The angle made by the line joining a given place on the earth's surface to the centre of the earth with the equatorial line is called the latitude $(\lambda)$ of that place.

$\therefore$ For the equator latitude $\lambda=0^{0}$ and for the poles latitude $\lambda=90^{\circ}$

As shown in figure the latitude of the place $P$ on the earth's surface is $\lambda=\angle \mathrm{POE}$. At this position consider a particle of mass $m$. Two forces acting on it.

$(1)$ Earth's gravitational force $m g$ is in $\overrightarrow{\mathrm{PQ}}$ direction.

$\ldots$ $(1)$

$(2)$ Earth has an acceleration due to rotational motion. So, this particle is in the accelerated frame of reference.

At this point the acceleration of the frame of reference is $\left(\frac{v^{2}}{r}\right)$ in $\overrightarrow{P M}$ direction. Hence, friction acceleration of particle is $\left(\frac{v^{2}}{r}\right) .$ It is in $\overrightarrow{P Q}$ direction. Hence, frictional centripetal force

$=\frac{m v^{2}}{r}$ $=m r \omega^{2}$

$=m r \omega^{2} \quad[\because v=r \omega]$

This force is in $PQ$ direction.

The component frictional centripetal force in the direction of $\overrightarrow{P R}=m r \omega^{2} \cos \lambda$

$\therefore$ From equation $(1)$ and $(2)$ effective force on $P, F$ $=m g-m r \omega^{2} \cos \lambda$

If $g^{\prime}$ is the effective gravitational acceleration of this particle then,

$m g^{\prime}=m g-m r \omega^{2} \cos \lambda$

$\therefore g^{\prime}=g-r \omega^{2} \cos \lambda$