Kinetic Energy Required for Object to Escape Earth Orbit

To find the kinetic energy of the object at the time of its ejection, we analyze its total mechanical energy required to just escape Earth's gravitational pull.

1. Orbital Speed of the Satellite:
   Let $M$ be the mass of the Earth and $r$ be the radius of the circular orbit of the satellite. The gravitational force on the satellite provides the required centripetal force for its circular motion:
   $\frac{G M m_{\text{sat}}}{r^2} = \frac{m_{\text{sat}} V^2}{r}$
   Solving for $V^2$, we get:
   $V^2 = \frac{G M}{r} \implies \frac{G M}{r} = V^2$

2. Gravitational Potential Energy of the Object:
   At the moment of ejection, the object of mass $m$ is at a distance $r$ from the center of the Earth. Its gravitational potential energy $U$ at this location is:
   $U = -\frac{G M m}{r}$

3. Condition for Just Escaping:
   For the object to just escape from the gravitational pull of the Earth, its total mechanical energy $E$ immediately after ejection must be zero ($E = 0$), so that it reaches infinity with zero residual kinetic energy.
   $E = K + U = 0$
   where $K$ is the kinetic energy of the object at the time of ejection.

4. Calculating the Kinetic Energy ($K$):
   $K - \frac{G M m}{r} = 0$
   $K = \frac{G M m}{r} = m \left(\frac{G M}{r}\right)$

   Substituting $\frac{G M}{r} = V^2$ into the expression above:
   $K = m V^2$

Thus, the kinetic energy of the object at the time of its ejection is $m V^2$.

Correct Option: (B)