Large meteors can strike the Earth with speeds of $80,000 \mathrm {~km} / \mathrm { h }$. Such a meteor hits the Earth (radius $6.37 \times 10 ^ { 6 } \mathrm {~m}$ ) at a point on the plane of the Earth's equator at $80,000 \mathrm {~km} / \mathrm { h }$, resulting in a change of the Earth's angular momentum of $80.0 \times 10 ^ { 12 } \mathrm {~kg} \cdot \mathrm { m } ^ { 2 } / \mathrm { s }$. The meteor's mass is
A) 13.0 kg.
B) 56.0 kg.
C) 126 kg.
D) 565 kg.
E) 637 kg.

Question

Quizplus · Accepted Answer

The angular momentum of the Earth-meteor system is conserved, so the change in the Earth's angular momentum is equal to the angular momentum of the meteor. The angular momentum of a point mass is $L = mvr$, where $m$ is the mass, $v$ is the speed, and $r$ is the distance from the axis of rotation. In this case, the distance from the axis of rotation is the radius of the Earth, $r = 6.37 \times 10^6 \mathrm{~m}$. The speed of the meteor is $v = 80,000 \mathrm{~km/h} = 22.2 \mathrm{~km/s}$. The change in the Earth's angular momentum is $\Delta L = 80.0 \times 10^{12} \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}$. So, the mass of the meteor is:$$m = \frac{\Delta L}{vr} = \frac{80.0 \times 10^{12} \mathrm{~kg} \cdot \mathrm{m}^2 / \mathrm{s}}{(22.2 \mathrm{~km/s})(6.37 \times 10^6 \mathrm{~m})} = 565 \mathrm{~kg}$$

Large Meteors Can Strike the Earth with Speeds Of 80,000 km/h80,000 \mathrm {~km} / \mathrm { h }80,000 km/h

Large Meteors Can Strike the Earth with Speeds Of $80,000 \mathrm {~km} / \mathrm { h }$