Finding the potential energy of a gravitational force?

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SUMMARY

The discussion focuses on calculating the potential energy function U(x) for two masses M1 and M2 under gravitational attraction, described by the force equation F(r) = GM1M2/r^2. The correct potential energy function is derived as U(x) = -GM1M2/r, which contrasts with the incorrect positive equation initially proposed. Additionally, the work required to increase the separation of the particles is linked to the gravitational force, emphasizing the need for a negative sign in the force equation to maintain clarity in sign conventions.

PREREQUISITES
  • Understanding of gravitational force equations, specifically F(r) = GM1M2/r^2
  • Knowledge of potential energy concepts in physics
  • Familiarity with calculus, particularly integration techniques
  • Basic grasp of sign conventions in physics
NEXT STEPS
  • Study the derivation of potential energy functions in gravitational systems
  • Learn about the work-energy theorem in the context of gravitational forces
  • Explore the implications of sign conventions in physics problems
  • Investigate advanced topics in gravitational physics, such as gravitational potential energy in multi-body systems
USEFUL FOR

Students studying physics, particularly those focusing on classical mechanics, as well as educators and tutors assisting with gravitational force and potential energy concepts.

CoreanJesus
Sorry if something is wrong... this is my first ever post
1. Homework Statement

The Question:
In one dimension, the magnitude of the gravitational force of attraction between a particle of mass M1, and one of mass M2 is given by:
F(r)=GM1M2/r^2
Where G is a constant and x is the distance between the particles.

a) What is the potential energy function U(x)? Assume that U(x) -> 0 as x -> infinity.
b) How much work is required to increase the separation of the particles from x=x1 to x=x1+d?

Homework Equations


F(r)=-dU/dx

The Attempt at a Solution


For a), The answer given is -GM1M2/r but I keep getting a positive equation...
∫-dU=∫GM1M2/r^2
-U=-GM1M2/r
U=GM1M2/r
For b) I just need a clue as to how to start...
 
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CoreanJesus said:
In one dimension, the magnitude of the gravitational force of attraction between a particle of mass M1, and one of mass M2 is given by:
F(r)=GM1M2/r^2
The force is attractive, i. e. towards smaller r. It would be better to add a minus sign here, otherwise the sign convention is confusing.
CoreanJesus said:
For b) I just need a clue as to how to start...
What is the potential energy at the described places?
 
mfb said:
The force is attractive, i. e. towards smaller r. It would be better to add a minus sign here, otherwise the sign convention is confusing.What is the potential energy at the described places?
No potential energy is given...
 
You just calculated it!
 
Oh :D Thank You!
 

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