Question

1. the moon has a mass of ( 7.35 \times 10^{22} ) kg, a radius of 1,738 km, and an angular velocity of 1 revolution every 27.3 days. determine the moons angular momentum.

( 2.36 \times 10^{29} \frac{\text{kg} cdot \text{m}^2}{\text{s}} )
( 5.91 \times 10^{29} \frac{\text{kg} cdot \text{m}^2}{\text{s}} )
( 8.35 \times 10^{40} \frac{\text{kg} cdot \text{m}^2}{\text{s}} )
( 3.34 \times 10^{40} \frac{\text{kg} cdot \text{m}^2}{\text{s}} )

Explanation:

Step1: Convert radius to meters

$r = 1738\ \text{km} = 1738 \times 10^3\ \text{m} = 1.738 \times 10^6\ \text{m}$

Step2: Convert angular velocity to rad/s

First, $1\ \text{revolution} = 2\pi\ \text{rad}$, $27.3\ \text{days} = 27.3 \times 24 \times 3600\ \text{s} = 2.35872 \times 10^6\ \text{s}$
$\omega = \frac{2\pi}{2.35872 \times 10^6} \approx 2.6617 \times 10^{-6}\ \text{rad/s}$

Step3: Calculate moment of inertia

For a solid sphere, $I = \frac{2}{5}mr^2$
$I = \frac{2}{5} \times 7.35 \times 10^{22} \times (1.738 \times 10^6)^2$
$I = \frac{2}{5} \times 7.35 \times 10^{22} \times 3.020644 \times 10^{12} \approx 8.87 \times 10^{34}\ \text{kg·m}^2$

Step4: Calculate angular momentum

$L = I\omega$
$L = 8.87 \times 10^{34} \times 2.6617 \times 10^{-6} \approx 2.36 \times 10^{29}\ \frac{\text{kg·m}^2}{\text{s}}$

Answer:

$2.36 \times 10^{29} \frac{\text{kg·m}^2}{\text{s}}$