A particle of mass 'm' is projected from the surface of earth with velocity`upsilon =2upsilon_(e)`, where `upsilon_(e)` is the value of escape velocity from the surface of earth . Find velocity of the particle on reaching to interstellar space (at infinity) in terms of `upsilon_(e)`.

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The particle is projected from the surface of the Earth with a velocity of \( v = 2v_e \), where \( v_e \) is the escape velocity from the surface of the Earth. We need to find the velocity of the particle when it reaches interstellar space (at infinity).

### Step-by-Step Solution:

1. **Understand Escape Velocity**:
The escape velocity \( v_e \) from the surface of the Earth is given by:
\[ v_e = \sqrt{\frac{GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Initial Kinetic Energy**:
The initial kinetic energy \( KE_i \) of the particle when it is projected is:
\[ KE_i = \frac{1}{2} m (2v_e)^2 = \frac{1}{2} m (4v_e^2) = 2mv_e^2 \]

3. **Initial Potential Energy**:
The initial potential energy \( PE_i \) at the surface of the Earth is:
\[ PE_i = -\frac{GMm}{R} \]

4. **Final Kinetic Energy at Infinity**:
At infinity, the potential energy \( PE_f \) is zero (since gravitational potential energy approaches zero as distance approaches infinity). Let the final velocity at infinity be \( v_a \). The final kinetic energy \( KE_f \) is:
\[ KE_f = \frac{1}{2} mv_a^2 \]

5. **Conservation of Mechanical Energy**:
According to the conservation of mechanical energy:
\[ KE_i + PE_i = KE_f + PE_f \]
Substituting the values we have:
\[ 2mv_e^2 - \frac{GMm}{R} = \frac{1}{2} mv_a^2 + 0 \]

6. **Simplifying the Equation**:
We can factor out \( m \) (assuming \( m \neq 0 \)):
\[ 2v_e^2 - \frac{GM}{R} = \frac{1}{2} v_a^2 \]

7. **Substituting for \( \frac{GM}{R} \)**:
From the escape velocity formula, we know:
\[ \frac{GM}{R} = v_e^2 \]
Therefore, we can substitute this into our equation:
\[ 2v_e^2 - v_e^2 = \frac{1}{2} v_a^2 \]
This simplifies to:
\[ v_e^2 = \frac{1}{2} v_a^2 \]

8. **Solving for \( v_a \)**:
Rearranging gives us:
\[ v_a^2 = 2v_e^2 \]
Taking the square root:
\[ v_a = \sqrt{2} v_e \]

9. **Final Result**:
Therefore, the velocity of the particle when it reaches interstellar space (at infinity) is:
\[ v_a = \sqrt{2} v_e \]