Let $O$ the circumcenter of $\triangle ABC$ and $G$ its centroid. Extend $OG$ until a point $P$ such that $OG/GP=1/2$ We'll prove that $P$ is the orthocenter $H$

Draw the median $AA'$ where $A'$ is the midpoint of $BC$ Triangles $OGA'$ and $PGA$ are similar, since $GP=2GO$ $AG=2A'G$ and $\angle OGA'=\angle PGA$ Then $\angle OA'G =\angle PGA$ and $OA'\parallel AP$ But $OA'\perp BC$ so $AP\perp BC$ that is, $AP$ is a height of the triangle.

Repeating the same argument for the other medians proves that $P$ lies on the three heights and therefore it must be the orthocenter $H$

The ratio is $OG/GH=1/2$ since we constructed it that way.