- #1

GwtBc

- 74

- 6

## Homework Statement

Consider a homogeneous thin ring of mass 2.5 x 1022 kg and outer radius 3.9 x 108 m (the figure).

**(a)**What gravitational attraction does it exert on a particle of mass 69 kg located on the ring's central axis a distance 3.7 x 108 m from the ring center?

**(b)**Suppose that, starting at that point, the particle falls from rest as a result of the attraction of the ring of matter. What is the speed with which it passes through the center of the ring?

## Homework Equations

U = - GMm/r

F = GMm/r^2

## The Attempt at a Solution

[/B]

Seeing as the ring is uniform and the mass is sitting on it's central axis, the first part is simple enough, I think, it's just:

F = GMm/r^2 cos(arcsin(R/r))

where r is the distance from the mass to each dm on the ring and R is the radius of the ring.

I'm not sure about the second part. I currently have:

##W = \int_{R}^{r_{i}} \frac{GMm}{r}\cos (\arcsin(\frac{R}{r})) dr##

## KE_{f} = W \Rightarrow \frac{1}{2}mv_{f}^2 = W##

Where ##r_{i}## is the initial distance of the mass from the ring (the ring itself, not it's centre of mass)

but using my numbers, this returns ## v_{f} = 8.03568 * 10^5## which seems a bit off.

any help is appreciated.