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Here's a heuristic argument that I've seen which supposedly shows that $m_{g} = m_{i}$:

Consider an arbitrary object, with inertial mass $m_{i}$ and (passive) gravitational mass $m_{g}$. Its acceleration during free-fall is proportional to the gravitational force it experiences, which in turn is proportional to $m_{g}$. But according to Newton's Second Law, it's also inversely proportional to $m_{i}$.

Thus its acceleration is proportional to $\frac{m_{g}}{{m_i}}$:

$$\ddot{x} = k \frac{m_{g}}{m_{i}} $$

where $k$ is a constant. But all objects fall with the same acceleration, so $ k \frac{m_{g}}{m_{i}}$ must be constant for any object, thus $m_{g} = m_{i}$.

But the ratio $\frac{m_{g}}{{m_i}}$ need not be unity; it could be $2.5$, say, and this would not affect the conclusion that all objects fall with the same acceleration. So is the above argument sufficient?

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  • $\begingroup$ I guess your argument is valid. You can absorb the constant in the gravitational constant. $\endgroup$ – Tony Feb 8 '17 at 10:50
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Your statement that "all objects fall with the same acceleration" assumes that the inertial and gravitational mass of an object are identical. The argument is therefore not sufficient.

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As Albert Einstein puts it (which I believe correctly so) :

A little reflection will show that the theorem of the equality of the inert and the gravitational mass is equivalent to the theorem that the acceleration imparted to a body by a gravitational field is independent of the nature of the body. For Newton's equation of motion in a gravitational field, written out in full, is

$$ (Inert \ mass) \cdot (Acceleration) = (Intensity \ of \ the \ gravitational \ field) \cdot (Gravitational \ mass) $$

It is only when there is numerical equality between the inert and gravitational mass that the acceleration is independent of the nature of the body.

Albert Einstein, The Meaning of Relativity https://en.wikisource.org/wiki/Page:The_Meaning_of_Relativity_-_Albert_Einstein_(1922).djvu/75

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  • $\begingroup$ About how it connects to your answer, the assumptions that "all objects fall with the same acceleration" and "gravitational mass is equal to inertial mass" are equivalent, so you cannot use one in the other's "proof". $\endgroup$ – mlg556 Nov 3 '17 at 3:57

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