Kepler's third law and conservation of angular momentum: apparent fallacy

It doesn't contradict Kepler's third law.

The equation $$ L = m\omega r^2 = const. $$ does not mean that $L$ is the same for all bodies orbiting the center. It means that $L = m\omega r^2$ of a single orbiting body is constant in time.

We can express $\omega$ as $$ \omega = \frac{\ell}{r^2}. $$ where $\ell = \omega r^2$ is angular momentum per unit mass, equal to twice the area speed (the area swept by the radius vector per unit time). This quantity, just as $L$, is constant in time. But it is not the same for all orbits. $\ell$ can (and does) depend on the radius $r$.

For circular orbits, where $\omega$ and $r$ are constant in time, we can express the orbital period as

$$ T = \frac{2\pi}{\omega} = 2\pi \frac{1}{\ell}r^2. $$

But this equation does not mean that $T$ is a quadratic function of the radius $r$. The expression depends also on $\ell$, which can (and does) depend on $r$. We can express this quantity as $$ \ell = 2\frac{\pi r^2}{T}. $$ In this, we can see that $\ell$ is indeed twice the area speed.

To find how it depends on the radius $r$, we'll use Kepler's third law.

Kepler's third law states that for two orbiting bodies 1 and 2, the ratio of their periods squared is the same as the ratio of their semimajor axes cubed. For circular orbits,

$$ \frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}. $$ For a single body, we have

$$ GM T^2 = r^3, $$ where $GM$ is some constant characterizing the central field, same for all orbits.

We can write

$$ \frac{r^4}{T^2} = GM r, $$ so $$ \frac{r^2}{T} = \sqrt{GM r~}. $$

We get the area speed

$$ \ell = 2\pi \sqrt{GM r~}, $$ confirming that $\ell$ depends on $r$.