1. From what do stars form?
2. What makes a star shine?
3. What makes a gas cloud contract to form stars?
4. What makes a star stop contracting?
5. When a star has stopped contracting and settled down, where is it in the HR diagram?
6. What fuel do stars on the main sequence burn?
7.Why does fuel "burn" only in a star's core?
8. What happens to the star's core as the hydrogen there is used up?
9. What happens to the star's outer layers as the fuel in its core is used up?
10. Why does the star's core get hotter as the core shrinks?
11. Why does a high mass star evolve differently than a low mass star?
12. What allows a high mass star to burn different fuels than a low mass star?
13. What determines how rapidly a star burns up its fuel?
14. What of the following most directly determines how much fuel a star has?
15. What kind of object does a main sequence star become on first using up its core hydrogen?
16. What quantities are plotted in an H-R diagram?
Suppose your car burns 2 gallons of gasoline per hour and your fuel tank holds 10 gallons.
17. How many hours can you drive your car before it uses up a full tank?
You can see from the above that the time to use up fuel is simply the amount of fuel divided by the burning rate.
Stars work the same way. They remain on the main sequence for a time equal to the amount of hydrogen fuel they have available divided by the rate at which they burn the
hydrogen. Thus, to figure out the star's main sequence lifetime, we need to know the amount of fuel it has and the fuel burning rate.
The amount of fuel available to a star turns out to be about 10% of its mass. The rate at which it burns fuel is set by the star's luminosity. That is, the star burns fuel at whatever rate it needs to supply its luminosity.
For example, the Sun has a total mass of 2x1030 kilograms. About 10% of this is available for burning, so that the Sun has an amount of fuel equal to
2x1030/10 = 2x1029 kilograms.
The Sun's luminosity in metric units is 4x1026 joules per second. It turns out that one kilogram of hydrogen releases 6.3x1014 joules when it is converted into helium. Thus, to supply 4x1026 joules per second, the Sun must convert 4x1026 joules per second/6.3x1014
joules per kilogram to get (4/6.3)x1012 kilograms per second.
That is, to supply 1 solar luminosity of power, the Sun must burn
6.3x1011 kilograms of hydrogen into helium per second.
We can now figure out how long the Sun will be able to power itself by burning hydrogen.
If the Sun burns 6.3x1011 kilograms of hydrogen into helium per second, then, because it has a mass of fuel equal to 2x1029 kilograms, it can burn for a time equal to 2x1029 kilograms/6.3x1011
kilograms/second = (2/6.3)x1018 seconds = 3.2x1017 seconds.
Our answer above is in seconds. To convert that to years, divide by the number of seconds in 1 year: about 3.16x107. Thus, the Sun can burn for 3.2x1017 seconds /3.16x107 seconds = about
We can extend this result to other stars by scaling. That is, if a star has 10 solar masses, multiply the above result by 10. If it has 100 solar luminosities, divided the above result
by 100. In other words the life time of a star is about 1010 M/L years.
18. Can the Sun burn all of the hydrogen it contains before leaving the main sequence?
Only the material in the core is hot enough to fuse and there is no very good way to mix the hydrogen from the surface inward. In fact, by the time a little over 10% of the mass is
burnt, the star will begin to turn into a red giant and leave the main sequence. This is why we use 10% of a star's mass in estimating its age.