorial inclination (#4) we can enjoy
the fact the the inertia and quadrupole moment tensors are diagonal in
our coordinate system, and their principal axes are along the coordinate
axes. Making assumptions # 1 - 5 allows us to find expressions for e and
f to leading order in the small parameters and conclude that:
e - E = (3/2)*((a/R)^3)*(m/M) and
f - F = (1/2)*((a/R)^3)*((3 + l)*(m/M) + l)
where l = ([omega]_s/[omega]_o)^2 is the square of the ratio of the
sidereal spin rate of body M divided by the orbital angular velocity
of the m - M system, and the parameters E and F are shape-dependent
distortion effects of the internal mass distribution of body M.  Here
E = (3/2)*(I_yy - I_xx)/(M*a^2) and F = (3/2)*(I_zz - I_xx)/(M*a^2) where
I_xx, I_yy, and I_zz are the principal moments of inertia along the
x, y, and z axes respectively.  By using our model for the mass
distribution in assumption #6 we can evaluate parameters E and F.  To
leading order in the small parameters E and F are directly proportional
e and f respectively with the same proportionality constant C (i.e.
E = e*C and F = f*C) where C = (3/5)*q + 3*(1 - q)/(n + 7), with q
defined as q = [rho]_s/[rho]_bar being the ratio of the surface density
divided by the mean density of body M.  This allows us to get
explicit formulae for e and f as:
e = (3/2)*((a/R)^3)*(m/M)/(1 - C) and
f = (1/2)*((a/R)^3)*((3 + l)*(m/M) + l)/(1 - C).
In order to make use of these formulae we still need to have decent
estimates for the internal density profile dependent parameters q and n.
We do this differently in each case at hand.

First, let's consider the case of the shape of the Earth (we take M to be
the Earth and m to be the Moon).  From observation we know that
m/M = 0.01230002, (a/R)^3 = 4.568 x 10^(-6), and l = 750.5661.  To
evaluate e (and f) we still need to find C.  Consider the case of the
lunar tidal induced bulge giving the e parameter.  For earth-tides most of
the tidal effect would seem to come from the oceans since the atmospheric
tides don't have enough mass to make much of a difference and the crust
and rocky interior would seem to respond too slowly adequately keep up
very well with the semi-diurnal frequency of the tidal dependence.  Thus
for this case we can assume that nearly all of the tidal distortion of
the Moon on the Earth induces a surface-only response in the Earth.  This
means that we then take n --> [infinity] in our model mass distribution
and take q = 1.02/5.515 = 0.185 = density of sea water/ mean density of
the Earth.  Using this model we get C = 0.111 using this parameter in 
(and the other observational parameters) our formula for e gives
e = 9.48 x 10^(-8).  Multiplying by a = 6.37814 x 10^6 m as the Earth's
equatorial radius gives a tidal bulge height difference of h = a*e =
0.60 m.  So *if* the Earth's ocean surface was in instantaneous
equilibrium with the tidal forces acting on it we would expect a tidal
bulge height of about 60 cm.  I do not know what the actual value of this
is but 60 cm seems reasonable to me.  (Jim Green will be satisfied, I
think, with only a zero value for this parameter.)  Since this bulge
would be expected to only exist in a relatively undistorted shape in deep
mid-ocean regions (far from complicating continental effects) I do not
know quite how measure it.  Probably using a GPS reciever with very
sensitive altitude resolution in a row boat in the middle of a calm
Pacific Ocean over a 12 1/2 hr time interval, or maybe consulting the
tidal records of some place like Kiritimati (Christmas) Island may do the
trick.

Now we consider the f parameter measuring the spin-induced equatorial
bulge for the Earth.  In this case we actually have a well known
experimentally measured value of f = 1/298.257.  Since we know the other
parameters in the expression for f we can use the experimental value of
f to work backwards to find the expressions for C and hence n for the
case of the spinning Earth (with a completely relaxed interior response
to the spin).  Working backwords from the measured f value gives a C
value of C = 0.4824.  Now since the whole plastic Earth responds to the
Earth's spin we can neglect the oceans and consider the crustal rock's
density to be the surface density of the Earth.  This allows us to
calculate the surface to mean density ratio q = 2.8/5.515 = 0.508.  Using
this value in our expression for C allow us to extract the
phenomenological parameter n = 1.3.  This reasonable value allows us to
accurately fit the shape of the Earth due to its spin if the anisotropy
parameter f decreases with depth according to: f_effective = f*t^1.3 for
t ranging: (center) 0 <= t <= 1 (surface).

Now we consider the *Moon* as our body M (& Earth as m) so we can answer
Dave Dockstader's original question.  In this case we have m/M = 81.3006,
l = 1 (since the Moon is tidally locked to the Earth), and
(a/R)^3 = 9.243 x 10^(-8).  If we assume that the Moon's crustal density
is similar to the crustal rocks of the Earth we get a q parameter of
q = 2.8/3.34 = 0.84 (since the mean density of the Moon is 3.34 g/cm^3).
Now sonce the Moon has less internal gravity with which to "fight off"
the tidal effects of the Earth than the Earth has to resist quadrupolar
deformations in its shape we expect that the anisotropy in the Moon's
density profile will tend to remain to greater depths inside the Moon
than the corresponding profile does inside the Earth.  IOW we expect that
the Earth's surfaces of constant density get 'rounder' with depth faster
than the Moon's surfaces of constant density do.  Since for the Earth's
interior we obtained n = 1.3 we would expect that for the Moon that its
effective n value is somewhat less (this is especially so considering
that the Moon is not supposed to have a large high density iron core like
the Earth does).  Therefore I will estimate that for the Moon we have an
n value something like n = 1.  (This would also allow the Moon's density
profile to smoothly and analytically "round out" as we go to its center.
Using n = 1 and q = 0.84  we obtain C = 0.564. Using this value in our
formulae for e and f gives e = 2.59 x 10^(-5), and f = 3.46 x 10^(-5).
Multiplying by the lunar radius we get a - b = 44.9 m and a - c = 60.1 m.

So if the Moon was perfectly deformable and possessed a shape in
equilibrium with the tidal/spin stresses induced in it, it would have a
semi-major axis toward the Earth about 45 m longer than the semi-middle
axis equatorial axis perpendicular to the Earth's direction, and some
60 m longer than its semi-minor polar axis. Since from the lunar
perspective the Earth stays (more or less) in the same place in the
lunar sky, the Earth-induced tidal bulges on the Moon would tend to stay
put except for the (already much discussed) librations of a few degrees
one way or the other over the lunar surface.  What about solar-induced
tidal bulges?  Since the tidal field gradient of the Sun is about 40%
of that of the Moon on the Earth this means we would expect a solar
tidal bulge on the Earth's mid-ocean regions of some 0.2 m in height.
On the Moon the solar-induced tidal bulge would come out to about 0.25 m
which is quite insignificant compared to the 45 m tidal bulge induced by
the Earth. On the moon this solar tidal bulge would not remain constant,
however, but would follow the sun going through a tidal cycle at a fixed
point on the Moon every 2 weeks.

Since we have just finished calculating the induced tidal effects on an
ideally deformable Moon, let me take this opportunity to correct my
posted calculation of 19 MAR 98 concerning the resonant libration
frequency of oscillations for an ideal Moon rocking in the Earth's
gravitational field gradient.  Recall from that post that the formula
for the period of such a rocking motion (whose rotation axis is
effectively along the Moon's spin axis) is given by:
T = 2*[pi]/sqrt((3/2)*f'*g') where g'is the magnitude of the Earth's
gravitational field gradient in the vicinity of the Moon and
f' == (I_yy - I_xx)/I_zz.  In the interest of preventing excess
confusion let me note the the parameter f' here is denoted as f in my
19 MAR post and in John Mallinckrodt's 17 MAR post. But since I have
already used the conventional symbol f for the polar flattening factor in
this post, I had to change the notation symbol to f' here.  In my 
previous post I estimated a value for the f' factor based on a cruder
picture of the behavior for the internal mass distribution.  In
particular, I had neglected the 1 - C factor that appears in the above
formuale for the e & f factors calculated in this work.
If we use the same model for the density as in given in assumption #6,
i.e. [rho] = [rho]_0 - ([rho]_0 - [rho]_s)*t^2 where t is given
implicitly in terms of:
r = a*t(1 - (t^n)*(e*sin^2([theta])*sin^2([phi]) + f*cos^2([theta]))),
then we obtain a value for the f' factor of:
f' = e*(q + 5*(1 - q)/(n + 7))*7/(5 + 2*q).
Now since we have calculated above for the Moon an e-value of
e = 2.59 x 10^(-5) using q = 0.84 and n = 1 we thus get a corrected value
for f' = 2.55 x 10^(-5).  This is some 2.7 times larger than the value I 
had calculated in my 19 MAR post.  Using this corrected value for f'
we estimate the resonant libration period to be T = 2.7 x 10^8 s = 8.6 yr
rather than the 14 yr result I gave previously.  This makes the libratory
driving frequencies about 115 times higher than the resonant frequency
(not the 185-190 times larger that I had earlier claimed).

Everything claimed so far in this post about tidal bulges induced in a
bound 2-body system was predicated on the validity of the 6 assumptions
given at the beginning of the post.  If it turns out that some of these
assumptions are very incorrect then the predictions made and the
conclusions drawn here will be, likewise, incorrect.  It turns out that
we have evidence to believe that assumption 5 is quite incorrect for the
case of the Earth-induced tidal bulge in the Moon.  (Jim Green would
also claim that it is a bad assumption for the Moon-induced tidal bulge
on the Earth as well.)  I found this evidence in the CRC Handbook:  On
pages_14_-2,3 (78th 1997-8 ed.) under section titled "PROPERTIES OF THE
SOLAR SYSTEM" is a table of potential coefficients for the harmonic
expansion of the gravitational potential of the various planets in terms
Legendre polymonials.  This expansion assumes that each body has axial
symmetry about its spin axis.  The l = 2 order expansion coefficient J_2
for the Moon has the experimental value of J_2 = 2.027 x 10^(-4).  We
can compare this result with the model given here because the value of
J_2 can be calculated from the model for the mass distribution given in
assumption 6 above, and it comes out that: J_2 = (2*f - e)*C/3.  If we
use the values of e, f and C calculated above for the Moon we get
J_2 = 8.14 x 10^(-6) and this is 24.9 times *smaller* than the published
experimental value.  Since the values of e, f, J_2, etc. are all
proportional to 1/R^3 we can determine that the published value for J_2
would agree with our equilibrium model here *if* the value of R was
decreased by a factor of about 2.9 (assuming that the moon remained
tidally locked to the earth).  This would put the Moon only 132,000 km
from the Earth.  We thus conlude that apparently the Moon is very
frozen and rigid and *quite* capable of withstanding (without yielding)
the changing tidal stress induced in it by the Earth.  It also seems that
the moon "hardened" in place a long long time ago when it was much 
closer to the Earth than it is now, and its present stiffness prevents it
from relaxing to the more spherical shape it would like to now have in
the weaker current terrrestrial field gradient.  Either that, or only
recently the moon was spinning much faster than it currently is now doing
and a miracle occurred which stopped it and tidally locked it to the
Earth such that there has not yet been enough time for the moon to
subsequently relax from its former more distorted shape.  Somehow this
second senario doesn't seem quite as elegant as the former one.

David Bowman
dbowman@gtc.georgetown.ky.us