Question

A particle is projected from the surface of the Earth with a velocity equal to twice the escape velocity. When the particle is far from the Earth, its speed will be:

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

Hint:

When a particle is projected with a velocity greater than escape velocity, use energy conservation to find the final velocity at infinity.

Answer 1

The Correct Option is C

Step 1: Using Energy Conservation The total mechanical energy of the particle is: \[ E = \frac{1}{2} m v^2 - \frac{GMm}{R} \] For escape velocity: \[ v_e = \sqrt{\frac{2GM}{R}} \] Step 2: Applying Given Condition The initial velocity given is \( v = 2v_e \): \[ E = \frac{1}{2} m (4 v_e^2) - \frac{GMm}{R} \] Substituting \( v_e^2 = \frac{2GM}{R} \): \[ E = 2GMm/R - GMm/R = GMm/R \] At infinity, kinetic energy remains: \[ \frac{1}{2} m v_{\infty}^2 = GMm/R \] Solving for \( v_{\infty} \): \[ v_{\infty} = \sqrt{3} v_e \] Thus, the correct answer is option (3).