Solutions to Homework 3

 

 

 

  1. Suppose you are a space explorer on board a starship which goes into orbit around a previously uncharted blue star.   Your spaceship is orbiting the star at a distance of 3A.U., and you deduce that it will take you half of an Earth year to orbit around the star once.  Furthermore, when you deploy a solar panel from the side of your space-craft, you collect 1000 times more energy than when the same solar panel is used to collect Sun-light on Earth. Using this information, deduce

 

    1. a rough estimate for the temperature of the star,

 

The temperature of the star is estimated using the color of the star.  Using the fact that this is a blue star, we estimate that the temperature is 20,000-30,000K.

 

Grading scheme:

q      1 point for stating that temperature comes from color

q      2 points for getting something in the appropriate range (20,000-30,000)

q      1 point for using correct unit (either K or degrees-C).

 

 

 

    1. the luminosity of the star (you can give you answer in terms of Sun luminosities… i.e., “this star is <your-answer> times more luminous than the Sun”).  Show your working.

 

You derive the luminosity of the star using two facts: the distance (d) and the observed flux of radiation from the star (F).  The luminosity is then given by

 

 

Let’s write this expression twice, once for the star (subscript *) and once for the Sun (subscript Sun):

 

 

Now divide these two expressions to get the ratio of the star’s and the Sun’s luminosity:

 

 

Finally, cancel the 4p’s and plug in the numbers relevant to the star and the Sun (remembering that the questions tells you that the ratio F*/FSUN=1000):

 

 

So, the star is 9000 times more luminous than the Sun.

 

Grading scheme:

q      2 points for correct formula (or for just recognizing general approach)

q      2 points for correct answer.

 

 

 

 

    1. the mass of the star (you can give your answer in terms of Sun Masses).  Show your working. [Hint – think about the discussion in class on how we can measure the mass of the Sun.  We discussed a formula that will be very useful for this question.]

 

From the lectures, the formula relating the mass of a central star (M) to the distance (d) and period (P) of an orbiting planet/spaceship is:

 

 

So, comparing the orbit of the spacecraft around the star and the Earth around the Sun, you deduce that the ratio of the mass of the star and the mass of the Sun is:

 

 

Cancel the constants, and plug the number in:

 

 

So, the star is 108 times the mass of the Sun.

 

Grading scheme

q      1 point for general effort

q      2 point for identifying correct formula (or otherwise demonstrating a clear understanding of what’s going on!)

q      2 points for correct answer

 

 

 

Do you think this star will suffer a supernova explosion at some point in the future?  Will it leave a White Dwarf, a Neutron Star or a Black Hole?

 

A star with such a large mass will certainly undergo a supernova explosion, leaving behind a black hole.

 

Grading scheme

q      1 point for stating that star will undergo supernova

q      1 point for stating that a black hole will be left behind

 

 

 

 

  1. The Sun has a total mass of 2´1030kg, of which initially (i.e., at the time of formation) 75% is hydrogen.  It “burns” (or more correctly, processes) hydrogen at a rate of about 6´1011kg per second.   

 

    1. If you suppose that the Sun processes all of its hydrogen while on the main sequence, calculate the “main-sequence lifetime” of the Sun.   Give you answer in terms of years.  [Hint – you will need to figure out how many seconds are in a year.]

 

The amount of hydrogen  initially in the Sun is (0.75)´(2´1030)kg=1.5´1030kg.  Thus, if it burns at a constant rate of 6´1011kg/s, it will take

 

 

to burn through all of this hydrogen.   Now, the number of seconds in a year is 60´60´24´365=31,536,000s.  So, dividing the number above by this, it will take 79.3 billion years to burn through all of the Sun’s hydrogen.

 

Grading scheme:

q      1 point for general effort

q      1 point for correctly using the “75%”

q      1 point for dividing the amount of hydrogen by the rate

q      1 point for dividing the life-time in seconds by number of seconds in a year.

q      1 points for doing these calculations correctly!

 

 

 

 

    1. The actual main-sequence lifetime of the Sun is 10 billion years.  What is the reason for the difference between this number and the one that you calculated in part (a) above.

 

 

[5 points] The above calculation is the time that it takes to burn through ALL of the Sun’s hydrogen.  But, the nuclear reactions only occur in the Sun’s core, so not all of the Sun’s hydrogen participates in the nuclear fusion.  In fact, only about 1/8th of the Sun’s mass participates… hence the lifetime is shorter by a factor of 8.