# Problem Involving Ring's gravitational force and velocity of distant partice

1. Aug 21, 2010

### Cocagola

1. The problem statement, all variables and given/known data
A uniform circular ring of radius R is fixed in place. A particle is placed on the axis of the ring
at a distance much greater than R and allowed to fall towards the ring under the influence of the ring’s gravity. The particle achieves a maximum speed v. The ring is replaced with one of the same(linear) mass density but radius 2R, and the experiment is repeated. What is the new maximum speed of the particle?

[a.] .5v [b.] v/sqrt(2) [c.] v [d.] sqrt(2)v [e.] 2v

(The answer is [c.] v. I don't know why)

2. Relevant equations

F = GMm/r^2 = m*dv/dt

3. The attempt at a solution

I don't know if I'm overcomplicating things but here's what I did.

I set up a differential equation, letting r equal the distance between the particle and the edge of the ring and M being the mass of the ring. (R is the radius of the ring, as stated in the problem)

so F = ma

GM/(R+r)^2 = dv/dt (the mass of the particle cancels)

I then used the chain rule to get rid of the variable t.

dv/dt = dv/dr *dr/dt. dr/dt = v, so dv/dt = vdv/dr.

GM(dr)/(R+r)^2 = vdv

I integrated both sides, the left side of the equation has limits of integration from 0 to D, the maximum distance between the particle and the ring. The right side of the equation has limits of integration from 0 to v, the maximum velocity of the particle.

After simplifying, I got an expression for v in terms of R.

v = sqrt((2D)/(RD + R^2)). (I took out the constants G and M, because they weren't important.)

D is just a constant, so when R changes by a factor of 2, v changes by a factor of 1/sqrt(2). However, the answer key says that the velocity stays constant regardless if R doubles. Can somebody please help me?

2. Aug 21, 2010

### Swap

ur answer seems to be overly complicated to me. Here is my solution:
-GMm/r=-GMm/R + 1/2 m v^2 . r being the distance between the particle and centre of the ring.
from here u will get v = sqrt. 2GM(1/R-1/r) which can be reduced to sqrt.2GM/R since R<< r. Then it easily follows the answer. because if u double the radius without changing the linear density then M will be doubled and same for the radius and hence they will cancel out and wont affect the answer.

3. Aug 22, 2010

### Cocagola

oh shoot! I forgot that the mass doubles as well as the radius. Yes, I was overcomplicating things by using kinematics rather than using conservation of energy...

But the thing that was messing me up was not doubling the mass as well as the radius.

4. Aug 22, 2010

Thanks!