A satellite is revolving near by earth of radius `=R_e`. If its velocity is increased `sqrt(3/2)v` where v is the orbital speed of satellite then find maximum distance of satellite from the center of the earth.

Options

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

Answer

To solve the problem of finding the maximum distance of a satellite from the center of the Earth after its velocity is increased, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The satellite is initially in a circular orbit at a radius \( R_e \) (the radius of the Earth).
- The orbital speed \( v \) of the satellite at this radius is given by the formula:
\[ v = \sqrt{\frac{GM}{R_e}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

2. **Increase Velocity**:
- The velocity of the satellite is increased to \( v' = \sqrt{\frac{3}{2}} v \).

3. **Conservation of Energy**:
- The total mechanical energy (kinetic + potential) is conserved when no external work is done.
- The potential energy \( U \) at a distance \( r \) from the center of the Earth is given by:
\[ U = -\frac{GMm}{r} \]
- The kinetic energy \( K \) of the satellite is given by:
\[ K = \frac{1}{2} mv'^2 \]

4. **Calculate Initial Energy**:
- At the initial position (radius \( R_e \)):
\[ U_1 = -\frac{GMm}{R_e} \]
\[ K_1 = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{GM}{R_e}\right) = \frac{GMm}{2R_e} \]
- Total energy at position 1:
\[ E_1 = U_1 + K_1 = -\frac{GMm}{R_e} + \frac{GMm}{2R_e} = -\frac{GMm}{2R_e} \]

5. **Calculate Final Energy**:
- At the maximum distance \( r \) from the center of the Earth:
\[ U_2 = -\frac{GMm}{r} \]
\[ K_2 = \frac{1}{2} m v'^2 = \frac{1}{2} m \left(\frac{3}{2} v\right)^2 = \frac{1}{2} m \left(\frac{3}{2}\right)^2 \left(\frac{GM}{R_e}\right) = \frac{9GMm}{8R_e} \]
- Total energy at position 2:
\[ E_2 = U_2 + K_2 = -\frac{GMm}{r} + \frac{9GMm}{8R_e} \]

6. **Set Energies Equal**:
- Since energy is conserved:
\[ E_1 = E_2 \]
\[ -\frac{GMm}{2R_e} = -\frac{GMm}{r} + \frac{9GMm}{8R_e} \]

7. **Solve for \( r \)**:
- Rearranging gives:
\[ -\frac{GMm}{2R_e} + \frac{9GMm}{8R_e} = -\frac{GMm}{r} \]
\[ \frac{9GMm}{8R_e} - \frac{4GMm}{8R_e} = -\frac{GMm}{r} \]
\[ \frac{5GMm}{8R_e} = -\frac{GMm}{r} \]
- Cross-multiplying gives:
\[ 5r = -8R_e \implies r = \frac{8R_e}{5} \]

8. **Final Result**:
- The maximum distance of the satellite from the center of the Earth is:
\[ r = 2R_e \]