GPE and gravitational force exerted by a ring

In summary: Make sense?In summary, the conversation discusses the calculation of gravitational attraction between a homogeneous thin ring and a particle located on its central axis. The first part involves using the equation F = GMm/r^2 cos(arcsin(R/r)) to calculate the gravitational attraction, while the second part involves finding the speed at which the particle passes through the center of the ring by finding the difference in gravitational potential energy between its initial and final locations.
  • #1
GwtBc
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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.
 
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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
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1. What is GPE and how is it related to gravitational force exerted by a ring?

GPE stands for gravitational potential energy, which is the energy an object possesses due to its position in a gravitational field. The gravitational force exerted by a ring refers to the force that a ring exerts on an object due to its mass and distance from the object. GPE and gravitational force are related because GPE is a measure of the potential energy that an object has due to its position in a gravitational field, and gravitational force is the force that causes this potential energy.

2. How is the gravitational force exerted by a ring calculated?

The gravitational force exerted by a ring can be calculated using the formula F = (GmM)/r^2, where G is the gravitational constant, m is the mass of the object, M is the mass of the ring, and r is the distance between the object and the ring. This formula is based on Newton's Law of Universal Gravitation.

3. Does the distance between the object and the ring affect the gravitational force exerted?

Yes, the distance between the object and the ring directly affects the gravitational force exerted. According to the formula F = (GmM)/r^2, as the distance (r) increases, the force (F) decreases. This means that the farther an object is from the ring, the weaker the gravitational force will be.

4. What is the difference between GPE and kinetic energy?

GPE and kinetic energy are both forms of energy, but they are related to different types of motion. GPE is the energy an object has due to its position in a gravitational field, while kinetic energy is the energy an object has due to its motion. GPE is dependent on the object's height and mass, while kinetic energy is dependent on its velocity and mass.

5. How does the mass of the ring affect the gravitational force exerted?

The mass of the ring directly affects the gravitational force exerted. According to the formula F = (GmM)/r^2, as the mass (M) of the ring increases, the force (F) increases as well. This means that a heavier ring will exert a stronger gravitational force on an object than a lighter ring, if the distance and object's mass are kept constant.

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