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:

Similar Questions

The depth $d$ below the surface of the earth where the value of acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface of the earth is ($R$ = radius of the earth).

If the radius of the Earth is $R$,then the height $h$ at which the value of $g$ becomes one-fourth is:

$A$ body weighs $300 \ N$ on the surface of the earth. How much will it weigh at a distance $\frac{R}{2}$ below the surface of the earth (in $N$)? ($R$ is the radius of the earth.)

$Assertion$ : In a free fall,weight of a body becomes effectively zero.
$Reason$ : Acceleration due to gravity acting on a body having free fall is zero.

Derive the equation for the variation of acceleration due to gravity $g$ with depth $d$ below the surface of the Earth.

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