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 = 1⋅5 × 10^{8} km.

#### Solution

Given

r = 6400 km = 6.4 × 10^{6} m;

R = 1⋅5 × 10^{8} km = 1⋅5 × 10^{11} m

About its axis, we have

T = 1 day = 86400 s;

\[\omega = \frac{2\pi}{T}\]

Angular momentum of the Earth about its axis,

\[L = I\omega\]

\[= \frac{2}{5}m r^2 \times \left( \frac{2\pi}{86400} \right)\]

About the Sun's axis,

*T* = 365 day = 365 × 86400 s

Angular momentum of the Earth about the Sun's axis,

\[L' = m R^2 \times \left( \frac{2\pi}{86400 \times 365} \right)\]

Ratio of angular momentums,

\[\frac{L}{L'} = \frac{2/5m r^2 \times \left( 2\pi/86400 \right)}{m R^2 \times 2\pi/\left( 86400 \times 365 \right)}\]

\[ = \frac{\left( 2 r^2 \times 365 \right)}{5 R^2} = \left( \frac{2 \times \left( 6 . 4 \times {10}^6 \right)^2 \times 365}{5 \times \left( 1 . 5 \times {10}^{11} \right)^2} \right)\]

\[ = 2 . 65 \times {10}^{- 7}\]