Question

Calculate the ratio of the angular momentum of the earth about its axis due to its spinning motion to that about the sun due to its orbital motion. Radius of the earth=$6400$km and radius of the orbit of the earth about the sun is equal to $1.5\times {10}^8$km.

Answer 1

Angular momentum=$I\omega$

$=I\left(\frac{2\pi }{T}\right)$

Angular Momentum of earth about its axis=$\frac{2}{5}m{r}^2\times \frac{2\pi }{T}$

$T=1$day i.e. $24\times 60\times 60=86400sec$

Angular Momentum of earth about the sun=$m{R}^2\times \frac{2\pi }{365T}$

Ratio of the two angular momentus is:

$=\frac{\frac{2}{5}m{r}^2\frac{2\pi }{T}}{m{R}^2\frac{2\pi }{365T}}=\frac{2\times 365r^2}{5R^2}=\frac{146r^2}{R^2}=\frac{146\times {\left(6400\right)}^2}{{(1.5\times {10}^8)}^2}=2.65785\times {10}^{-7}$