8. A grain silo can be roughly regarded as a frustum of a cone, with the radius of the upper base being 3 meters, the radius of the lower base being 6 meters, and the height being $\sqrt{7}$ meters. A well-fed mouse is exercising and plans to start from point $A$ on the circumference of the lower base, run around the side of the frustum, and return to point $A$. To make the journey as short as possible, the mouse needs to run at least $\qquad$ meters.
8. $8 \sqrt{3}+\frac{10 \pi}{3}$.

Unfolding the lateral surface of the frustum, we get a part of an annulus, as shown in Figure 2. In the unfolded diagram, arcs $\overparen{A D}$ and $\overparen{E F}$ correspond to the lower and upper bases of the frustum, respectively, and point $A(D)$ is the starting point of the mouse.

Assume the inner radius of the annulus is $r$, the outer radius is $R$, and the central angle is $\alpha$. Clearly,
$$
\begin{array}{l}
\alpha r=2 \pi \times 3, \alpha R=2 \pi \times 6, \\
(6-3)^{2}+(\sqrt{7})^{2}=(R-r)^{2} .
\end{array}
$$

Solving these equations, we get $r=4, R=8, \alpha=\frac{3 \pi}{2}$.
Draw $A B$ tangent to arc $\overparen{E F}$ at point $B$, and $D C$ tangent to arc $\overparen{E F}$ at point $C$. From the diagram, we know that $A \rightarrow B \rightarrow C \rightarrow D$ is the shortest path, where the segment $B \rightarrow C$ lies on the arc.

Calculating, we get $A B=C D=\sqrt{R^{2}-r^{2}}=4 \sqrt{3}$, and it is easy to see that $\angle A O B=\angle C O D=\frac{\pi}{3}$, $\angle B O C=\alpha-\angle A O B-\angle C O D=\frac{5 \pi}{6}$. Therefore, the length of arc $\overparen{B C}$ is $\frac{5 \pi}{6} \times 4=\frac{10 \pi}{3}$. Thus, the length of the shortest path is $4 \sqrt{3}+\frac{10 \pi}{3}+4 \sqrt{3}=8 \sqrt{3}+\frac{10 \pi}{3}$ (meters).