Question

A particle of mass $m$ is thrown upwards from the surface of the earth, with a velocity $u$ . The mass and the radius of the earth are, respectively, $M$ and $R. G$ is gravitational constant and $g$ is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is

• A. $ \sqrt{ \frac{ 2GM}{R^2}}$
• B. $ \sqrt{ \frac{ 2GM}{R}}$
• C. $ \sqrt{ \frac{ 2gM}{R^2}}$
• D. $\sqrt{2g R^2 }$

Answer 1

The Correct Option is B

To solve this problem, we need to determine the minimum velocity \( u \) that allows a particle to escape the Earth's gravitational field without returning. This is essentially the escape velocity problem.

1. To find the escape velocity, we use the formula for escape velocity: v_e = \sqrt{\frac{2GM}{R}}. Here, G is the gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.
2. The escape velocity is derived from the conservation of energy principle, where the total mechanical energy (sum of kinetic and potential energy) must be zero or more for the particle to escape:
   • The initial kinetic energy is given by \frac{1}{2}mu^2.
   • The initial potential energy relative to infinity is -\frac{GMm}{R}.
   • At infinity, both kinetic energy and potential energy become zero.
3. Setting the total mechanical energy equal to zero gives: \frac{1}{2}mu^2 - \frac{GMm}{R} = 0.
4. Solving for u, we get: u^2 = \frac{2GM}{R}
5. Thus, the minimum velocity \( u \) for the particle to not return to Earth (escape velocity) is: u = \sqrt{\frac{2GM}{R}}.

Therefore, the correct minimum value of u so that the particle does not return back to Earth is: \sqrt{\frac{2GM}{R}}. This corresponds to option 2.