Find the imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero. (take `g = 10 m//s^(2)` for the acceleration due to gravity, if the earth were at rest and radius of earth equal to 6400 km and `phi = 60^(@)`)

Explanation

Acceleration due to gravity,
`g'=g-omega^(2)R cos^(2)phi`
`0=g-omega^(2)Rcos^(2)60" "(because g'=0)`
`0=g-(omega^(2)R)/(4)`
`omega=2 sqrt((g)/(2))=2 sqrt((10)/(6400xx1000))`
`rArr omega=(1)/(400)=2.5 xx10^(-3) "rad/s"`.