Question

A planet with a mass of 5.51x10²⁴kg and a radius of 6.36x10⁶m orbits its sun, a distance of 139.4x10⁹m away, in a uniform circular motion with velocity 3.42x10⁴m/s. Assuming the planet is a uniform solid sphere spinning on its axis and has the same length of day as Earth, what is the orbital angular momentum of this planet (in kg⋅m²/s)?

Answer 1

The orbital angular momentum is

$$L=I\omega=\bigg(\frac25mr^2+ma^2\bigg)\frac va,\\\space\\ L=2.63·10^{40}\text{ kg}·\text{m}^2\text{/s}$$