A satellite is moving with a constant speed 'V' in a circular orbit about the earth. An object of mass 'm' is ejected from the satellite such that it just escapes form the gravitational pull of the earth. At the tme of its ejection, the kinetic energy of the object is

Options

A. `(1)/(2)mV^2`
B. `mV^2`
C. `(3)/(2)mV^2`
D. `2mV^2`

Explanation

To find the kinetic energy of an object of mass 'm' ejected from a satellite moving in a circular orbit around the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Orbital Velocity**:
The satellite is moving in a circular orbit with a constant speed 'V'. The orbital velocity \( V \) is given by the formula:
\[ V = \sqrt{\frac{GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the distance from the center of the Earth to the satellite.

2. **Escape Velocity**:
The escape velocity \( V_e \) from the Earth's gravitational field is given by:
\[ V_e = \sqrt{2gR} \]
where \( g \) is the acceleration due to gravity at the surface of the Earth. We can also express escape velocity in terms of the orbital velocity:
\[ V_e = \sqrt{2} \cdot V \]

3. **Kinetic Energy at Ejection**:
When the object of mass 'm' is ejected from the satellite, it must have enough kinetic energy to escape the gravitational pull of the Earth. The kinetic energy (KE) of the object at the time of ejection is given by:
\[ KE = \frac{1}{2} m V_e^2 \]

4. **Substituting Escape Velocity**:
Now, substituting the expression for \( V_e \) in terms of \( V \):
\[ KE = \frac{1}{2} m \left(\sqrt{2} \cdot V\right)^2 \]
Simplifying this gives:
\[ KE = \frac{1}{2} m \cdot 2V^2 = mV^2 \]

5. **Final Result**:
Therefore, the kinetic energy of the object at the time of its ejection is:
\[ KE = mV^2 \]

### Conclusion:
The kinetic energy of the object of mass 'm' at the time of its ejection from the satellite is \( mV^2 \).