= Solution
\b[False.] The <singular isothermal sphere> is infinite in extent and mass. The isolated finite-system <virial theorem> cannot be applied to its global divergent energies without boundary terms.
A direct <Spherical Jeans equation> counterexample is enough. For isotropic <velocities> with constant one-dimensional dispersion $\sigma^2$, its density slope gives
$$
\frac d{dr}(\rho\sigma^2)=-\frac{2\rho\sigma^2}{r}=\rho\psi'=-\frac{\rho v_0^2}{r}.
$$
Thus $\sigma^2=v_0^2/2$ and the <circular and isotropic speed comparison in a singular isothermal sphere> is
$$
\boxed{\langle v^2\rangle_{\mathrm{isotropic}}=\frac32v_0^2,\qquad v_{\mathrm{rms}}=\sqrt{\frac32}\,v_0.}
$$
This is realized by a positive isotropic distribution $F\propto\exp(E/\sigma^2)$, which gives a <Maxwell-Boltzmann distribution> of <velocities> and $\rho\propto\exp(\psi/\sigma^2)\propto r^{-2}$. <Circular orbits> have instead $\langle v_r^2\rangle=0$ and $\langle v_\theta^2\rangle=\langle v_\phi^2\rangle=v_0^2/2$, so their total mean square speed is $v_0^2$. The isotropic model contains eccentric orbits except for a measure-zero subset.
The <boundary virial term of a singular isothermal sphere> explains the different energy budgets of a finite spherical region. At radius $R$, its radial-pressure surface term is $4\pi R^3\rho(R)\sigma_r^2$. It is nonzero in the isotropic model and zero in the circular-orbit model. Equality of the density therefore does not enforce equality of the <velocity> energy through an incorrectly omitted surface term.
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