Question

The speed of earth's rotation about its axis is $\omega$. Its speed is increased to $x$ times to make the effective acceleration due to gravity equal to zero at the equator, then the value of $x$ is $(g=10{\text{ms}}^{-2};R=6400\text{km})$.

• A. $1$
• B. $8.5$
• C. $17$
• D. $34$

Answer 1

The correct option is C $17$

The acceleration due to gravity at the Equator is given by $$g^{\prime }=(g-{\omega }^{2}R),$$ where $\omega$ is the angular velocity of the Earth and $R$ the radius of the Earth.

Since, the acceleration due to gravity of the body becomes zero. i.e. $g^{\prime }=0$
and ${\omega }^{\prime }=x\omega$

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

$\Rightarrow g=(x\omega {)}^{2}R$

$\Rightarrow x=\frac{1}{\omega }\times \left(\sqrt{\frac{g}{R}}\right)$

Given,
$g=10{\text{m/s}}^2;R=64\times {10}^5\text{m}$ and

Earth rotates about its own axis in $24\text{hours}$ So, the time period, $T=24\times 60\times 60\text{sec}$. Thus, angular speed of the earth about its own axis
$\omega =\frac{2\pi }{T}=\frac{2\pi }{24\times 60\times 60}\text{rad/sec}$

$\Rightarrow x=\frac{24\times 60\times 60}{2\pi }\times \sqrt{\left(\frac{10}{64\times {10}^5}\right)}$

$\therefore x\approx 17$

Hence, option (c) is the correct answer.

The acceleration due to gravity at the Equator is given by $$g^{\prime }=(g-{\omega }^{2}R),$$ where $\omega$ is the angular velocity of the Earth and $R$ the radius of the Earth. Why this question? To familiarize students with the change in acceleration due to gravity due to the rotation of the earth.