5. Let $M$ and $N$ be the points of intersection of the left and right directrices of the ellipse $G$ with the major axis. If there exists a point $P$ on the ellipse $G$ such that $\angle M P N=120^{\circ}$, then the range of the eccentricity of the ellipse $G$ is $\qquad$
5. $\left(0, \frac{\sqrt{3}}{3}\right]$.

Let the ellipse $G: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
Then $M\left(-\frac{a^{2}}{c}, 0\right), N\left(\frac{a^{2}}{c}, 0\right)$.
Also, $P\left(x_{0}, y_{0}\right)\left(y_{0}>0\right), \angle P M N=\alpha$, $\angle P N M=\beta$, then
$\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1$,
$\tan \alpha=\frac{y_{0}}{\frac{a^{2}}{c}+x_{0}}$,
$\tan \beta=\frac{y_{0}}{\frac{a^{2}}{c}-x_{0}}$.
And $\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=\tan 60^{\circ}=\sqrt{3}$,

Substituting equations (1), (2), and (3) and simplifying, we get
$$
\frac{2 y_{0} a^{2} b^{2} c}{a^{2} b^{4}+c^{4} y_{0}^{2}}=\sqrt{3} \text {. }
$$

Therefore, $\frac{2 y_{0} a^{2} b^{2} c}{2 \sqrt{a^{2} b^{4} c^{4} y_{0}^{2}}} \geqslant \sqrt{3}$.
Thus, $e=\frac{c}{a} \leqslant \frac{\sqrt{3}}{3}$.