Calculate angular velocity of the earth so that acceleration due to gravity at `60^(@)` latitude becomes zero (radius of the earth = 6400 km, gravitational acceleration at poles = `10 m//s^(2) , cos60^(@) = 0.5`)

Options

A. `7.8 xx 10^(-2)` rad/s
B. `0.5 xx 10^(-3)` rad/s
C. `1 xx 10^(-3)` rad/s
D. `2 . 5 xx 10^(-3)` rad/s

Answer

D

Explanation

New acceleration due to gravity g' is given by
`g' g - R_(e) omega^(2) cos^(2) lambda `
`0 = 10 - 6.4 xx 10^(6) omega^(2) cos^(2) 60^(@)`
`implies omega^(2) = (10)/(6.4 xx 10^(6) [0.5]^(2)) = 2.5 xx 10^(-3) rad//s`