Answer
Using our expression for $U,$ the change in potential energy as the particle falls to the
center is
$$
\Delta U=-G M m\left(\frac{1}{R}-\frac{1}{\sqrt{x^{2}+R^{2}}}\right)
$$
By conservation of mechanical energy, this must "turn into" kinetic energy,
$\Delta K=-\Delta U=m v^{2} / 2 .$ We solve for the speed and obtain
$$
\frac{1}{2} m v^{2}=G M m\left(\frac{1}{R}-\frac{1}{\sqrt{x^{2}+R^{2}}}\right) \Rightarrow v=\sqrt{2 G M\left(\frac{1}{R}-\frac{1}{\sqrt{x^{2}+R^{2}}}\right)}
$$
Work Step by Step
Using our expression for $U,$ the change in potential energy as the particle falls to the
center is
$$
\Delta U=-G M m\left(\frac{1}{R}-\frac{1}{\sqrt{x^{2}+R^{2}}}\right)
$$
By conservation of mechanical energy, this must "turn into" kinetic energy,
$\Delta K=-\Delta U=m v^{2} / 2 .$ We solve for the speed and obtain
$$
\frac{1}{2} m v^{2}=G M m\left(\frac{1}{R}-\frac{1}{\sqrt{x^{2}+R^{2}}}\right) \Rightarrow v=\sqrt{2 G M\left(\frac{1}{R}-\frac{1}{\sqrt{x^{2}+R^{2}}}\right)}
$$
