asteroid traveling

Anonymous

2013-07-12 5:44 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.

What would have to be the mass of this asteroid, in terms of the earth's mass M
, for the day to become 25.0% 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.

回答 (1)

天同

2013-07-12 7:26 am

✔ 最佳答案

Moment of inertia of earth = (2/5)MR^2
where R is the radius of the earth.

Initial angular momentum of earth = [(2/5)MR^2]. w
where w is the rotational speed of the earth, which is equal to 2.pi/T (T being the rotational period of the earth)
Hence, initial angular momentum of earth = (2MR^2/5).(2.pi/T)

Let m be the mass of the asteroid.
After the asteroid is buried into earth surface, the total moment of inertia of the earth and asteroid = 2MR^2/5 + mR^2
Hence, final angular momentum of earth and asteriod
= (2MR^2/5 + mR^2).(2.pi/T')
where T' is the new period of earth rotation, which is equal to 1.25T

Using Law of Conservation of Angular Momentum before and after collision of the asteriod with earth,
(2MR^2/5).(2.pi/T) = (2MR^2/5 + mR^2).(2.pi/1.25T)
After simplification,
2M/5 = (2M/5 + m).(1/1.25)
solve for m gives m = 0.1M

The mass of the asteriod is 10% of that of the earth.

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