Question

The sun radiates energy at the rate $3.8\times {10}^{26}\mathrm{W}$. The source of this energy is fusion, a nuclear reaction in which mass is transformed into energy. The mass of the sun is $2.0\times {10}^{30}\mathrm{kg}$.

a. How much mass does the sun lose each year?

b. What percent is this of the sun’s total mass?

c. Fusion takes place in the core of a star, where the temperature and pressure are highest. A star like the sun can sustain fusion until it has transformed about $0.10\%$of its total mass into energy, then fusion ceases and the star slowly dies. Estimate the sun’s lifetime, giving your answer in billions of years.

Short answer

a. mass of the sun used in one year is $1.33\times {10}^{17}\mathrm{kg}.$

b. Percentage is $6.65\times {10}^{-12}\%.$

c. Sun lifetime is$1.5\times {10}^{13}\mathrm{years}.$

Step-by-step solution

Part (a) Step 1: Given information

We have given,

power =$3.8\times {10}^{26}\mathrm{W}$

Mass of sun =$2.0\times {10}^{30}\mathrm{kg}$

We have to find the mass used in fusion in one year.

Simplify

The Energy obtain by sun in one year time,

$$\begin{gathered} \mathrm{E}=\mathrm{Pt} \\ \mathrm{E}=3.8\times {10}^{26}\mathrm{W}\times 31536000\mathrm{s} \\ \mathrm{E}=1.2\times {10}^{34}\mathrm{J} \end{gathered}$$

Then, mass will be,

$$\begin{gathered} \mathrm{E}={\mathrm{mc}}^2 \\ 1.2\times {10}^{34}\mathrm{J}=\mathrm{mc} \\ \mathrm{m}=1.33\times {10}^{17}\mathrm{kg} \end{gathered}$$

Part (b) Step 1: Given information

We have given,

Mass of the sun =$2.0\times {10}^{30}\mathrm{kg}$

We have to find the percentage of mass used.

Simplify

$$\begin{gathered} \%=\frac{{\mathrm{m}}_{\mathrm{used}}}{{\mathrm{m}}_{\mathrm{total}}}\times 100 \\ \%=\frac{1.33\times {10}^{17}\mathrm{kg}}{2\times {10}^{30}\mathrm{kg}}\times 100 \\ \%=6.65\times {10}^{-12} \end{gathered}$$

Part (c) Step 1: Given information

We have to find the life time of the sun.

Simplify 

$$\begin{gathered} \mathrm{T}=\frac{\mathrm{total}\mathrm{mass}}{\mathrm{rate}\mathrm{of}\mathrm{fusion}} \\ \mathrm{T}=\frac{2\times {10}^{30}}{1.33\times {10}^{17}} \\ \mathrm{T}=1.5\times {10}^{13}\mathrm{years}. \end{gathered}$$