From: zowie@daedalus.stanford.edu (Craig "Powderkeg" DeForest)
Subject: Re: NASA Coverup
Organization: Stanford Center for Space Science and Astrophysics
Date: 8 Nov 92 18:27:45

In a rather misleading article, gary@ke4zv.uucp (Gary Coffman) writes:
   In article <foo> snarfy@cruzio.santa-cruz.ca.us writes:
   > Dillon Pyron writes:
   >> To  the point.  Your calculations assume that the earth and moon have the
   >> same density, and that it is homogenous.
   >Beiser states , on page 118 , "A spherical object behaves  gravitationally
   >as  if  it's  mass  were concentrated at it's center"

   Beiser's assumption holds up as a close approximation when the two bodies 
   are far enough apart that their radius is an insignificant part of their 
   separation distance, say Earth-Sun distance. But they fail miserably when 
   the separation is less than a few body radiuses. The mascons will warp the
   orbit of the satellite a measurable amount.

Actually, Beiser's assumption holds up extremely well, for snarfy's purposes.
The gravitational force from, say, the Earth on a small test mass near it
is the integral over the entire volume of the Earth of the pull from each 
little chunk of mass inside the Earth.  With the assumption that the Earth
is homogenous, the integral can be done in closed form, easily.  One finds
that the pull is the same as that of a point mass at the center of the Earth.

If the mascons had a significant effect (say, 10%) on the local gravity 
somewhere in the Earth's neighborhood, we'd find significant `mountains' in
the middle of the ocean.  While mascons do cause small (I seem to recall
order of 50') changes in sea level in certain places, a 10% shift in local
gravity at the surface would result in a change in sea level of the order
of (R-sub-earth) * (10%)^2, or about 50 miles, much higher than Mt. Everest!

Since we don't see 50-mile humps in the ocean (the CRC lists Earth's minimum
radius [the polar one] at 6357 miles, and maximum [equatorial] at 6379 miles),
we can assume that, to within better than 10%, the Earth is a homogeneous
sphere.  QED.

   The existance of tides is a direct result of the effect
   of differentials in gravitational potential across the diameter of a 
   body. Using Beiser's simplifing assumption, there could be no tides on
   Earth. Since we can easily observe that there are, his assumption is
   invalid for bodies as close together as the Earth and the Moon.

This is plain wrong.  Using Beiser's simplifying assumption, there could be
tides on Earth.  The tides are due to the fact that the parts of the Earth
closer to the Moon are, well, closer to the Moon and therefore attracted more
strongly than those farther from it.  This is true for every point mass in
the Earth's vicinity, including Gary Coffman, who (if he lives in the
United States) attracts the Empire State Building considerably more than he
does the Opera House in Sydney.

   The other assumption Beiser uses is that of spherical objects. Neither
   the Earth nor the Moon are spheres. The Earth is an oblate spheriod, and
   the Moon is somewhat pear shaped with the greatest mass on the side facing 
   Earth. This is again significant when the separation distance is a few 
   planetary radiuses.

It's not significant unless the separation distance is less than a few times
the deviation from a sphere, considerably less than a planetary radius!

   The most damning evidence against your theory that the Moon has a 
   gravity of .6 G is that we know the orbital period of the Moon to
   a great accuracy, and we know the mass of the Earth and the distance
   to the Moon. With those three numbers, we can calculate exactly how
   much centrifigual force is in the system, and thus how much gravitational
   force is required to counterbalance it. So if the Moon stays in orbit,
   and it does, we can state it's gravitational pull to a high degree of
   accuracy.

This is semi-right, but I'm not sure whether Mr. Coffman understands exactly
what he's saying.  If we assume that the Earth is much more massive than the
Moon, then knowing the distance to the moon and the mass of the Earth tells us
nothing about the mass of the moon.  The reason is that, with Me >> Mm, the
Moon could be made out of lead or out of papier-mache, and it would orbit with
the same period: the heavier the Moon is, the harder the Earth must pull it
to keep it in orbit; but the harder the Earth *does* pull it.

This is one of the great puzzles of Physics -- that gravitational `charge' 
should be exactly the same quantity as inertial `mass', so that the mass 
of an object is irrelevant to its trajectory through a fixed gravitational
field.  It is one of the reasons why most physicists believe Einstein's 
theory of General Relativity, which unifies the ideas of gravity and inertia
on a macroscopic level.

If we start looking at the system more carefully, we note that the Moon doesn't
revolve around the exact center of the Earth, but that both the Earth and Moon
revolve around a point somewhere above the center of the Earth (but, in this
case, within the Earth).  The more complicated problem can be simplified by 
modelling the Moon as orbiting around a large, `reduced mass' at the center
of the Earth.  If Snarfy really wants to prove his point, he should calculate
the orbital period of the Moon based upon the reduced mass implicit in his
claimed lunar mass.  [I'd do it here but (a)I have real work to do, and 
(b) I believe the figures in my CRC.]

[BTW, one of my pet peeves is people who say "it's", short for "it is", when 
they really mean "its", which is the proper form of the singular neuter 
possesive pronoun in English.  So there!]
--
Craig DeForest -- astrophysicist for hire.   DoD#314159; PhD#271828
Newsgroups: alt.conspiracy