Question

The angular velocity of the earth with which it has to rotate so that the acceleration due to gravity on ${60}^0$ latitude zero is:

• A. $\omega =2.48\times {10}^{-3}rad/s$
• B. $\omega =3.58\times {10}^{-3}rad/s$
• C. $\omega =6.46\times {10}^{-6}rad/s$
• D. $\omega =4.25\times {10}^{-5}rad/s$

Answer 1

The correct option is A $\omega =2.48\times {10}^{-3}rad/s$

we know that the variation of the acceleration due to gravity with rotation and latitude of earth is given by

$g^{\prime }=g-R{\omega }^{2}co{s}^2\varphi$

where

R = radius of the earth = 6400000 m

$\varphi$ = ange of latitude = ${60}^{\circ }$ (given)

then,

$0=9.8-6400000\times {\omega }^{2}co{s}^{2}60$

$9.8=6400000\times {\omega }^{2}co{s}^{2}60$

$\omega =2.48\times {10}^{-3}rad/s$