Originally Posted by nutant gene 71

Inertial-mass is always equivalent to its gravitational-mass. Let me illustrated it thus:

Take a one kilogram mass (same cube or rod or ball or whatever, did not change its composition in any way) from Earth's 1 G and move it to Jupiter's (hypo) 5 G. Place this kilogram of mass on a balance scale, and it balances against another kilogram mass. Now try lifting this kilogram off Jupiter's surface: it will take 5 times the force to do so. I interpret this as 5F = 5m*a; while you interpret this as 5F = m*5a; this is our fundamental difference. But Jupiter is not 5 times the mass it was before, same mass. If Jupiter were 5 times larger mass, than we operate in 1G, where 5F = m* 5a; but if Jupiter is still the same mass, we operate in 5G, where 5F = 5m*a, which means Jupiter's "kilogram" is one-fifth Earth's kilogram.. Now take a one "kilogram" mass from Jupiter, which is one-fifth the size of our previous mass, and take it to Earth's 1 G. Place that 0.2 mass on a balance scale with Earth's one kilogram mass, and the Earthian kilogram wins hands down. This is why Jupiter (if it really is in 5 G) is a far smaller planet core than we realize. Because this core is hidden by its incredibly vast atmosphere, we can't see it, though Voyager reported that it is about two or three Earth masses, which is still very small. Yet Jupiter's small core is operating in 5 G, so it balances on a scale with Earth 1 G mass, but it would take ten to fifteen Earths to do so. If you tried to move 10 to 15 Earths rather than two to three Earths, you would need a great deal more force, and that is why the inertial mass and gravitational mass are equivalent.