Gravitation takes a toll

Since we neglect the interaction between each gas particle, we can modeled the cloud as a particle Free Falling with height ${{r}_{0}}$ , but the acceleration will be varies depends on the distance from the center of cloud. In this case of Free Fall, we shall use Keppler’s 3rd law with eccentricity of the orbit $e=\infty$ (which appears as line). $\frac{{{T}^{2}}}{{{a}^{3}}}=\frac{4{{\pi }^{2}}}{GM}$ Where $a$ is the semi-major axis, in this case $a$ equals to $\frac{{{r}_{0}}}{2}$ $\Rightarrow \frac{{{T}^{2}}}{{{\left( \frac{{{r}_{0}}}{2} \right)}^{3}}}=\frac{4{{\pi }^{2}}}{GM}$ ${{T}^{2}}=\frac{4{{\pi }^{2}}{{\left( \frac{{{r}_{0}}}{2} \right)}^{3}}}{GM}$ ${{T}^{2}}=\frac{4{{\pi }^{2}}r_{0}^{3}}{8GM}$ And mass of the cloud $M={{\rho }_{0}}.V={{\rho }_{0}}\frac{4}{3}\pi r_{0}^{3}$ $\Rightarrow {{T}^{2}}=\frac{4{{\pi }^{2}}r_{0}^{3}}{8G{{\rho }_{0}}\frac{4}{3}\pi r_{0}^{3}}$ $\Rightarrow T=\sqrt{\frac{3\pi }{8G{{\rho }_{0}}}}$ Where $T$ is the period of the particle. Let the time needed for the cloud to collapse entirely be $P$ , where $P$ is given by $P=\frac{T}{2}$ $P=\frac{1}{2}\sqrt{\frac{3\pi }{8G{{\rho }_{0}}}}$ $\Rightarrow P=\sqrt{\frac{3\pi }{32G{{\rho }_{0}}}}$ Now, we can solve for $k$ $kt=\sqrt{\frac{3\pi }{32G{{\rho }_{0}}}}$ $k\sqrt{\frac{\pi }{96G{{\rho }_{0}}}}=\sqrt{\frac{3\pi }{32G{{\rho }_{0}}}}$ $k=\sqrt{\frac{288\pi G{{\rho }_{0}}}{32\pi G{{\rho }_{0}}}}$ $\Rightarrow k=\sqrt{\frac{288}{32}}=\sqrt{9}$ $\therefore k=3$ Yes, 3 is a rational number :)