4. As shown in Figure 5, given $\triangle A B C$ is inscribed in $\odot O$, chord $A F \perp B C$ at point $H, G$ is the midpoint of $B F$. Then $\frac{A C}{O G}=$
4. 2 .

Connect $F O$. Then
$$
\begin{array}{l}
\text { Rt } \triangle A B H \backsim \text { Rt } \triangle O \\
\Rightarrow \frac{A H}{B H}=\frac{O G}{F G} . \\
\text { Also } \triangle A C H \backsim \triangle B F H \\
\Rightarrow \frac{A H}{B H}=\frac{A C}{B F} .
\end{array}
$$

Therefore $A C=2 O G \Rightarrow \frac{A C}{O G}=2$.