Question

If Earth is supposed to be a sphere of radius $R$, if $g_{{30}^o}$ is value of acceleration due to gravity at latitude of ${30}^{\circ }$ and $g$ at the equator, the value of $g-g_{{30}^o}$ is

• A. $\frac{1}{4}{\omega }^{2}R$
• B. $\frac{3}{4}{\omega }^{2}R$
• C. ${\omega }^{2}R$
• D. $\frac{1}{2}{\omega }^{2}R$

Answer 1

Answer: (B)

$\frac{3}{4}{\omega }^{2}R$

Acceleration due to gravity at latitude '$\lambda$' is given by
$g_{\lambda }=g-R{\omega }^{2}co{s}^2\lambda$
At ${30}^{\circ },g_{30}=g-R{\omega }^{2}co{s}^{2}30=g-\frac{3}{4}R{\omega }^2$
or, $g-g_{30}=\frac{3}{4}R{\omega }^2$