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Build the spherical system by bringing successive thin spherical shells in from infinity. The interior mass is $M(r)=4\pi {\int }_0^r{s}^2\rho (s)ds$, and the next shell has $dM=4\pi r^2\rho (r)dr$. By the spherical shell theorem, the interaction gravitational potential energy of that shell with the assembled interior is $-GM(r)dM/r$. Each pair of mass elements is counted exactly once in this construction. Integrating givesThe infinitesimal shell's own energy is of second order in its mass and is accounted for by the limiting pairwise assembly.

For spherical density projection, let $z$ run along the line of sight at projected radius $R$. Spherical symmetry givesOn $z>0$, use $r=\sqrt{R^2+z^2}$, so $dz=rdr/\sqrt{r^2-R^2}$. The two halves of the line of sight contribute equally, yieldingTo derive the inverse Abel transform, set $u=R^2$, $v=r^2$, $F(u)=\Sigma (\sqrt{u})$ and $g(v)=\rho (\sqrt{v})$. The projection becomes $F(u)={\int }_u^{\infty }g(v)(v-u{)}^{-1/2}dv$. Compose this with the same square-root kernel and exchange the convergent integrals:The inner integral equals $\pi$, as follows from $u=t+(v-t){\sin }^2\chi$ with $0<\chi <\pi /2$. Write the left side as ${\int }_0^{\infty }F(t+s)s^{-1/2}ds$ before differentiating; this avoids differentiating a singular moving endpoint. Under sufficient regularity and decay,Finally put $t=r^2,u=R^2$. Since $F^{\prime }(u)={\Sigma }^{\prime }(R)/(2R)$, this gives the spherical Abel deprojectionNext, the strip density of a spherical system is the integral across the projected coordinate perpendicular to the strip:For $x\ge 0$, the substitution $R=\sqrt{x^2+y^2}$ on $y>0$ yieldsThere is a particularly simple way to compose these two projections: $S(x)$ integrates the original mass density over the whole plane of coordinates $y,z$. Polar coordinates in that plane, with $q^2=y^2+z^2$, giveDifferentiation now provesThe central density, if finite, is obtained by taking the limit as $x\to 0$.

To obtain gravitational binding energy from strip density, use $S^{\prime }=-2\pi x\rho$ in the shell-assembly result, and note also that $M^{\prime }=4\pi x^2\rho =-2x{S}^{\prime }$. Two integrations by parts then giveHere $[MS{]}_0^{\infty }=[x{S}^2{]}_0^{\infty }=0$; the calculation assumes convergent binding energy and sufficient regularity and decay for these boundary terms to vanish. ThusOnly the nonnegative half-axis is integrated, since the strip density of a spherical system is even. Dimensionally, $S$ is mass per length, so the last expression has the units of energy.