1. 1. We factor the constant energy
      $$\begin{eqnarray*} E &=& \frac{1}{2}\ mr^2 (t)\, H^2(t)\ \left(\ 1 - \frac{\rho(... ... \frac{H^2(t)}{\rho_c(t)}\ \left(\ \rho_c(t) - \rho(t)\ \right) \end{eqnarray*}$$
      
      
      
      
      
      and note that
      
      $$\frac{H^2(t)}{\rho_c (t)} = \frac{8\pi G}{3}\ \ .$$
      
      
      is constant in time. We thus have
      
      $$r^2(t)\ \ (\ \rho_c(t) - \rho(t)\ ) = \frac{3E}{4\pi Gm}$$
      
      
      constant.
   2. From part (a), we have $r^2(t)\,\rho_c(t)\ \ (\ 1 - \Omega(t)\ )$ constant, with $\Omega (t) \equiv \rho(t)/\rho_c(t)$. Thus
      
      $$r^2(t_1)\,\rho_c(t_1)\ \ (\ 1 - \Omega(t_1)\ ) = r^2(t_2)\,\rho_c(t_2)\ \ (\ 1 - \Omega(t_2)\ )$$
      
      
      which implies that
      $$\begin{eqnarray*} \frac{1 - \Omega(t_1)}{ 1- \Omega(t_2)} &=& \frac{ r^2(t_2)\,\... ...quad = \quad \left(\ \frac{r(t_1)}{r(t_2)}\ \right)^{n-2} \ \ , \end{eqnarray*}$$
      
      
      
      
      
      using $\rho \propto r^{-n}$.
   3. We neglect the recent shift from a matter-dominated to a $\Lambda$-dominated universe. Thus, from $10^{12}$ s to now ($10^{18}$ s), we use the matter-dominated scaling $n=3$. Thus
      
      $$\frac{1 - \Omega(t_o)}{ 1- \Omega(t_{MD})} = \left(\ \frac{... ...D})}\ \right) =\left(\ \frac{t_o}{t_{MD}}\ \right)^{2/3} \ \ ,$$
      
      
      since $r \propto t^{2/3}$ in the matter-dominated era. This gives
      
      $${ 1- \Omega(t_{MD})} = (\ 1 - \Omega(t_o)\ )\ \left(\ \frac{t_o}{t_{MD}}\ \right)^{-2/3} = 0.9 \ \times\ 10^{-4}\ ,$$
      
      
      using the given values for $\Omega_o$, $t_o,$ and $t_{MD}$.
   4. During the time span given (between $10^{12}$ s and $10^{2}$ s), the universe was radiation-dominated, so we use the above scaling with $n=4$ and $r \propto t^{1/2}$. This gives
      $$\begin{eqnarray*} {1 - \Omega(t_{RD})}& =& (\ 1- \Omega(t_{MD})\ )\ \left(\ \fr... ...\Omega(t_{MD})\ )\ \left(\ \frac{t_{RD}}{t_{MD}}\ \right) \ \ . \end{eqnarray*}$$
      
      
      
      
      
      Using the given values for $1- \Omega_{MD}$, $t_{MD},$ and $t_{RD}$ this gives
      
      $${1 - \Omega(t_{RD})} = 0.9 \ \times\ 10^{-4}\ \cdot 10^{-10} = 0.9 \ \times\ 10^{-14}\ \ .$$