A satellite of mass m is launched vertically upwards with an initial speed U from the surface of the Earth. After it reaches height 2 R (R = radius of Earth) it ejects a rocket of mass `(m)/(5)` So, that subsequently the satellite moves in a circular orbit. The kinetic energy of rocket is (M is mass of Earth)

Options

A. `(m)/(10)(25U^(2) - (84 GM)/(3 R))`
B. `(m)/(10) (15U^(2) - (74 GM)/(3R) )`
C. `(m)/(8) ( U sqrt((10 GM)/(3R)))`
D. `(m)/(6) (25U^(2) - (64 GM)/(3R))`

Explanation

To solve the problem step by step, we will follow the principles of conservation of energy and momentum.

### Step 1: Initial Energy Calculation
The satellite is launched from the surface of the Earth with an initial speed \( U \). We will calculate the total mechanical energy (kinetic + potential) when the satellite is at the surface of the Earth.

- **Kinetic Energy (KE)** at the surface:
\[ KE_1 = \frac{1}{2} m U^2 \]

- **Potential Energy (PE)** at the surface:
\[ PE_1 = -\frac{GMm}{R} \]

Thus, the total energy at the surface \( E_1 \) is:
\[ E_1 = KE_1 + PE_1 = \frac{1}{2} m U^2 - \frac{GMm}{R} \]

### Step 2: Energy at Height \( 2R \)
When the satellite reaches a height of \( 2R \) (which means its distance from the center of the Earth is \( 3R \)), we calculate the total energy again.

- **Kinetic Energy (KE)** at height \( 2R \):
Let the velocity at height \( 2R \) be \( V \).
\[ KE_2 = \frac{1}{2} m V^2 \]

- **Potential Energy (PE)** at height \( 2R \):
\[ PE_2 = -\frac{GMm}{3R} \]

Thus, the total energy at height \( 2R \) \( E_2 \) is:
\[ E_2 = KE_2 + PE_2 = \frac{1}{2} m V^2 - \frac{GMm}{3R} \]

### Step 3: Conservation of Energy
According to the conservation of energy, the total energy at the surface must equal the total energy at height \( 2R \):
\[ \frac{1}{2} m U^2 - \frac{GMm}{R} = \frac{1}{2} m V^2 - \frac{GMm}{3R} \]

### Step 4: Simplifying the Energy Equation
Rearranging the equation gives:
\[ \frac{1}{2} m U^2 + \frac{GMm}{3R} = \frac{1}{2} m V^2 + \frac{GMm}{R} \]
\[ \frac{1}{2} m U^2 = \frac{1}{2} m V^2 + \frac{GMm}{R} - \frac{GMm}{3R} \]
\[ \frac{1}{2} m U^2 = \frac{1}{2} m V^2 + \frac{2GMm}{3R} \]

### Step 5: Solving for \( V^2 \)
Now, we can isolate \( V^2 \):
\[ \frac{1}{2} m U^2 - \frac{2GMm}{3R} = \frac{1}{2} m V^2 \]
\[ U^2 - \frac{4GM}{3R} = V^2 \]

### Step 6: Ejecting the Rocket
After reaching height \( 2R \), the satellite ejects a rocket of mass \( \frac{m}{5} \). The remaining mass of the satellite is \( \frac{4m}{5} \).

### Step 7: Conservation of Momentum
Using conservation of momentum in the tangential direction:
\[ 0 = \left(\frac{4m}{5}\right) V_{orbital} - \left(\frac{m}{5}\right) V_t \]
This gives:
\[ V_t = 4 V_{orbital} \]

### Step 8: Finding Orbital Velocity
The orbital velocity \( V_{orbital} \) at radius \( 3R \) is given by:
\[ V_{orbital} = \sqrt{\frac{GM}{3R}} \]

### Step 9: Kinetic Energy of the Rocket
The kinetic energy of the rocket is given by:
\[ KE_{rocket} = \frac{1}{2} \left(\frac{m}{5}\right) V_t^2 + \frac{1}{2} \left(\frac{m}{5}\right) V_r^2 \]

Using \( V_t = 4V_{orbital} \) and \( V_r = 5V_{orbital} \):
\[ KE_{rocket} = \frac{1}{2} \left(\frac{m}{5}\right) (4V_{orbital})^2 + \frac{1}{2} \left(\frac{m}{5}\right) (5V_{orbital})^2 \]
\[ = \frac{m}{10} (16V_{orbital}^2 + 25V_{orbital}^2) = \frac{m}{10} (41V_{orbital}^2) \]

### Step 10: Substituting \( V_{orbital} \)
Substituting \( V_{orbital} = \sqrt{\frac{GM}{3R}} \):
\[ KE_{rocket} = \frac{m}{10} (41 \cdot \frac{GM}{3R}) \]

### Final Answer
Thus, the kinetic energy of the rocket is:
\[ KE_{rocket} = \frac{41mGM}{30R} \]