Chapter 23, Problem 55Q

(a)

To determine

The total number of intracluster cluster gas atoms in the Coma Cluster, if it is given that this gas is made of Hydrogen atom. The mass of the intracluster gas in the Coma Cluster is ${10}^{13}{{\rm M}}_{\odot }$.

Answer to Problem 55Q

Solution:

$1.2\times {10}^{70}$ atoms.

Explanation of Solution

Given data:

The Coma Cluster contains approximately ${10}^{13}{{\rm M}}_{\odot }$ of intracluster gas.

The gas is made up of hydrogen atoms.

Formula used:

The total number of the intracluster gas atoms can be given as:

$n=\frac{\text{mass of the }{10}^{13}{\text{M}}_{\text{e}}\text{ atoms}}{\text{mass of the one hydrogen atom}}$

Explanation:

The mass of hydrogen atom is $1.67\times {10}^{-27}\text{ kg}$ and the mass of $1{{\rm M}}_{\odot }$ is $1.989\times {10}^{30}\text{}\text{kg}$.

Then, the mass of the ${10}^{13}{{\rm M}}_{\odot }$ is,

$\begin{array}{c}\text{Mass of the }{10}^{13}{\text{M}}_{\text{e}}\text{ atoms}={10}^{13}\left(1.989\times {10}^{30}\text{}\text{kg}\right)\\ =1.989\times {10}^{43}\text{ kg}\end{array}$

Recall the expression for the total number of the intracluster gas atoms.

$n=\frac{\text{mass of the }{10}^{13}{\text{M}}_{\text{e}}\text{ atoms}}{\text{mass of the one hydrogen atom}}$

Substitute $1.989\times {10}^{43}\text{ kg}$ for $\text{the mass of the }{10}^{13}{\text{M}}_{\text{e}}\text{ atoms}$ and $1.67\times {10}^{-27}\text{ kg}$ for the $\text{mass of the one hydrogen atom}$.

$\begin{array}{c}n=\frac{1.989\times {10}^{43}\text{ kg}}{1.67\times {10}^{-27}\text{ kg}}\\ =1.2\times {10}^{70}\text{ atoms}\end{array}$

Conclusion:

Therefore, the total number of intracluster cluster gas atoms in the coma cluster is $1.2\times {10}^{70}$ atoms.

(b)

To determine

The number of intracluster gas atoms per cubic centimeter in the Coma Cluster if it is given that the gas fills the cluster uniformly. The Coma Cluster is roughly spherical in shape with a radius of about $3\text{ Mpc}$. The mass of intracluster gas in the Coma Cluster is ${10}^{13}{{\rm M}}_{\odot }$.

Answer to Problem 55Q

Solution:

$3.6\times {10}^{-6}$ atoms.

Explanation of Solution

Given data:

The Coma Cluster contains about ${10}^{13}{{\rm M}}_{\odot }$ of intracluster gas.

The radius of Coma Cluster is $3\text{ Mpc}$.

Formula used:

The volume of the sphere is written as:

$V=\frac{4}{3}\pi r^3$

Here, V is the volume of the sphere and r is the radius of the sphere.

Explanation:

Recall the formula of volume of Coma Cluster, considering it to be a perfect sphere.

$V=\frac{4}{3}\pi r^3$

Substitute $3\text{ Mpc}$ for r.

$\begin{array}{c}V=\frac{4}{3}\pi {\left(3\text{ Mpc}\right)}^3\\ =\frac{4}{3}\pi {\left(3\text{ Mpc}\left(\left(\frac{{10}^6\text{ pc}}{1\text{ Mpc}}\right)\left(\frac{3.086\times {10}^{16}\text{ m}}{1\text{ pc}}\right)\left(\frac{100\text{ cm}}{1\text{ m}}\right)\right)\right)}^3\\ =\frac{4}{3}\pi {\left(9.3\times {10}^{24}\text{ cm}\right)}^3\\ =3.3\times {10}^{75}{\text{ cm}}^3\end{array}$

The density $\left(\rho \right)$ is the atoms per unit volume. So,

$\rho =\frac{\text{number of atoms}}{\text{Volume}}$

Substitute $1.2\times {10}^{70}\text{ atoms}$ for number of atoms (from part (a)) and $3.3\times {10}^{75}{\text{ cm}}^3$ for Volume.

$\begin{array}{c}\rho =\frac{1.2\times {10}^{70}\text{ atoms}}{3.3\times {10}^{75}{\text{ cm}}^3}\\ =3.6\times {10}^{-6}{\text{ atoms/cm}}^{\text{3}}\end{array}$

Conclusion:

Therefore, the atoms per centimeter cube in the Coma Cluster are $3.6\times {10}^{-6}$ atoms.

(c)

To determine

The comparison of the intracluster gas in the Coma Cluster with the gas present in Earth’s atmosphere, which has a temperature of 300 K and $3\times {10}^{19}$ molecules per cubic centimeter. A gas cloud within our own galaxy has a temperature of 50 K and a few hundred molecules per cubic centimeter. The corona of the Sun has a temperature of ${10}^6\text{ K}$ and ${10}^5$ atom per cubic centimeter. The mass of the intracluster gas in the Coma Cluster is ${10}^{13}{{\rm M}}_{\odot }$.

Answer to Problem 55Q

Solution:

The density of Coma Cluster is lower than our atmosphere, a gas cloud, and corona of the Sun.

Explanation of Solution

Given data:

The temperature of our atmosphere is 300 K and the molecules per cubic centimeter are $3\times {10}^{19}$.

The temperature of the corona of the Sun is ${10}^6\text{ K}$ and the molecules per cubic centimeter are ${10}^5$.

The temperature of a typical cloud in our own galaxy is 50 K or less, and the molecules per cubic centimeter are in a few hundred.

Formula used:

The volume of the sphere is written as:

$V=\frac{4}{3}\pi r^3$

Here, V is the volume of the sphere and r is the radius of the sphere.

Explanation:

The density of the Coma Cluster is $3.6\times {10}^{-6}{\text{ atoms/cm}}^{\text{3}}$ (from part (b)).

The temperature of our atmosphere is 300 K and the molecules per cubic centimeter are $3\times {10}^{19}$.

The temperature of corona of the Sun is ${10}^6\text{ K}$ and the molecules per cubic centimeter are ${10}^5$.

The temperature of a typical cloud in our own galaxy is 50 K or less, and the molecules per cubic centimeter are in a few hundred.

So, from the above values, it can be derived that the density of the Coma Cluster is much lower than the density of our atmosphere.

Conclusion:

Hence, the density of the Coma Cluster is lower than our atmosphere.