The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60° latitude becomes zero is (Radius of earth = 6400 km. At the poles g = 10 ms–2) :
Answer & explanation
Correct answer: option 1
Effective acceleration due to gravity due to rotation of earth
\(\Rightarrow 0 = g - \omega^2 R cos^2 60^o\)
\(\Rightarrow \frac{\omega^2 R}{4} = g\)
\(\Rightarrow \omega = \sqrt{\frac{4g}{R}}\)
\(\Rightarrow \omega = 2\sqrt{\frac{g}{R}}\)
\(\Rightarrow \omega = \frac{2}{800} rad \text{ } sec^{-1}\)
\(g' = 0 \text{ and } \lambda = 60^o\)
\(\Rightarrow \omega = \frac{1}{400} \)
\(\Rightarrow \omega = 2.5 \text{ x } 10^{-3} rad \text{ } sec^{-1}\)