Chapter 25, Problem 41Q

(a)

To determine

The critical density of the matter in the universe $\left({\rho }_c\right)$, if the value of Hubble constant is $50\text{km/s/Mpc}$.

Answer to Problem 41Q

Solution:

$4.7\times {10}^{-27}{\text{ kg/m}}^3$

Explanation of Solution

Given data:

The Hubble constant is $50\text{km/s/Mpc}$.

Formula used:

The relation between the Hubble constant and the critical density is:

${\rho }_c=\frac{3{\left(H_0\right)}^2}{8\pi G}$

Here, $H_0$ is the Hubble constant and G is universal gravitational constant having a value of $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Explanation:

The conversion process of a unit of Hubble constant from $\text{km/s/Mpc}$ to ${\text{s}}^{\text{-1}}$ is:

$\begin{array}{c}{H}_0=50\text{ km/s/Mpc}\\ =5\times {\text{10}}^4\frac{\text{m}}{\text{s}\cdot \text{Mpc}}\left(\frac{1\text{ Mpc}}{3.09\times {10}^{22}\text{ m}}\right)\\ =1.62\times {10}^{-18}{\text{ s}}^{-1}\end{array}$

Recall the relation between the Hubble constant and the critical density.

${\rho }_c=\frac{3{\left(H_0\right)}^2}{8\pi G}$

Substitute $1.62\times {10}^{-18}{\text{ s}}^{-1}$ for $H_0$ and $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for G.

$\begin{array}{c}{\rho }_c=\frac{3{\left(1.62\times {10}^{-18}{\text{ s}}^{-1}\right)}^2}{8\pi \left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)}\\ =\frac{7.87\times {10}^{-36}}{1.67\times {10}^{-9}}\\ =4.7\times {10}^{-27}{\text{ kg/m}}^3\end{array}$

Conclusion:

Therefore, the critical density of matter is $4.7\times {10}^{-27}{\text{ kg/m}}^3$, if the value of Hubble constant is $50\text{km/s/Mpc}$.

(b)

To determine

The critical density of the matter in the universe $\left({\rho }_c\right)$ if the value of Hubble constant is $100\text{km/s/Mpc}$.

Answer to Problem 41Q

Solution:

$1.9\times {10}^{-26}{\text{ kg/m}}^3$

Explanation of Solution

Given data:

Hubble constant is $100\text{km/s/Mpc}$.

Formula used:

The relation between Hubble constant and the critical density is:

${\rho }_c=\frac{3{\left(H_0\right)}^2}{8\pi G}$

Here, $H_0$ is the Hubble constant and G is universal gravitational constant having a value of $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$.

Explanation:

The conversion process of unit of Hubble constant from $\text{km/s/Mpc}$ to ${\text{s}}^{\text{-1}}$ is:

$\begin{array}{c}{H}_0=100\text{ km/s/Mpc}\\ {\text{=10}}^5\frac{\text{m}}{\text{s}\cdot \text{Mpc}}\left(\frac{1\text{ Mpc}}{3.09\times {10}^{22}\text{ m}}\right)\\ =3.24\times {10}^{-18}{\text{ s}}^{-1}\end{array}$

Recall the relation between the Hubble constant and the critical density.

${{\rho }^{\prime }}_c=\frac{3{\left({H^{\prime }}_0\right)}^2}{8\pi G}$

Here, ${H^{\prime }}_0$ is the Hubble constant $\left(100\text{km/s/Mpc}\right)$ and ${{\rho }^{\prime }}_c$ is the critical density for ${H^{\prime }}_0$.

Substitute $3.24\times {10}^{-18}{\text{ s}}^{-1}$ for ${H^{\prime }}_0$ and $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for G.

$\begin{array}{c}{{\rho }^{\prime }}_c=\frac{3{\left(3.24\times {10}^{-18}{\text{ s}}^{-1}\right)}^2}{8\pi \left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)}\\ =\frac{3.14\times {10}^{-35}}{1.67\times {10}^{-9}}\\ =1.9\times {10}^{-26}{\text{ kg/m}}^3\end{array}$

Conclusion:

Therefore, the critical density of matter is $1.9\times {10}^{-26}{\text{ kg/m}}^3$, if the value of Hubble constant is $100\text{km/s/Mpc}$.