Question

How much (in times) faster than its normal rate should the earth be rotating about its axis so that the weight of the body at the equator becomes zero $?$ $(R=6.4\times {10}^6\text{m},g=10$${\text{m/s}}^2)$

• A. $\text{nearly}17$
• B. $\text{nearly}12$
• C. $\text{nearly}10$
• D. $\text{nearly}14$

Answer 1

The correct option is A $\text{nearly}17$

If $m$ is the mass of the body, the weight at the equator is given by $$m{g}^{\prime }=m(g-{\omega }^{2}R)...\left(1\right)$$ Let the weight of the body becomes zero, when the angular speed becomes $n$ times of the initial angular speed.i.e, ${\omega }^{\prime }=n\omega$

From equation $\left(1\right)$,

$\Rightarrow 0=m(g-n^2{\omega }^{2}R)$

$\Rightarrow n=\frac{1}{\omega }\sqrt{\frac{g}{R}}$

Substituting, $g=10{\text{m/s}}^2$

$R=6.4\times {10}^6\text{m}$

$\omega =\frac{2\pi }{T}=\frac{2\pi }{24\text{hrs}}=\frac{2\pi }{24\times 60\times 60}$

$\Rightarrow n=\frac{86400}{2\pi }\sqrt{\frac{10}{6.4\times {10}^6}}$

$\Rightarrow n=17.19$

So, $n$ nearly equal to $17$. Hence, option (a) is the correct answer.