# Momentum of an asteroid / Earth

1. Nov 15, 2006

### canvas01

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?

2. Nov 15, 2006

### OlderDan

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)