PHYSICS UNIVERSITY( ROTATION

DAMON

2013-10-25 5:47 am

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.

Q What would have to be the mass of this asteroid, in terms of the earth's mass M {\rm M} , for the day to become 26.0% 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

m=_____M ?

回答 (1)

天同

2013-10-25 7:39 am

✔ 最佳答案

Let T be the original period of earth rotation
Hence, new period T' = 1.26T after asteroid impact

Moment of inertial of earth = (2/5)MR^2 where R is radius of earth
Initial angular momentum of earth = (2/5)MR^2.(2.pi/T)

After impact, new moment of inertia of earth and asteroid = I
new angular momentum of earth and asteroid
= I.(2.pi/T')
= I.(2.pi/1.26T)

By conservation of angular momentum,
(2/5)MR^2.(2.pi/T) = I.(2.pi/1.26T)
(2/5)MR^2 = I/1.26
I = 0.504MR^2

But I = (2/5)MR^2 + mR^2 where m is the mass of asteroid
i.e. 0.504MR^2 = (2/5)MR^2 + mR^2
m = 0.104M

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