Problem #1: 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 flux and a luminosity, and we're asked for a distance. Unfortunately, this is an equation hunter's dream, since there exists an equation relating flux, distance, and luminosity. I will try to avoid playing into the equation hunters' hands so completely in the future.
Nonetheless, we can use the distance dimming equation:
and solving for distance, we get,
where (x)1/2 means "the square root of x" (I don't know how to make square root signs in HTML). Plugging in the numbers from the problem, we get
Solution -- 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,
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 (x)1/2.
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 (250)1/2 higher than the old distance estimate, or 15.8 times bigger. So the new distance is
Problem #2: A Hydrogen atom in an interstellar cloud makes a transition from its 54th excited state to its 43rd excited state, and in doing so, lowers it energy by 4.31 x 10-22 Joules and launches a photon.
a) What is the frequency of the photon emitted by this process (according to the atoms in the cloud)?
b) If you are in a spaceship traveling toward this cloud at as speed of 10% of the speed of light, what frequency light do you measure?
Solution -- Part a): Energy doesn't just disappear, so if a hydrogen atom lowers its energy by 4.31 x 10-22 Joules, then it has to go somewhere. Where? Into the photon it creates.
So the energy of the photon is 4.31 x 10-22 Joules. Can we use that information to get the frequency of the light? Sure, using the handy-dandy energy frequency relation for photons:
Solution -- Part b): The answer above is the frequency of the light emitted by the cloud as the cloud sees it, and since the cloud isn't moving with respect to itself, this is the frequency anyone who is at rest with respect to the cloud will see. That is, 6.5 x 1011 Hz is the rest frequency.
If we're moving toward the cloud in our spaceship, we won't see the same thing someone who's not moving with respect to the cloud will see, because of the Doppler effect. Because of our motion, we will see waves coming toward us "bunched up" and the frequency of the waves will be greater.
The Doppler relation tells us how much the frequency will change:
Doppler formula: speed of emitter change in frequency
------------------- = --------------------
speed of wave rest frequency
In this case, we know the rest frequency (from part a), and the speed
of the waves (they're light waves). We also know the speed of the
emitter, because if we're on the spaceship traveling toward the cloud
at a speed of 10% of the speed of light, it will appear that the cloud
is coming toward us at a speed of 10% of the speed of
light. Therefore, we can solve for the change in frequency:
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 of the Sun's total mass would be converted into energy? Assume that the Sun is 80% hydrogen by mass. (Note: I am not asking how much hydrogen gets used up, but instead how the total mass of the Sun (i.e., hydrogen and helium) changes through the fusion process.)
b) How much energy would be produced?
c) 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
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. 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":
Solution -- Part b): 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
where the last step comes from the definition of the energy unit, the Joule.
Solution -- Part c): 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,
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,
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). 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.