Using conservation of energy, the escape velocity from a body of mass $M$ and radius $R$ is:

1. On the surface, total energy $E_1 = \dfrac{1}{2}mv_{esc}^2 - \dfrac{GMm}{R}$. 2. At infinity both kinetic and potential energy are zero, so $E_2 = 0$. 3. Conservation gives $E_1 = E_2$: $\dfrac{1}{2}mv_{esc}^2 - \dfrac{GMm}{R} = 0$. 4. So $v_{esc}^2 = \dfrac{2GM}{R}$, hence $v_{esc} = \sqrt{\dfrac{2GM}{R}}$. 5. Dropping the 2 or squaring $R$ in the denominator gives the wrong options. _Source: Balbharati (Maharashtra Board) Class 10 Science & Technology, Ch 1 "Gravitation", p.23_