= Solution
For an isolated smooth mass distribution with <spherical symmetry>, assemble each shell $dM$ at radius $r$ against the <Newtonian gravitational potential> of its interior mass. The <spherical shell theorem> gives
$$
W=-G\int_0^\infty\frac{M(r)}r\,dM(r)=-\frac G2\int_0^\infty\frac1r\,d[M(r)^2].
$$
Assume finite total mass, finite binding energy and no central point-mass self-energy, so $M(r)^2/r$ vanishes at both endpoints. <Integration by parts> then gives
$$
\boxed{W=-\frac G2\int_0^\infty\frac{M(r)^2}{r^2}\,dr.}
$$
For example, $M(r)\propto r^3$ near a regular center guarantees the central boundary condition, while finite total mass guarantees the one at infinity. A divergent central binding energy would fall outside this formula's finite-energy hypotheses.
For the discrete system use the <centre of mass> frame, let $N$ be the particle number, and write $M_{\rm tot}=Nm$. The printed $M$ in the particle-count factor and the subsequent $n$ must be understood consistently as $N$. Define
$$
K=\frac12\sum_i m|\dot{\mathbf r}_i|^2,\qquad
W=-Gm^2\sum_{i<j}\frac1{r_{ij}},\qquad
I=\sum_i m|\mathbf r_i|^2.
$$
By <Newton's second law>, differentiation gives
$$
\frac12\ddot I=2K+\sum_i\mathbf r_i\cdot\mathbf F_i.
$$
Each pair contributes
$$
\mathbf r_i\cdot\mathbf F_{ij}+\mathbf r_j\cdot\mathbf F_{ji}
=(\mathbf r_i-\mathbf r_j)\cdot\left[-Gm^2\frac{\mathbf r_i-\mathbf r_j}{r_{ij}^3}\right]
=-\frac{Gm^2}{r_{ij}}.
$$
Therefore the exact dynamical <virial theorem> is $\ddot I/2=2K+W$. In a stationary statistical state, or on a bounded-motion time average for which the mean $\ddot I$ vanishes, it becomes
$$
\boxed{2\langle K\rangle+\langle W\rangle=0.}
$$
There is no external pressure, tidal field or boundary work in this isolated point-particle derivation. For a time-averaged system, the following $W$ and <gravitational radius> refer to that same averaged binding energy.
Using $W=-G(Nm)^2/R_g$ and $2K=Nm\langle v_{\rm eq}^2\rangle$, <virial equilibrium> gives
$$
\boxed{\langle v_{\rm eq}^2\rangle=\frac{GNm}{R_g},\qquad v_{{\rm eq},{\rm rms}}=\sqrt{\frac{GNm}{R_g}}.}
$$
This is the three-dimensional mean squared speed about the <centre of mass>. For an isotropic system each one-dimensional <velocity dispersion> squared is one third of it.
Set the <Newtonian gravitational potential> to zero at infinity and exclude each particle's own field. The local <escape velocity> in the instantaneous frozen potential obeys $v_{{\rm esc},i}^2=-2\Phi_i$, where $\Phi_i=-Gm\sum_{j\ne i}r_{ij}^{-1}$. Since every pair occurs twice, $\sum_i m\Phi_i=2W$. Averaging over the identical particles proves the <gravitational rms escape and collapse speeds> relation
$$
\boxed{\langle v_{\rm esc}^2\rangle=-\frac{4W}{Nm}=\frac{4GNm}{R_g}=4\langle v_{\rm eq}^2\rangle,\qquad v_{{\rm esc},{\rm rms}}=2v_{{\rm eq},{\rm rms}}.}
$$
An actual encounter in a changing potential can exchange energy, so this is the usual local escape-speed definition, not a claim about every many-body trajectory.
Finally, dissipation-free collapse from rest at infinite separation has total energy zero. <Conservation of energy> at the specified binding energy therefore gives $K_{\rm coll}=-W$, twice the equilibrium kinetic energy at the same $W$. Hence
$$
\boxed{\langle v_{\rm coll}^2\rangle=\frac{2GNm}{R_g}=2\langle v_{\rm eq}^2\rangle,\qquad v_{{\rm coll},{\rm rms}}=\sqrt2\,v_{{\rm eq},{\rm rms}}.}
$$
This mean squared speed includes coherent infall; a local random <velocity dispersion> after subtracting the infall velocity need not obey that ratio.
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