A satellite of mass 'M' is projected radially from surface of earth with speed 'u'. When is reaches a height equal to radius of earth, it ejects a rocket of mass `( M)/(10 )`and it itself starts orbiting the earth in circular path of radius 2R, find the kinetic energy of rocket.

Options

A. `5M(u^(2)-(119GM_(e))/(200R))`
B. `5M(u^(2)-(113GM_(e))/(200R))`
C. `(M)/(20)(u^(2)-(119GM_(e))/(200R))`
D. `(M)/(20)(u^(2)-(113GM_(e))/(200R))`

Answer

To solve the problem step by step, we will analyze the motion of the satellite and the rocket, applying the principles of conservation of energy and conservation of momentum.

### Step 1: Understand the Initial Conditions
The satellite of mass \( M \) is projected radially from the surface of the Earth with speed \( u \). It reaches a height equal to the radius of the Earth, which means it is at a distance of \( 2R \) from the center of the Earth (where \( R \) is the radius of the Earth).

### Step 2: Apply Conservation of Energy
At the surface of the Earth (point 1), the total mechanical energy is given by:
\[ E_1 = \text{K.E.} + \text{P.E.} = \frac{1}{2} M u^2 - \frac{G M M_e}{R} \]
where \( M_e \) is the mass of the Earth.

At height \( R \) (point 2), the total mechanical energy is:
\[ E_2 = \text{K.E.} + \text{P.E.} = \frac{1}{2} M v^2 - \frac{G M M_e}{2R} \]

By conservation of energy, we have:
\[ \frac{1}{2} M u^2 - \frac{G M M_e}{R} = \frac{1}{2} M v^2 - \frac{G M M_e}{2R} \]

### Step 3: Rearranging the Energy Equation
Rearranging the equation gives:
\[ \frac{1}{2} M u^2 + \frac{G M M_e}{2R} = \frac{1}{2} M v^2 + \frac{G M M_e}{R} \]
This simplifies to:
\[ \frac{1}{2} M u^2 = \frac{1}{2} M v^2 + \frac{G M M_e}{2R} \]

### Step 4: Solve for \( v \)
Rearranging gives:
\[ v^2 = u^2 - \frac{G M_e}{R} \]

### Step 5: Ejecting the Rocket
When the satellite ejects the rocket of mass \( \frac{M}{10} \), the remaining mass of the satellite is \( \frac{9M}{10} \). We will use conservation of momentum to find the velocities of both the rocket and the satellite after the ejection.

### Step 6: Apply Conservation of Momentum
Let \( v_r \) be the velocity of the rocket after ejection. The momentum before ejection is:
\[ M v \]
The momentum after ejection must equal the momentum before:
\[ M v = \frac{M}{10} v_r + \frac{9M}{10} v_s \]
where \( v_s \) is the velocity of the satellite after ejection.

### Step 7: Determine the Velocities
Using the fact that the satellite moves in a circular orbit of radius \( 2R \), we know:
\[ v_s = \sqrt{\frac{G M_e}{2R}} \]
Substituting this into the momentum equation gives:
\[ M v = \frac{M}{10} v_r + \frac{9M}{10} \sqrt{\frac{G M_e}{2R}} \]

### Step 8: Solve for \( v_r \)
Rearranging gives:
\[ v_r = 10\left(v - \sqrt{\frac{G M_e}{2R}}\right) \]

### Step 9: Calculate the Kinetic Energy of the Rocket
The kinetic energy \( K_r \) of the rocket is given by:
\[ K_r = \frac{1}{2} \left(\frac{M}{10}\right) v_r^2 \]
Substituting \( v_r \) into the kinetic energy equation gives:
\[ K_r = \frac{1}{20} \left(10\left(v - \sqrt{\frac{G M_e}{2R}}\right)\right)^2 \]

### Step 10: Final Expression for Kinetic Energy
After substituting \( v \) and simplifying, we can express \( K_r \) in terms of \( u \) and \( G M_e \).

### Final Answer
The final expression for the kinetic energy of the rocket is:
\[ K_r = \frac{M}{20} \left(100 u^2 - 119 \frac{G M_e}{R}\right) \]