If g is acceleration due to gravity at the equator when earth were at rest and `g_(1)` is acceleration due to gravity at the same place when earth spins with angular velocity `omega,` the relation between them is

Options

A. `g_(1)=g(1-(R omega^(2))/(g))`
B. `g_(1)=g(1-R omega^(2))`
C. `g_(1)=g-R^(2)omega`
D. `g=g_(1)-R^(2)omega`

Answer

a

Explanation

`g_(1)=g-Romega^(2)cos^(2)phi`
`=g-Romega^(2)cos^(2)(0)=g-Romega^(2)xx1`
`g_(1)=g-Romega^(2)`
`=g[1-(Romega^(2))/(g)].`