= Solution
By the shell theorem, only the mass inside the particle's radius contributes:
$$
M(r)=\frac{4\pi}{3}\rho r^3.
$$
The conserved Newtonian energy per unit particle mass is
$$
\frac12\dot r^2-\frac{GM}{r}=E.
$$
Writing $r=a(t)r_0$ and using conservation of enclosed mass, so that $\rho a^3$ is constant, gives
$$
\boxed{\left(\frac{\dot a}{a}\right)^2
=\frac{8\pi G}{3}\rho-\frac{k}{a^2}},
\qquad
k=-\frac{2E}{r_0^2}.
$$
This is the first <Friedmann equation> without a cosmological constant.
The derivation has two central defects. It uses instantaneous Newtonian gravity and omits the relativistic gravitational effect of pressure, so it is not valid for general cosmological matter. It also starts from a finite ball with a preferred centre and boundary, contrary to exact homogeneity and isotropy; the assumption $r\ll L$ only hides that defect locally.
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