Conservation of Angular Momentum of an asteroid

[ASUengineer16]

Homework Statement

Suppose that an asteroid traveling straight toward the center of the Earth were to collide with our planet at the equator and bury itself just below the surface.

Now The question asks: What would have to be the mass of this asteroid, in terms of the Earth's mass {\rm M} , for the day to become 28.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the Earth and that the Earth is uniform throughout.

Homework Equations

In the book it mentions this is a conservation of angular momentum problem, so:
I think that I1w1 + mvl = I2w2, where I1 is the inertia of the Earth and I2 is the inertia of the Earth and asteroid together; and w1 is angular velocity of Earth originally and w2 is angular velocity of the Earth and asteroid combined. I am not sure how to go about solving this one and how to relate the length of day being 28% longer to angular momentum. Any help would be greatly appreciated.

 

[salmon]

You have to conserve angular momentum about the center of the earth. Since the asteroid is headed straight for the center, your equation becomes:

I1 w1 = I2 w2, where:

I1 = (2/5) M(earth) R^2;
I2 = (2/5) M(earth) R^2 + m(ast) R^2

There's no mvl term since the asteroid doesn't hit the Earth tangentially.

For the w business, use T = 24 hrs initially => w1 = 2 pi / T etc.

 

[ASUengineer16]

Thank you that worked perfectly!

Came out to be .112M as the mass of the asteroid. Thanks again, I'm not a big fan of this angular moment stuff :(