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 asteroid (m = 3.19·1022 kg) hits the Earth while traveling at a speed of 3.49·103 m/s (assume the asteroid is a point mass compared to the radius of the Earth) in each of the following cases:

a) The asteroid hits the Earth dead center.

b) The asteroid hits the Earth nearly tangentially in the direction of Earth's rotation.

c) The asteroid hits the Earth nearly tangentially in the direction opposite of Earth's rotation.

Explanation / Answer

Earth angular speed wo = 7.272x10^-5 rad/s

m =3.19x10^22 kg , Radius of earth R = 6.371x10^6 m

mass of earth M = 5.98x10^24 kg

Moment of inertia of earth :

earth is solid sphere
Io = 2MR² / 5 = 2 * 5.98x1024 * (6.371x106)² / 5 = 9.71x1037 kg·m²

Initial angular momentum Lo = Io * o =

Lo = (9.71x1037 )(7.272x10-5 ) =7.06x1033 kg·m²/s

a) The asteroid adds to the moment of inertia, but does not contribute an angular momentum component of its own. Assuming it remains near the earth's surface,
new I = Io+ mR2

= (9.71x1037 ) + [(3.19x1022 )(6.371x106)² ]= 9.84x1037 kg·m²

From conservaion of angular momentum

Iowo=Iw

w = Iowo/I

= (9.71x1037 x7.272x10-5 )/(9.84x1037)

w =7.174x10-5 rad/s

b) Now the asteroid contributes L = mvR = (3.19x1022)(3490)(6.371x106)

L = 0.7093x1033 kg·m²/s

Lnew = Learth+Lasteroid
so Lnew = (7.06 + 0.7093)x1033= 7.7693x1033kg·m²/s

Lnew =Inew* wnew

Then new = Lnew / Inew = 7.7693x1033 / 9.84x1037

w = 7.896x10-5 rad/s

c) Now the asteroid contributes -0.7093x1033 kg·m²/s, so

so Lnew = (7.06 - 0.7093)x1033= 6.3507x1033kg·m²/s

L =Iw

new = 6.3507x1033 / 9.84x1037

w =6.454x10-5 rad/s

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