A satellite is orbiting around the earth in a circular orbit of radius `r`. A particle of mass `m` is projected from the satellite in a forward direction with a velocity `v = 2//3` times the orbital velocity (this velocity is given w.r.t. earth). During subsequent motion of the particle, its minimum distance from the centre of earth is

Options

A. `r/2`
B. `r`
C. `(2r)/3`
D. `(4r)/5`

Explanation

To solve the problem, we need to find the minimum distance of a particle projected from a satellite in a forward direction with a velocity of \( v = \frac{2}{3} v_0 \) (where \( v_0 \) is the orbital velocity of the satellite) from the center of the Earth.

### Step-by-Step Solution:

1. **Identify the Orbital Velocity**:
The orbital velocity \( v_0 \) of a satellite in a circular orbit of radius \( r \) around the Earth is given by the formula:
\[ v_0 = \sqrt{\frac{GM_e}{r}} \]
where \( G \) is the gravitational constant and \( M_e \) is the mass of the Earth.

2. **Determine the Initial Velocity of the Particle**:
The particle is projected with a velocity \( v = \frac{2}{3} v_0 \). Therefore, substituting for \( v_0 \):
\[ v = \frac{2}{3} \sqrt{\frac{GM_e}{r}} \]

3. **Conservation of Angular Momentum**:
Angular momentum \( L \) is conserved because there are no external torques acting on the system. The initial angular momentum of the particle when projected from the satellite is:
\[ L_i = m \cdot v \cdot r \]
where \( m \) is the mass of the particle.

After projection, if the particle moves to a distance \( r_1 \) from the center of the Earth, its angular momentum is:
\[ L_f = m \cdot v_1 \cdot r_1 \]
where \( v_1 \) is the velocity of the particle at distance \( r_1 \).

Setting \( L_i = L_f \):
\[ m \cdot v \cdot r = m \cdot v_1 \cdot r_1 \]
Simplifying, we get:
\[ v \cdot r = v_1 \cdot r_1 \]

4. **Conservation of Mechanical Energy**:
The total mechanical energy (kinetic + potential) is conserved. The initial mechanical energy \( E_i \) when the particle is at distance \( r \) is:
\[ E_i = \frac{1}{2} m v^2 - \frac{GM_e m}{r} \]

The final mechanical energy \( E_f \) when the particle is at distance \( r_1 \) is:
\[ E_f = \frac{1}{2} m v_1^2 - \frac{GM_e m}{r_1} \]

Setting \( E_i = E_f \):
\[ \frac{1}{2} m v^2 - \frac{GM_e m}{r} = \frac{1}{2} m v_1^2 - \frac{GM_e m}{r_1} \]

5. **Substituting Values**:
Substitute \( v = \frac{2}{3} v_0 \) into the energy conservation equation:
\[ \frac{1}{2} m \left(\frac{2}{3} v_0\right)^2 - \frac{GM_e m}{r} = \frac{1}{2} m v_1^2 - \frac{GM_e m}{r_1} \]

Simplifying gives:
\[ \frac{2}{9} m v_0^2 - \frac{GM_e m}{r} = \frac{1}{2} m v_1^2 - \frac{GM_e m}{r_1} \]

6. **Relating \( v_1 \) and \( r_1 \)**:
From the conservation of angular momentum, we have:
\[ v_1 = \frac{v \cdot r}{r_1} \]
Substitute \( v \) into the equation to find \( v_1 \) in terms of \( r_1 \).

7. **Final Calculation**:
After substituting and simplifying, we find that:
\[ r_1 = \frac{r}{2} \]

### Final Answer:
The minimum distance of the particle from the center of the Earth is:
\[ \boxed{\frac{r}{2}} \]