Chapter 23, Problem 55Q

To determine

(a)

The total number of interstellar gas atoms in the Coma Cluster.

Answer to Problem 55Q

The number of interstellar gas atoms in the Coma Cluster is $1.2\times {10}^{70}\text{atoms}$.

Explanation of Solution

Given:

The mass of the Coma Cluster is, $M={10}^{13}{\text{M}}_{\odot }$

Formula used:

The number of atoms in the Coma Cluster is given by,

$\begin{array}{c}N=\frac{M}{M_H}\\ =\frac{{10}^{13}{M}_{\odot }}{M_H}\end{array}$

Calculation:

The mass of Sun is, ${\text{M}}_{\odot }=1.98\times {10}^{30}\text{kg}$

The mass of one atom of hydrogen is, $M_{\text{H}}=1.66\times {10}^{-27}\text{kg}$

The number of atoms in the Coma Cluster is calculated as,

$\begin{array}{c}N=\frac{{10}^{13}{M}_{\odot }}{M_H}\\ =\frac{{10}^{13}\left(1.98\times {10}^{30}\text{kg}\right)}{1.66\times {10}^{-27}\text{kg}}\\ =\frac{1.98\times {10}^{43}\text{kg}}{1.66\times {10}^{-27}\text{kg}}\\ =1.2\times {10}^{70}\end{array}$

Conclusion:

The number of interstellar gas atoms in the Coma Cluster is $1.2\times {10}^{70}\text{atoms}$.

To determine

(b)

The total number of intracluster gas atoms per cubic centimeters in the Coma Cluster.

Answer to Problem 55Q

The number of atoms per centimeter cube is $3.56\times {10}^{-6}\text{atoms}/{\text{cm}}^3$ .

Explanation of Solution

Given:

The radius of the Coma Cluster is, $r=3\text{Mpc}$.

Formula used:

The volume of the cluster is given by,

$V=\frac{4}{3}\pi {\left(r\right)}^3$

The number of atoms per centimeter cube is given by,

$n=\frac{N}{V}$

Calculation:

The volume of cluster is calculated as,

$\begin{array}{c}V=\frac{4}{3}\pi {\left(r\right)}^3\\ =\frac{4}{3}\pi \left(3\text{Mpc}\times \frac{31\times {10}^{23}\text{cm}}{1\text{Mpc}}\right)\\ =\frac{4}{3}\pi {\left(9.3\times {10}^{24}\text{cm}\right)}^3\\ =3.37\times {10}^{75}{\text{cm}}^3\end{array}$

The number of atoms per centimeter cube is calculated as,

$\begin{array}{c}n=\frac{N}{V}\\ =\frac{1.2\times {10}^{70}\text{atoms}}{3.37\times {10}^{75}{\text{cm}}^3}\\ =3.56\times {10}^{-6}\text{atoms}/{\text{cm}}^3\end{array}$

Conclusion:

The number of atoms per centimeter cube is $3.56\times {10}^{-6}\text{atoms}/{\text{cm}}^3$.

To determine

(c)

The comparison between the intracluster gas in the Coma Cluster with the gas in the atmosphere, a typical gas cloud in our own galaxy and the corona of the Sun.

Answer to Problem 55Q

The number of molecules in the Earth’s atmosphere per centimeter cube is $8.4\times {10}^{24}$ times the number of molecules on Coma Cluster.The number of molecules in the Milky Way galaxy per centimeter cube is $1.12\times {10}^8$ times the number of molecules on Coma Cluster. The number of molecules in the corona of Sun per centimeter cube is $2.8\times {10}^{10}$ times the number of molecules on Coma Cluster.

Explanation of Solution

Given:

The number of molecules per centimeter cube in Earth’s atmosphere is, $m_{\text{e}}=3\times {10}^{19}{\text{cm}}^{-3}$.

The number of molecules in the typical gas cloud in the Milky way galaxy is, $m_{\text{g}}=400{\text{cm}}^{-\text{3}}$.

The number of molecules in the corona of the Sun is, $m_{\text{s}}={10}^5{\text{cm}}^{-3}$.

Calculation:

The ratio of the number of molecules in Coma Cluster and the molecules in the Earth’s atmosphere is calculated as,

$\begin{array}{c}\frac{m_{\text{e}}}{n}=\frac{3\times {10}^{19}{\text{cm}}^{-\text{3}}}{3.56\times {10}^{-6}{\text{cm}}^{-3}}\\ m_{\text{e}}=\left(8.4\times {10}^{24}\right)n\end{array}$

The ratio of the number of molecules in Coma Cluster and the molecules in the Milky Way galaxy is calculated as,

$\begin{array}{c}\frac{m_{\text{g}}}{n}=\frac{400{\text{cm}}^{-\text{3}}}{3.56\times {10}^{-6}{\text{cm}}^{-3}}\\ m_{\text{g}}=\left(1.12\times {10}^8\right)n\end{array}$

The ratio of the number of molecules in Coma Cluster and the molecules in the corona of the Sun is calculated as,

$\begin{array}{c}\frac{m_{\text{s}}}{n}=\frac{{10}^5{\text{cm}}^{-\text{3}}}{3.56\times {10}^{-6}{\text{cm}}^{-3}}\\ m_{\text{s}}=\left(2.8\times {10}^{10}\right)n\end{array}$

Conclusion:

The number of molecules in the Earth’s atmosphere per centimeter cube is $8.4\times {10}^{24}$ times the number of molecules on Coma Cluster.The number of molecules in the Milky Way galaxy per centimeter cube is $1.12\times {10}^8$ times the number of molecules on Coma Cluster. The number of molecules in the corona of the Sun per centimeter cube is $2.8\times {10}^{10}$ times the number of molecules on Coma Cluster.