2. As shown in Figure 1, in the obtuse triangle $\triangle ABC$, $BC=1, \angle A = 30^{\circ}, D$ is the midpoint of side $BC$, and $G$ is the centroid of $\triangle ABC$. If $B$ and $C$ are fixed points, when point $A$ moves, the range of the length of segment $GD$ is ( ).
(A) $0 < GD \leqslant \frac{\sqrt{13}}{6}$
(B) $\frac{1}{6} < GD < \frac{\sqrt{13}}{6}$
(C) $0 < GD \leqslant \frac{2+\sqrt{3}}{6}$
(D) $\frac{1}{6} \leqslant GD \leqslant \frac{2+\sqrt{5}}{6}$
2.B.

Since $\angle A=30^{\circ}$, and $\triangle A B C$ is an obtuse triangle, point $A$ is on the arc $\overparen{A_{1} B}$ or $\overparen{A_{2} C}$ (excluding the endpoints) as shown in Figure 4 (where $A_{1} B \perp B C, A_{2} C \perp B C$).

Let $G_{1}$ be the centroid of $\mathrm{Rt} \triangle A_{1} B C$, then
$$
\frac{1}{3} B D < G D < G_{1} D \text{. }
$$

It is easy to know that $A_{1} B=\sqrt{3}$, then
$G_{1} D=\frac{1}{3} A_{1} D=\frac{1}{3} \sqrt{A_{1} B^{2}+B D^{2}}=\frac{\sqrt{13}}{6}$.
Also, $\frac{1}{3} B D=\frac{1}{6}$, so, $\frac{1}{6}<G D<\frac{\sqrt{13}}{6}$.