The GUESS iSS PRiNCiPLE of EQUiVALENT SYNONYMs (La PoEsy)
G-FORCE F = G*M1*m1 / r1^2 ..FRAME of REFERENCE:
= m1*(n - 1)*a_1
= m1*(n - 1)*v1^2 / r1
= 4*(pi)^2*m1*(n - 1)*r1 / t1^2
= 4*(pi)^2*m1*(n - 1)*(ORBiT radius) /(PERiOD sec)^2
= 4*(pi)^2*m2*(n - 1)*lbob / tbob^2
= m2*(n - 1)*vbob^2 / lbob
= m2*(n - 1)*a_3
= m3*(n - 1)*g_eotvos
= m3*(n - 1)*a_4
= LATERAL FORCE = m4*(DiSTANCE) / (DURATiON second)^2
= m4*a_2.

CONCLUSiONs:
1. All, mass m1 = m2 = m3 = m4 ..are equal iDENTiCALLY.
2. EQUAL accelerations a_1, a_2, a_3 & a_4 DiFFERENTLY.
3. Let forces F1 = F2 = F3 = F4 be equal iDENTiCALLY.
4. LATERAL FORCE F is NOT affected by (n - 1) LiFT.
5. Uncle's TEST is ONLY for m4*g_Eotvos = m4*a_4.
6. Let mass m1, m2, m3 & m4 be PENDULUM CLOCKs.
7. A ROSE is still a ROSE by any other name.

May the FORCEs be WiTH you.
VERY Sincerely c,
```Brian
p.s.
ANGULAR momentum pA = planet*radius*velocity:
(There's NO quantity for CENTRAL MASS there.)


-- 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.
[```SNiP```]
What if the inertial mass is greater than the gravitational
mass, and the binding energy is tied to inertial mass?

Bilge wrote:
Ben Rudiak-Gould: [```SNiP```]
Your argument seems to assume a particular Newtonian model
with unequal inertial and gravitational masses.
[```SNiP```]
Newtonian gravity is not correct, and there is no analogue
in GR of differing inertial and gravitational masses.
[```SNiP```]
get a null a result such that the difference between inertial
and gravitational masses cancel through some miracle.
[```SNiP```]
One can imagine anything.