Question

At what angular speed the Earth should rotate, so that effective acceleration due to gravity at ${45}^{\circ }$ latitude becomes half of effective acceleration due to gravity at pole? (Radius of earth is $R$ and acceleration due to gravity without taking variation of Earth’s rotation at surface is $‘g^{\prime }$)

• A. $\omega =\sqrt{\frac{g}{R}}$
• B. $\omega =\sqrt{\frac{g}{2R}}$
• C. $\omega =\sqrt{\frac{2g}{R}}$
• D. $\omega =\sqrt{\frac{3g}{4R}}$

Answer 1

The correct option is A $\omega =\sqrt{\frac{g}{R}}$


$g^{\prime }=g-R{w}^2{\cos }^2\varphi$

given $g_{{45}^{\circ }}=\frac{1}{2}\times g_{{90}^{\circ }}$


$g_{{45}^{\circ }}=g-R{\omega }^2\times \frac{1}{2}$

$g_{{90}^{\circ }}=g$

$⟹g-\frac{{\omega }^{2}R}{2}=\frac{1}{2}\times g$

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

Hence $\omega =\sqrt{\frac{g}{R}}$