Question 28.5: How long will the Sun shine? According to the present models......
How long will the Sun shine?
According to the present models of stellar structure, the central parts of stars, which contain about 10% of the star’s mass, have conditions favorable for fusion. Assuming the Sun is made of pure hydrogen, how much time will it take the Sun to radiate the energy released when 10% of its hydrogen has fused into helium? The mass of the Sun is about 2.00 \times 10^{30} \mathrm{~kg} . The total luminosity L (also power, or energy/second) emitted by the Sun is L=3.85 \times 10^{26} \mathrm{~W} .
Sketch and translate Imagine the Sun as an object that simultaneously releases energy through fusion reactions and then radiates it as electromagnetic radiation.
Simplify and diagram Assume that during a fusion reaction in the Sun’s interior, two protons and two neutrons join together to form a helium nucleus. The amount of energy released is equal to approximately 0.008 of the rest energy of the participating particles (see Exercise 28.4). Further, assume that there are no other sources of energy, so that the rate at which energy is produced inside the Sun by fusion reactions is the same as the rate at which the energy is radiated.
Represent mathematically The total energy released by fusion reactions throughout the lifetime of the Sun is equal to 0.8% of the rest energy of the 10% of the Sun’s mass energy that is available for fusion:
E_{\text {released }}=(0.008)(0.10) m_{\text {Sun }} c^2
This equals the luminosity L of the Sun times the time interval Δt during which the Sun will shine, or
E_{\text {released }}=L \Delta t
Solve and evaluate Setting the above two expressions equal to each other, solving forΔt, and inserting the appropriate values, we estimate the expected lifetime of the Sun:
\Delta t=\frac{E_{\text {released }}}{L}
=\frac{(0.008)(0.10)\left(2.00 \times 10^{30} \mathrm{~kg}\right)\left(3.00 \times 10^8 \mathrm{~m} / \mathrm{s}\right)^2}{3.85 \times 10^{26} \mathrm{~W}}
=3.7 \times 10^{17} \mathrm{~s}
=3.7 \times 10^{17} \mathrm{~s}\left(\frac{1 \text { year }}{3.16 \times 10^7 \mathrm{~s}}\right)=1.2 \times 10^{10} \text { years }
This result is about 2.5 times the age of Earth. Nuclear fusion is a possible mechanism for powering the Sun for billions of years.
Try it yourself: Studying the energy emitted by stars of different masses, astronomers found that the energy that a typical star emits every second is proportional to its mass raised to the power of 3.5, that is, L \propto m^{3.5} . Will a 10 solar mass star take more or less time than the Sun to exhaust its nuclear fuel, assuming that both the star and the Sun have 10% of its mass in hydrogen available for fusion?
Answer: The star will take significantly less time—about 0.003 times the time for the Sun:
\frac{\Delta t_{\text {star }}}{\Delta t_{\text {Sun }}}=\frac{\left(\frac{E_{\text {released }}}{L}\right)_{\text {star }}}{\left(\frac{E_{\text {released }}}{L}\right)_{\text {Sun }}}=\frac{\frac{(0.008)(0.10)\left(10 m_{\text {Sun }}\right) c^2}{L_{\text {star }}}}{\frac{(0.008)(0.10) m_{\text {Sun }} c^2}{L_{\text {Sun }}}}
=\frac{10 L_{\text {Sun }}}{1 L_{\text {Star }}}=\frac{10 m_{\text {Sun }}^{3.5}}{1 m_{\text {Star }}^{3.5}}=\frac{10}{1}\left(\frac{m_{\text {Sun }}}{10 m_{\text {Sun }}}\right)^{3.5}
= 0.0032