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Time for ring of equidistant particles to collapse (gravity)

  • #1
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(I assume that the three section headings below form the template referred to below)

1. Homework Statement

n identical equi-distant particles are distributed equi-distantly around the circumference of a ring of radius r in space. Each particles is of mass m, so the total mass of the ring is n*m. External gravitational fields are negligible and there is no other mass within the ring. The particles are initially at rest and the ring is not rotating. How long will it take the particles to meet at the center of the ring under gravitational attraction between the particles?


Homework Equations



"Time to meet function" for two particles at rest to meet under the force of gravity:

t = time to meet function = pi * (8 * G)^(-1/2) * x^(3/2) * (m1 + m2)^(-1/2)[/B]

where

t is the time to meet
G is the universal gravitational constant
x is the initial distance between the two particles
m1 and m2 are the masses of the two particles


Ref: Answers to Physics Forums question

The Attempt at a Solution




radius of ring = r
n particles of mass m
total mass n*m
the centre of mass of the ring is at the physical center of the ring
there are n^2 unique pairs of particles where each pair includes two particles attracting one another
the mutual gravitational attraction of each unique pair is independent of every other pair, i.e. the mutual attractions can be combined linearly and additively

Result for two particles is pi * (8 * G)^(-1/2) * (2r)^(3/2) * (m + m)^(-1/2)
Result for three particles is:[/B]

Side of an equilateral triangle inscribed in a circle of radius 1 is approx 1.732
There are three identical pairs each 1.732r, but the infall direction is towards the center of the ring not along the straight line between the particles, and it is hard to see how to use the time to meet function in this situation and also to generalize to n particles
 

Answers and Replies

  • #2
haruspex
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there are n^2 unique pairs of particles
No, less than half that. Each pair consists of choosing 2 from n.
the mutual attractions can be combined linearly
Yes, but as vectors, so there will be cancellation. By symmetry, you only care about the forces pulling the particles to the common centre.

Just consider one particle A attracted to another, B, k steps around the circle.
Each step around the circle subtends what angle at the centre?
What angle do these two particles subtend?
How far is it between these two particles?
What is the force on A from B?
What is the component of that force acting towards the common centre?
 
  • #3
67
0
No, less than half that. Each pair consists of choosing 2 from n.

Yes, but as vectors, so there will be cancellation. By symmetry, you only care about the forces pulling the particles to the common centre.

Just consider one particle A attracted to another, B, k steps around the circle.
Each step around the circle subtends what angle at the centre?
What angle do these two particles subtend?
How far is it between these two particles?
What is the force on A from B?
What is the component of that force acting towards the common centre?
Dear Haruspex, thanks so much for your help, I really appreciate it very much, and I think I can do it given your excellent assistance, and thanks to Berkeman too :-)
 

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