Re: Challenge to Jim Scotti

Article: <[email protected]> 
Subject: Re: Challenge to Jim Scotti
Date: 12 May 1998 13:33:08 GMT

In article <[email protected]> Chris Franks writes:
> Nancy wrote:
>> 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? 
>
> MASS of the orbiting object is in both the gravity equation
> and in the energy equation.  If you double the mass of the
> Moon, its gravitational attraction would double, but so 
> would its balancing energy double.  All orbiting objects 
> at the same distance will orbit with the same period, 
> regardless of mass. 

I see mass, or gravity, on ONE side of the equations, with velocity on
the other in Paul Campbell's equations below.  And it appears that the
period increases as the mass, or gravity pull, get larger, in Joshua's
formula.  Is that not so?  As the mass increases, the object must move
faster to maintain it's current position, per your math.  This seems to
be intrinsic to your orbital mechanics equations.  Coming closer,
moving faster, at a distance, moving slower, etc. 

So, I repeat my question.  How was the mass of the Moon CALCULATED!  To
match the velocity we see, as the opposite side of an orbital mechanics
equation, or by some other means?  If the latter, then we have the
explaination of why such a heavy object is up there, sedately drifting.
If one were to calculate the weight of the moon, taking it's dimension
and assuming solid rock, as it no longer has a molten core, then what
WOULD that weight be?  And if one were to plug THAT into the equations
that Greg Neil proffered, would the Moon still be orbiting, according
to human orbital mechanics formulas?

In article <[email protected]> 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. 

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.