GPE and gravitational force exerted by a ring

Click For Summary

Homework Help Overview

The problem involves a homogeneous thin ring and its gravitational effects on a particle located along its central axis. Participants are tasked with calculating the gravitational attraction exerted by the ring on the particle and determining the particle's speed as it falls through the center of the ring.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply gravitational force equations and work-energy principles to solve the problem. Some participants suggest alternative methods, such as calculating gravitational potential energy at different locations. There is a discussion about the complexity of the integral involved in the work calculation.

Discussion Status

Participants are actively exploring different approaches to the problem. Some guidance has been offered regarding the use of gravitational potential energy instead of directly integrating the work done. There is recognition of the challenges posed by the integral, and participants are questioning the setup and calculations presented.

Contextual Notes

Participants note potential issues with the original poster's calculations and assumptions, particularly regarding the integration process and the treatment of gravitational potential energy. There is an emphasis on ensuring that the correct expressions are used for the gravitational force and potential energy.

GwtBc
Messages
74
Reaction score
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.
 
Physics news on Phys.org
We'd need to see how you carried out the work integral. It looks like it could be nasty.

Perhaps it would be easier to find the gravitational potential energy of the system at the initial and final locations and take the difference?
 
gneill said:
We'd need to see how you carried out the work integral. It looks like it could be nasty.

Perhaps it would be easier to find the gravitational potential energy of the system at the initial and final locations and take the difference?
It is nasty, but my calculator doesn't mind that. Also the denominator 'r' terms is supposed to be 'r^2' in the integral.

Aren't I essentially finding the initial and final GPEs and then taking the difference? How else can I find the GPE?
 
Since PE is a scalar you don't have to worry about vector components or integrating the contributions. Instead, you know that all the dm's are the same distances from the given location in each case. So you should be in a position to write an expression for the GPE at each location by inspection. For example, at the ring's center all the ring's mass is located at the same distance r. So the GPE there must be GMm/r.
 
  • Like
Likes   Reactions: GwtBc

Similar threads

  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
7
Views
2K
Replies
1
Views
1K
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K