Question

How much faster than its normal rate should the earth rotate about its axis so that the weight of the body at the equator becomes zero ? (Radius of the earth$=6.4\times {10}^{6}m$ and $g=9.8m/s^2$)

• A. Nearly 17 times
• B. Nearly 12 times
• C. Nearly 10 times
• D. Nearly 14 times

Answer 1

The correct option is A Nearly 17 times

The angular speed($\omega$) of Earth is approximately $7.3\times {10}^{-5}{s}^{-1}$.
$\omega =7.3\times {10}^{-5}{s}^{-1}$.
For the weight to be zero at the equator, value of gravity should be zero. i.e
$\Rightarrow g^{\prime }=g-R{\omega }^{\prime 2}=0$
$\Rightarrow {\omega }^{\prime }=\sqrt{\frac{g}{R}}=123.7\times {10}^{-5}{s}^{-1}$
$\Rightarrow \frac{{\omega }^{\prime }}{\omega }=\left(123.7/7.3\right)=16.95$