7. As shown in Figure 2, given points $A, B, C$ on the ellipse
$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)
$$

$A B$ passes through the origin $O, A C$ passes through the right focus $F$. If
$$
B F \perp A C \text {, and } \overrightarrow{A F}=2 \overrightarrow{F C} \text {, }
$$

then the eccentricity of the ellipse is $\qquad$
7. $\frac{\sqrt{5}}{3}$.

Let the left focus of the ellipse be $F^{\prime}$.
By symmetry, we know $A F^{\prime} \perp A C$, and $\left|A F^{\prime}\right|=|B F|$.
Let $\left|A F^{\prime}\right|=|B F|=m,|A F|=n$.
Then $|C F|=\frac{n}{2}$.
Thus, $\left\{\begin{array}{l}m+n=2 a, \\ m^{2}+n^{2}=4 c^{2}, \\ m^{2}+\left(n+\frac{n}{2}\right)^{2}=F^{\prime} C^{2}, \\ F^{\prime} C+\frac{n}{2}=2 a\end{array}\right.$
$\Rightarrow m=\frac{4}{\sqrt{5}} c, n=\frac{2}{\sqrt{5}} c$
$\Rightarrow \frac{4}{\sqrt{5}} c+\frac{2}{\sqrt{5}} c=2 a \Rightarrow e=\frac{\sqrt{5}}{3}$.