Physics - Gravitation Question with Solution | TestHub

As,$g^{\prime }=g-{\omega }^2 R{\cos }^2\lambda$

Now, for poles$\lambda =90°$and for equation$\lambda =0°$

(i) For$\lambda ={90}^{ }$

$g_{pole}^{\prime }=g-{\omega }^2 R{\cos }^{2}90°=g$

$g^{\prime }=g$

There is no effect on$g$at the poles.

(ii) For$\lambda =0^{ }$

$g_{\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}}=g-{\omega }^2 R{\cos }^{2}0°$

$= g-{\omega }^2 R$

The effect of the rotation of Earth on the value of$g$at the equator is maximum.