Re: Challenge to Jim Scotti

Article: <[email protected]> 
Subject: Re: Challenge to Jim Scotti
Date: 22 May 1998 13:30:15 GMT

In article <[email protected]> Paul Campbell
writes:
>> 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, is 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?
>
> I would love to see your definition of weight, ours is F=ma.

(Begin ZetaTalk[TM])
Fine.  We have an equation from you, but without, of course, an
application to orbital mechanics.  Per your fellow, M.C. Harrison, the
force of gravity is equal to the force BETWEEN two objects, and one
must take them both into consideration.  Take the weight of a square
foot of granite, on the surface of your Earth, and use that as a
constant for mass, as some constant must exist.  Calculate the weight,
then, of the Earth, and likewise the Moon, and thus produce F, the
force of gravity between them in a manner the common man can relate to.
Two masses of solid granite, or nearly so, pulling toward each other.

===>This gives us M1, M2, r, and thus F, per M.C. Harrison, below.
(End ZetaTalk[TM])

In article <[email protected]> M.C. Harrison
writes:
> F=M*A works just fine for photons. M is zero so acceleration
> to infinite speed requires no force. It's wrong, because 
> relativity is more useful when dealing with stuff on that scale.

In article <[email protected]> M.C. Harrison
writes.
> The force of gravity is an inverse square law, which means
> a mass will experience a force due to another mass according to 
> the equation F=M1*M2/r^2 where M1 is one mass, M2 is the 
> other mass, and r is the separation of the masses. ... 

(Begin ZetaTalk[TM])
Now we deal with velocity, which you can calculate nicely as you have
your Moon at a determinate distance from the Earth, and it rounds in a
short ellipse every 28 days or so.  What is that speed, in miles per
hour, something the common man can relate to!  Or is your math not REAL
and meant to reflect the world around you.  Do you only deal in
abstracts?

===>This gives us v, per M.C. Harrison, below<===

And he states that the change in velocity, V, would equal the old
velocity plus or minus a factor of acceleration and time.  We now have
the Moon, putting along out there at X miles per hour, with F, the
force of gravity, pulling on it every minute.  Let us state that we
will calculate this little drama for a minute, t equaling a single
minute.

===>This gives us t, time, per M.C. Harrison, below<===
(End ZetaTalk[TM])

In article <[email protected]> M.C. Harrison
writes.
> Motion of an 
> object is to proceed in a straight line at a constant velocity, but 
> with an acceleration, it will go faster in the direction of the force,
> V=v+At, where V is the new velocity and v is the old velocity, A 
> is the acceleration and t is the number of seconds that have passed. 
> A simple Basic program which moves an object according to this
> inverse square law can then be written, which by varying the rate
> and starting position of the object, will demonstrate the circular 
> path in question. Similarily, ellipses can be produced. 

(Begin ZetaTalk[TM])
Since we know v, the existing straight line velocity, and t, a minute,
we now can calculate what ACCELERATION toward the earth that the pull
of gravity must be exerting to balance out to the same rate of
velocity, V, for the Moon!  If you make V, new velocity, and v, old
velocity, the same, then the draw must be zero in order for the Moon to
maintain its pace!  Surely this could not be the case, that the pull of
all those tons of granite toward each other would be nothing!

Now, if we state, as many have argued, that the Moon is FALLING toward
Earth, and this is accelerating its rate rather than slowing it, then A
is SOMETHING, and per M.C. Harrison is "M1*M2/r^2 where M1 is one mass,
M2 is the other mass, and r is the separation of the masses."  This
would mean, for the Moon to maintain an even keel, that the force of
its existing velocity  must be a something that is equal to F, the
force of gravity.  You have x tons of granite, being pushed AWAY from
the Earth by its astonishing velocity, v, to arrive at V, the new
velocity of the continuing steady pace.  

So what does it take to lift x tons off the face of the Earth?  And how
much would that force be diminished by the distance between the Earth
and the Moon?  Pardon us for placing your math together in one lump, a
situation that is avoided, because it simply DOES NOT WORK!
(End ZetaTalk[TM])