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 need to determine the final velocity of an object that is projected vertically upwards from the surface of the Earth with a velocity three times the escape velocity.

### Given:
- Escape velocity from the Earth's surface, \( v_e = \sqrt{\frac{2GM}{R}} \)
- Initial velocity of the object, \( v = 3v_e \)

### Steps to Solve:

1. **Calculate the Escape Velocity**:
The escape velocity 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 Kinetic Energy**:
The initial kinetic energy (KE) of the object when it is projected is:
\[ KE_i = \frac{1}{2} m v^2 = \frac{1}{2} m (3v_e)^2 = \frac{1}{2} m (9v_e^2) = \frac{9}{2} mv_e^2 \]

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

4. **Total Mechanical Energy at the Surface**:
The total mechanical energy (E) at the surface of the Earth is the sum of the initial kinetic energy and potential energy:
\[ E = KE_i + PE_i = \frac{9}{2} mv_e^2 - \frac{GMm}{R} \]

5. **Final Potential Energy at Infinity**:
As the object escapes the gravitational pull of the Earth, its potential energy at infinity becomes zero:
\[ PE_f = 0 \]

6. **Final Kinetic Energy at Infinity**:
The total mechanical energy must remain constant. Therefore, at infinity:
\[ E = KE_f + PE_f \]
\[ E = KE_f + 0 = KE_f \]
Thus, we have:
\[ KE_f = \frac{9}{2} mv_e^2 - \frac{GMm}{R} \]

7. **Setting Up the Equation**:
We know that \( v_e^2 = \frac{2GM}{R} \), so we can substitute this into our equation:
\[ KE_f = \frac{9}{2} mv_e^2 - \frac{GMm}{R} = \frac{9}{2} m \left(\frac{2GM}{R}\right) - \frac{GMm}{R} \]
\[ KE_f = \frac{9GMm}{R} - \frac{GMm}{R} = \frac{8GMm}{R} \]

8. **Final Velocity**:
The final kinetic energy can also be expressed in terms of the final velocity \( v_f \):
\[ KE_f = \frac{1}{2} mv_f^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{2} mv_f^2 = \frac{8GMm}{R} \]
Dividing both sides by \( m \) and multiplying by 2:
\[ v_f^2 = \frac{16GM}{R} \]
Taking the square root:
\[ v_f = \sqrt{\frac{16GM}{R}} = 4v_e \]

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