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 motion of a satellite that is initially in a low circular orbit around the Earth and then has its speed increased. Let's break it down step by step.

### Step 1: Understand the Initial Conditions
The satellite is in a circular orbit at a radius equal to the radius of the Earth, \( R_e \). The orbital speed \( v_0 \) of the satellite can be calculated using the formula for circular motion:
\[ v_0 = \sqrt{\frac{GM_e}{R_e}} \]
where \( G \) is the gravitational constant and \( M_e \) is the mass of the Earth.

**Hint:** Remember that the gravitational force provides the centripetal force required for circular motion.

### Step 2: Speed Increase
The speed of the satellite is increased by a factor of \( \sqrt{\frac{3}{2}} \):
\[ v = v_0 \cdot \sqrt{\frac{3}{2}} \]

### Step 3: Conservation of Angular Momentum
Since the satellite is moving in a circular orbit, we can apply the conservation of angular momentum. The initial angular momentum \( L_i \) and final angular momentum \( L_f \) can be expressed as:
\[ L_i = m \cdot v_0 \cdot R_e \]
\[ L_f = m \cdot v \cdot R \]
Setting these equal gives:
\[ m \cdot v_0 \cdot R_e = m \cdot v \cdot R \]
Cancelling \( m \) from both sides, we have:
\[ v_0 \cdot R_e = v \cdot R \]

### Step 4: Substitute for \( v \)
Substituting \( v = v_0 \cdot \sqrt{\frac{3}{2}} \) into the angular momentum equation:
\[ v_0 \cdot R_e = (v_0 \cdot \sqrt{\frac{3}{2}}) \cdot R \]
Dividing both sides by \( v_0 \) (assuming \( v_0 \neq 0 \)):
\[ R_e = \sqrt{\frac{3}{2}} \cdot R \]

### Step 5: Solve for \( R \)
Rearranging gives:
\[ R = \frac{R_e}{\sqrt{\frac{3}{2}}} = R_e \cdot \sqrt{\frac{2}{3}} \]

### Step 6: Find the Farthest Distance
The farthest distance \( R \) from the center of the Earth that the satellite reaches is:
\[ R = R_e \cdot \sqrt{\frac{2}{3}} \]

### Conclusion
Thus, the value of \( R \) is \( R_e \cdot \sqrt{\frac{2}{3}} \).

### Final Answer
\[ R = \frac{R_e}{\sqrt{\frac{3}{2}}} \]