The acceleration due to gravity at pole and equator can be related as

The correct option is gₚ > gₑ.

To understand the relationship between the acceleration due to gravity at the poles ($g_p$) and at the equator ($g_e$), we must consider two main factors: the shape of the Earth and the rotation of the Earth.

1. Shape of the Earth:
The Earth is not a perfect sphere; it is an oblate spheroid. Due to its rotation, it bulges at the equator and is flattened at the poles. Consequently, the polar radius ($R_p$) is shorter than the equatorial radius ($R_e$).
$R_p<R_e$
The basic formula for acceleration due to gravity ($g$) on the surface of a planet of mass $M$ and radius $R$ is:
$g=\frac{GM}{R^2}$
This formula shows that acceleration due to gravity is inversely proportional to the square of the radius of the Earth ($g\propto \frac{1}{R^2}$). Since the distance to the Earth's center is shorter at the poles ($R_p<R_e$), the gravitational pull is stronger there:
$g_p>g_e$

2. Rotation of the Earth:
As the Earth rotates on its axis, it generates an outward centrifugal force. This force is maximum at the equator and zero at the poles. The effective acceleration due to gravity ($g\text{'}$) at a latitude $\theta$ is given by the relation:
$g\text{'}=g-{\omega }^{2}R{\cos }^2\theta$
where $\omega$ is the angular velocity of the Earth.
At the equator ($\theta =0$), we have:
$g_e=g-{\omega }^2{R}_e$
At the poles ($\theta =90°$), the centrifugal effect is zero:
$g_p=g$
Therefore, the rotation of the Earth also reduces the effective gravity at the equator relative to the poles.

Combining both factors, the acceleration due to gravity is significantly greater at the poles than at the equator:
$g_p>g_e$