Gravity due to a uniform ring of mass

In summary: The gravitational attraction between the particle and the small element of the ring is F=GMm/R^2where m=(rdθ)M/(πr2)
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Homework Statement


Several planets (Jupiter, Saturn, Uranus) are encircled by rings, perhaps composed of material that failed to form a satellite. In addition, many galaxies contain ring-like structures. Consider a homogeneous thin ring of mass 2.1 x 1022 kg and outer radius 4.3 x 108 m (the figure). (a) What gravitational attraction does it exert on a particle of mass 76 kg located on the ring's central axis a distance 4.5 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


Kepler's 3rd Law: T^2=(4*π^2/GM)*R^3
F=GmM/R^2

The Attempt at a Solution


I haven't really been able to attempt this question, I have absolutely no idea how to establish a relation between the mass on the axis and the ring. Please enlighten me with your genius ideas!
 

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  • #2
The problem should remind you of the similar one involving electric force: a ring of charge acting on a point charge along the axis of the ring.
 
  • #3
gneill said:
The problem should remind you of the similar one involving electric force: a ring of charge acting on a point charge along the axis of the ring.
I haven't previously studied electric force, I'm not sure what you meant.
 
  • #4
1. In what direction will the force from the whole ring be?
2. Consider a small segment of the ring, length rdθ, say. What force does it exert at the given point on the axis? What component of that acts in the direction in (1)?
 
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  • #5
Mentor's note:
Please note that the the thread title has been changed to: Gravity due to a uniform ring of mass.

The original title, "Challenging planetary problem", was too vague and non-descriptive of the actual problem.
 
  • #6
haruspex said:
1. In what direction will the force from the whole ring be?
2. Consider a small segment of the ring, length rdθ, say. What force does it exert at the given point on the axis? What component of that acts in the direction in (1)?
I'm guessing for 1) the force would be towards the centre, and 2) I don't see how r*dθ would exert a force, pls instruct me on how to analyse this system, thanks
 
  • #7
gneill said:
Mentor's note:
Please note that the the thread title has been changed to: Gravity due to a uniform ring of mass.

The original title, "Challenging planetary problem", was too vague and non-descriptive of the actual problem.
Okay I'll be more specific next time
 
  • #8
i_hate_math said:
I don't see how r*dθ would exert a force
The ring as a whole has a given mass M, so it has a mass per unit length density. If the ring has radius r, and a small piece of it subtends angle dθ at the centre of the ring then it has a mass (rdθ)M/(2πr). That will exert a gravitational force on the particle.
 
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  • #9
haruspex said:
The ring as a whole has a given mass M, so it has a mass per unit length density. If the ring has radius r, and a small piece of it subtends angle dθ at the centre of the ring then it has a mass (rdθ)M/(πr2). That will exert a gravitational force on the particle.
I am starting to understand. Is this related to Kepler's Second Law? I will need to go thru the textbook again if it is
 
  • #10
i_hate_math said:
I am starting to understand. Is this related to Kepler's Second Law? I will need to go thru the textbook again if it is
No, it's Newton's law of gravitation, which you quoted up front. What is the gravitational attraction between the particle and the small element of the ring?
 
  • #11
haruspex said:
No, it's Newton's law of gravitation, which you quoted up front. What is the gravitational attraction between the particle and the small element of the ring?
Right, F=GMm/R^2
where m=(rdθ)M/(πr2)
 
  • #12
i_hate_math said:
Right, F=GMm/R^2
where m=(rdθ)M/(πr2)
I made an error in my post #8, which you have blindly copied. Trust no-one!
Also, be careful with the M's - you have them mixed up. I suggest fixing on M for the mass of the ring and m for the mass of the particle.
 
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  • #13
haruspex said:
I made an error in my post #8, which you have blindly copied. Trust no-one!
Also, be careful with the M's - you have them mixed up. I suggest fixing on M for the mass of the ring and m for the mass of the particle.
I have managed to got the correct answer for this question, thank you very much for the help!
 

What is the concept of gravity due to a uniform ring of mass?

The concept of gravity due to a uniform ring of mass refers to the gravitational force experienced by an object placed near or within a circular ring with a constant distribution of mass. This force is a result of the sum of all the individual gravitational forces exerted by each small section of the ring on the object.

How is the gravitational force calculated for a uniform ring of mass?

The gravitational force on an object due to a uniform ring of mass is calculated using the formula F = G * M * m / d^2, where G is the universal gravitational constant, M is the mass of the ring, m is the mass of the object, and d is the distance between the object and the center of the ring.

What is the direction of the gravitational force for a uniform ring of mass?

The direction of the gravitational force on an object due to a uniform ring of mass is always towards the center of the ring. This is because the gravitational force exerted by each small section of the ring is directed towards the center, and the total force is the sum of all these individual forces.

How does the mass and distance of the object affect the gravitational force of a uniform ring?

The gravitational force exerted by a uniform ring of mass on an object is directly proportional to the mass of the object and inversely proportional to the square of the distance between the object and the center of the ring. This means that a heavier object or a smaller distance will result in a stronger gravitational force.

Can the gravitational force of a uniform ring be negative?

No, the gravitational force of a uniform ring can never be negative. The force is always attractive and directed towards the center of the ring. However, the net force on an object can be negative if there are other forces acting on the object in addition to the gravitational force from the ring.

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