Free fall of stones 
I suppose if the objects were large, they would begin to move towards each other (D).
However, the objects are small and this renders them weightless. I would suppose that in this case both A and B would continue to be at rest relative to the cabin (A).
The answer key, however, says that the correct option is B (that A moves slowly upward and B moves slowly downward relative to the cabin). The only reasoning I can provide for it is that B moves down due to the gravitational force (but in my opinion, B is too small for g to have any effect on it)
Can someone please explain how this works? Thanks.
 A: The correct answer can be deduced when you think about the fact that gravity is a function of distance (inverse square law). The object that is closer to Earth feels a slightly larger pull of gravity than the object that is further away. In the frame of reference of the cabin, that means the higher object falls more slowly (up) and the lower one falls faster (down). This leads to answer B.
Assuming that the distance between the objects is $x$, and that you are a distance $R$ from the center of the Earth; then the difference in gravitational acceleration is
$$\begin{align}
dg &= \frac{GM}{R^2}-\frac{GM}{(R+x)^2}\\
&=\frac{GM}{R^2}\left(1-\frac{1}{\left(1+\frac{x}{R}\right)^2}\right)\\
&=\frac{GM}{R^2}\left(1-(1-\frac{2x}{R})\right)\\
&=\frac{2GMx}{R^3}\\
&=\frac{2gx}{R}
\end{align}$$
So in the frame of reference of the cabin, this is the relative acceleration of the two stones. If they start out 3.2 m apart (1/2,000,000th of the earth's radius), their relative acceleration will be about $10^{-5}~\rm{m/s^2}$
A: The correct answer should be A, regardless of the mass of the stones (I am obviously neglecting effects such as air resistance). This is because at the moment the stones are released, their velocity relative to earth is the same as the velocity of the cabin (because they're at rest relative to the cabin). Thus, having the same velocity and acceleration (-g), the stones will exactly follow the cabin's motion and stay at rest relative to it.   The only case I can think of where A may not be true is when the cabin's height is so ridiculously big that the change in g between its top and bottom needs to be taken into account. Maybe you should look into that.
A: The answer is D and the reason is because the cabin is described as "narrow but tall" and the question is careful to mention the center of mass of the cabin.
First off, everything is in free-fall so the gravity of earth can be ignored.  The key to this question is in the gravitational forces that the cabin exerts on the stones and, to a lesser extent, that the stones exert on one another.
The cabin, being very massive compared to the stones, will generate a gravitational pull on the stones and, roughly speaking, that will pull them toward the cabin's center of mass.  Now, if the cabin were a perfect sphere, the net gravitational force on something inside it would be zero (this is one of the interesting quirks of spheres and gravity).
But the cabin being "narrow and tall" indicates that when you're "much above the center of mass" you're going to be pulled downward.  Most of the cabin's mass is below you, and because the cabin is narrow, most of it is focused directly below you.  So stone A will be pulled downward toward the center of mass of the cabin.  B will be pulled upward for the same reason.  So the answer is D.
And of course, the stones will exert very weak gravitational attractions on one another that will contribute (probably negligibly) to their attraction to one another.
