Question

A planet with a mass of 5.86x10²⁴kg and a radius of 6.87x10⁶m orbits its sun, a distance of 144.87x10⁹m away, in a uniform circular motion with velocity 3.37x10⁴m/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 spin angular momentum of this planet (in kg⋅m²/s)?

Answer 1

$$L=\frac{0.8\pi MR^2}{T}\\L=\frac{0.8\pi (5.86\cdot10^{24})(6.87\cdot10^{6})^2}{24(3600)}=8.05\cdot10^{33}\frac{kgm^2}{s}$$