From: gary@ke4zv.uucp (Gary Coffman)
Newsgroups: alt.conspiracy,sci.space
Subject: Re: NASA Coverup
Date: 9 Nov 92 17:41:33 GMT
Article-I.D.: ke4zv.1992Nov9.174133.2968
Reply-To: gary@ke4zv.UUCP (Gary Coffman)
Organization: Gannett Technologies Group

In article <ZOWIE.92Nov8182745@daedalus.stanford.edu> zowie@daedalus.stanford.edu (Craig "Powderkeg" DeForest) writes:
>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.

Sorry, pulling a 10% number out of the air has no bearing on the discussion.
It's a strawman argument. The Earth, and the Moon, are both inhomogeneous
bodies. The average pull is not 100% symetrical about a center. No claim
of a 10% asymetry was made by anyone but you as a strawman to convienently
knock down. Note that a probe trajectory from Earth with a 50 foot course
error caused by a mascon at engine cutoff translates into a 500 mile error 
at Lunar distance. That's nasty for a Ranger probe.

>   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.

Well that's exactly what I said. The Earth isn't a point mass, different
parts of it are attracted with different force by the Moon. And the reverse
is also true, different parts of the Earth attract the Moon by differing
amounts. Treating the Earth as a point means there's no gradient, and that's
observably false. 


>   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!

If you insist on a value of 10% to indicate significance, then it's not
significant. I make no such demand. This was in response to the Ranger
"misses" spoken of by Snarfy. As I noted with mascons above, a small
error here can make a large difference there.


>   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.

But of course we *are* assuming that the Moon is within a couple of orders
of magnitude of the same mass as the Earth. The correct value is about
.01 Earth and Snarfy needs it to be a little over .1 Earth. Thus we have
a two body problem with the bodies revolving around a common center. We
know the orbital motion of the Moon very precisely, and we know the distance 
to the Moon. From the motion of the two bodies around the common center
we can establish a ratio of the masses of the two bodies. Since I postulated
that we know the mass of the Earth, the Moon's mass is then trivially
calculated. 

Ok, I stated it very badly the first time. What I was trying to say 
up there is that if we assume the Moon's gravity is .6 G as Snarfy 
claims, then we know the combined pull of the two bodies, and we know 
their revolution rates. The centripedal force caused by the motion has 
to exactly balance the combined gravitational pull of the two bodies.
Knowing the mass of the Earth, we should then be able to see if the
revolution rate makes sense with a .6 G Moon. It won't, it's too
slow, the Moon should be spiraling in to the Earth.

>[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!]

Oooh, a spelling flame, I'm crushed.

Gary
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