https://en.wikipedia.org/wiki/Effective_potential
In this wiki link, They write
$$ E = \frac{1}{2} mv^2 + \frac{1}{2} mr^2 \omega^2 - \frac{GMm}{r}$$
Then they sub
$$ L= mr\omega$$
Which turns the energy expression into,
$$ E = \frac{1}{2} mv^2 + \frac{L^2}{2mr^2} - \frac{GMm}{r}$$
Now, at the end wiki writes that this turns a two variable problem into a single variable one but how?
We still have velocity and radius as varying parameters..
In this particular type of problem, the mechanical energy, E, is constant because no external work is done on the system.
What you seem to quote ("They write ...") is not an accurate copy. Rather, they write $$E = \frac{1}{2}m\left(\dot{r}^2+r^2\dot{\phi}^2\right)-\frac{GmM}{r}.$$
That makes all the difference in the world because, when you change your $v$ to $\dot{r}$ in your last equation, you now have a first-order non-linear differential equation in one variable, $r$.
