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. `7.8xx120^(-2)rad s^(-1)`
B. `0.5xx10^(-3)rad s^(-1)`
C. `1xx10^(-3) rad s^(-1)`
D. `2.5xx10^(-3) rad s^(-1)`

Answer

D

Explanation

New acceleartin due to gravity g is given by
`g' =g-R_(e )omega^(2) cos^(2) lambda`
`0=10 -6.4 xx10^(6) omega^(2)cos^(2) 60^(@)`
`rarr omega^(2)=(10)/(6.4xx10^(6)[0.5]^(2))=2.5xx10^(-3) rad s^(-1)`