I'm not sure what that means.
> If you ever find where Young worked out the equivalency to P.E. then you'll find he had m v² = 2 m g h.
I don't get it. mv² is not equal to 2mgh. That makes no sense.
I think I now know what you mean. You mean that a mass will gain kinetic equal to 1/2 mv² by the time it hits the ground from a height of h. This doesn't prove your claim. Using my own derivation of kinetic energy, I get mgh = mv². Here's why:
PE = mgh is the same thing as Newton's Work equation, W = Fd = mad. Why? Because g is acceleration (a) and h is distance (d).
"proportional to mv²" also means "proportional to 1/2 mv²" and "proportional to 123.45 mv²".
Assuming your gravitational and inertial masses are supposed to be the same, then 1/2 m v² is, by Noether's theorem, a consequence of conservation of energy. https://en.wikipedia.org/wiki/Noether%27s_theorem#Example_1:... .
Yes. This is my point. The error has never been found experimentally because everything eventually gets resolved to Newton's E = Fd. Dividing every result by 2 makes no difference because the proportions do not change.
I still don't see why gravitational mass is a problem but I'll keep looking into it.