Is this argument sufficient to demonstrate the equality between passive gravitational mass and inertial mass?

Question

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?

Answer 1

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.

Answer 2

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