Chapter 19, Problem 50Q

To determine

Thetime period for which the Sun will be a red giant.

Answer to Problem 50Q

The Sun will be a red giant for $35\text{million}\text{years}$.

Explanation of Solution

Given:

The luminosity of the Sun if it becomes a red giant is, $L=2000L_{\odot }$.

Formula Used:

The luminosity of the star is given by,

$L=\frac{E}{t}$ …… (I)

Here, $E$ is the total energy that the star emits and $t$ is the time.

Energy is emitted by the star by converting a fraction of total mass of the star and is given by,

$E=fM{c}^2$

Here, $f$ is the fraction of the total mass $M$ and $c$ is the speed of light.

Thus, the luminosity of the star is given by,

$\begin{array}{l}L=\frac{fM{c}^2}{t}\\ L\propto \frac{M}{t}\end{array}$

The ratio of the luminosity of a star for two different cases is given by,

$\begin{array}{c}\frac{L}{L^{\prime }}=\frac{\left(\frac{M}{t}\right)}{\left(\frac{M^{\prime }}{t^{\prime }}\right)}\\ =\frac{M}{M^{\prime }}\cdot \frac{t^{\prime }}{t}\end{array}$

Calculations:

Consider figure 19-9 of the book “Universe: Stars and Galaxies”; the main sequence lifetime of the Sun is $1.2\times {10}^{10}\text{years}$.

The Sun consumes hydrogen at the rate of $600\text{million}\text{tons}/\text{second}$.

Therefore, the mass of hydrogen consumed by the Sun in $1.2\times {10}^{10}\text{years}$ is calculated as,

$\begin{array}{c}{M}_{\text{1}}=\left(\text{Rate}\text{of}\text{consumption}\right)\left(\text{time}\right)\\ =\left(600\text{million}\text{tons}/\text{second}\right)\left(1.2\times {10}^{10}\text{years}\right)\\ =\left\{\begin{array}{l}\left(600\text{million}\text{tons}/\text{second}\times \frac{{10}^6\text{tons}}{1\text{million}\text{tons}}\times \frac{{10}^3\text{kg}}{1\text{ton}}\right)\\ \left(1.2\times {10}^{10}\text{years}\times \frac{365\text{days}}{1\text{year}}\times \frac{24\text{hr}}{1\text{day}}\times \frac{60\text{min}}{1\text{hr}}\times \frac{60\text{sec}}{1\text{min}}\right)\end{array}\right\}\\ =\left(6\times {10}^{11}\text{kg}/\text{sec}\right)\left(3.78\times {10}^{17}\text{sec}\right)\end{array}$

Solve further,

$M_1=2.27\times {10}^{29}\text{kg}$

The standard value of the mass of the Sun is, $M_{\odot }=1.99\times {10}^{30}\text{kg}$.

Assume the amount of hydrogen in the Sun at present is $74\%$ by mass of the Sun. Therefore, the mass of the amount of hydrogen in the Sun at present is calculated as,

$\begin{array}{c}{M}_{\text{p}}=\left(74\%\right)M_{\odot }\\ =\left(\frac{74}{100}\right)\left(1.99\times {10}^{30}\text{kg}\right)\\ =1.47\times {10}^{30}\text{kg}\end{array}$

The Sun is considered to be $5\text{billion}\text{years}$ old. Therefore, the amount of hydrogen consumed by the Sun in $5\text{billion}\text{years}$ is calculated as,

$\begin{array}{c}{M}_{\text{2}}=\left(\text{Rate}\text{of}\text{consumption}\right)\left(\text{time}\right)\\ =\left(600\text{million}\text{tons}/\text{second}\right)\left(5\text{billion}\text{years}\right)\\ =\left\{\begin{array}{l}\left(600\text{million}\text{tons}/\text{second}\times \frac{{10}^6\text{tons}}{1\text{million}\text{tons}}\times \frac{{10}^3\text{kg}}{1\text{ton}}\right)\\ \left(5\text{billion}\text{years}\times \frac{{10}^9\text{years}}{1\text{billion}\text{years}}\times \frac{365\text{days}}{1\text{year}}\times \frac{24\text{hr}}{1\text{day}}\times \frac{60\text{min}}{1\text{hr}}\times \frac{60\text{sec}}{1\text{min}}\right)\end{array}\right\}\\ =\left(6\times {10}^{11}\text{kg}/\text{sec}\right)\left(1.58\times {10}^{17}\text{sec}\right)\end{array}$

Solve further,

$M_2=9.48\times {10}^{28}\text{kg}$

The initial amount of hydrogen present in the Sun is calculated as,

$\begin{array}{c}{M}_{\text{i}}=M_2+M_{\text{p}}\\ =9.48\times {10}^{28}\text{kg}+1.47\times {10}^{30}\text{kg}\\ =1.56\times {10}^{30}\text{kg}\end{array}$

The mass of hydrogen left after the main sequence lifetime of the Sun is calculated as,

$\begin{array}{c}{M}_{\text{r}}=M_{\text{i}}-M_1\\ =1.56\times {10}^{30}\text{kg}-2.27\times {10}^{29}\text{kg}\\ =\text{1}\text{.33}\times {10}^{30}\text{kg}\end{array}$

The time period for which the Sun will be a red giant is calculated as,

$\begin{array}{c}\frac{t}{t^{\prime }}=\frac{M}{M^{\prime }}\cdot \frac{L^{\prime }}{L}\\ \frac{t}{1.2\times {10}^{10}\text{years}}=\frac{\text{1}\text{.33}\times {10}^{30}\text{kg}}{2.27\times {10}^{29}\text{kg}}\cdot \frac{L_{\odot }}{2000L_{\odot }}\\ t=\left(35.15\times {10}^6\text{years}\times \frac{1\text{million}\text{years}}{{10}^6\text{years}}\right)\\ \approx 35\text{million}\text{years}\end{array}$

Conclusion:

Thus, for $35\text{million}\text{years}$ the Sun will be a red giant.