For the acceleration of an object $m_2$ falling towards a much more massive object $m_1$, we have something like the following:
$$F=G\frac{m_1 m_2}{r^2} = \left\{ G\frac{m_1}{r^2}\right\} m_2=m_2a_2$$
where: $$ a_2=\left\{ G\frac{m_1}{r^2}\right\}$$
But we also have: $$F=G\frac{m_1 m_2}{r^2} = \left\{ G\frac{m_2}{r^2}\right\} m_1=m_1a_1$$
where: $$ a_1=\left\{ G\frac{m_2}{r^2}\right\}$$
Then, since $m_1$ is is so much more massive than $m_2$, the force exerted results in $a_2>>a_1$ and we simplify $a_1=0m/s^2$. This results in a constant acceleration of gravity for all small objects towards the more massive object $m_1$.
This results in the the good'ol "heavier objects fall at the same rate as lighter objects.
But staring at the equations, it seems constant acceleration of gravity independent of mass isn't true when the objects are of comparable masses thus resulting in non-negligible accelerations towards each other. I'm trying to think about where $a_1$ and $a_2$ are being measured. It doesn't seem to make sense that the acceleration of one object is measured with respect to the other object since they can be different and if you measured things this way the accelerations must necessarily be equal and opposite.
1. The only choice left is to measure the accelerations of the two objects with respect to a third, fixed point, correct?
2. And this would result in the objects accelerating towards each other at $|a_1|+|a_2|$?
3. And this in turn would result in a more massive objects falling faster than less massive objects? (use whichever mass you want as the reference, aka the "planet" that the other object falls into)
