= Solution
The stated relation follows for an isolated uniform sphere ending at $r_0$, with the <gravitational potential> zero at infinity. Uniform density in a core alone does not fix the <escape velocity>: exterior shells add a constant interior <gravitational potential> while leaving the interior force unchanged.
Under the isolated-sphere assumption, the enclosed <mass> and <circular speed> are
$$
m(r)=\frac{4\pi}3\rho_0r^3,\qquad V_c^2(r)=\frac{Gm(r)}r=\frac{4\pi G\rho_0}3r^2.
$$
The boundary value is $\Phi(r_0)=-Gm(r_0)/r_0$. Integrating $\Phi'=Gm(r)/r^2$ inwards gives
$$
\Phi(r)=-\frac{2\pi G\rho_0}3(3r_0^2-r^2),\qquad v_e^2(r)=-2\Phi(r).
$$
Thus the <escape and circular speeds inside an isolated uniform sphere> satisfy
$$
\boxed{v_e^2(r)=V_c^2(r)\left(\frac{3r_0^2}{r^2}-1\right),\qquad 0<r\leq r_0.}
$$
The finite central limit is $v_e^2(0)=4\pi G\rho_0r_0^2$, although $V_c(0)=0$. Adding an exterior spherical shell of <mass> $m_s$ at radius $s>r_0$ raises $v_e^2$ by $2Gm_s/s$ and is a direct counterexample to deducing the boxed relation from a uniform core alone.
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