Chapter 16, Problem 10Q

To determine

The fraction of the Sun’s mass that will be converted into helium over the next 5 billion years, if it is assumed that the current rate of the hydrogen fusion remains constant. Also, discuss its effect on the overall chemical composition of the Sun.

Answer to Problem 10Q

Solution:

$6.83\times {10}^{26}\text{ kg}$, after the next 5 billion year, 30% of the Sun’s mass will be helium and 69% will be hydrogen.

Explanation of Solution

Given data:

The current rate of the hydrogen fusion in the Sun remains constant.

The time is 5 billion year.

Formula used:

The expression for the mass-energy equation is written as:

$E=m{c}^2$

Here, E is the total energy radiated, m is the mass of object, and c is the speed of light having value $3\times {10}^8\text{ m/s}$.

Explanation:

The Sun’s luminosity is $3.9\times {10}^{26}\text{ J/s}$.

The total energy radiated by the Sun in 5 billion year is calculated as,

$\begin{array}{c}E=\left(3.9\times {10}^{26}\text{ J/s}\right)\left(5\text{ billion yr}\right)\\ =\left(3.9\times {10}^{26}\text{ J/s}\right)\left(5\text{ billion yr}\left(\frac{{10}^9\text{ yr}}{1\text{ billion yr}}\right)\left(\frac{3.154\times {10}^7\text{ s}}{1\text{ yr}}\right)\right)\\ =6.15\times {10}^{43}\text{J}\end{array}$

Recall the expression of the mass-energy equation.

$E=m{c}^2$

Rearrange the above expression for $m$,

$m=\frac{E}{c^2}$

Substitute $6.15\times {10}^{43}\text{ J}$ for E and $3\times {10}^8\text{ m/s}$ for c.

$\begin{array}{c}m=\frac{6.15\times {10}^{43}\text{ J}}{{\left(3\times {10}^8\text{ m/s}\right)}^2}\\ =\frac{6.15\times {10}^{43}}{9\times {10}^{16}}\\ =6.83\times {10}^{26}\text{ kg}\end{array}$

Only .7% of the mass is transformed into energy at the time of conversion of the hydrogen into helium. So, the amount of hydrogen used in Sun’s energy production over the next 5 billion year is,

$\begin{array}{c}{m}^{\prime }=\frac{6.83\times {10}^{26}\text{ kg}}{.007}\\ =9.7\times {10}^{28}\text{ kg}\end{array}$

The mass of the Sun is $1.99\times {10}^{30}\text{ kg}$. So, the percent of amount of hydrogen used of the Sun’s mass is,

$\begin{array}{c}{m^{\prime }}^{\prime }=\frac{9.77\times {10}^{28}\text{ kg}}{1.99\times {10}^{30}\text{ kg}}\\ =.049\\ =4.9\%\text{ of sun's mass}\end{array}$

The present composition of the Sun is 74% hydrogen, 25% helium, and 1% heavy elements.

Thus, the mass of each component in Sun is,

The mass of hydrogen is,

$\begin{array}{c}{m}_{\text{Hydrogen}}=0.74\left(1.99\times {10}^{30}\text{ kg}\right)\\ =1.47\times {10}^{30}\text{ kg}\end{array}$

The mass of helium is,

$\begin{array}{c}{m}_{\text{Helium}}=0.25\left(1.99\times {10}^{30}\text{ kg}\right)\\ =4.98\times {10}^{29}\text{ kg}\end{array}$

The mass of heavy elements is,

$\begin{array}{c}{m}_{\text{Heavy element}}=0.01\left(1.99\times {10}^{30}\text{ kg}\right)\\ =1.99\times {10}^{28}\text{ kg}\end{array}$

Over 5 billion year the composition of the Sun is as:

The composition of hydrogen is,

$\begin{array}{c}{\left(m_{\text{Hydrogen}}\right)}_{\text{after 5 billion year}}=m_{\text{Hydrogen}}-m^{\prime }\\ =\left(1.47\times {10}^{30}\text{ kg}\right)-\left(9.77\times {10}^{28}\text{ kg}\right)\\ =1.37\times {10}^{30}\text{ kg}\end{array}$

The fraction of the sun’s mass that will be hydrogen in 5 billion year is,

$\begin{array}{c}{\left(\%m_{\text{Sun}}\right)}_{\text{in Hydroen, after 5 billion yr}}=\frac{{\left(m_{\text{Hydrogen}}\right)}_{\text{after 5 billion year}}}{m_{\text{Heavy element}}}\\ =\frac{1.37\times {10}^{30}\text{ kg}}{1.99\times {10}^{30}\text{ kg}}\\ =.688\\ \approx 69\%\end{array}$

It is 69% of the Sun’s mass.

Now, the composition of helium is,

$\begin{array}{c}{\left(m_{\text{Helium}}\right)}_{\text{after 5 billion year}}=m_{\text{Helium}}+m^{\prime }-0.007\left(m^{\prime }\right)\\ =\left(4.98\times {10}^{29}\text{ kg}\right)+\left(9.77\times {10}^{28}\text{ kg}\right)-0.007\left(9.77\times {10}^{28}\text{ kg}\right)\\ =\left(5.957\times {10}^{29}\text{ kg}\right)-\left(6.839\times {10}^{26}\text{ kg}\right)\\ =5.95\times {10}^{29}\text{ kg}\end{array}$

The fraction of the sun’s mass that will be helium in 5 billion year is,

$\begin{array}{c}{\left(\%m_{\text{Sun}}\right)}_{\text{in Helium, after 5 billion yr}}=\frac{5.95\times {10}^{29}\text{ kg}}{1.99\times {10}^{30}\text{ kg}}\\ =0.2989\\ \approx 30\%\end{array}$

It is 30% of the Sun’s mass.

Conclusion:

Therefore, the mass converted into energy in 5 billion year is $6.83\times {10}^{26}\text{ kg}$. And, the chemical composition of helium will change in small quantity, but not a negligible quantity.