# How to derive Kepler's third law of planetary motion using angular momentum of Earth around the Sun?

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?

Clearly, $$L$$ varies with $$r$$ somehow. So, let's work from first principles; in fact, let's not restrict ourselves to circular orbits. The Binet equation proves closed orbits are elliptical with semi-latus rectum $$\ell=\frac{h^2}{G\mu}$$, with $$h$$ the orbiting body's specific angular momentum and $$\mu$$ the orbiting and central bodies' reduced mass. This implies $$L\propto h$$ is proportional to lengths' square roots. Indeed, $$L\propto r^{1/2}$$ is just what you need to fix your problem.
As a bonus, I'll give the full proof of Kepler's third law. Area is swiped out at rate $$\frac12r^2\dot{\theta}=\frac12h$$. Let $$a,\,b$$ denote the major and minor semi-axes, so the area is $$\pi ab$$, and the orbital period is $$T=\frac{2\pi ab}{h}$$. If $$e$$ is the orbit's eccentricity then$$a=\frac{\ell}{1-a^2},\,b=\frac{\ell}{\sqrt{1-a^2}}\implies b^2=a\ell\implies T^2=\frac{4\pi a^2b^2}{G\mu\ell}=\frac{4\pi a^3}{G\mu}.$$
The first derivation mentioned in the question is valid. The starting formula, equating the expression of the modulus of acceleration in a uniform circular motion to the acceleration due to Newton's force, is valid for all the circular motions, each corresponding to a different pair $$r$$,$$T$$.
The second cannot work. The angular momentum is a constant in a circular motion of radius $$r$$ and period $$T$$, but it is a different constant in a circular motion of radius $$r'$$ and period $$T'$$. Therefore, it is not possible to use the conservation of angular momentum to derive a relation between $$r$$ and $$T$$.
Said in another way, the third Kepler's law is not a consequence of the angular momentum conservation, valid for all the central forces, but depends on the $$1/r^2$$ character of the force law.