A small block of superdense material has a mass of 3 × 10^ 24 kg ....

Let the mass of the block be $m = 3 \times 10^{24} \text{ kg}$ and the mass of the earth be $M = 6 \times 10^{24} \text{ kg}$. We have $M = 2m$.

Since no external forces act on the earth-block system, the center of mass remains at rest. By conservation of linear momentum:
$M v_1 = m v_2$
$2m v_1 = m v_2 \Rightarrow v_1 = \dfrac{v_2}{2}$
where $v_1$ and $v_2$ are the speeds of the earth and the block, respectively.

The height of the block from the earth's surface decreases from $h$ to $h/2$, meaning the relative distance between them decreases by $h/2$. Since $h$ is much smaller than the earth's radius $R$, the gravitational force remains approximately constant:
$F = \dfrac{GMm}{R^2} = mg$

The loss in gravitational potential energy of the system is:
$\Delta U = F \times \Delta r = mg$\left(\dfrac{h}{2}\right)

This loss in potential energy is converted into the kinetic energy of both the earth and the block. By conservation of energy:
$\Delta K = \Delta U$
\dfrac{1}{2}$M v_1^2 +$\dfrac{1}{2}$m v_2^2 = mg$\dfrac{h}{2}

Substituting $M = 2m$ and $v_1 =$\dfrac{v_2}{2}:
\dfrac{1}{2} ($2m) \left(\dfrac{v_2}{2}\right)^2 +$\dfrac{1}{2}$m v_2^2 = mg$\dfrac{h}{2}

\dfrac{1}{4}$m v_2^2 +$\dfrac{1}{2}$m v_2^2 = mg$\dfrac{h}{2}

\dfrac{3}{4}$m v_2^2 = mg$\dfrac{h}{2}

$v_2^2 =$\dfrac{2gh}{3}

$v_2 =$\sqrt{\dfrac{2gh}{3}}

Answer: \sqrt{\dfrac{2gh}{3}}