The common understanding is that, setting air resistance aside, all objects dropped to Earth fall at the same rate. This is often demonstrated through the thought experiment of cutting a large object in half—the halves clearly don't fall more slowly just from having been sliced into two pieces.
However, I believe the answer is that when two objects fall together, attached or not, they do "fall" faster than an object of less mass alone does. This is because not only does the Earth accelerate the objects toward itself, but the objects also accelerate the Earth toward themselves. Considering the formula:
$$
F_{\text{g}} = \frac{G m_1 m_2}{d^2}
$$
Given $F = ma$ thus $a = F/m$, we might note that the mass of the small object doesn't seem to matter because when calculating acceleration, the force is divided by $m$, the object's mass. However, this overlooks that a force is actually applied to both objects, not just to the smaller one. An acceleration on the second, larger object is found by dividing $F$, in turn, by the larger object's mass. The two objects' acceleration vectors are exactly opposite, so closing acceleration is the sum of the two:
$$
a_{\text{closing}} = \frac{F}{m_1} + \frac{F}{m_2}
$$
Since the Earth is extremely massive compared to everyday objects, the acceleration imparted on the object by the Earth will radically dominate the equation. As the Earth is $\sim 5.972 \times {10}^{24} \, \mathrm{kg} ,$ a falling object of $5.972 \times {10}^{1} \, \mathrm{kg}$ (a little over 13 pounds) would accelerate the Earth about $\frac{1}{{10}^{24}}$ as much, which is one part in a trillion trillion.
Thus, in everyday situations we can for all practical purposes treat all objects as falling at the same rate because this difference is so small that our instruments probably couldn't even detect it. But I'm hoping not for a discussion of practicality or what's measurable or observable, but of what we think is actually happening.
Am I right or wrong?
What really clinched this for me was considering dropping a small Moon-massed object close to the Earth and dropping a small Earth-massed object close to the Moon. This thought experiment made me realize that falling isn't one object moving toward some fixed frame of reference, and treating the Earth as just another object, "falling" consists of multiple objects mutually attracting in space.
Clarification: one answer points out that serially lifting and dropping two objects on Earth comes with the fact that during each trial, the other object adds to the Earth's mass. Dropping a bowling ball (while a feather waits on the surface), then dropping the feather afterward (while the bowling ball stays on the surface), changes the Earth's mass between the two experiments. My question should thus be considered from the perspective of the Earth's mass remaining constant between the two trials (such as by removing each of the objects from the universe, or to an extremely great distance, while the other is being dropped).
Best Answer
Using your definition of "falling," heavier objects do fall faster, and here's one way to justify it: consider the situation in the frame of reference of the center of mass of the two-body system (CM of the Earth and whatever you're dropping on it, for example). Each object exerts a force on the other of
$$F = \frac{G m_1 m_2}{r^2}$$
where $r = x_2 - x_1$ (assuming $x_2 > x_1$) is the separation distance. So for object 1, you have
$$\frac{G m_1 m_2}{r^2} = m_1\ddot{x}_1$$
and for object 2,
$$\frac{G m_1 m_2}{r^2} = -m_2\ddot{x}_2$$
Since object 2 is to the right, it gets pulled to the left, in the negative direction. Canceling common factors and adding these up, you get
$$\frac{G(m_1 + m_2)}{r^2} = -\ddot{r}$$
So it's clear that when the total mass is larger, the magnitude of the acceleration is larger, meaning that it will take less time for the objects to come together. If you want to see this mathematically, multiply both sides of the equation by $\dot{r}\mathrm{d}t$ to get
$$\frac{G(m_1 + m_2)}{r^2}\mathrm{d}r = -\dot{r}\mathrm{d}\dot{r}$$
and integrate,
$$G(m_1 + m_2)\left(\frac{1}{r} - \frac{1}{r_i}\right) = \frac{\dot{r}^2 - \dot{r}_i^2}{2}$$
Assuming $\dot{r}_i = 0$ (the objects start from relative rest), you can rearrange this to
$$\sqrt{2G(m_1 + m_2)}\ \mathrm{d}t = -\sqrt{\frac{r_i r}{r_i - r}}\mathrm{d}r$$
where I've chosen the negative square root because $\dot{r} < 0$, and integrate it again to find
$$t = \frac{1}{\sqrt{2G(m_1 + m_2)}}\biggl(\sqrt{r_i r_f(r_i - r_f)} + r_i^{3/2}\cos^{-1}\sqrt{\frac{r_f}{r_i}}\biggr)$$
where $r_f$ is the final center-to-center separation distance. Notice that $t$ is inversely proportional to the total mass, so larger mass translates into a lower collision time.
In the case of something like the Earth and a bowling ball, one of the masses is much larger, $m_1 \gg m_2$. So you can approximate the mass dependence of $t$ using a Taylor series,
$$\frac{1}{\sqrt{2G(m_1 + m_2)}} = \frac{1}{\sqrt{2Gm_1}}\biggl(1 - \frac{1}{2}\frac{m_2}{m_1} + \cdots\biggr)$$
The leading term is completely independent of $m_2$ (mass of the bowling ball or whatever), and this is why we can say, to a leading order approximation, that all objects fall at the same rate on the Earth's surface. For typical objects that might be dropped, the first correction term has a magnitude of a few kilograms divided by the mass of the Earth, which works out to $10^{-24}$. So the inaccuracy introduced by ignoring the motion of the Earth is roughly one part in a trillion trillion, far beyond the sensitivity of any measuring device that exists (or can even be imagined) today.