Question 44.7: Critical density of the universe. Use energy conservation an......
Critical density of the universe. Use energy conservation and escape velocity (Section 8-7) to estimate the critical density of the universe.
APPROACH At the critical density, \rho_{\mathrm{c}}, any given galaxy of mass m will just be able to “escape” away from our Galaxy. As we saw in Section 8-7, escape can just occur if the total energy E of the galaxy satisfies
E=K+U=\frac{1}{2} m v^{2}-G \frac{m M}{R}=0
Here R is the distance of that galaxy m from us. We approximate the total mass M that pulls inward on m as the total mass within a sphere of radius R around us (Appendix D). If we assume the density of galaxies is roughly constant, then M=\frac{4}{3} \pi \rho_{\mathrm{c}} R^{3}.
Substituting this M into the equation above, and setting v=H R (Hubble’s law, Eq. 44-4), we obtain
v=H d . (44-4)
\frac{G M}{R}=\frac{1}{2} v^{2}
or
\frac{G\left(\frac{4}{3} \pi \rho_{\mathrm{c}} R^{3}\right)}{R}=\frac{1}{2}(H R)^{2}
We solve for \rho_{\mathrm{c}} :
rho_{\mathrm{c}}=\frac{3 H^{2}}{8 \pi G} \approx \frac{3\left[(22 \mathrm{~km} / \mathrm{s} / \mathrm{Mly})\left(1 \mathrm{Mly} / 10^{19} \mathrm{~km}\right)\right]^{2}}{8(3.14)\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right)} \approx 10^{-26} \mathrm{~kg} / \mathrm{m}^{3} .