Q. The mass of the earth is increasing at the rate of 1 part in $5 \times 10^{19}$ per day by the attraction of meteors falling normally on the earth’s surface. Assuming that the density of earth is uniform, the rate of change o the period of rotation of the earth is
A) $2.0 \times 10^{- 20}$
B) $2.66 \times 10^{- 19}$
C) $4.33 \times 10^{- 18}$
D) $5.66 \times 10^{- 17}$
Explanation
(a) As angular momentum is conserved in the absence of a torque, therefore
$I_{0}\omega_{0} = I\omega$
$\left( \frac{2}{3}MR^{2} \right)\left( \frac{2\pi}{T_{0}} \right) = \left\lbrack \frac{2}{5}MR^{2} + \frac{2}{5}\frac{MR^{2}}{5 \times 10^{19}}\ \right\rbrack\frac{2\pi}{T}$
$\frac{T}{T_{0}} = 1 + \frac{1}{5 \times 10^{19}}$
$\frac{T}{T_{0}} - 1 = \frac{1}{5 \times 10^{19}} = 2 \times 10^{- 20}$
$\frac{T - T_{0}}{T_{0}} = 2 \times 10^{- 20}\ or\ \frac{\mathrm{\Delta}T}{T_{0}} = 2 \times 10^{- 20}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.