Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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. - Physics

Sum

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.

#### Solution

Given

r = 6400 km = 6.4 × 106 m;

R = 1⋅5 × 108 km = 1⋅5 × 1011 m

T = 1 day = 86400 s;

$\omega = \frac{2\pi}{T}$

Angular momentum of the Earth about its axis,

$L = I\omega$

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

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,

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

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

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

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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 10 Rotational Mechanics
Q 48 | Page 198