= Solution
Differentiate the Friedmann equation and use
$$
\dot\rho=-3H(\rho+P),
\qquad
\dot H=\frac{\ddot a}{a}-H^2.
$$
After cancelling the curvature term with the original Friedmann equation, one obtains the <Friedmann acceleration equation>
$$
\boxed{
\frac{\ddot a}{a}
=-\frac{4\pi G}{3c^2}(\rho+3P)
+\frac{\Lambda c^2}{3}}.
$$
The Newtonian derivation does not explain why <pressure> gravitates through the $3P$ term; that input enters only through a relativistic-looking continuity law. It also treats space and time as Euclidean and absolute, so it cannot supply the spacetime meaning of curvature $k$, horizons, or the global geometry of the cosmological model.
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