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Re: Challenge to Jim Scotti

Article: <6j5019$> 
Subject: Re: Challenge to Jim Scotti
Date: 10 May 1998 19:43:37 GMT

In article <35533323.6D43@my.signature> Joshua Hewitt writes:
> Paul Campbell wrote:
>> V^2=GM/r = GM*1/r for all objects going in a circle.
>> V^2=GM*1/r some how relates to V^2=GM*(2/r-1/a)
>> V^2=GM*(2/r-1/a) but for a circle a=r then
>> V^2=GM*(2/r-1/r) which equals V^2=GM*(1/r)
> Didn't quite follow that, but the period of an elliptical orbit 
> is given by
> T=[(4*pi^2*a^3/G*(m_1+m_2)]^1/2
> The above equation wasn't just pulled out of a hat, any 
> book on orbital mechanics will derive it from first principles. 

(Begin ZetaTalk[TM])
Fine, now take those very equations for that quarter of the ellipse
when the Moon is pointed every so slightly TOWARD the Earth, rather
than away.  At any given point in time, which is the way the Moon is
dealing with this situation, it has the pull of gravity at an inverse
square rule, per you.  At this same point in time, it has its motion,
which in this case is moving TOWARD the Earth rather than away.  

Now why, in this situation, would the Moon not INCREASE in its
direction, pointing toward the Earth, as it INCREASES is its speed as
it does so?  It increases its speed, it is pointing more toward the
Earth than away, with an angle less than 90 degrees, and then .. and
then .. it pulls AWAY from the Earth, pointing slightly more AWAY than

(End ZetaTalk[TM])

The Zetas wish to emphasize that the above is a retorical question,
they darn well know the answer, etc.  

By the way, in the above equations, there is a balance so that if the
gravity is greater but the speed remains the same, the orbit becomes a
plummet, right?  Or if the gravity lessens but the speed remains the
same, it becomes a trajectory out into space, right?  So, have you
calculated the weight of the Moon, its mass, based on its speed?  Since
you don't know its composition, how else did you get its weight?  Isn't
this cheating?  You're saying that your equations balance, but then
only balance because you've CALCULATED the weight of the Moon using the
orbital mechanics formula, right?  So if you would enter any other
weight for the Moon into those equations, then the Moon either plummets
or ejects into space?  So .. how about estimating the weight of the
moon, based on it being solid rock, for instance, which it probably is
without a molten core.  What is its weight REALLY?  We do know a bit
more about its composition now that in earlier times, right?  Having
picked up Moon rocks and all.