Astronomy 110 | PRINT Name __________________________ |

Fall 2005 Section 006 | |

Homework 7
: Sunshine & Sunspots |
(Due Thursday, Oct 27, 2005) |

This homework asks you to do some simple calculations related to
the production of energy in the Sun. Recall that the Sun produces
energy by the fusion of hydrogen into helium in its core, and that in
the process, it loses mass as some mass is converted into energy. The
formula, Energy = Mass times (speed of light)^{2} gives the
conversion between mass and energy. Answer the following questions.
[To get the right answer, it's important to express the quantities in
**consistent units**. If you are uncertain about what units to
use, refer to Appendix A of the textbook, especially page A8.]

**How long will the Sun shine**? Recall that we are led to the idea of fusion in the Sun being it's energy source by the fact that no other energy source could power the Sun for its long history of 4.5 billion years. Will fusion do the job? You can calculate this as follows:

The amount of mass available in the Sun to be used as "fuel" is the mass in the core of the Sun, which is approximately 10% of its mass. (The mass of the Sun is 2 x 10^{30}kg.) Recall that, each time four hydrogen nuclei fuse into a helium nucleus, 0.7% of the mass of hydrogen is converted into energy (where E = mc^{2}).

You can put these numbers together to figure out the total amount of**energy**that the Sun can produce in its hydrogen-burning lifetime, i.e., as long as there is hydrogen available to "burn".**(2 pts.)**

**In a fusion reaction in the Sun 0.7% of the mass involved is converted to energy, and if 10% of the mass of the Sun is available for fusion, then the total mass that can be converted to energy is: 0.1*0.007*2*10**^{30}=1.4*10^{27}kg. In order to use this in the equation E=mc^{2}we need to know the speed of light, which is c=3*10^{8}m/s (we need the units to be meters per second for the units to work out). E=(1.4*10^{27}kg)*(3*10^{8}m/s)^{2}=__1.26*10__^{44}Joules.Then, to figure out how long this lifetime is, you need to kow the rate at which the Sun is

**losing**energy. We know this from the Sun's**luminosity**, which is 3.78 x 10^{26}Joules/second (a Joule is a unit of energy, see page A8).

Can you put this all together to figure out**how long the Sun can shine (i.e., lose energy) at its present rate**?**(2 pts.)**

**The Sun loses energy at rate of 3.78*10**^{26}Joules/second, and if it has 1.26*10^{44}Joules of available energy, then we can divide the two to determine how long the Sun can shine: (3.78*10^{26}Joules/second)/(1.26*10^{44}Joules)=3.33*10^{17}seconds. We should convert this to more apropriate units of years, 3.33*10^{17}seconds/(60 sec/min*60 min/hr*24 hr/day*365 day/year)=1.05*10^{10}years or__10.5 billion years.__- How does the time you've just calculated compare with the present
age of the Sun? How much longer will the Sun shine as it does now?
**(1 pt.)**

**As noted above, the Sun is 4.5 billion years old, so with our calculations from problem one we can estimate that the Sun can shine for 6 billion more years before running out of energy.** - We know that the Sun loses 3.78 x 10
^{26}Joules of energy every second (this is the Sun's luminosity). By using E = mc^{2}, figure out how much mass this corresponds to. That is,**how much mass does the Sun lose every second**? Express your answer in metric tons. (A metric ton = 1000 kg.)**(2 pts.)**

**To figure out how much mass is lost every second we need to rewrite the equation as m=E/c**^{2}. Plugging in the numbers, we have: m=(3.78*10^{26}Joules/sec)/(3*10^{8}m/s)^{2}=4.21*10^{9}kg/sec. Or converting this to metric tons, the Sun loses__4.21 million tons per second.__ - Sunspots look dark because they are at a temperature that is
typically 1500 K cooler than the surrounding photosphere, whose
temperature is 5780 K. Using the Stefan-Boltzmann law, compare the
surface brightness of a dark sunspot to that of the surrounding
photosphere. The surface brightness is the energy per unit time
radiated from each square meter of the surface, either of the spot or
of the photosphere.
**(3 pts.)****Pages 119-120 in our textbook explain the Stefan-Boltzmann law. We can use this law to figure how much energy per second is emitted by a one square-meter sized patch of the Sun, which is called the flux. This is determined by the temperature of the patch taken to the fourth power. We also need to know the value of the Stefan-Boltzmann constant, represented by the Greek letter "sigma" =5.67*10**^{-8}Joules/(second*meter^{2}*Kelvin^{4}). For a 5780 Kelvin patch of the photosphere: Flux=5.67*10^{-8}*5780^{4}=__6.33*10__For a 4280 Kelvin patch of a sunspot: Flux=5.67*10^{7}Joules per second per square-meter.^{-8}*4280^{4}=__1.90*10__This means that the slightly hotter photosphere is about three times brighter than a sunspot.^{7}Joules per second per square-meter.

Better yet, you can make the comparison without having to know the flux from either the sunspot or the surrounding photosphere separately, and therefore without having to look up the value of the constant, "sigma". You can do this by noting that the flux is__proportional to__the fourth power of the temperature, so if you__divide__the flux from the photosphere by the flux from the sunspot you get that the__ratio__of the fluxes is = sigma*5780^{4}/sigma*4280^{4}= (5780/4280)^{4}= 1.35^{4}= 3.33, just as before. Notice that the constant, sigma, cancels out. (This is the easy way to do the calculation.)