= Solution
Work in the centre-of-mass frame and define $I=\sum_i m|\mathbf r_i|^2$, $T=\tfrac12\sum_i m|\mathbf v_i|^2$. Differentiation gives
$$
\frac12\ddot I=2T+\sum_i\mathbf r_i\cdot\mathbf F_i.
$$
For each gravitational pair, the two contributions combine to
$$
(\mathbf r_i-\mathbf r_j)\cdot\mathbf F_{ij}
=-\frac{Gm^2}{|\mathbf r_i-\mathbf r_j|}.
$$
Summing pairs therefore gives the exact scalar <virial theorem>
$$
\boxed{\frac12\ddot I=2T+W,\qquad
W=-\sum_{i<j}\frac{Gm^2}{r_{ij}}.}
$$
In a stationary equilibrium, or a bounded long-time average with vanishing average $\ddot I$, $2T=-W$. With $M=Nm$, the stated <gravitational radius> gives $W=-GM^2/R_g$, hence
$$
\boxed{\langle v_{\rm eq}^2\rangle=\frac{2T}M=\frac{GNm}{R_g}.}
$$
The rms speed itself is $[GNm/R_g]^{1/2}$. This is a three-dimensional mean-square speed; for an isotropic system the one-component <velocity dispersion> is smaller by a factor of three in its square. No assumption about a uniform spatial density was used.
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