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. \bigcirc $2.36 \times 10^{29} \frac{\text{kg} \cdot \text{m}^2}{\text{s}}$ \bigcirc $5.91 \times 10^{29} \frac{\text{kg} \cdot \text{m}^2}{\text{s}}$ \bigcirc $8.35 \times 10^{40} \frac{\text{kg} \cdot \text{m}^2}{\text{s}}$ \bigcirc $3.34 \times 10^{40} \frac{\text{kg} \cdot \text{m}^2}{\text{s}}$

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, convert days to seconds: $27.3\ \text{days} = 27.3 \times 24 \times 3600\ \text{s} = 2.35872 \times 10^6\ \text{s}$
Angular velocity: $\omega = \frac{2\pi}{T} = \frac{2\pi}{2.35872 \times 10^6\ \text{s}} \approx 2.6617 \times 10^{-6}\ \text{rad/s}$

Step3: Calculate moment of inertia

For a solid sphere, $I = \frac{2}{5}mr^2$
$I = \frac{2}{5} \times 7.35 \times 10^{22}\ \text{kg} \times (1.738 \times 10^6\ \text{m})^2$
$I = \frac{2}{5} \times 7.35 \times 10^{22} \times 3.020644 \times 10^{12}\ \text{kg·m}^2$
$I \approx 8.93 \times 10^{34}\ \text{kg·m}^2$

Step4: Calculate angular momentum

$L = I\omega$
$L = 8.93 \times 10^{34}\ \text{kg·m}^2 \times 2.6617 \times 10^{-6}\ \text{rad/s} \approx 2.36 \times 10^{29}\ \frac{\text{kg·m}^2}{\text{s}}$

Answer:

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