5. Given the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ intersects with the line $x+y=1$ at points $M, N$, and $O M \perp O N$ (where $O$ is the origin), when the eccentricity of the ellipse $e \in\left[-\frac{3}{3}, \overline{2}\right]$, the range of the length of the major axis of the ellipse is $\qquad$
5. $[\sqrt{5}, \sqrt{6}]$

Analysis: Solving the system $\left\{\begin{array}{l}\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \\ x+y=1\end{array}\right.$, we get $\left(a^{2}+b^{2}\right) x^{2}-2 a^{2} x+a^{2}-a^{2} b^{2}=0$. Let $M\left(x_{1}, y_{1}\right)$, $N\left(x_{2}, y_{2}\right)$, then by $O M \perp O N$ we have $x_{1} x_{2}+y_{1} y_{2}=0$, which simplifies to $2 x_{1} x_{2}-\left(x_{1}+x_{2}\right)+1=0$. Substituting $x_{1}+x_{2}=\frac{2 a^{2}}{a^{2}+b^{2}}, x_{1} x_{2}=\frac{a^{2}-a^{2} b^{2}}{a^{2}+b^{2}}$ gives $\frac{1}{a^{2}}+\frac{1}{b^{2}}=2$. Given $\frac{\sqrt{3}}{3} \leq \frac{c}{a} \leq \frac{\sqrt{2}}{2}$, we have $\frac{1}{3} \leq \frac{b^{2}}{a^{2}} \leq \frac{2}{3}$, thus $\frac{3}{2} \leq a^{2}\left(2-\frac{1}{a^{2}}\right) \leq 2$, solving for $a$ yields $\sqrt{5} \leq 2 a \leq \sqrt{6}$, meaning the range of the major axis length is $[\sqrt{5}, \sqrt{6}]$.