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 apply the principles of conservation of energy and momentum.

### Step 1: Apply Conservation of Energy
Initially, the satellite has kinetic energy due to its initial speed \( U \) and potential energy due to its position at the surface of the Earth.

The initial kinetic energy (KE_initial) is:
\[ KE_{\text{initial}} = \frac{1}{2} m U^2 \]

The initial potential energy (PE_initial) is:
\[ PE_{\text{initial}} = -\frac{GMm}{R} \]

At a height of \( 2R \), the total potential energy (PE_final) is:
\[ PE_{\text{final}} = -\frac{GMm}{3R} \]

The final kinetic energy (KE_final) at height \( 2R \) is:
\[ KE_{\text{final}} = \frac{1}{2} m v^2 \]

Using conservation of energy:
\[ KE_{\text{initial}} + PE_{\text{initial}} = KE_{\text{final}} + PE_{\text{final}} \]

Substituting the expressions:
\[ \frac{1}{2} m U^2 - \frac{GMm}{R} = \frac{1}{2} m v^2 - \frac{GMm}{3R} \]

### Step 2: Simplify the Equation
Rearranging the equation gives:
\[ \frac{1}{2} m U^2 + \frac{GMm}{3R} = \frac{1}{2} m v^2 + \frac{GMm}{R} \]

Multiplying through by 2 to eliminate the fraction:
\[ m U^2 + \frac{2GMm}{3R} = m v^2 + \frac{2GMm}{R} \]

Now, simplifying gives:
\[ U^2 + \frac{2GM}{3R} = v^2 + \frac{2GM}{R} \]

Rearranging further:
\[ v^2 = U^2 - \frac{4GM}{3R} \]

### Step 3: Ejecting the Rocket
When the satellite ejects the rocket of mass \( \frac{m}{5} \), we need to consider momentum conservation. Let \( V_t \) be the tangential velocity of the rocket and \( V_r \) be the radial velocity.

Using conservation of momentum in the tangential direction:
\[ 0 = \frac{4}{5} m V_0 - \frac{m}{5} V_t \]
From this, we find:
\[ V_t = 4 V_0 \]

### Step 4: Finding the Orbital Velocity
The orbital velocity \( V_0 \) at height \( 3R \) is given by:
\[ V_0 = \sqrt{\frac{GM}{3R}} \]

### Step 5: Finding the Rocket's Velocity
From the momentum conservation in the radial direction:
\[ m v = \frac{4}{5} m V_0 + \frac{m}{5} V_r \]
This leads to:
\[ V_r = 5v \]

### Step 6: Kinetic Energy of the Rocket
The kinetic energy of the rocket is given by:
\[ KE_{\text{rocket}} = \frac{1}{2} \left(\frac{m}{5}\right) V_r^2 + \frac{1}{2} \left(\frac{m}{5}\right) V_t^2 \]

Substituting \( V_r = 5v \) and \( V_t = 4V_0 \):
\[ KE_{\text{rocket}} = \frac{1}{2} \left(\frac{m}{5}\right) (5v)^2 + \frac{1}{2} \left(\frac{m}{5}\right) (4V_0)^2 \]

Calculating:
\[ = \frac{m}{10} (25v^2) + \frac{m}{10} (16V_0^2) \]

### Step 7: Substitute for \( v^2 \) and \( V_0^2 \)
Substituting \( v^2 \) and \( V_0^2 \) into the equation gives:
\[ KE_{\text{rocket}} = \frac{m}{10} \left(25\left(U^2 - \frac{4GM}{3R}\right) + 16\left(\frac{GM}{3R}\right)\right) \]

### Final Result
After simplifying, we get:
\[ KE_{\text{rocket}} = \frac{m}{10} \left(25U^2 - \frac{84GM}{3R}\right) \]