First Friedmann equation

Fix a comoving test galaxy at comoving radius $r_0$ from an arbitrary origin, so its physical distance is $R(t)=a(t)r_0$ and its physical velocity is $\dot R = \dot a\, r_0$. By the Newtonian shell theorem, the gravitational pull it feels comes only from the mass $M$ enclosed within radius $R$, and since no comoving matter crosses the comoving sphere of radius $r_0$ as the universe expands, $M = \dfrac{4}{3}\pi (a r_{0})^{3}\rho$ is constant in time (for pressureless matter). Treat the test galaxy, of mass $m$, as a projectile: its total mechanical energy $\dfrac12 \dot a^{2} r_{0}^{2} - \dfrac{GM m}{a r_{0}} = E$ is conserved.

Substitute $M$: $\dfrac12 \dot a^2 r_0^2 - \dfrac{G m}{a r_0}\cdot\dfrac{4}{3}\pi (ar_0)^3\rho = E$, which simplifies to $\dfrac12\dot a^2 r_0^2 - \dfrac{4}{3}\pi G m \rho\, a^2 r_0^2 = E$. Divide through by $\tfrac12 m a^2 r_0^2$: $\left(\dfrac{\dot a}{a}\right)^2 - \dfrac{8\pi G}{3}\rho = \dfrac{2E}{m a^2 r_0^2}$.

The right-hand side must be independent of the arbitrary radius $r_0$ chosen (the equation has to hold for every comoving observer, and $a(t)$ itself does not depend on $r_0$), so $\dfrac{2E}{mr_0^2}$ is a constant of the test galaxy's orbit that can only depend on the fixed comoving coordinate structure — define this constant by $\dfrac{2E}{m r_{0}^{2}} \equiv -kc^{2}$, giving $H^{2} = \dfrac{8\pi G}{3}\rho - \dfrac{kc^{2}}{a^{2}}$. A separate thermodynamic argument (treating the cosmological constant as a fluid with constant energy density $\rho_\Lambda = \Lambda c^2/(8\pi G)$ and substituting it into this same formula) restores the $\Lambda c^2/3$ term, giving exactly $H^{2} = \left(\dfrac{\dot a}{a}\right)^{2} = \dfrac{8\pi G}{3}\rho - \dfrac{kc^{2}}{a^{2}} + \dfrac{\Lambda c^{2}}{3}$ — and remarkably, the full general-relativistic calculation from Einstein's field equations produces this identical equation, with $k$ now properly identified as the actual sign of spatial curvature rather than just an integration constant.