7. Given that $A$ and $B$ are two points on the ellipse $\Gamma$ with eccentricity $e$:
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

$F_{1}$ and $F_{2}$ are the two foci of the ellipse $\Gamma$, $O$ is the origin, and $O A \perp O B, \overrightarrow{A F_{1}} \cdot \overrightarrow{A F_{2}}+\overrightarrow{B F_{1}} \cdot \overrightarrow{B F_{2}}=0$. Then the range of $e$ is . $\qquad$
7. $\sqrt{2-\sqrt{2}} \leqslant e<1$.

Let $O A=r_{1}, O B=r_{2}, O A \perp O B$. Then
$$
\begin{array}{l}
\frac{1}{r_{1}^{2}}+\frac{1}{r_{2}^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}} . \\
\text { By } \overrightarrow{A F_{1}} \cdot \overrightarrow{A F_{2}}+\overrightarrow{B F_{1}} \cdot \overrightarrow{B F_{2}} \\
=O A^{2}+O B^{2}-2 c^{2} \\
=r_{1}^{2}+r_{2}^{2}-2 c^{2}=0 \\
\Rightarrow r_{1}^{2}+r_{2}^{2}=2 c^{2} . \\
\text { Hence }\left(\frac{1}{r_{1}^{2}}+\frac{1}{r_{2}^{2}}\right)\left(r_{1}^{2}+r_{2}^{2}\right) \\
=\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right)\left(2 c^{2}\right) \geqslant 4 \\
\Rightarrow\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}\right) c^{2}=e^{2}+\frac{e^{2}}{1-e^{2}} \geqslant 2 \\
\Rightarrow \sqrt{2-\sqrt{2}} \leqslant e<1 .
\end{array}
$$