Question

Let the value of acceleration due to gravity at poles and equator of earth $g_{p}$ and $g_{e}$ respectively. Assuming the earth to be a sphere of radius $R$ rotating about its axis with angular speed $\omega$ then $g_{p}-g_{e}$ is given by:

• A. $\frac{\omega^{2}}{R}$
• B. $R \omega^{2}$
• C. $R^{2} \omega^{2}$
• D. $\frac{\omega^{2}}{R^{2}}$

Answer 1

Answer: (B)

Effective value of acceleration due to gravity $\left(g_{\phi}\right)$ is given by $g_{\phi}=g\left(1-\frac{\omega^{2} R}{g} \cos ^{2} \phi\right)$
where $\omega$ is angular velocity.
At poles: $\phi=90^{\circ}$
$\therefore g_{\phi}=g_{p}=g_{\ldots \text { (i) }}$
At equator $\phi=0 g_{\phi}=g_{e}=g\left(1-\frac{\omega^{2} R}{g}\right)=g-\omega^{2} R \ldots$ (ii)
From Eqs. (i) and (ii), we get
$\therefore g_{p}-g_{e}=g-\left(g-\omega^{2} R\right)=\omega^{2} R$