7. As shown in the figure, in tetrahedron $ABCD$, face $ABC$ and face $BCD$ form a dihedral angle of $60^{\circ}$, the projection of vertex $A$ on face $BCD$ is the orthocenter $H$ of $\triangle BCD$, and $G$ is the centroid of $\triangle ABC$. If $AH=4, AB=AC$, then $GH=$ $\qquad$
Given, $A H \perp$ plane $B C D$, connect $D H$ intersecting $B C$ at $M$, then $H M \perp B C \Rightarrow A M \perp B C \Rightarrow M$ is the midpoint of $B C$.
In $\triangle A M H$, $A H=4, \angle A M H=60^{\circ}$
$\Rightarrow H M=\frac{4}{\sqrt{3}}, A M=\frac{8}{\sqrt{3}} \Rightarrow G M=\frac{8}{3 \sqrt{3}}$,
Thus, $G H^{2}=\frac{16}{3}+\frac{64}{27}-\frac{32}{9}=\frac{112}{27} \Rightarrow G H=\frac{4 \sqrt{21}}{9}$.