The acceleration due to gravity at the pole ( | vedclass

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(D) The acceleration due to gravity $g$ at a point on the surface of the Earth is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitati

Vedclass · Accepted Answer

$g_p > g_e$

Vedclass · Answer

(D) The acceleration due to gravity $g$ at a point on the surface of the Earth is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is the distance from the center of the Earth.Because the Earth is not a perfect sphere but is flattened at the poles and bulges at the equator,the radius at the equator $(R_e)$ is greater than the radius at the poles $(R_p)$.Since $g \propto \frac{1}{R^2}$,a smaller radius at the poles results in a higher value of acceleration due to gravity $(g_p > g_e)$.Therefore,the correct option is $D$.

The acceleration due to gravity at the pole $(g_p)$ and at the equator $(g_e)$ are related as:

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