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 = 1⋅5 × 108 km.

Answer 1

The ratio of angular momentum is 2.65 × 10`-7

Radius of the earth = 6400 km = 6.4 × 10`6m  (Given)

Radius of the sun = 1.5 10`8km = 1.5  × 10`11m  (Given)

About the axis it will be -  

Time = 1 day = 86400 seconds

w = 2π/T

Angular momentum of earth about its axis = L = Iw

= 2/5 mr² ( 2π/86400)

About the sun's it will be -  

Time = 365 days = 365  × 86400 seconds

Angular momentum of earth about the sun's axis = L = mR²  

= mR² ( 2π / 86400 . 365)

Ratio of angular momentum -  

= 2r² × 365 / 5R²  

= 2 × (6.4 × 10`6)² . 365/ 5 ( 1.5 × 10`11)²

= 2.65 × 10`-7

Thus, the ratio of angular momentum is 2.65 × 10`-7

Answer 2

The angular momentum of the earth is 2.65 x 10-7.

Explanation:

It is given that

Radius of the Earth, r = 6400 km

Radius of the Sun, $R = 1.5 \times 10^{8} km$

Time = 1 Day = 86400 seconds

The angular momentum of the earth about its axis $L=I \omega$

Where $\omega=\frac{2 \pi}{T}$  and $I = \frac{2}{5} mr^2$

$L=\frac{2}{5} mr^{2} \times \frac{2\pi }{85400}$  

About sun’s axis, the angular moment is  $= mr^{2} \times \frac{2\pi }{85400 \times 365}$

This is because$I = mr^2$

Hence, angular momentum’s ratio =  $\frac{\frac{2}{5} mr^{2} \times \frac{2\pi }{85400}}{ mr^{2} \times \frac{2\pi }{85400 \times 365}}$                       =$\frac{ (2r^{2} \times 365)}{5R^2}$

= $\frac{2.990 \times 10^{10}}{1.125 \times 10^{17}}$

Angular momentum's ratio =  $2.65 \times 10^{-7}$

Therefore the angular momentum's ratio of the earth about its axis due to its spinning motion to that about the sun due to its orbital axis  $2.65 \times 10^{-7}$