Answer
The angular momentum depends on $~~\sqrt{r}$
Work Step by Step
We can find an expression for the speed:
$\frac{GMm}{r^2} = \frac{mv^2}{r}$
$v^2 = \frac{GM}{r}$
$v = \sqrt{\frac{GM}{r}}$
We can find an expression for the angular momentum:
$L = mvr$
$L = (m)(\sqrt{\frac{GM}{r}})(r)$
$L = m~\sqrt{GMr}$
The angular momentum depends on $~~\sqrt{r}$
