Two Spheres Attracted by Gravitational Force

Click For Summary
SUMMARY

The discussion focuses on the gravitational interaction between two spheres, each with a radius of 14m and mass of 1420kg, initially separated by a distance of 56m (4R). The conservation of energy principle is applied to determine their speed upon collision. The initial kinetic energy is zero, and the gravitational potential energy is calculated using the formula U(g) = -G*m1*m2/R. The correct final speed, accounting for both spheres' kinetic energy, is derived to be v = √(G*m/2R), which requires correction from the initial attempt of 5.8172e-5 m/s.

PREREQUISITES
  • Understanding of gravitational potential energy and kinetic energy concepts
  • Familiarity with the conservation of energy principle
  • Knowledge of gravitational constant G and its application in physics
  • Basic algebra for manipulating equations and solving for variables
NEXT STEPS
  • Study the derivation of gravitational potential energy equations
  • Learn about the conservation of energy in multi-body systems
  • Explore the implications of kinetic energy in collision scenarios
  • Investigate the effects of varying mass and distance on gravitational attraction
USEFUL FOR

Physics students, educators, and anyone interested in understanding gravitational interactions and energy conservation in multi-body systems.

maximoanimo
Messages
6
Reaction score
0

Homework Statement


Two spheres have equal radius R = 14m and mass m = 1420kg. Their centers are separated a distance 4R. The spheres are released from rest. What will be their speed when they collide? Answer in m/s.

Homework Equations



U(g) = -G*m1*m2/R

The Attempt at a Solution



We know from the conservation of energy that U(i) + K(i) = U(f) + K(f). Since the spheres are initially at rest, K(i) = 0. Initially, the two spheres are 4R apart. Upon collision, they must be 2R apart (the sum of their radii when they are touching). Knowing this information, we can plug in:

-Gm^2/4R = -Gm^2/2R + mv^2/2
-Gm^2/4R = -2Gm^2/4R + 2Rmv^2/4R
Gm^2 = 2Rmv^2
v^2 = Gm^2/2Rm = Gm/2R

Plugging in the values given, I found v = 5.8172e-5 m/s. However, it says this solution is incorrect. Can someone help me find the flaw in my solution?

Thanks so much,
Steven
 
Physics news on Phys.org
Remember that each of the spheres are going to have a value kinetic energy, so you'll have to take that into account when balancing the initial and final energies. Fix that and the rest should follow in a similar manner as in your derivation.
 
Ah, it took me a while to see what was wrong with the derivation in Post #1. Cider is correct; there are two spheres of course.
 

Similar threads

Replies
8
Views
2K
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 23 ·
Replies
23
Views
4K
Replies
335
Views
16K