Question

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

• A. $2.5\times {10}^{-3}rad/s$
• B. $1.5\times {10}^{-3}rad/s$
• C. $4.5\times {10}^{-3}rad/s$
• D. $0.5\times {10}^{-3}rad/s$

Answer 1

The correct option is A $2.5\times {10}^{-3}rad/s$

The variation of acceleration due to gravity with Earth's latitude is given by:
$g^{{}^{\prime }}=g-R{\omega }^{2}co{s}^2\lambda$

Now if $g^{{}^{\prime }}=0$, then

$g=R({\omega }^{{}^{\prime }}{)}^{2}co{s}^2\lambda =R({\omega }^{{}^{\prime }}{)}^{2}co{s}^{2}60=\left(R/4\right)({\omega }^{{}^{\prime }}{)}^2$

$\Rightarrow {\omega }^{{}^{\prime }}=\sqrt{\frac{4g}{R}}=2.47\times {10}^{-3}rad/s$