An objects is projected vertically upwards from the surface of the earth with a velocity `3` times the escape velocity `v_(e )` from earth's surface. What will be its final velocity after escape from the earth's gravitational pull ?

Options

A. `sqrt(2)v_(e )`
B. `2sqrt(2)v_(e )`
C. `2v_(e )`
D. `(v_(e ))/(sqrt(2))`

Answer

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy (kinetic + potential) at the surface of the Earth will equal the total mechanical energy at infinity (where the potential energy is zero).

### Step-by-Step Solution:

1. **Define the Escape Velocity**:
The escape velocity \( v_e \) from the surface of the Earth is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

2. **Initial Conditions**:
The object is projected with a velocity \( v = 3v_e \). The initial kinetic energy \( KE_i \) and gravitational potential energy \( PE_i \) at the surface of the Earth are:
\[ KE_i = \frac{1}{2} m (3v_e)^2 = \frac{9}{2} mv_e^2 \]
\[ PE_i = -\frac{GMm}{R} \]

3. **Final Conditions at Infinity**:
At infinity, the potential energy \( PE_f \) is zero, and the final kinetic energy \( KE_f \) is:
\[ KE_f = \frac{1}{2} mv^2 \]

4. **Conservation of Energy**:
According to the conservation of mechanical energy:
\[ KE_i + PE_i = KE_f + PE_f \]
Substituting the expressions we have:
\[ \frac{9}{2} mv_e^2 - \frac{GMm}{R} = \frac{1}{2} mv^2 + 0 \]

5. **Substituting Escape Velocity**:
We know that \( v_e^2 = \frac{2GM}{R} \), so substituting this into the equation:
\[ \frac{9}{2} m \left(\frac{2GM}{R}\right) - \frac{GMm}{R} = \frac{1}{2} mv^2 \]
Simplifying the left side:
\[ \frac{9GMm}{R} - \frac{GMm}{R} = \frac{1}{2} mv^2 \]
\[ \frac{8GMm}{R} = \frac{1}{2} mv^2 \]

6. **Isolating \( v^2 \)**:
Dividing both sides by \( m \):
\[ \frac{8GM}{R} = \frac{1}{2} v^2 \]
Multiplying both sides by 2:
\[ \frac{16GM}{R} = v^2 \]

7. **Finding Final Velocity \( v \)**:
Taking the square root of both sides:
\[ v = \sqrt{\frac{16GM}{R}} = 4\sqrt{\frac{GM}{R}} = 4v_e \]

### Final Answer:
The final velocity of the object after escaping the Earth's gravitational pull is \( 4v_e \).