An items reaction to a gravitational pull inside a hollow cylinder of infinite mass?by thesweeneyman Tags: gravity, mass, theoratical phyiscs 

#1
Mar3011, 03:29 PM

P: 3

My applied maths teacher proposed this question while we were having a discussion about gravitational effects. We do not know the answer.
Given a hollow cylinder with a mass large enough to have its own gravitational effects, placed in zero gravity. If you propel an object into to an open end of the cylinder, what path will it take? (a) continue straight downward? (b) move to the centre and continue straight downward? (c) move to the nearest side? (d) spiral downward? the attachment with this thread is my bad diagram of the cylinder. We are extremely curious as to what the answer is so we would be extremely grateful if it could be figured out by better minds. 



#2
Mar3011, 03:58 PM

P: 47

Interesting question.
http://www.jstor.org/pss/2397270 Apparently someone else had the same question. 



#3
Mar3111, 04:05 PM

P: 3

i did not know that page existed thanks




#4
Mar3111, 04:26 PM

P: 47

An items reaction to a gravitational pull inside a hollow cylinder of infinite mass?
I just wanted to place a guess.
From the following (probably faulty) reasoning. Imagine a planet orbiting a large body. I think that if the planet follows a stable orbit, the large body does not undergo a displacement over the period of the orbit. So now, let's image the planet breaks up for forms a belt. The large body is not accelerated by the belt, and for the orbit to be stable, it must be an ellipse. Your cylinder is not an ellipse, it's a circle with the object off center. To make an ellipse a circle, you must pull some parts away from the object, which results in a weaker pull toward the center, and the object will deflect toward the outside of the cylinder. I would be interested to know what the paper says. 



#5
Mar3111, 05:02 PM

P: 288

I think it would continue straight through, because for any point in a 2D ring the gravitational force is 0, so if you integrated over dz you would still get a net force of 0N, so its position wouldn't change.
Edit: Hm actually I'm not sure now, deriving it from a small area of uniform surface density, it suggests that F1/F2=R2/R1 which suggests it moves to the nearest side. 



#6
Apr311, 06:21 AM

P: 3

@Sciurus interesting theory but there is no rotation in the cylinder like there is in a belts orbit.... P.s i cant access that paper
@Vagan fair point but the radius may be shorter on the near side which povides a large force but there is a lot more mass on the opposite side 


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