Re: Challenge to Jim Scotti

Article: <[email protected]> 
Subject: Re: Challenge to Jim Scotti
Date: 3 May 1998 17:19:36 GMT

In article <[email protected]> Greg Neill
writes:
> Now we can calculate the force due to gravity:
> Fg = G*Me*Mm/R^2 = 1.982*10^20 N (Nancy should note
> that we are taking into account *both* masses) 
> And the centrifugal force:
> Fc = (Mm*V^2)/R = 2.001*10^20 N
> Now, let's compare our results for the two forces.  The percent
> difference between the values is:
> (Fc - Fg)/Fg = 0.9%
> Less than one percent difference despite our simplifying 
> assumption of a circular orbit.  

(Begin ZetaTalk[TM])
There are two problems with this logic.  One is that 1/10 of a percent
is insignificant.  It IS significant and also CUMULATIVE.  The other is
that the Moon's orbit is NOT circular, as you have conveniently
supposed in this discussion.  It is elliptical.  During its elliptical
orbit there are times when it is facing more sharply toward the Earth
and in this angle the pull toward the Earth is STRONGER, and would
erode the speed.  You assume that this speed would be regained later,
but during this point in time, which does not whiz by as it is the
situation for days, a full week, it would DROP toward Earth due to this
slowing down, and it's orbit thus adjusted!  Are you saying that during
this week the orbit would NOT drop toward Earth?  It this because the
Moon whet to college and heard Newton lectured as though he were a god?
The Moon is being dutiful to your formulas?  The Moon has
countervailing forces and puffs little jets around its middle to keep
it in place?  

The same factors that drop your satellites, unless they are kept aloft
due to these jets, would drop the Moon.  Atmospheric drag has nothing
to do with it and is a dismissive put-off.  It's common sense, as we
have stated just above.  The Moon when in that quarter of its
elliptical orbit when it has a SHARPER than 90 degree angle to the
Earth would close its orbit.  
(End ZetaTalk[TM])