The value of acceleration due to gravity $g$ is maximum at
The value of acceleration due to gravity $g$ is maximum at
poles
centre
equator
surface of earth
Solution
The acceleration due to gravity on the surface of earth is given as
$
g^{\prime}=g-\omega^2 R_e \cos ^2 \lambda
$
where, $\lambda$ is the angle of lattitude.
At poles, $\lambda=90^{\circ}$
$\therefore \quad \cos 90^{\circ}=0$
From Eq. (i),
$
g^{\prime}=g
$
Hence, at poles, the acceleration due to gravity is maximum