At a mantle radius $R_1\leq r\leq R_p$, the same integration gives
$$
\boxed{P(r)=P_0+G\rho_2\left[
C\left(\frac1r-\frac1{R_p}\right)
+\frac D2(R_p^2-r^2)
\right]}.
$$
Inside the core, $M(r)=4\pi\rho_1r^3/3$, so matching to $P_b$ gives
$$
\boxed{P(r)=P_b+\frac{2\pi G}{3}\rho_1^2(R_1^2-r^2),
\qquad0\leq r\leq R_1}.
$$
The central pressure is therefore
$$
\boxed{P_c=P_b+\frac{2\pi G}{3}\rho_1^2R_1^2}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 315 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-315/2/c/solution