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I've imagined this little scenario to help me conceptualize things.

Let's say we have a doughnut-shaped object with a hole whose diameter is greater than that of a sphere. Let's say that the sphere is vertically aligned with the center of the doughnut and is horizontally gravitating towards it according to Newton's law of universal gravitation. What will happen?

1) The sphere will pass through the doughnut, travel a certain distance (but how far) and accelerate back towards the doughnut (and oscillate back and forward).

2) The sphere will stop as it reaches the center of the doughnut. The reasoning behind this is that the distance between the two objects will be zero and hence the acceleration which is inversely proportional to the distance will be infinity (but in both directions?). This doesn't sound right.

3) Something else

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  • $\begingroup$ Is your doughnut fixed in place? $\endgroup$
    – BMS
    Apr 7, 2014 at 20:19
  • $\begingroup$ I assumed it was, but will it change things if it wasn't? $\endgroup$
    – Simeon
    Apr 7, 2014 at 20:27
  • $\begingroup$ At the center of the torus acceleration is zero. With mass symmetricaly distributed about the sphere feels an equal tug from each side. The torus mass to the left would cancel the torus mass to the right. $\endgroup$
    – HopDavid
    Apr 7, 2014 at 20:46
  • $\begingroup$ @dfan: Your comment made me think for a while and created doubts in myself. I will to try look for that problem (I have deleted the post now), if possible I will reply. Anyway thank you for the comment, I learnt something new. $\endgroup$
    – Sensebe
    Apr 8, 2014 at 1:02

4 Answers 4

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1) is correct. The wrong reasoning about 2) is that what you have in mind is probably Newtons Law for point masses. When the sphere is close to the doghnut the gravitational force will be more complicated, but still point towards the centre of the doughnut due to the symmetries in the situation. It will however stay finite, because all points of the sphere still have a finite distance to all points of the doughnut.

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  • $\begingroup$ If they were point masses and could slide past each other instead of colliding which of the cases would hold true? $\endgroup$
    – Simeon
    Apr 8, 2014 at 5:45
  • $\begingroup$ Well, you get a divergent acceleration then. But it doesn't make sense to consider this situation. The acceleration will increase to positive infinity when the point masses approach and then instantaneously change to negative infinity. For a physically reasonable situation you always have to assume some finite extent of your masses. A point mass is just an approximation. $\endgroup$
    – André
    Apr 8, 2014 at 8:37
  • $\begingroup$ Just to clarify my last sentence: Elementary particles, as for example electrons, are (at least as far as we know today) indeed point particles. But when they approach you still cannot consider them as classical point particles but have to take quantum mechanics into account. $\endgroup$
    – André
    Apr 8, 2014 at 8:45
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Your answer (1) is the correct one. It is actually quite simple if you think in terms of conservation of energy. What you have described is a simplified version of a two body problem. Note that strictly speaking, both the doughnut $(D)$ and the ball $(B)$ will move towards each other. But without outside influence, their combined center of mass should be fixed (or moving at constant velocity). However, the solution is the same if we think of $D$ as being fixed and only $B$ is moving (essentially we are just using $D$ as our frame of reference).

We look at what happens at each of the following to answer the question:

  1. the initial time when $B$ is $d$ distance away from $D$ and 'released' with $0$ initial velocity
  2. when $B$ reaches the center $c$ of $D$
  3. after $B$ passes though $D$

1) the energy on $B$ is purely potential (gravitational pull exerted on it by $D$). It has $0$ kinetic energy since velocity at that initial time is $0$.

2) the gravitational force on $B$ becomes $0$ only when it reaches the center of $D$. Note that by symmetry the perpendicular (to it's path) component of the net force acting on $B$ is zero at all times. So it will at least move towards $c$ until it reaches it. But the gravitational potential energy on $B$ cannot just disappear, it must have completely converted into kinetic energy. So the velocity of $B$ at $c$ cannot be zero; $B$ shoots through $c$.

3) $B$ will stop when the kinetic energy it carried at $c$ is completely converted into potential gravitational energy. By symmetry of the set up, this will happen when $B$ has traveled a distance $d$ away from $c$.

So we see that the ball $B$ will oscillate back and forth perpetually.

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when your ball is far from a donut, there's a potential energy of the gravitational field. when the ball reaches the donut, that potential energy must be converted into something else, it can't disappear. do it must be that it converted into the kinetic energy of movement, i.e. the ball must be moving at this point, hence, it'll pass the center at some speed, it won't stop still

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Newton's law of universal gravitation holds for point like masses. For spatial masses you would have to integrate of infinitely small parts of that mass. However it does turns out that the result of this for a sphere of constant density at a given radius does yield that simple formula, see Gauss's law.

To get a better understanding of this, without having to derive the analytical answer yourself, you can play with Algodoo (a free 2D physics simulator). Due to axial-symmetry you can approximate this problem in 2D by two fixed circles between which a third circle can move freely.
Do keep in mind that this problem is not stable. With this I mean that if the third circle is slightly closer to one of the circle, it will be pulled closer over time.

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