Momentum of an asteroid / Earth

Click For Summary
SUMMARY

The discussion focuses on the relationship between the angular momentum of Earth and the linear momentum of an asteroid striking it tangentially at the equator. The asteroid, despite being non-rotating, possesses angular momentum relative to Earth's rotation axis. The Earth’s angular momentum is calculated using the formula Iω, where I is the moment of inertia (0.4mr²) and ω is the angular velocity derived from the Earth's circumference and rotation period. The key question is how to relate the asteroid's linear momentum (mv) to the Earth's angular momentum to determine the percentage change in Earth's angular speed post-impact.

PREREQUISITES
  • Understanding of angular momentum and linear momentum concepts
  • Familiarity with the moment of inertia and its calculation
  • Knowledge of angular velocity and its relation to linear speed
  • Basic principles of collision mechanics in physics
NEXT STEPS
  • Study the conservation of angular momentum in collision scenarios
  • Learn how to calculate the moment of inertia for various shapes
  • Explore the relationship between linear and angular momentum
  • Investigate the effects of external forces on angular velocity
USEFUL FOR

Physics students, educators, and professionals interested in mechanics, particularly those studying angular momentum and collision dynamics in astrophysics.

canvas01
Messages
1
Reaction score
0
Please help me to understand this question: I'm paraphrasing it. A speed (relative to earth) and mass for an asteroid are given, and the asteroid strikes the Earth tangentially in the direction of the Earth's rotation at the equator. The question tells us to use angular momentum to find the percentage change in the angular speed of the earth.

My approach: Because the asteroid is not rotating, it has no angular momentum, just linear momentum represented as mv. The Earth however, has no linear momentum, just angular momentum which is equal to Iw where I = .4mr^2 and w = v/r where v = circumference of the earth/time of one complete rotation of the earth. I'm confused as to where to go from here? How are the angular momentum of the Earth and the linear momentum of the asteroid related?
 
Physics news on Phys.org
canvas01 said:
Please help me to understand this question: I'm paraphrasing it. A speed (relative to earth) and mass for an asteroid are given, and the asteroid strikes the Earth tangentially in the direction of the Earth's rotation at the equator. The question tells us to use angular momentum to find the percentage change in the angular speed of the earth.

My approach: Because the asteroid is not rotating, it has no angular momentum, just linear momentum represented as mv. The Earth however, has no linear momentum, just angular momentum which is equal to Iw where I = .4mr^2 and w = v/r where v = circumference of the earth/time of one complete rotation of the earth. I'm confused as to where to go from here? How are the angular momentum of the Earth and the linear momentum of the asteroid related?
The asteroid does have angular momentum relative to the rotation axis of the earth. What is the definition of angular momentum? (It is not Iω; that is derived from the definition applied to a rotating rigid object)
 

Similar threads

Replies
335
Views
17K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
67
Views
5K
Replies
28
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
11
Views
988
Replies
9
Views
3K
Replies
18
Views
7K