|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.]
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 1026 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*1026 Joules/second, and if it has 1.26*1044 Joules of available energy, then we can divide the two to determine how long the Sun can shine: (3.78*1026 Joules/second)/(1.26*1044 Joules)=3.33*1017 seconds. We should convert this to more apropriate units of years, 3.33*1017 seconds/(60 sec/min*60 min/hr*24 hr/day*365 day/year)=1.05*1010 years or 10.5 billion years.
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"
Joules/(second*meter2*Kelvin4). For a 5780
Kelvin patch of the photosphere:
Joules per second per square-meter. For a 4280 Kelvin patch of a
1.90*107 Joules per second per square-meter. This
means that the slightly hotter photosphere is about three times
brighter than a sunspot.
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*57804/sigma*42804 = (5780/4280)4 = 1.354 = 3.33, just as before. Notice that the constant, sigma, cancels out. (This is the easy way to do the calculation.)