Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The effect of rotation of the earth on the value of acceleration due to gravity is

UPSEEUPSEE 2009Gravitation

Solution:

As $g^{'}=g-\omega^{2} R \cos ^{2} \lambda$
The latitude at a point on the surface of the earth is defined as the angle, which the line joining that point to the centre of earth makes with equitorial plane. It is denoted by $\lambda$. For the poles $\lambda=90^{\circ}$ and for equator $\lambda=0^{\circ}$
(i) Substituting $\lambda=90^{\circ}$ in the above expression, we get
$g_{\text {pole }}=g-\omega^{2} \,R \,\cos ^{2} \,90^{\circ}$
$\therefore g_{\text {pole }}=g$
ie, there is no effect of rotational motion of the earth on the value of $g$ at the poles.
(ii) Substituting $\lambda=0^{\circ}$ in the above expression, we get
$g_{\text {equator }}=g-\omega^{2}\, R \,\cos ^{2} \,0^{\circ}$
$\therefore \,\,g_{\text {equator }}=g-\omega^{2}\, R$
ie, the effect of rotation of earth on the value of $g$ at the equator is maximum.