Question

What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes $\dfrac{3}{5}$ of the present weight at the equator. Equatorial radius of earth is 6400km.
$\begin{aligned} & \text{A}\text{. 8}\text{.7}\times \text{1}{{\text{0}}^{-7}}rad/s \\\ & \text{B}\text{. 7}\text{.8}\times \text{1}{{\text{0}}^{-4}}rad/s \\\ & \text{C}\text{. 6}\text{.7}\times \text{1}{{\text{0}}^{-4}}rad/s \\\ & \text{D}\text{. 7}\text{.4}\times \text{1}{{\text{0}}^{-3}}rad/s \\\ \end{aligned}$

Answer 1

To solve this question we will use the concept of apparent weight. The weight we feel in the equator of earth’s surface can be expressed in terms of the original weight and the centripetal force experienced by the body due to rotation of earth. Obtain the mathematical expression and put the given values to find the required answer.

Complete step-by-step answer:
Earth is rotating about its own axis. So, any object in the equator of the earth will experience a centripetal force. This centripetal force is due to the gravitational force of earth.
Let the mass of the person is m.
So, the weight of the person at the equator will be,
$W=mg$
Where, g is the acceleration due to gravity.
Now, the observed weight at the equator is,
${W}'=m{g}'=\dfrac{3}{5}mg$
Now, when the earth rotates centripetal force is acting at the equator. So, the apparent weight of the person at the equator can be given by the equation,

& {W}'=W-{{F}_{c}} \\\ & \Rightarrow m{g}'=mg-\dfrac{m{{v}^{2}}}{R} \\\ & \Rightarrow \dfrac{3}{5}mg=mg-\dfrac{m{{v}^{2}}}{R} \\\ & \Rightarrow \dfrac{3}{5}g=g-\dfrac{{{v}^{2}}}{R} \\\ & \Rightarrow \dfrac{{{v}^{2}}}{R}=g-\dfrac{3}{5}g \\\ & \Rightarrow \dfrac{{{v}^{2}}}{R}=\dfrac{2}{5}g \\\ \end{aligned}$$ Now, the linear velocity v can be expressed in terms of the angular velocity as, $v=\omega R$ Where, $\omega $ angular velocity of the rotation of earth and R is the radius of earth at the equator. Putting this in the above equation, $\begin{aligned} & \Rightarrow \dfrac{{{\omega }^{2}}{{R}^{2}}}{R}=\dfrac{2}{5}g \\\ & \Rightarrow {{\omega }^{2}}=\dfrac{2g}{5R} \\\ & \Rightarrow \omega ={{\left( \dfrac{2g}{5R} \right)}^{\dfrac{1}{2}}} \\\ \end{aligned}$ Acceleration due to gravity have a value of $g=9.8m{{s}^{2}}$ at the surface of earth and the radius of earth is $R=6400km=6.4\times {{10}^{6}}m$ Putting these values, $$\omega ={{\left( \dfrac{2\times 9.8}{5\times 6.4\times {{10}^{6}}} \right)}^{\dfrac{1}{2}}}=7.8\times {{10}^{-4}}rad/s$$ So, the velocity of rotation of earth due to rotation of earth about its own axis will be $$7.8\times {{10}^{-4}}rad/s$$. **So, the correct answer is “Option B”.** **Note:** Centripetal force is a force which acts on the body moving in a circular path and the force is directed toward the centre of rotation. For an object at the equator of earth, the centripetal force is given by the gravitational force of earth. Since this centripetal force depends on the velocity of the rotation, the apparent weight of objects will change depending on the velocity.