The earth has an angular speed of 7.272*10^-5 rad/s in its rotation. Find the ne

Question

The earth has an angular speed of 7.272*10^-5 rad/s in its rotation. Find the new angular speed if an asteriod (m=1.22*10^22kg) hits the earth while traveling at a speed of 1.45*10^3 m/s.(assume the asteriod is a point mass compared to the radius of the earth) in each of the following cases.

A. The asteriod hits the earth dead center along the radial line from the Sun through the Earth's center. Answer is given 7.24*10^-5rad/s

B. The asteriod hits the Earth nearly tangentially in the direction of Earth's rotation. Answer is given 7.35*10^-5rad/s

C. The asteroid hits the Earth nearly tangentially in the direction opposite to Earth's rotation. Answer is given 7.12*10^-5 rad/s

Explanation / Answer

A.

w = 7.272*10^-5 rad/s (there is no change here because asteroid hits the Earth dead center).

B.

Formula to use:   lw + mvr = lw'

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(2/5*5.97E24*6.37E6^2)*(7.272E-5) + 1.22*10^22*1.45*10^3 *6.37E6 = (2/5*5.97E24*6.37E6^2)*(w)

w = (2/5*5.97E24*6.37E6^2)*(7.272E-5) + 1.22*10^22*1.45*10^3 *6.37E6 / (2/5*5.97E24*6.37E6^2)

w = 7.35*10^-5rad/s

C. Formula to use: lw - mvr = lw'

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(2/5*5.97E24*6.37E6^2)*(7.272E-5) - 1.22*10^22*1.45*10^3 *6.37E6 = (2/5*5.97E24*6.37E6^2)*(w)

w = (2/5*5.97E24*6.37E6^2)*(7.272E-5) - 1.22*10^22*1.45*10^3 *6.37E6 / (2/5*5.97E24*6.37E6^2)

w = 7.12*10^-5 rad/s

HOPE THIS HELPS, PLEASE RATE!

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