If the earth were to stop rotating, the apparent value of g on its...

The apparent acceleration due to gravity at a latitude $\lambda$ on the surface of the earth is given by:

$g' = g - R\omega^2 \cos^2 \lambda$

where $R$ is the radius of the earth and $\omega$ is its angular velocity.

If the earth stops rotating, $\omega = 0$, and the apparent acceleration due to gravity becomes $g' = g$.

At the poles ($\lambda = 90^{{}^{{}^{\circ}}}$), $\cos \lambda = 0$, so $g' = g$ both before and after the earth stops rotating. The value of $g$ remains unchanged at the poles.

At the equator ($\lambda = 0^{{}^{{}^{\circ}}}$), $\cos \lambda = 1$, so $g'$ changes from $g - R\omega^2$ to $g$. The value of $g$ increases.

At all other latitudes ($0^{{}^{{}^{\circ}}} < \lambda < 90^{{}^{{}^{\circ}}}$), $g'$ increases as the subtractive term $R\omega^2 \cos^2 \lambda$ becomes zero.

Thus, the apparent value of $g$ increases at some locations and remains unchanged at others.