Will the universe continue to expand forever? To attack this question, assume that the theory of dark energy is in error and that the recessional speed$\mathit{v}$ of a galaxy a distance$\mathit{r}$ from us is determined only by the gravitational interaction of the matter that lies inside a sphere of radius$\mathit{r}$ centered on us. If the total mass inside this sphere in$\mathit{M}$,the escape speed ${\mathit{v}}_{\mathbf{e}}$from the sphere is ${\mathit{v}}_{\mathbf{e}}\mathbf{=}\sqrt{\mathbf{2}\mathbf{G}\mathbf{M}\mathbf{/}\mathbf{r}}$(Eq.13.28).(a) Show that to prevent unlimited expansion, the average density inside the sphere must be at least equal to

.$\mathit{\rho }\mathbf{=}\frac{\mathbf{3}{\mathbf{H}}^{\mathbf{2}}}{\mathbf{8}\mathbf{\pi }\mathbf{G}}$

(b) Evaluate this “critical density” numerically; express your answer in terms of hydrogen atoms per cubic meter. Measurements of the actual density are difficult and are complicated by the presence of dark matter

Assume that the theory of dark energy is in error and that the recessional speed$v$ of a galaxy a distance$r$ from us is determined only by the gravitational interaction of the matters that lies inside a sphere of radius $r$centered on us.

The total mass inside the sphere in,$M$the escape speed from the sphere is .

$v_e=\sqrt{2GM/r}$

a)Give formula of the escape velocity.

${\mathit{v}}_{\mathbf{e}}\mathbf{=}\sqrt{\mathbf{2}\mathbf{G}\mathbf{M}\mathbf{/}\mathbf{r}}$…… (1)

b)Give the formula of density.

$\mathit{\rho }\mathbf{=}\frac{\mathbf{3}{\mathbf{H}}^{\mathbf{2}}}{\mathbf{8}\mathbf{\pi }\mathbf{G}}$ ……. (2)

(a)

Compare the speed of the Sphere and the escape speed as follows,

$\begin{array}{l}v\left(r\right)=Hr\le v_e=\sqrt{2GM/r}\\ Hr\le \sqrt{2GM/r}\end{array}$

Square on both sides,

$\begin{array}{l}{\left(Hr\right)}^2\le {\left(\sqrt{2GM/r}\right)}^2\\ H^2{r}^2\le 2GM/r\\ H^2\le {2GM/r}^3\end{array}$

$M/r^3\ge H^2/2G$

Thus, solve further as:

$\rho =\frac{M}{4\pi r^2/3}$

$\frac{3M}{4\pi r^3}\ge \frac{3H^2}{8\pi G}$...... (3)

Therefore, From the Equation (3) it has been shown that, to prevent unlimited expansion, the average density $\rho$inside the sphere must be at least equal to .

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

b)

Consider that the density is being expressed in$\text{H-atoms}/{\text{m}}^{\text{3}}$ is equivalent as follows:

${\rho }_0=\frac{mH}{m^3}=1.67\times {10}^{-27}\text{kg}/{\text{m}}^{\text{3}}$

The critical density is calculated as follows:

$\begin{array}{l}\rho =\frac{3H^2}{8\pi G{\rho }_0}(H-atoms/m^3)\\ \rho =\frac{3{(0.0218\text{m}/\text{s.ly})}^2{(1.00ly/9.460\times {10}^{15}\text{m})}^2\left({\text{H-atoms/m}}^{\text{3}}\right)}{8\pi (6.67\times {10}^{-11}{\text{m}}^{\text{3}}/{\text{kg.s}}^{\text{2}})(1.67\times {10}^{-27}\text{kg}/{\text{m}}^{\text{3}})}\end{array}$

Therefore, the critical density is.$\rho =5.7{\text{H-atoms/m}}^{\text{3}}$