Question

There is a moon orbiting an Earth-like planet. The mass of the moon is 5.49 × 1022 kg, the center-to-center separation of the planet and the moon is 2.32 × 105 km, the orbital period of the moon is 21.2 days, and the radius of the moon is 1710 km. What is the angular momentum of the moon about the planet?

Answer 1

To solve this problem, it is necessary to apply the concepts related to angular momentum and express these units in a way related to the values we have. Mathematically and initially it can be expressed as,

$L = mvr$

Here,

m = mass

v = Linear velocity

r = Radius

We know that linear speed is the product of angular speed times radius, so the expression can be converted

$L = mr^2\omega$

At the same the expression representing the angular velocity in terms of the period,

$L = mr^2 (\frac{2\pi}{T})$

$L = \frac{(5.49*10^{22})(2.32*10^8)^2(2 \pi)}{(21.2)(24)(3600)}$

$L = 1.01363*10^{34} kg \cdot m^2/s$

Therefore the angular momentum of the moon about the planet is  $1.013*10^{34} kg\cdot m^2/s$