#### Answer

$v_0=\sqrt {\dfrac{G M}{r_0}}$
and $\sqrt {\dfrac{G M}{r_0}} \lt v_0 \lt \sqrt {\dfrac{2G M}{r_0}}$
and $v_0=\sqrt {\dfrac{2 G M}{r_0}} $
and $v_0 \gt \sqrt {\dfrac{2 G M}{r_0}} $

#### Work Step by Step

The eccentricity can be written as: $e=\dfrac{r_0^2v_0^2}{G M}-1$
We know that the orbit will be a circle when $e=0$
So, $v_0=\sqrt {\dfrac{G M}{r_0}}$
We know that the orbit will be an ellipse when $ 0 \lt e \lt 1$
So, $\sqrt {\dfrac{G M}{r_0}} \lt v_0 \lt \sqrt {\dfrac{2G M}{r_0}}$
We know that the orbit will be a parabola when $ e=1$
So, $v_0=\sqrt {\dfrac{2 G M}{r_0}} $
We know that the orbit will be a hyperbola when $ e \gt 1$
So, $v_0 \gt \sqrt {\dfrac{2 G M}{r_0}} $