1. Given: The body is released from a height equal to the radius of the earth, so $h=R$ Hence its initial distance from the center of the earth is $r_i=R+h=2R$ and final distance at the surface is $r_f=R.$

2. Use conservation of mechanical energy.

   Initial kinetic energy: $K_i=0$ since the body is released from rest.

   Gravitational potential energy at distance $r$ from earth's center is $U=-\frac{GMm}{r}.$

   Therefore, $E_i=-\frac{GMm}{2R}$ and at the earth's surface, $E_f=\frac12 mv^2-\frac{GMm}{R}.$

3. Equate initial and final energies: $-\frac{GMm}{2R}=\frac12 mv^2-\frac{GMm}{R}$

   Rearranging, $\frac12 mv^2=\frac{GMm}{R}-\frac{GMm}{2R}=\frac{GMm}{2R}$

   So, $v^2=\frac{GM}{R}.$

4. **Use the relation between $g$ and $GM$: ** At the earth's surface, $g=\frac{GM}{R^2}$ so $GM=gR^2.$

   Substitute into the expression for $v^2$: $v^2=\frac{gR^2}{R}=gR$

   Hence, $v=\sqrt{gR}.$

5. Check options:

   • A: $\sqrt{\frac{gR}{2}}$ ❌
   • B: $\sqrt{4gR}$ ❌
   • C: $\sqrt{2gR}$ ❌
   • D: $\sqrt{gR}$ ✅

Therefore, the correct option is D.