The answers are: a. 5.82 x 10–5 rad/s b. ME/10 or 6 x 10^23 kg c. 2.1 x 10^29 J

Question

The answers are:

a. 5.82 x 10–5 rad/s

b. ME/10 or 6 x 10^23 kg

c. 2.1 x 10^29 J

Please show me the procedures.

P10: Suppose that an asteroid travelling straight towards the center of the Earth were to collide with our planet at the equator and bury itselfjust below the surface. (R,-6.4 x 106 m) a) If due to the impact the day becomes 25% longer, what is the new angular speed of Earth? b) What would have to be the mass of the asteroid (in terms of the mass of the Earth, ME-597 x 1024 kg) for the day to become 25% longer? c) What is the kinetic energy of the system after the impact?

Explanation / Answer

a)We know that

w = 2 pi /T = 2 x 3.14/86400 = 7.27 x 10^-5 rad/s

T' = 25% longer 24 = 30 hrs = 108000

w' = 2 x 3.14/108000 = 5.82 x 10^-5 rad/s

Hence, w' = 5.82 x 10^-5 rad/s

b)Ii = 2/5 M R^2

after asteriods interaction

2/5 M R^2 x 1.25 = 2/5 M R^2 + m M R^2

M R^2 both sides get cancelled

0.5 = 0.4 = k

k = 0.5 - 0.4 = 0.1

m = 0.1 M = 1/10 M = 0.1 x 6 x 10^24 = 6 x 10^23

Hence, m = 6 x 10^23 kg

c)KE = 1/2 I w^2

I = 2/5 x 6 x 10^24 x (6.4 x 10^6)^2 + 6 x 10^23 x (6.4 x 10^6)^2 = 1.23 x 10^38

KE = 0.5 x 1.23 x 10^38 x (5.82 x 10^-5)^2 = 2.1 x 10^29 J

Hence, KE = 2.1 x 10^29 J

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