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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.

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  • $\begingroup$ Near the Earth's surface suggests that the strength of gravity is constant. $\endgroup$ – sammy gerbil Oct 30 '17 at 2:00
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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 $10^{-5}~\rm{m/s^2}$

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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.

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  • $\begingroup$ Nope. $g$ is a function of height. $\endgroup$ – Floris Oct 29 '17 at 5:15
  • $\begingroup$ Please read the last paragraph of my answer before you say that. $\endgroup$ – Sahand Tabatabaei Oct 29 '17 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 '17 at 5:27
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    $\begingroup$ @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. $\endgroup$ – Chappo Oct 29 '17 at 11:18
  • $\begingroup$ Yes I guess you're right. My estimation on the significance of that was a bit off. $\endgroup$ – Sahand Tabatabaei Oct 29 '17 at 14:19
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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.

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  • $\begingroup$ 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? $\endgroup$ – Sahand Tabatabaei Oct 29 '17 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 '17 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! $\endgroup$ – Matthew Kuzma Oct 30 '17 at 12:49

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