$10 . B$ ship is at a position $45^{\circ}$ north of west from $A$ ship, the two ships are $10 \sqrt{2} \mathrm{~km}$ apart. If $A$ ship sails west, and $B$ ship sails south at the same time, and the speed of $B$ ship is twice that of $A$ ship, then the closest distance between $A$ and $B$ ships is $\qquad$ $\mathrm{km}$.
$10.2 \sqrt{5}$
As shown in the figure, after $t$ hours, ships $A$ and $B$ have sailed to positions $A_{1}$ and $B_{1}$, respectively. Let $A A_{1}=x$. Therefore, $B B_{1}=2 x$. Given that $A B = 10 \sqrt{2}$, we have
$$
\begin{aligned}
A C & =B C=10 . \\
\therefore A_{1} C & =|10-x|, B_{1} C=|10-2 x| \\
\therefore A_{1} B_{1} & =\sqrt{|10-x|^{2}+|10-2 x|^{2}} \\
& =\sqrt{5(x-6)^{2}+20 .}
\end{aligned}
$$

When $x=6$, $A_{1} B_{1}=2 \sqrt{5}$ is the minimum.