While I was trying to derive Kepler's third law of planetary motion, I tried the gravitational force for the Earth method which goes something like this:

$$\frac{mv^2}{r}=\frac{GMm}{r^2}$$ $$\Rightarrow\frac{4\pi r^2}{T^2}=\frac{GM}{r}$$ From further rearrangement of the equation, we can figure out that $T^2\propto r^3$.

Now, I am trying to derive the same thing using angular momentum (assuming no external force disrupts the Sun-Earth system). So, what I did goes as follows: $$L=mvr$$ $$\Rightarrow L=m\times \frac{2\pi r}{T}\times r$$ Now since, $\frac{dL}{dt}=0$ (angular momentum is conserved), we can treat $L$ like a constant (from what I think). $$\Rightarrow T=\frac{2\pi r^2 m}{L}$$ $$\therefore T\propto r^2$$ But this is not what Kepler's third law of planetary motion says. Now I am sure my angular momentum method has gone wrong somewhere, I am just not sure where. Can someone please point out my error?