The minimum velocity required for a particle to escape the Earth's gravitational field is called the escape velocity ($v_e$). It is derived using the principle of conservation of energy. The total mechanical energy (Kinetic + Potential) at the surface must equal the total energy at infinity (which is zero for the minimum condition).

$K.E. + P.E. = 0$ $\frac{1}{2}mu^2 - \frac{GmM}{R} = 0$ $\frac{1}{2}mu^2 = \frac{GmM}{R}$ $u^2 = \frac{2GM}{R}$ $u = \sqrt{\frac{2GM}{R}}$

Alternatively, using $g = \frac{GM}{R^2}$, the expression can be written as $\sqrt{2gR}$. Since $\sqrt{2gR}$ is not explicitly listed (Option 3 has $R^2$), the correct form in terms of $G$ and $M$ is the one in Option A.