An object is allowed to fall from a height R above the earth, where...

As gravitational field is a conservative field and therefore mechanical energy will be conserved. Therefore,
Loss in PE = Gain in KE

$\left(- \dfrac{GMm}{2R}\right) - \left(- \dfrac{GMm}{R}\right) = \dfrac{1}{2}m{v}^{2}$

$⇒ {v}^{2} = \dfrac{GM}{R} = gR\left(\text{as}g = \dfrac{GM}{{R}^{2}}\right)$

$⇒ v = \sqrt{gR}$