7. In tetrahedron $ABCD$, $AB \perp BC$, $CD \perp BC$, $BC=2$, and the angle between the skew lines $AB$ and $CD$ is $60^{\circ}$. If the radius of the circumscribed sphere of tetrahedron $ABCD$ is $\sqrt{5}$, then the maximum volume of tetrahedron $ABCD$ is $\qquad$
Answer: $2 \sqrt{3}$
Analysis: Consider the triangular prism $A B D_{1}-A_{1} C D$, where $B C=2, \angle \triangle S D_{1}=60^{\circ}$. Then, the tetrahedron $A B C D$ satisfies the given conditions, and the circumscribed sphere of the tetrahedron $A B C D$ is the same as that of the triangular prism $A B D_{1}-A_{1} C D$. Let the circumradius of the base triangle $\triangle A B D_{1}$ of the triangular prism be $r$, then $r^{2}+\left(\frac{B C}{2}\right)^{2}=5 \Rightarrow r=2$. In $\triangle A B D_{1}$, by the Law of Sines, $\frac{A D_{1}}{\sin \angle A B D_{1}}=2 r \Rightarrow A D_{1}=2 \sqrt{3}$; by the Law of Cosines, $A D_{1}^{2}=A B^{2}+B D_{1}^{2}-2 A B \cdot B D_{1} \cdot \cos \angle A B D_{1} \Rightarrow A B^{2}+B D_{1}^{2}-A B \cdot B D_{1}=12$, thus by the AM-GM inequality, $A B \cdot B D_{1} \leq 12$, so $V_{A B C D}=\frac{1}{3} V_{A B D_{1}-A_{1} C D}=\frac{1}{3} \cdot \frac{1}{2} A B \cdot B D_{1} \cdot \sin \angle A B D_{1} \cdot B C \leq 2 \sqrt{3}$. When the triangular prism $A B D_{1}-A_{1} C D$ is a regular triangular prism, the equality holds, hence the maximum volume of the tetrahedron $A B C D$ is $2 \sqrt{3}$.