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. `(GMm)/(11R)`
B. `(GMm)/(10R)`
C. `(mgR)/(11G)`
D. `(10GMm)/(11R)`

Answer

d

Explanation

`g_(d)=g-Romega^(2)cos^(2)phi,0=g-R omega^(2)cos^(2)phi,`
`Romega^(2)cos^(2)phi=g`
`omega^(2)=(g)/(Rcos^(2)phi)`
`omega=sqrt((g)/(Rcos^(2)phi))`
`=sqrt((10)/(6.4xx10^(6)xx(1)/(4)))=sqrt((100)/(16xx10^(6)))`
`=(10)/(4)xx10^(-3)`
`omega=2.5xx10^(-3)"rad/s"`