Question

A body is projected vertically upwards from earth’s surface with velocity \( 2v_e \), where \( v_e \) is escape velocity from earth’s surface. The velocity when body escapes the gravitational pull is

• A. \( \sqrt{7} v_e \)
• B. \( \sqrt{3} v_e \)
• C. \( \sqrt{5} v_e \)
• D. \( 2 v_e \)

Hint:

When a body is projected upwards with velocity greater than the escape velocity, use energy conservation principles to solve for the final velocity at infinity.

Answer 1

The Correct Option is B

Step 1: Escape velocity and energy considerations.
The escape velocity is the minimum velocity required for a body to escape the gravitational pull of the Earth. The total mechanical energy at the Earth's surface is: \[ E = \frac{1}{2} m v_e^2 - \frac{GMm}{R} \] where \( G \) is the gravitational constant, \( R \) is the radius of the Earth, and \( m \) is the mass of the body. For escape, the total energy at infinity is zero, so we set the total energy equal to zero.

Step 2: Energy after projection.
After projecting with velocity \( 2v_e \), the total energy at the Earth's surface is: \[ E = \frac{1}{2} m (2v_e)^2 - \frac{GMm}{R} = 2m v_e^2 - \frac{GMm}{R} \] Since the body escapes the Earth, the total energy at infinity is zero, so we equate this energy to zero: \[ E = \frac{1}{2} m v_f^2 = 2 m v_e^2 - \frac{GMm}{R} \] Solving for the final velocity \( v_f \), we get: \[ v_f = \sqrt{3} v_e \]
Step 3: Conclusion.
The velocity of the body when it escapes the Earth's gravitational pull is \( \sqrt{3} v_e \), which corresponds to option (B).