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 = 4.99·1022 kg) hits the Earth while traveling at a speed of 3.23·103 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.
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We need to find the angular momentum of earth:
Io = 2mr² / 5 = 2 * 5.98e24kg * (6.371e6m)² / 5 = 9.71e37 kg·m²
ωo = 2π rads/day * 1day/24hr * 1hr/3600s = 7.272e-5 rad/s
Initial angular momentum Lo = Io * ωo = 7.06e33 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 = 9.71e37kg·m² + 4.99e22kg * (6.371e6m)² = 9.91e37 kg·m²
Then ω = 7.272e-5 rad/s * 9.71 / 9.91 = 7.123e-5 rad/s
b) Now the asteroid contributes L = mvr = 4.99e22kg * 3230m/s * 6.371e6m,
or L = 1.027e33 kg·m²/s
so Lnew = (7.06 + 1.027)e33 kg·m²/s = 8.087e33 kg·m²/s
Then new ω = Lnew / Inew = 8.087e33kg·m²/s / 9.91e37kg·m² = 8.16e-5 rad/s
c) Now the asteroid contributes -1.027e33 kg·m²/s, so
Lnew = 6.033e33 kg·m²/s, so
new ω = 6.033kg·m²/s / 9.91e37kg·m² = 6.088e-5 rad/s