Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 315 / 3 / a

For a spherical planet, combine the hydrostatic pressure support equation with $dm/dr=4\pi r^2\rho$:Neglect surface pressure. Since $r(m)\le R$, the hydrostatic lower bound on planetary central pressure isFor 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,Uniform mass density attains the sharper bound; direct integration with $m(r)=M{r}^3/R^3$ gives $P_c=3G{M}^2/(8\pi R^4)$. Real planets are centrally concentrated through compression and dense cores, so their central pressure is higher.

Using $M_{\oplus }=5.97\times 10^{24}\mathrm{kg}$, $R_{\oplus }=6.37\times 10^6\mathrm{m}$, $M_J=1.90\times 10^{27}\mathrm{kg}$, and $R_J=7.15\times 10^7\mathrm{m}$, the universal bounds are approximately $5.75\times 10^{10}\mathrm{Pa}$ for Earth and $3.66\times 10^{11}\mathrm{Pa}$ for Jupiter. The uniform-density estimates are respectively $1.72\times 10^{11}\mathrm{Pa}$ and $1.10\times 10^{12}\mathrm{Pa}$, or $1.72$ and $11.0$ megabars. Either comparison gives a Jupiter minimum about 6.4 times the Earth minimum.