Chapter 25, Problem 21Q

To determine

(a)

The age of the universe.

Answer to Problem 21Q

The age of the universe is $\text{19}\text{.542}\text{billion}\cdot \text{year}$.

Explanation of Solution

Given:

The value of the Hubble’s constant $1$ is $H_1=50\text{km}/\text{s}/\text{Mpc}$.

The value of the Hubble’s costant $2$ is $H_2=75\text{km}/\text{s}/\text{Mpc}$.

The value of the Hubble’s costant $3$ is $H_3=100\text{km}/\text{s}/\text{Mpc}$.

Formula used:

The life of the universe is given by,

$T_1=\frac{1}{H_1}$

Calculation:

The life of the universe is calculated by,

$\begin{array}{c}{T}_1=\frac{1}{H_1}\\ =\frac{1}{50\text{km}/\text{s}/\text{Mpc}}\\ =\frac{1}{50\text{km}/\text{s}/\left(\text{Mpc}\right)\left(\frac{{10}^6\text{pc}}{\text{Mpc}}\right)}\\ =\frac{1}{50\text{km}/\text{s}/\left({10}^6\text{pc}\right)}\end{array}$

Solve further,

$\begin{array}{c}{T}_1=\frac{1}{50\text{km}/\text{s}/\left(3.08568\times {10}^{19}\text{km}\right)}\\ =\frac{\left(3.08568\times {10}^{19}\text{km}\right)}{50\text{km}/\text{s}}\\ =6.17136\times {10}^{17}\text{s}\left(\frac{1\text{year}}{\left[\left(3600\right)\left(24\right)\left(365.25\right)\right]\text{s}}\right)\\ =1.9542\times {10}^{10}\text{year}\end{array}$

Solve further,

$\begin{array}{c}{T}_1=1.9542\times {10}^{10}\text{year}\left(\frac{1\text{billion}\cdot \text{year}}{{10}^9\text{year}}\right)\\ =\text{19}\text{.542}\text{billion}\cdot \text{year}\end{array}$

Conclusion:

The age of the universe is $\text{19}\text{.542}\text{billion}\cdot \text{year}$.

To determine

(b)

The age of the universe.

Answer to Problem 21Q

The age of the universe is $\text{13}\text{.004}\text{billion}\cdot \text{year}$.

Explanation of Solution

Formula used:

The life of the universe is given by,

$T_2=\frac{1}{H_2}$

Calculation:

The life of the universe is calculated by,

$\begin{array}{c}{T}_2=\frac{1}{H_2}\\ =\frac{1}{75\text{km}/\text{s}/\text{Mpc}}\\ =\frac{1}{75\text{km}/\text{s}/\left(\text{Mpc}\right)\left(\frac{{10}^6\text{pc}}{\text{Mpc}}\right)}\\ =\frac{1}{75\text{km}/\text{s}/\left({10}^6\text{pc}\right)}\end{array}$

Solve further,

$\begin{array}{c}{T}_2=\frac{1}{75\text{km}/\text{s}/\left({10}^6\text{pc}\right)}\\ =\frac{1}{75\text{km}/\text{s}/\left({10}^6\text{pc}\left(\frac{206265\text{AU}}{1\text{pc}}\right)\right)}\\ =\frac{1}{75\text{km}/\text{s}/\left(206265\times {10}^6\text{AU}\right)}\\ =\frac{1}{75\text{km}/\text{s}/\left(206265\times {10}^6\text{AU}\right)\left(\frac{149597871\text{km}}{\text{1}\text{AU}}\right)}\end{array}$

Solve further,

$\begin{array}{c}{T}_2=\frac{1}{75\text{km}/\text{s}/\left(3.08568\times {10}^{19}\text{km}\right)}\\ =\frac{\left(3.08568\times {10}^{19}\text{km}\right)}{75\text{km}/\text{s}}\\ =4.10666\times {10}^{17}\text{s}\left(\frac{1\text{year}}{\left[\left(3600\right)\left(24\right)\left(365.25\right)\right]\text{s}}\right)\\ =1.30043\times {10}^{10}\text{year}\end{array}$

Solve further,

$\begin{array}{c}{T}_2=1.30043\times {10}^{10}\text{year}\left(\frac{1\text{billion}\cdot \text{year}}{{10}^9\text{year}}\right)\\ =\text{13}\text{.004}\text{billion}\cdot \text{year}\end{array}$

Conclusion:

The age of the universe is $\text{13}\text{.004}\text{billion}\cdot \text{year}$.

To determine

(c)

The age of the universe and how the age of globular cluster is used to place the limit on the maximum value of Hubble’s constant.

Answer to Problem 21Q

The age of the universe is $9.7779\text{billion}\cdot \text{year}$.

Explanation of Solution

Formula used:

The life of the universe is given by,

$T_3=\frac{1}{H_3}$

Calculation:

The life of the universe is calculated by,

$\begin{array}{c}{T}_3=\frac{1}{H_3}\\ =\frac{1}{100\text{km}/\text{s}/\text{Mpc}}\\ =\frac{1}{100\text{km}/\text{s}/\left(\text{Mpc}\right)\left(\frac{{10}^6\text{pc}}{\text{Mpc}}\right)}\\ =\frac{1}{100\text{km}/\text{s}/\left({10}^6\text{pc}\right)}\end{array}$

Solve further,

$\begin{array}{c}{T}_2=\frac{1}{100\text{km}/\text{s}/\left({10}^6\text{pc}\right)}\\ =\frac{1}{100\text{km}/\text{s}/\left({10}^6\text{pc}\left(\frac{206265\text{AU}}{1\text{pc}}\right)\right)}\\ =\frac{1}{100\text{km}/\text{s}/\left(206265\times {10}^6\text{AU}\right)}\\ =\frac{1}{100\text{km}/\text{s}/\left(\frac{149597871\text{km}}{\text{1}\text{AU}}\right)\left(206265\times {10}^6\text{AU}\right)}\end{array}$

Solve further,

$\begin{array}{c}{T}_2=\frac{1}{100\text{km}/\text{s}/\left(3.08568\times {10}^{19}\text{km}\right)}\\ =\frac{\left(3.08568\times {10}^{19}\text{km}\right)}{100\text{km}/\text{s}}\\ =3.08568\times {10}^{17}\text{s}\left(\frac{1\text{year}}{\left[\left(3600\right)\left(24\right)\left(365.25\right)\right]\text{s}}\right)\\ =9.7779\times {10}^9\text{year}\end{array}$

Solve further,

$\begin{array}{c}{T}_2=9.7779\times {10}^9\text{year}\left(\frac{1\text{billion}\cdot \text{year}}{{10}^9\text{year}}\right)\\ =9.7779\text{billion}\cdot \text{year}\end{array}$

Conclusion:

The age of the universe is $9.7779\text{billion}\cdot \text{year}$.

The age of the universe is decreasing as the value of the Hubble’s constant is increasing. Since. the age of the globular cluster cannot be less than the universe, as it is not possible that consist of the universe is older than the universe. So the age of the universe is more than the globular cluster and it apply the maximum limit on the value of the Hubble’s constant.