= Solution
Work in an inertial <centre of mass> frame and write $I=\sum_i m_i|\mathbf r_i|^2$. Differentiation and <Newton's second law> give
$$
\frac12\ddot I=2T+\sum_i\mathbf r_i\cdot\mathbf F_i.
$$
The pairwise Newtonian <gravitational potential energy> is
$$
\Omega=-\sum_{i<j}\frac{Gm_im_j}{|\mathbf r_i-\mathbf r_j|}.
$$
For each pair, combining the two force contributions gives $-Gm_im_j/|\mathbf r_i-\mathbf r_j|$. Therefore $\sum_i\mathbf r_i\cdot\mathbf F_i=\Omega$, and
$$
\frac12\ddot I=2T+\Omega.
$$
For a stationary configuration, or after a long-time average of a bounded system whose endpoint term vanishes, the <virial theorem> gives
$$
\boxed{2\langle T\rangle+\langle\Omega\rangle=0.}
$$
The brackets may be suppressed when $T,\Omega$ denote equilibrium averages. A collapsing cloud or an eccentric <Kepler orbit> need not obey $2T+\Omega=0$ instantaneously. Confining surface pressure, magnetic stress or other external forces add their own virial terms; the displayed result concerns an isolated gravitational particle system.
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