= Solution
To distinguish the moment quantity from the <gravitational constant>, write it as $\mathcal G=\sum_i\boldsymbol p_i\cdot\boldsymbol r_i$. Newton's equations give $\dot{\boldsymbol p}_i=\boldsymbol F_i$ and $\dot{\boldsymbol r}_i=\boldsymbol p_i/m_i$. Differentiating proves
$$
\boxed{\frac{d\mathcal G}{dt}=\sum_i\frac{p_i^2}{m_i}+\sum_i\boldsymbol F_i\cdot\boldsymbol r_i
=2T+\sum_i\boldsymbol F_i\cdot\boldsymbol r_i.}
$$
Average from $0$ to $\tau$, divide by $\tau$, and apply the supplied long-time boundary condition. The result is
$$
0=2\langle T\rangle+\left\langle\sum_i\boldsymbol F_i\cdot\boldsymbol r_i\right\rangle,
\qquad\boxed{2\langle T\rangle=-\left\langle\sum_i\boldsymbol F_i\cdot\boldsymbol r_i\right\rangle.}
$$
The PDF's intermediate formula omits this minus sign. It cannot follow from the preceding derivative identity: an attractive inverse-square circular orbit already has $\boldsymbol F\cdot\boldsymbol r<0$ and $T>0$, contradicting the printed positive-sign version. The later homogeneous-potential relation is consistent with the corrected sign.
Interpret $W$ as the total pair <potential energy>, with each pair counted once, for example $W=\sum_{i<j}C|\boldsymbol r_i-\boldsymbol r_j|^{p+1}$. The printed pair sum leaves one particle index unspecified; an ordered-pair convention with the corresponding factor absorbed into $C$ has the same degree of homogeneity. Under a common dilation of all positions,
$$
W(\lambda\boldsymbol r_1,\ldots,\lambda\boldsymbol r_N)=\lambda^{p+1}W(\boldsymbol r_1,\ldots,\boldsymbol r_N).
$$
Differentiating with respect to $\lambda$ at one is the <Euler theorem for homogeneous functions>. Since $\boldsymbol F_i=-\nabla_iW$, it gives
$$
\sum_i\boldsymbol F_i\cdot\boldsymbol r_i=-(p+1)W.
$$
Substitution into the correctly signed averaged identity proves the <virial theorem for a homogeneous pair potential>:
$$
\boxed{2\langle T\rangle-(p+1)\langle W\rangle=0.}
$$
For $p\ne-1$, the conserved total <energy> can be evaluated using these averages:
$$
E_{\rm tot}=\langle T\rangle+\langle W\rangle
=\frac{p+3}{p+1}\langle T\rangle.
$$
Use the kinetic <temperature> convention $\langle T\rangle=\kappa\Theta$, where $\kappa=3Nk_B/2>0$ for three-dimensional thermalized particles, or the same positive coefficient as a kinetic-temperature definition. At fixed particle number and interaction parameters,
$$
\frac{dE_{\rm tot}}{d\Theta}=\kappa\frac{p+3}{p+1}.
$$
Thus the equilibrium <heat capacity> is negative precisely when
$$
\boxed{-3<p<-1,}
$$
subject to existence of the bounded equilibrium sequence assumed by the <virial theorem>. At $p=-3$ the virial energy is zero and this derivative vanishes. At $p=-1$ the written power-law potential is constant, its force vanishes, and the bounded-motion virial condition excludes nonzero kinetic energy; division by $p+1$ is not permissible. The Newtonian case $p=-2$ recovers $E_{\rm tot}=-\langle T\rangle$.
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