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Build the spherical system by adding shells from the centre outwards. The shell theorem says that a shell of mass $dM$ at radius $r$ interacts with the already assembled interior mass through potential $-GM(r)/r$. Thus its contribution is $dW=-GM(r)dM(r)/r$. Each pair is counted when its outer member is added, so no further factor of one half is needed:Integration by parts givesFor an isolated finite-mass system without a singular central point mass, the endpoint term vanishes provided the central cusp has finite binding energy. ConsequentlyThis is the spherical binding-energy mass integral. If a central singularity makes the endpoint or integral divergent, the finite-energy assumption must be checked rather than silently discarding that term.

A thin uniform spherical shell of fixed mass $M_s$ and radius $R$ has Newtonian gravitational potential energyThis follows from the spherical binding-energy mass integral: the enclosed mass is zero inside $R$ and $M_s$ outside. The radial generalized force, positive outward, is consequently $-{\partial }_R{W}_{\mathrm{self}}=-G{M}_s^2/(2R^2)$, taking the derivative at fixed $M_s$. It is an inward resistance to uniform expansion; individual surface forces cancel as vectors by spherical symmetry.

If an interior spherical component has mass $M_c$, the spherical shell theorem adds inward load $G{M}_c{M}_s/R^2$. The total inward load isFor a swept gas fraction $f$ of an original enclosed mass $M$, with $M_s=fM$ and $M_c=(1-f)M$, this becomes $G{M}^{2}f(1-f/2)/R^2$. Omitting shell self-gravity would give $f(1-f)$ instead. In an accreting shell, evaluate the instantaneous load at its current mass; differentiating an assigned $M_s(R)$ in the fixed-mass energy formula would also include the energy of newly acquired material.