Question

A body of mass 'M' and radius 'R', situated on the surface of the earth becomes weightless at its equator when the rotational kinetic energy of the earth reaches a critical value 'K'. The value of 'K' is given by [Assume the earth as a solid sphere, g = gravitational acceleration on the earth's surface]

• A. $\frac{1}{2} MgR$
• B. $\frac{1}{3} MgR$
• C. $\frac{1}{4} MgR$
• D. $\frac{1}{5} MgR$

Hint:

To minimize steps, link translational and rotational energy definitions. Since $K = \frac{1}{2}I\omega^2$, replacing $I$ with $\frac{2}{5}MR^2$ collapses the expression into $K = \frac{1}{5}M(v_{\text{eq}})^2$. Since weightlessness implies the orbital condition $v^2 = gR$, substitute $gR$ directly to arrive at $\frac{1}{5}MgR$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the critical rotational kinetic energy $K$ of the Earth required to make an object placed at the equator completely weightless. Weightlessness means that the apparent acceleration due to gravity at the equator becomes exactly zero.

Step 2: Key Formula or Approach:
The apparent acceleration due to gravity $g'$ at the equator due to Earth's rotation is:
$$g' = g - \omega^2 R$$ For a body to experience weightlessness, $g' = 0$, meaning the outward centrifugal force balances the inward gravitational pull:
$$\omega^2 R = g \implies \omega^2 = \frac{g}{R}$$ The rotational kinetic energy of a spinning rigid body is:
$$K = \frac{1}{2} I \omega^2$$ For a uniform solid sphere like the Earth, the central moment of inertia is $I = \frac{2}{5}MR^2$.

Step 3: Detailed Explanation:
Substitute the moment of inertia for a solid sphere into the rotational kinetic energy formula:
$$K = \frac{1}{2} \left(\frac{2}{5} MR^2\right) \omega^2 = \frac{1}{5} MR^2 \omega^2$$ Now, substitute the critical weightless condition $\omega^2 = \frac{g}{R}$ into this equation:
$$K = \frac{1}{5} MR^2 \left(\frac{g}{R}\right)$$ One factor of $R$ cancels out perfectly from the numerator and denominator:
$$K = \frac{1}{5} MgR$$

Step 4: Final Answer:
The value of the critical kinetic energy is $\frac{1}{5} MgR$, which matches option (D).