= Solution
For a spherical planet, combine the <hydrostatic pressure support equation> with $dm/dr=4\pi r^2\rho$:
$$
\frac{dP}{dm}=-\frac{Gm}{4\pi r^4}.
$$
Neglect surface <pressure>. Since $r(m)\le R$, the <hydrostatic lower bound on planetary central pressure> is
$$
\boxed{P_c=\int_0^M\frac{Gm}{4\pi r(m)^4}\,dm\ge\frac{GM^2}{8\pi R^4}.}
$$
For a physically usual <mass density> decreasing outward, the mean interior density exceeds the global mean. Hence $r(m)\le R(m/M)^{1/3}$, giving the sharper minimum within this class,
$$
\boxed{P_c\ge\frac{3GM^2}{8\pi R^4}.}
$$
Uniform <mass density> attains the sharper bound; direct integration with $m(r)=Mr^3/R^3$ gives $P_c=3GM^2/(8\pi R^4)$. Real planets are centrally concentrated through compression and dense cores, so their central <pressure> is higher.
Using $M_\oplus=5.97\times10^{24}\,\mathrm{kg}$, $R_\oplus=6.37\times10^6\,\mathrm m$, $M_J=1.90\times10^{27}\,\mathrm{kg}$, and $R_J=7.15\times10^7\,\mathrm m$, the universal bounds are approximately $5.75\times10^{10}\,\mathrm{Pa}$ for <Earth> and $3.66\times10^{11}\,\mathrm{Pa}$ for <Jupiter>. The uniform-density estimates are respectively $1.72\times10^{11}\,\mathrm{Pa}$ and $1.10\times10^{12}\,\mathrm{Pa}$, or $1.72$ and $11.0$ megabars. Either comparison gives \b[a Jupiter minimum about 6.4 times the Earth minimum].
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