A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of earth's radius `R_e`. By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it becomes `sqrt(3/2)` times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is R. Value of R is :

Options

A. `2R_e`
B. `3R_e`
C. `4R_e`
D. `2.5 R_e`

Explanation

To solve the problem, we need to analyze the situation where a satellite is in a low circular orbit around the Earth and then has its speed increased. We will use the principles of conservation of angular momentum and conservation of energy to find the maximum distance \( R \) from the center of the Earth that the satellite can reach after the speed increase.

### Step-by-Step Solution:

1. **Initial Conditions**:
- The satellite is initially in a circular orbit at a radius \( R_e \) (the radius of the Earth).
- The initial speed of the satellite \( V_0 \) can be calculated using the formula for orbital velocity:
\[ V_0 = \sqrt{\frac{GM}{R_e}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

2. **Speed Increase**:
- The speed of the satellite is increased to \( V = \sqrt{\frac{3}{2}} V_0 \).

3. **Conservation of Angular Momentum**:
- The angular momentum before and after the speed increase must be conserved. The initial angular momentum \( L_i \) is given by:
\[ L_i = m V_0 R_e \]
- The final angular momentum \( L_f \) after the speed increase is:
\[ L_f = m V R \]
- Setting these equal gives:
\[ m V_0 R_e = m V R \]
- Canceling \( m \) from both sides, we have:
\[ V_0 R_e = V R \]
- Substituting \( V = \sqrt{\frac{3}{2}} V_0 \):
\[ V_0 R_e = \sqrt{\frac{3}{2}} V_0 R \]
- Dividing both sides by \( V_0 \) (assuming \( V_0 \neq 0 \)):
\[ R_e = \sqrt{\frac{3}{2}} R \]
- Rearranging gives:
\[ R = \frac{2}{\sqrt{3}} R_e \]

4. **Conservation of Energy**:
- The total mechanical energy (kinetic + potential) must also be conserved. The initial total energy \( E_i \) is:
\[ E_i = \frac{1}{2} m V_0^2 - \frac{GMm}{R_e} \]
- The final total energy \( E_f \) is:
\[ E_f = \frac{1}{2} m V^2 - \frac{GMm}{R} \]
- Setting \( E_i = E_f \):
\[ \frac{1}{2} m V_0^2 - \frac{GMm}{R_e} = \frac{1}{2} m \left(\sqrt{\frac{3}{2}} V_0\right)^2 - \frac{GMm}{R} \]
- Simplifying gives:
\[ \frac{1}{2} m V_0^2 - \frac{GMm}{R_e} = \frac{3}{4} m V_0^2 - \frac{GMm}{R} \]
- Canceling \( m \) and rearranging:
\[ \frac{1}{2} V_0^2 - \frac{3}{4} V_0^2 = \frac{GM}{R} - \frac{GM}{R_e} \]
- This simplifies to:
\[ -\frac{1}{4} V_0^2 = \frac{GM}{R} - \frac{GM}{R_e} \]

5. **Substituting \( V_0^2 \)**:
- Substitute \( V_0^2 = \frac{GM}{R_e} \):
\[ -\frac{1}{4} \cdot \frac{GM}{R_e} = \frac{GM}{R} - \frac{GM}{R_e} \]
- Multiplying through by \( R_e R \) gives:
\[ -\frac{1}{4} GM R = GM R_e - GM R_e \]
- Rearranging gives:
\[ R = 3 R_e \]

### Final Result:
Thus, the farthest distance \( R \) from the center of the Earth that the satellite reaches is:
\[ R = 3 R_e \]