Question

If ‘R’ is the radius of the earth and $\omega$ is the angular velocity with which it rotates about its axis, the variation in the value of ‘g’ at ${45^ \circ }$ latitude of the earth stops its rotation will be
A) Increases by $\dfrac{{R{\omega ^2}}}{2}$
B) Decreases by $\dfrac{{R{\omega ^2}}}{2}$
C) Increases by $\dfrac{{R{\omega ^2}}}{4}$
D) Decreases by $\dfrac{{R{\omega ^2}}}{4}$

Answer 1

The earth latitude of a point is the angle $\phi$ between the equatorial plane and line joining that point to the center of the earth. Due to the rotational motion of the earth about its axis, the body experiences a centrifugal force.

Formula Used: We will be using the formula of the variation of acceleration due to gravity with a Latitudinal position on earth is $g' = g - {\omega ^2}R{\cos ^2}\phi$.

Complete step by step solution:
At equator, the Latitude $\phi = 0$, the acceleration due to gravity is minimum. At poles $\phi = {90^ \circ }$ and so the acceleration due to gravity is maximum.

A body of mass m at a point P with latitude $\phi$ and g’ be the acceleration due to gravity at point P. The expression for acceleration due to gravity at a point P on the surface of the earth having latitude $\phi$ is defined as the difference between the two forces is equal to the weight of that body at that point.
$mg' = mg - mr{\omega ^2}\cos \phi$
Therefore, $r = R\cos \phi$
$g' = g - {\omega ^2}R{\cos ^2}\phi$
Given: The variation in the value of ‘g’ at ${45^ \circ }$ latitude of the earth stops. To find whether the rotation is increasing or decreases.
We already know the relation of the acceleration due to gravity due to earth’s rotation
$g' = g - {\omega ^2}R{\cos ^2}\phi$
According the given latitude ${45^ \circ }$,
$\phi = {45^ \circ }$ Acceleration due to gravity at ${45^ \circ }$ latitude
$g' = g - {\omega ^2}R{\cos ^2}{45^ \circ }$
$g' = g - \dfrac{{{\omega ^2}R}}{2}$
When earth stops rotating, that is $\omega = 0$, then the acceleration due to gravity at ${45^ \circ }$ latitude will increases by $\dfrac{{R{\omega ^2}}}{2}$

Option (A), Increases by $\dfrac{{R{\omega ^2}}}{2}$ is the answer.

Note: The acceleration due to gravity is defined as the acceleration gained by an object due to gravitational force. The standard value of g on the surface of the earth is $9.8m{s^{ - 2}}$. The acceleration due to gravity is constant but it varies from place to place.