Question 31.6: The Critical Density of the Universe Estimate the critical m...

Figure 31.20 shows a large section of the Universe with radius R, containing galaxies with a total mass M. Let us apply the isolated system model to an escaping galaxy and the section of the Universe; a galaxy of mass m and speed v at R will just escape to infinity with zero speed if the sum of its kinetic energy and the gravitational potential energy of the system is zero. The Universe may be infinite in extent, but a theorem such as the gravitational form of Gauss’s law implies that only the mass inside the sphere contributes to the gravitational potential energy of the system of the sphere and the galaxy.
Therefore,

$\begin{aligned}&E_{\text {total }}=0=K+U=\frac{1}{2} m v^{2}-\frac{G m M}{R} \\&\frac{1}{2} m v^{2}=\frac{G m \frac{4}{3} \pi R^{3} \rho_{c}}{R}\end{aligned}$

(1)    $v^{2}=\frac{8 \pi G}{3} R^{2} \rho_{c}$

Because the galaxy of mass m obeys the Hubble law, v = HR, (1) becomes

$H^{2}=\frac{8 \pi G}{3} \rho_{c}$

(2)    $\rho_{c}=\frac{3 H^{2}}{8 \pi G}$

Using $H=17 \times 10^{-3} m /( s \cdot ly )$, where 1 ly = $9.46 \times 10^{12}$ km, and G = $6.67 \times 10^{-11} N \cdot m ^{2} / kg ^{2}$ yields the critical density $\rho_{c}=6 \times 10^{-30} g / cm ^{3}$. Because the mass of a hydrogen atom is $1.67 \times 10^{-24}$ g, the value calculated for $\rho_{c}$ corresponds to $3 \times 10^{-6}$ hydrogen atoms per cubic centimeter or 3 atoms per cubic meter.