Question 354: Gravitation

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 surfacde]

Accepted Answer

To determine the rotational kinetic energy at which a body becomes weightless at the equator, we need to consider the balance between the centrifugal force and the gravitational force acting on the body.
LetÃ¢ÂÂs start by analyzing the forces acting on the body at the equator. For a body to become weightless, the centrifugal force must equal the gravitational force:
Centrifugal Force:
$$F_{	ext{centrifugal}} = M \omega^2 R$$
Gravitational Force:
$$F_{	ext{gravitational}} = Mg$$
For weightlessness:
$$M \omega^2 R = Mg$$
Solving for $$\omega^2$$:
$$\omega^2 R = g \implies \omega^2 = \frac{g}{R}$$
Next, the rotational kinetic energy of the Earth (considering it a solid sphere) is:
$$K = \frac{1}{2} I \omega^2$$
where $$I$$ is the moment of inertia of the Earth. For a solid sphere, the moment of inertia is given by:
$$I = \frac{2}{5} MR^2$$
Substituting the value of $$I$$ and $$\omega^2$$:
$$K = \frac{1}{2} \left( \frac{2}{5} MR^2 ight) \left( \frac{g}{R} ight)$$
Simplifying:
$$K = \frac{1}{2} \cdot \frac{2}{5} MR^2 \cdot \frac{g}{R}$$
$$K = \frac{1}{2} \cdot \frac{2}{5} \cdot M \cdot R \cdot g$$
$$K = \frac{1}{5} MgR$$
Thus, the value of the critical rotational kinetic energy $$K$$ is:
$$\frac{1}{5} \mathrm{MgR}$$
The correct option is:
Option D:
$$\frac{1}{5} \mathrm{MgR}$$

Answer

$$
\frac{1}{2} \mathrm{MgR}
$$

Answer

$$
\frac{1}{3} \mathrm{MgR}
$$

Answer

$$
\frac{1}{4} \mathrm{MgR}
$$

Answer

$$
\frac{1}{5} \mathrm{MgR}
$$

Gravitation

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