Question

If earth were to rotate on its own axis such that the weight of a person at the equator becomes half the weight at the poles, then its time period of rotation is: ($g$=acceleration due to gravity near the poles and $R$ is the radius of earth) (Ignore equatorial bulge)

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

Answer 1

Answer: (D)

$2\pi \sqrt{\frac{2R}{g}}$

Let weight of the person be $mg$ and the angular speed of the earth be $\omega$.

Thus weight of the person at poles=$mg$

Weight of the person at equator=$mg-m{\omega }^{2}R$

$⟹mg=2(mg-m{\omega }^{2}R)$

$⟹{\omega }^{2}R=\frac{g}{2}$

$⟹\omega =\sqrt{\frac{g}{2R}}$

Thus time period of earth's rotation=$\frac{2\pi }{\omega }$

$=2\pi \sqrt{\frac{2R}{g}}$