= Solution
For a <spherically symmetric potential>, the <Poisson equation for Newtonian gravity> becomes
$$
\frac1{r^2}\frac{d}{dr}\left(r^2\phi'(r)\right)=4\pi G\rho(r).
$$
Integrate from the center, excluding any additional point mass there. The boundary condition is $r^2\phi'(r)\to0$ as $r\to0$, so
$$
r^2\phi'(r)=4\pi G\int_0^r\rho(s)s^2\,ds=GM(r),\qquad \mathbf g(r)=-\frac{GM(r)}{r^2}\widehat{\mathbf r}.
$$
This derives the enclosed-mass form of the <spherical shell theorem> from the differential equation. Outside the mass, integrating $\phi'=GM/r^2$ gives $\phi=-GM/r+C$; the conventional zero at infinity sets $C=0$.
For the dyadic shells, multiply their constant <mass density> by their spherical volumes. The $n$th shell contributes
$$
\begin{aligned}
M_n&=\frac{4\pi}{3}\rho_0\,2^{n-1}\left(2^{3-3n}-2^{-3n}\right)\\
&=\frac{14\pi\rho_0}{3}\,4^{-n}.
\end{aligned}
$$
Summing the <geometric series> yields
$$
\boxed{M=\sum_{n=1}^\infty M_n=\frac{14\pi\rho_0}{9}.}
$$
At radius $r=2^{-N}$, the enclosed shells are exactly those with $n\geq N+1$; a boundary itself contributes no mass. Thus
$$
M(2^{-N})=\frac{14\pi\rho_0}{3}\sum_{n=N+1}^\infty4^{-n}=\frac{14\pi\rho_0}{9}\,4^{-N}.
$$
Dividing by $r^2=4^{-N}$ gives
$$
\boxed{\mathbf g(2^{-N})=-\frac{14\pi G\rho_0}{9}\widehat{\mathbf r}\qquad(N\geq1).}
$$
These <dyadic spherical shells with inverse-radius density scaling> have the same field magnitude at each shell boundary, although the field varies inside an individual shell. Finally, outside the planet,
$$
\boxed{\phi(r)=-\frac{14\pi G\rho_0}{9r}\qquad(r>1),}
$$
with zero potential at infinity; without that convention an arbitrary additive constant remains.
Back to article page
