enter image description here

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.

  • $\begingroup$ Near the Earth's surface suggests that the strength of gravity is constant. $\endgroup$ Oct 30, 2017 at 2:00

3 Answers 3


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 $10^{-5}~\rm{m/s^2}$


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.

  • $\begingroup$ Nope. $g$ is a function of height. $\endgroup$
    – Floris
    Oct 29, 2017 at 5:15
  • $\begingroup$ Please read the last paragraph of my answer before you say that. $\endgroup$ Oct 29, 2017 at 5:17
  • $\begingroup$ You are already told B is the right answer... and "tall" and "close observation" suggests they are looking for small effects. To answer the question you need to explain why B is correct, not come up with a reason why it is not (that doesn't help the OP, who doesn't understand why it would be B). $\endgroup$
    – Floris
    Oct 29, 2017 at 5:27
  • 1
    $\begingroup$ @SahandTabatabaei: your last paragraph identifies that the difference in g between top and bottom is the key. Perhaps you've been distracted by the scale of the difference: it will, of course, be quite tiny. But measurable nonetheless, especially over longer periods. $\endgroup$ Oct 29, 2017 at 11:18
  • $\begingroup$ Yes I guess you're right. My estimation on the significance of that was a bit off. $\endgroup$ Oct 29, 2017 at 14:19

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.

  • $\begingroup$ Interesting remark. But I have a feeling that for the acceleration caused by the gravitational attraction of the cabin and stones to become significant, its mass must be ridiculously large, don't you think? $\endgroup$ Oct 29, 2017 at 4:59
  • $\begingroup$ Inside a massive sphere there is no gravitational field due to the sphere. That may not be quite true for a cabin, but the effect will certainly be MUCH weaker than when you are outside. Did you try to estimate that? The fact that things are in free-fall does NOT mean gravity of earth can be ignored: it is a function of height, and I suspect that's the point of this question. $\endgroup$
    – Floris
    Oct 29, 2017 at 5:17
  • $\begingroup$ Good question. Rough estimate, assuming the cabin is a 727 and 90% of it is one one side of you, the net acceleration induced on the stone mass will be roughly $10^{-10}$ smaller than earth gravity, making it smaller than the effect you describe in your answer. Nice one! $\endgroup$ Oct 30, 2017 at 12:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.