Bilge wrote:
The the entire point of the experiment is discover whether
or not the `m' in Force = ma is the same `m' in Gm/r^2. --

You CONFUSE because your "Gm/r^2" is NOT a FORCE, Dimwit.!!
But, CLEANing up here, let notation for F = m2*a_2 ;

THEN: G FORCE F = m1*g1 = m1*v1^2 / r1
= G*M1*m1 / r1^2
= 4*(pi)^2*m1*r1 / t1^2
= 4*(pi)^2*m1*a_1
= m2*a_2.
CONCLUSiON:
GiVEN the mass m = m1 ..as per the Equivalence Principle;
AND ALSO the GiVEN FORCEs F are BOTH equivalent, as well;
THEN the ACCELERATiON a_2 ..is CLEARLY = to 4*(pi)^2*a_1.

The ACCELERATiON a_2, is CLEARLY equal to 4*(pi)^2*a_1.!!

CONVERSELY if a_1 gets "PEGGED" as equal to a_2, HOWEVER,
THEN mass m1 will HAVE to be EQUAL to mass 4*(pi)^2*m2.!!

GUESS, iNHERENTLY, does NOT have this 4*(pi)^2 PROBLEM.!!
Sincerely c,
```Brian


-- A non-null
result (which is what uncleal would like to get) means those
two `m's are not the same. It shouldn't be much of a surprise
that one can take advantage of the difference.

I.e. is the binding
energy of the system dependent on gravitational or intertial mass?

The binding energy depends on the electromagnetic forces.
Besides, I _dont_ need to use both isomers. It's sufficient to
just disassemble the molecule, transport the constituent atoms
and resassemble it.

In this gedanken, the gravitational mass of step (1) is greater than
after step (3).

Not necessarily. I made no assumption about which isomer had
a greater mass.
If it costs energy to put the object into the step (3)
configuration there is no problem with energy conservation. What if
the inertial mass is greater than the gravitational mass, and the
binding energy is tied to inertial mass?

What if the sky falls? Don't you think any of these things occured to
me before I ever posted that? Seriously, rather than just saying
``what if,'' without having any idea whether or not I already considered
these things, find a particular ``what if'' and make some numerical
estimates within the constraints we already know from experimental data.