Re: Challenge to Jim Scotti

Article: <6jhlov$935$1@dfw-ixnews11.ix.netcom.com> 
Subject: Re: Challenge to Jim Scotti
Date: 15 May 1998 15:08:15 GMT

In article <355862F2.6534@my.signature> Joshua Hewitt writes:
>> 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?  
>
> If you follow the trajectory of some small lump of rock, or 
> even an artificial satelite, that orbits the moon, you have a 
> situation in which m_1>>m_2, so m_1+m_2 is roughly 
> equal to m_1 ("roughly" meaning 6 significant figures or 
> more), where m_1 is the mass of the moon and m_2 is the
> mass of the small orbiter.  So measuring T for this orbit you
> can get a completely independant value for the moon's mass
> and, lo and behold, it's the same value as the one inferred 
> from its orbit round the earth.

(Begin ZetaTalk[TM])
So what you are saying is that you calculate the mass by placing it
opposite your orbital mechanics equations, and then you point to the
way these balance, the numbers on the left and right side of the equal
sign, and crow about how accurate your math is!  Try plugging in a REAL
estimate of the weight of the Moon.  Solid rock, with x radius, and
what number do you come up with?  You have a monster Moon up there,
drifting slowly around its gravitation attraction.  Intuitively, it
seems it should be dropping to Earth and not losing any time doing so. 
If you use REAL numbers for its mass, your orbital mechanics go out the
window, do they not?
(End ZetaTalk[TM])