Asteroid of mass 10^16 kg is travelling in the equatorial plane of Earth. It hits the surface at 45°. After the impact the day shortens by 1% (15 mins). How fast was the asteroid - comment? Neglect effects of atmosphere. Consider only inelastic collision.

Key idea - total angular momentum of the system is conserved. Before the collision: angular momentum of earth is given by LE=IwE, here wE is the angular velocity of Earth and I is its moment of innertia which is given by I=2/5MER2. R is the radius of Earth and ME is its mass. Angular momentum of the asteroid is LA=mvd. Here m is asteroid's mass, v its velocity and d the perpendicular distance of its trajectory from the centre of the Earth given by d=sin(45°)R (see the whiteboard).
After the collision (inelastic) the asteroid is stuck to the surface and the whole system rotates at different angular velocity wF. Total angular momentum is therefore given by LF=I
wF + m
R2wF.
By equating the two angular momenta we obtain an equation from which we can express v - the asteroid's velocity. By neglecting the asteroids mass compared to the mass of the Earth (factor of 108) we obtain an approximate expression v=(I
(wF-wE))/(m*sin(45°)R) = 15.910^8 i.e. faster than the speed of light!! This means that our classical approximations break down and we need to treat the whole problem relativistically which is well beyond the scope of A level Physics.

Answered by Jiri K. Physics tutor

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