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

[Cocagola]

Homework Statement

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)

Homework Equations

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

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?

 

[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 won't affect the answer.

 

[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.

 

[Cocagola]

Thanks!

 

[rwisz]

It seems like you have approached the problem correctly, but there may be a mistake in your integration or simplification. The correct answer is indeed [c.] v, meaning that the maximum speed of the particle remains constant regardless of the change in radius of the ring. This can be explained by the conservation of energy in the system. As the particle falls towards the ring, its gravitational potential energy is converted into kinetic energy, resulting in an increase in speed. When the ring is replaced with one of twice the radius, the gravitational force on the particle decreases by a factor of 4, but the distance it falls also increases by a factor of 4. This results in the same amount of potential energy being converted into kinetic energy, leading to the same maximum speed. I would suggest double-checking your integration and simplification steps to find where the mistake may have occurred. Additionally, it may be helpful to try and solve the problem using conservation of energy instead of a differential equation approach.