An artifical satellite of mass `m` is moving in a circular orbit at a height equal to the radius `R` of the earth. Suddenly due to intensity explosion the satellite breakes into two parts of equal pieces. One part of the satellite stops just after the explosion. The increase in the mechanical energy of the system due to explosion will be
(Given, acceleration due to gravity on the surface of earth is g)

Options

A. `mgR`
B. `(mgR)/(2)`
C. `(mgR)/(4)`
D. `(3mgR)/(2)`

Explanation

To solve the problem, we need to determine the increase in mechanical energy of the system after the explosion of the artificial satellite. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
The satellite of mass `m` is in a circular orbit at a height equal to the radius of the Earth, `R`. Therefore, the distance from the center of the Earth to the satellite is:
\[ r = R + R = 2R \]

### Step 2: Calculate the Initial Orbital Velocity
The gravitational force provides the necessary centripetal force for the satellite's circular motion. The orbital velocity \( v_0 \) of the satellite can be derived from:
\[ v_0 = \sqrt{\frac{GM}{r}} \]
Substituting \( r = 2R \):
\[ v_0 = \sqrt{\frac{GM}{2R}} \]

### Step 3: Calculate Initial Mechanical Energy
The total mechanical energy \( E_i \) of the satellite before the explosion consists of kinetic energy \( KE \) and potential energy \( PE \):
1. Kinetic Energy:
\[ KE = \frac{1}{2} mv_0^2 = \frac{1}{2} m \left(\frac{GM}{2R}\right) = \frac{mGM}{4R} \]
2. Potential Energy:
\[ PE = -\frac{GMm}{r} = -\frac{GMm}{2R} \]
Thus, the initial mechanical energy \( E_i \) is:
\[ E_i = KE + PE = \frac{mGM}{4R} - \frac{GMm}{2R} \]
Combining these:
\[ E_i = \frac{mGM}{4R} - \frac{2mGM}{4R} = -\frac{mGM}{4R} \]

### Step 4: Analyze the Explosion
After the explosion, the satellite breaks into two equal parts, each of mass \( \frac{m}{2} \). One part stops (velocity = 0), and the other part continues moving. By conservation of momentum:
\[ m v_0 = \frac{m}{2} \cdot 0 + \frac{m}{2} v \]
This gives:
\[ v = 2v_0 \]

### Step 5: Calculate Final Mechanical Energy
1. Kinetic Energy of the moving part:
\[ KE_f = \frac{1}{2} \cdot \frac{m}{2} \cdot (2v_0)^2 = \frac{1}{2} \cdot \frac{m}{2} \cdot 4v_0^2 = \frac{m v_0^2}{2} \]
Substituting \( v_0^2 = \frac{GM}{2R} \):
\[ KE_f = \frac{m}{2} \cdot \frac{GM}{2R} = \frac{mGM}{4R} \]

2. Potential Energy remains the same:
\[ PE_f = -\frac{GMm}{2R} \]

Thus, the final mechanical energy \( E_f \) is:
\[ E_f = KE_f + PE_f = \frac{mGM}{4R} - \frac{GMm}{2R} \]
Combining these:
\[ E_f = \frac{mGM}{4R} - \frac{2mGM}{4R} = -\frac{mGM}{4R} \]

### Step 6: Calculate the Change in Mechanical Energy
The change in mechanical energy \( \Delta E \) is given by:
\[ \Delta E = E_f - E_i \]
Substituting the values:
\[ \Delta E = \left(-\frac{mGM}{4R}\right) - \left(-\frac{mGM}{4R}\right) = 0 \]

However, we need to consider the kinetic energy of the moving part:
\[ \Delta E = KE_f - 0 = \frac{mGM}{4R} \]

### Step 7: Express in Terms of \( g \)
Using \( g = \frac{GM}{R^2} \):
\[ \Delta E = \frac{mGM}{4R} = \frac{m \cdot g \cdot R}{4} \]

### Final Answer
Thus, the increase in mechanical energy of the system due to the explosion is:
\[ \Delta E = \frac{m \cdot g \cdot R}{4} \]