Question

Because of the friction between the water in oceans with the earth's surface the rotational kinetic energy of the earth is continuously decreasing. If the earth's angular speed decreases by 0⋅0016 rad/day in 100 years find the average torque of the friction on the earth. Radius of the earth is 6400 km and its mass is 6⋅0 × 10²⁴ kg.

Answer 1

Rate of change of angular velocity, i.e., angular acceleration,

\[α = \left( \frac{0 . 0016}{100} \right)\text{ rad/day}\]

\[\Rightarrow \alpha = \left\{ \frac{0 . 0016}{\left( 86400 \right)^2 \times 100 \times 365} \right\}  ..........\left[1 \text{ year }= 365\text{ days }= 365 \times 86400\text{ sec} \right]\]

Torque produced by the ocean water in decreasing the Earth's angular velocity,

\[\tau = I\alpha = \frac{2}{5}m r^2 \alpha\]

\[   = \frac{2}{5} \times 6 \times  {10}^{24}  \times  \left( 64 \times {10}^5 \right)^2  \times \left\{ \frac{0 . 0016}{{86400}^2 \times 100 \times 365} \right\}\]

\[   = 5 . 8 \times  {10}^{20}   N - m\]