= Solution
For <inside-out collapse of a singular isothermal sphere>, the expansion front is at $r_f=a_Tt$. The original <mass> swept by this front is
$$
M_{\rm swept}=\frac{2a_T^2r_f}{G}=\frac{2a_T^3t}{G}.
$$
Thus the only time-independent accretion scale is $a_T^3/G$. The front-sweeping estimate has coefficient two, but <gas> takes time to fall from the front to the object; it is not an exact central accretion coefficient. Write
$$
\boxed{\dot M_\star=m_0\frac{a_T^3}{G}\propto\left(\frac{k_BT}{\mu m_H}\right)^{3/2}\propto T^{3/2},}
$$
for fixed mean particle <mass> $\mu m_H$. Determining the dimensionless coefficient $m_0$ requires the collapse solution; it is not fixed by front propagation alone and is not needed for the requested <temperature> scaling.<Self-Similar Collapse of Isothermal Spheres and Star Formation by Frank H. Shu>[1977 expansion-wave solution: central accretion and the free-falling envelope]{ref}
Well inside the front, where the point <mass> dominates and infall is approximately pressureless and locally steady, $|v_r|\simeq\sqrt{2GM_\star/r}$. Conserving the radial <mass> flux gives the <free-fall density profile around a growing point mass>:
$$
\boxed{\rho(r,t)=\frac{\dot M_\star}{4\pi r^2|v_r|}\simeq\frac{\dot M_\star}{4\pi\sqrt{2GM_\star(t)}}r^{-3/2}.}
$$
The inner slope is $-3/2$, versus $-2$ in the undisturbed outer cloud. For constant $\dot M_\star$ and negligible initial object <mass>, $M_\star\propto t$ and the inner normalization varies as $t^{-1/2}$. These are inner asymptotes, not an assertion of an abrupt exact power-law join at the front.
\Image[/past-exam-of-the-formation-of-structure-in-the-universe-course-of-the-university-of-cambridge/2015/paper-1-inside-out-collapse.png]
{title=An inside-out collapse has a freely falling inner density slope distinct from the undisturbed isothermal envelope}
{source=paper-1-inside-out-collapse.py}
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