Chapter 19, Problem 50Q

To determine

The time period for which the Sun will be a red giant. Assuming that its luminosity is because of fusion of remaining hydrogen at its core and will be 2000 times greater than the present time.

Answer to Problem 50Q

Solution:

35 million years.

Explanation of Solution

Introduction:

Rate of consumption of hydrogen can be defined as the mass of hydrogen combusted per second to produce energy and helium.

$\text{Rate of hydrogen combustion = }\frac{\text{mass of hydrogen combusted}}{\text{time in seconds}}$

The rate of hydrogen consumption is the same hence, the rate for its main sequence lifetime and red giant phase can be equated.

Explanation:

Obtain the value of main sequence lifetime of the Sun from the refered figure 19-9, as $1.2\times {10}^{10}\text{ years}$.

Understand that the amount of hydrogen consumed by the Sun per second is 600 millioin tons of hydrogen, which in kilogram is $6\times {10}^{11}\text{ kg}$.

Since, there are $3.16\times {10}^7\text{ seconds}$ in a year.

Thus, calculate the amount of hydrogen consumed by the Sun in 12 billion years.

$\begin{array}{c}{M}_{\text{12billion}}=6\times {10}^{11}\text{ kg/s}\left(12\times {10}^9\text{ yr}\left(\frac{3.16\times {10}^7\text{ s}}{1\text{ yr}}\right)\right)\\ =2.275\times {10}^{29}\text{ kg}\\ \approx 2.28\times {10}^{29}\text{ kg}\end{array}$

Consider, the standrad value for the mass of the Sun as, $1.99\times {10}^{30}\text{ kg}$.

Understand that the mass of hydrogen present in the present-Sun, assuming 74% of the Sun is hydrogen. Thus,

$\begin{array}{c}{M}_{\text{p}}=\left(1.99\times {10}^{30}\text{ kg}\right)\left(74\%\right)\\ =\left(1.99\times {10}^{30}\text{ kg}\right)\left(\frac{74}{100}\right)\\ =\left(1.99\times {10}^{30}\text{ kg}\right)\left(0.74\right)\\ \text{=1}\text{.47}\times {\text{10}}^{30}\text{ kg}\end{array}$

Here, $M_{\text{p}}$ is the mass of the present-Sun.

Assume, the Sun is 5 billion years old. Then, the hydrogen converted in 5 billion years is given by,

$\begin{array}{c}{M}_{\text{5billion}}=6\times {10}^{11}\text{ kg/s}\left(5\times {10}^9\text{ yr}\left(\frac{3.16\times {10}^7\text{ s}}{1\text{ yr}}\right)\right)\\ =9.48\times {10}^{28}\text{ kg}\\ \approx 0.09\times {10}^{30}\text{ kg}\end{array}$

Thus, the initial mass of hydrogen present in the Sun was,

$\begin{array}{c}{M}_{\text{total}}=M_{\text{p}}+M_{\text{5billion}}\\ =1.47\times {10}^{30}\text{ kg}+0.09\times {10}^{30}\text{ kg}\\ \text{=1}\text{.56}\times {\text{10}}^{30}\text{ kg}\end{array}$

Subtract the mass of hydrogen, consumed throughout the main-sequence life, from the total mass of hydrogen present in the Sun to obtain the mass of hydrogen remaining in the Sun. So, the mass of hydrogen remaining after main-sequence lifetime is over.

$\begin{array}{c}{M}_r=M_{total}-M_{12billion}\\ =1.56\times {10}^{30}\text{ kg}-0.228\times {10}^{30}\text{ kg}\\ =1.332\times {10}^{30}\text{ kg}\end{array}$

To calculate time form a proportion, use the unitary method.

Assume that the rate, by which the Sun consumes hydrogen, is constant throughout its main-sequence lifetime.

Since, the time required to finish $2.28\times {10}^{29}\text{ kg}$ of hydrogen is $12\times {10}^9\text{ years}$.

Thus, the time required to finish $1.332\times {10}^{30}\text{kg}$ of hydrogen will be ‘t’ years, which is calculared as,

$\begin{array}{c}\frac{2.28\times {10}^{29}\text{ kg}}{12\times {10}^9\text{ years}}=\frac{1.332\times {10}^{30}\text{ kg}}{t\text{ years}}\\ t=\frac{\left(12\times {10}^9\right)\left(1.33\times {10}^{30}\right)}{2.28\times {10}^{29}}\text{ years}\\ =\text{7}\times {\text{10}}^9\text{ years}\end{array}$

At 2000 times, the rate of consumption remaining life would be,

$\begin{array}{c}{t}_{\text{new}}=\frac{7\times {10}^{10}\text{ years}}{2000}\\ =35\text{ million years}\end{array}$

Conclusion:

The time, for the red giant phase to finish the conversion of hydrogen, is 35 million years.