Astronomy 102 Problem Set #3 Solutions

Problem #1: a) Calculate the distance to a star whose parallax is 0.012 arcseconds (NOTE: 1 arcsecond = 1/3600^th of a degree).
b) If this star has the same luminosity as the Sun, what is its flux (in W/m²) when viewed from the Earth?

Solution -- Part a): For this problem, you need to know what the parallax effect is, and more specifically, what is meant by the parallax of a star. You can learn about the former from your class notes, the book, or the Special Web Page on this topic. The latter is simply a definition, and can be found in all three places.

The parallax of a star is the angle through which a star appears to move when viewed from two positions separated by 1 Astronomical Unit (A.U.). It is the precise definition of the baseline (1 A.U.) which allows us to make a direct relationship between the angle and the distance to the star. With this information we can use some simple trigonometry to determine the distance.

For this problem, alpha is the parallax, or 0.012 arcseconds, w is half the baseline, or 1 A.U., and R is what we're looking for, the distance to the star. Now before we can plug all of this into the definition of a sin, we need to pay attention to units.

sin (alpha) = w/R

alpha must have units of degrees. We know that 1 degree = 3600 arcseconds, so

alpha = 0.012 arcseconds x (1 degree/3600 arcseconds)
= 3.33 x 10^-6 degrees

Now the above result becomes

sin alpha = w/R
sin (3.33 x 10^-6 degrees) = 1A.U./R
5.81 x 10^-8 = 1 A.U./R

and therefore

R = 1 A.U./5.81 x 10^-8
= 1.72 x 10⁷ A.U.

Note that our answer has units of A.U. because we used units of A.U. to express w. We didn't have to, but since we were given w in units of A.U., it was easier, and the answer is just as valid as it would be expressed in meter of parsecs.

We can convert this answer to meters by multiplying by the appropriate conversions factor. Since 1 A.U. = 1.5 x 10⁸ m

R = 1.72 x 10⁷ A.U. x (1.5 x 10¹¹ m/ 1 A.U.)
= 2.58 x 10¹⁸ m

We can also use parsec, as in your textbook FETU, and since 1 pc = 2.06 x 10⁵ A.U.

R = 1.72 x 10⁷ A.U. x (1 A.U./ 2.06 x 10⁵ pc)
= 83 pc.

Note: Your book provides you with a shortcut for getting from the parallax of a star to its distance in parsecs:

parallax (in arcseconds) = 1/distance (in parsecs)

This is a perfectly legitimate way of determining the distance from the parallax, but beware: you must use units of arcseconds for the parallax, and you must use units of parsecs for the distance. I'm not going to put this formula on your exam, and I just might ask you to compute the distance to a star whose parallax I give you in degrees. Using this shortcut formula is correct but dangerous. Make sure you understand how it works if you chose to use it.

Part b): The intensity of a luminous object is directly related to that object's luminosity and the distance between the observer and emitter as follows:

I = L/(4 pi R²)

where I is the intensity, L is the luminosity(power!), and R is the distance between the observer and emitter. Note that R is not the radius of the star in this case.

We're told that the luminosity of this star is the same as that of the Sun, i.e., 3.8 x 10²⁶ J/s, or W. We've just calculated the distance above, so let's put these quantities into our relation

I= L/(4 pi R²)
= 3.8 x 10²⁶ W/ (4 x 3.14 x (2.58 x 10¹⁸ m)²
= 4.5 x 10^-12 W/m²

Note that in order to get an answer in units of W/m², you need to use a distance on meters. If you didn't calculate the distance in meters in the first part of this problem, you'll need to do some converting.

Problem #2:

You observe that a K star in a star cluster has a flux of 6.23 x 10-14 W/m2.

a) Assuming that this K star is a Main Sequence star and that Main Sequence K stars have a luminosity of 0.4 Lo, calculate the
distance to the star cluster.

b) Further observations shows that this particular K star is actually a red giant star (i.e., it's not on the Main Sequence). Since the
luminosity of K giants is more like 100 Lo, what is the true distance to this cluster?

Solution --

Part a): In this problem, we're given a intensity and a luminosity, and we're asked for a distance. Unfortunately, this is
an equation hunter's dream, since there exists an equation relating intensity, distance, and luminosity. I will try to avoid playing into the equation hunters' hands so completely in the future, especially not after a problem #1 that uses the same equation...

Nonetheless, we can use the distance dimming equation:

intensity = luminosity/(4 pi distance2)

and solving for distance, we get,

distance = SQRT{(luminosity/(4 pi flux))}

where SQRT(x) means "the square root of x". Plugging in the numbers from
the problem, we get

     distance = ((0.4 x 3.8 x 10^26 W)/SQRT(4 x 3.14 x 6.23 x 10^-14 W/m2))
     = SQRT(1.94 x 10^38 m^2)
     = 1.39 x 10^19 m, or 451 pc.

Part b): There are two ways to deal with this part of the problem. The brute force method would be to redo the
whole calculation with the new luminosity. This will work just fine, but it's the long way to go.

A shorter way is to look at how the distance depends on the luminosity of the star. The distance relation that we derived above,

distance = SQRT(luminosity/(4 pi flux))

shows that the distance varies with the square root of the luminosity; that is, if the flux remains constant and the luminosity
increases by some factor x, then the distance to the source must have increased by a factor of SQRT(x).

In this problem, we find that the new luminosity for the source is 250 times higher than the old luminosity (100 Lo/0.4 Lo = 250);
therefore the new distance estimate must be SQRT(250) higher than the old distance estimate, or 15.8 times bigger. So the new
distance is

     new distance = 15.8 x old distance
     = 15.8 x 1.39 x 10^19 m
     = 2.20 x 10^20 m, or 7120 pc.

Problem #3:

Assume for a moment that all of the hydrogen in the Sun is converted into helium via nuclear fusion (this doesn't
really happen).

a) How much energy would be produced in the sun?

Remember that a fusion process some of the mass of hydrogen is converted into helim, and some is converted into energy.
Assume that the Sun is 80% hydrogen by mass.

b) How long would it take for the Sun to radiate this much energy at its current luminosity?

Solution --

Part a): OK, let's start by figuring out how much "raw material" we have available. For hydrogen fusion, the raw
material is hydrogen, and 80% of the Sun is hydrogen, so

     mass of H available = 0.8 x (mass of Sun)
     = 0.8 x 1.99 x 10^30 kg
     = 1.59 x 10^30 kg.

That's how much H was have available to us. If we were to convert all of the H into He, we would "lose" 0.7% of the mass in the
conversion process. Look up the periodic table (one of the appendixes in your book has it, with atomic mass units).

Here is how we get the 0.7%: Given that the mass of hydrogen is 1.008 atomic mass units, and helium is 4.0025 atomic mass units, and that 4 hydrogens are converted into helium, the percentage of mass lost is:

4 x 1.008 - 4.0025
-------------------- x 100 = 0.7 %
4.0025

Where does that mass go? It's converted into energy, which then heats the core and makes the Sun radiate.
But we're getting ahead of ourselves. Let's first figure out how much mass is "lost":

     mass "lost" = 0.007 x (mass of H available)
     = 0.007 x 1.59 x 10^30 kg
     = 1.11 x 10^28 kg.

Now that we know how much mass is "lost," we can figure out how much energy is generated. Why?
Because the reason this mass is "lost" is that it has been converted to energy. Energy and mass are related by Einstein's famous
equation E = mc2, where m is the mass that has been converted to energy, and c is the speed of light. So

     Energy produced = (mass "lost") x c2
     = 1.11 x 10^28 kg x (3 x 10^8 m/s)2
     = 9.99 x 10^44 kg m^2/s^2
     = 9.99 x 10^44 J.

where the last step comes from the definition of the energy unit, the Joule.

Solution -- Part b): Now we know how much energy is contained in the H-fusing potential of the Sun. How long will it last?
Well, that depends on how much energy we've got (we know this from Part b) and how fast we're losing it.

Here's an everyday analog: If you've got a bathtub containing 200 gallons of water, and the water passes through a drain in the
bottom of the tub at a rate of 5 gallons per minute, how many minutes will it take to empty the tub? Well, in one minute you'll
remove 5 gallons, and in two minutes you'll remove 10 gallons. You can calculate how long it will take by dividing the amount of
water you've got, 200 gallons, by the rate at which the water leaves, 5 gallons/minute,

     time to empty tub = (amount in tub)/(rate at which it leaves tub)
     = (200 gallons)/(5 gallons/minute)
     = 40 minutes.

The game is exactly the same for the Sun. You've got a supply of energy, 9.99 x 1044 J and it's "leaking" out of the Sun at a rate
of 3.8 x 1026 J/s. How do I know this? Because the Sun is radiating (or "giving away") energy at a rate equal to its luminosity.
After all, luminosity is nothing more than energy per second, or more explicitly, Joules per second. So, following the bathtub
analogy,

     time to give away all of the Sun's energy = (amount of energy)/(rate at which energy leaves the Sun)
     = (9.99 x 10^44 J)/(3.8 x 10^26 J/s)
     = 2.63 x 10^18 s.

Note that this time (80 billion years) is much longer than the time the Sun would radiate if it only had its gravitational energy
available (30 million years. To be mentioned again soon...). This is why it's so important to the Sun have H-fusion as its energy source.

In reality, the Sun will fuse hydrogen into helium for only 10 billions years or so. That's because only the hydrogen in the core will
be hot enough to fuse. The rest of the hydrogen in the Sun hangs out in the outer parts of the Sun, where it's too cool for fusion.
Only 10% or so of the Sun's hydrogen is in the core, so only that part of its hydrogen supply is fused, and therefore the Sun sits on
the Main sequence for only 10 billion years.