Question

A body of mass $m$ is thrown vertically upward with speed $\sqrt{3}v_e^2$, where $v_e$ is the escape velocity of a body from earth surface. The final velocity of the body is

• A. 0
• B. 2ve
• C. √3ve
• D. √2ve

Answer 1

The Correct Option is D

1. Given: A body of mass m is projected vertically upward with speed \( \sqrt{3}v_e \), where \( v_e \) is the escape velocity.
2. Escape velocity: This is the minimum velocity required to escape the Earth's gravitational field, and is given by:
   \[ v_e = \sqrt{\frac{2GM}{R}} \] where \( G \) is the gravitational constant, \( M \) is the Earth's mass, and \( R \) is Earth's radius.
3. Total mechanical energy at projection point:
   \[ E_{\text{initial}} = \frac{1}{2}m(\sqrt{3}v_e)^2 - \frac{GMm}{R} = \frac{3}{2}mv_e^2 - \frac{GMm}{R} \] But \( v_e^2 = \frac{2GM}{R} \), so: \[ E_{\text{initial}} = \frac{3}{2}m\left(\frac{2GM}{R}\right) - \frac{GMm}{R} = 3\frac{GMm}{R} - \frac{GMm}{R} = \frac{2GMm}{R} \]
4. At infinity: Potential energy = 0, so total energy = kinetic energy:
   \[ E_{\text{final}} = \frac{1}{2}mv^2 \] Set this equal to initial energy: \[ \frac{1}{2}mv^2 = \frac{2GMm}{R} \Rightarrow v^2 = \frac{4GM}{R} \Rightarrow v = \sqrt{4GM/R} = \sqrt{2}v_e \]
5. Final Answer: \( \boxed{\sqrt{2}v_e} \)