Obtain an expression for the variation in effective gravitational acceleration $g'$ with latitude due to the Earth's rotation.

(N/A) The angle made by the line joining a given place on the Earth's surface to the center of the Earth with the equatorial plane is called the latitude $(\lambda)$ of that place.
For the equator,latitude $\lambda = 0^{\circ}$ and for the poles,latitude $\lambda = 90^{\circ}$.
As shown in the figure,the latitude of the place $P$ on the Earth's surface is $\lambda = \angle POE$. Consider a particle of mass $m$ at this position. Two forces act on it:
$(1)$ The Earth's gravitational force $mg$ acting towards the center $O$ along the line $PO$.
$(2)$ Due to the Earth's rotation,the particle experiences a centrifugal force $m r \omega^2$ acting outwards along the radius of the circular path $PM$ (where $r$ is the radius of the circle of latitude).
From the geometry of the figure,$r = R_e \cos \lambda$,where $R_e$ is the radius of the Earth.
The component of the centrifugal force $m r \omega^2$ acting along the line $PO$ (towards the center) is $m r \omega^2 \cos \lambda$.
Substituting $r = R_e \cos \lambda$,the component is $m (R_e \cos \lambda) \omega^2 \cos \lambda = m R_e \omega^2 \cos^2 \lambda$.
The effective force $F$ on the particle towards the center is $F = mg - m R_e \omega^2 \cos^2 \lambda$.
If $g'$ is the effective gravitational acceleration,then $mg' = F = mg - m R_e \omega^2 \cos^2 \lambda$.
Therefore,the expression for effective gravitational acceleration is $g' = g - R_e \omega^2 \cos^2 \lambda$.