1. Interpret the height

   The object is released from a height $R$ above the earth's surface.

   Since the radius of earth is also $R$, the initial distance from the center of earth is $r_i = R + R = 2R$ and the final distance when it reaches the surface is $r_f = R.$

2. Use conservation of mechanical energy

   Initial kinetic energy is zero because the object is allowed to fall from rest.

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

   So,

   • Initial energy: $E_i = 0 - \frac{GMm}{2R}$
   • Final energy: $E_f = \frac{1}{2}mv^2 - \frac{GMm}{R}$

   By conservation of energy, $-\frac{GMm}{2R} = \frac{1}{2}mv^2 - \frac{GMm}{R}.$

3. Solve for $v$

   Bring the potential terms together: $\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R}.$

   Therefore, $\frac{1}{2}mv^2 = \frac{GMm}{2R}$ $v^2 = \frac{GM}{R}.$

4. Use the relation $g = \dfrac{GM}{R^2}$

   Hence, $GM = gR^2.$

   Substituting into $v^2 = \dfrac{GM}{R}$: $v^2 = \frac{gR^2}{R} = gR.$

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

5. Check options

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

Correct option: B