5. Five concentric circles $\Gamma_{0}, \Gamma_{1}, \Gamma_{2}, \Gamma_{3}, \Gamma_{4}$ have radii that form a geometric sequence with common ratio $q$ in the given order. Find the maximum value of $q$ such that a closed broken line $A_{0} A_{1} A_{2} A_{3} A_{4}$ can be drawn, with each segment of equal length, and points $A_{i} (i=0,1, \cdots, 4)$ lying on circle $\Gamma_{i}$.
5. $q=\frac{\sqrt{5}+1}{2}$.

We can assume $q \geqslant 1$. Let the radius of $\Gamma_{i}$ be $r_{i}=r q$.
Choose some positive number $l$, start from a point $A_{0} \in \Gamma_{0}$, and attempt to construct a broken line with each segment length being $l$ as required by the problem.

Assume $A_{i} \in \Gamma_{i}$ has been chosen. The distance from it to a point on the circle $\Gamma_{i+1}$ fills the interval $\left[r_{i+1}-r_{i}, r_{i+1}+r_{i}\right]$, which is $\left[r q^{i}(q-1), r q^{i}(q+1)\right]$. A point $A_{i+1}$ can be found if and only if $l$ belongs to this interval. This indicates that the broken line can be constructed if and only if
$$
r q^{i}(q-1) \leqslant l \leqslant r q^{i}(q+1)(i=0,1,2,3) \text {. }
$$

Since $q \geqslant 1$, these four inequalities indicate
$$
r q^{3}(q-1) \leqslant l \leqslant r(q+1) \text {. }
$$

Therefore, such an $l$ exists if and only if
$$
\begin{array}{l}
q^{3}(q-1) \leqslant q+1 \\
\Leftrightarrow q^{4}-q^{3}-q-1 \leqslant 0 \\
\Leftrightarrow\left(q^{2}-q-1\right)\left(q^{2}+1\right) \leqslant 0 .
\end{array}
$$

Thus, the maximum value of $q$ that satisfies the above inequality is $\frac{\sqrt{5}+1}{2}$.