Assemble a uniform-density sphere from shells. Since $m(r)=M(r/R)^3$ and $dm=3Mr^2dr/R^3$,
$$
U=-\int_0^R\frac{Gm(r)}r\,dm
=-\frac{3GM^2}{R^6}\int_0^Rr^4dr
=\boxed{-\frac{3GM^2}{5R}}.
$$
Hydrostatic equilibrium gives its central pressure
$$
\boxed{P_c=\frac{3GM^2}{8\pi R^4}}.
$$
Thus, relative to the same uniform-density estimate for Earth,
$$
\frac{P_{c,p}}{P_{c,\oplus}}
=\left(\frac{M_p}{M_\oplus}\right)^2
\left(\frac{R_p}{R_\oplus}\right)^{-4}.
$$
Using $(M,R)_{\rm N}\simeq(17.1,3.88)$ gives $P_{c,\rm N}/P_{c,\oplus}\simeq1.3$, while $(M,R)_{\rm J}\simeq(318,11.2)$ gives $P_{c,\rm J}/P_{c,\oplus}\simeq6.4$. Real central pressures differ because all three planets are compressible and compositionally stratified.


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 315 / 3 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-315/3/b/solution