Astronomy Assignment #01: The Sun
***** Solutions *****
Answer the following Review Questions:
1. What are the two main gases in the Sun? How does the Sun's mass and size compare with Jupiter?
The two main gases in the Sun are hydrogen (H, about 75%) and helium (He, about 24%). The mass of
the Sun is 1,048 times the mass of Jupiter (and 333,166 Earth masses). The diameter is the Sun is 9.73
times the diameter of Jupiter (and 109.1 Earth diameters). If we were to compare the volume of the Sun
to the volume of Jupiter we would find that the volume of the Sun is equivalent to 921Jupiter volumes
(and 1.3 million Earth volumes). A rough comparison of the sizes of the Earth, Jupiter and the Sun can
be seen if you compare a golf ball to a basketball to a ball the diameter of the circle at the ”top of the key
on a basketball court.
2. What goes on in the core, Envelope (radiative zone, and convection zone) of the Sun?
The core is the sole location of energy production via the proton-proton chain. As described in the TwoLayer Model of the Sun (As presented on Hot Tips), the core occupies only 1% of the Sun’s volume,
contains 2/3rd’s the mass of the Sun, has a density of around 100 g/cm3 and a temperature of 15 million
Kelvins. In spite of the small physical size of the Sun’s core, all the energy the Sun produces occurs in the
core. Because the proton-proton chain occurs in the core of the Sun, the amount of hydrogen in the core
is decreasing as helium is produced by the p-p chain. Currently, stellar models suggest that the core is
34% helium – a significant increase from the overall composition of 24% helium.
The rest of the Sun is called the envelope (in the Two-Layer Model) and is traditionally divided into an
inner radiative zone and an outer convective zone. Energy flows through the envelope. In the radiative
zone energy flows by photons scattering off the atoms in the envelope. In the convective zone energy
flows by creating large convective currents that carry the energy toward the surface much like the
warmed wax in a “Lava Lamp” rises to the surface.
3. How does nuclear fusion produce energy?
Nuclear energy produces energy by converting mass into energy following Einstein’s famous formula E =
mc2. In the case of the p-p chain a small amount of mass (0.047×10-27 kg) disappears with each complete
of the chain of reactions. This little bit of mass lost reappears as new gamma rays (i.e. energy).
Both fusion and fission nuclear reactions produce energy the same way – converting mass into energy.
4. Why does nuclear fusion need high temperatures and densities?
To fuse nuclei, you have to bring the individual nuclei very close together (on the order of 10-15 m
separation) so the nuclear strong force can bind the two nuclei together. Bringing two nuclei this close
together is difficult to accomplish because the nuclei are both positively charged and “like charges repel
each other”. To overcome this charge barrier (called the Coulomb barrier after August Coulomb a 17th
century French military engineer who experimented with electricity and discovered the law that describes
the repulsive force between two like charges, called Coulomb’s Law.) – back on topic, to overcome the
Coulomb barrier the nuclei must be moving very fast (about 610 km/s, fast enough to travel from NY to
LA in 8.2 seconds!). To get nuclei moving that fast, you need to heat them up to very high temperatures
(millions of Kelvins). So the high temperatures are needed for fusion to get the nuclei moving fast enough
for the strong force to bind the nuclei together.
The high densities are needed for fusion to increase the chances of collisions. Even if nuclei are moving
fast enough, if they are spread out (as in a low density gas) they are not likely to collide. No collision –
no fusion. By increasing the density to VERY high values, the fast moving nuclei are more likely to
engage in a collision and fuse.
5. Why is it so hard to develop nuclear fusion as a dependable power source on Earth?
It so hard to develop nuclear fusion as a dependable power source on Earth because the high
temperature and high densities required are very difficult to maintain. For example, any container you
but the high temperature hydrogen in melts! There is no material that can withstand the millions of
Kelvins of temperature that the gas has. Scientists and engineers are working on containing the hot gas
in a closed magnetic field – a magnetic bottle, as it were. Progress is being made, but stable fusion is not
yet within reach. So far the only way we can create fusion is in an uncontained nuclear fusion – with
catastrophic consequences!
6. Why will chemical reactions or gravitational contraction not work for powering the Sun?
Chemical reactions cannot power the Sun, because there are no known reactions that can produce the
extraordinary amount of energy the Sun produces each second for the length of time that the Sun has
been producing that energy. The Sun has had roughly (within 15%) the same luminosity for over the last
4 billion years. Any source of chemical energy would have “burned up” all the material being reacted.
Here are some interesting numbers: Combusting hydrogen gas releases about 1.5×108 Joules of energy
per kilogram of Hydrogen consumed. Nuclear fusion produces about 9×1016 Joules of energy per
kilogram of mass consumed. So… nuclear fusion produces about 600 million times more energy per unit
mass than chemical combustion reactions. If the Sun produced its energy by chemical reactions it would
have to “burn” through its mass 600 million times faster and, thus, have a lifetime 600 million times
shorter – too short to have been around for a long as the geologic record indicates.
7. What is the net result of the nuclear fusion chain process? Why does nature use the complicated chain process
instead of a one-step fusion procedure?
The net result of the proton-proton chain process is to transform four hydrogen atoms into one helium
atom, two neutrinos and half a dozen gamma rays, leaving the Sun slightly less massive than it was
before. This net result is accomplished through a series of nuclear reactions. This more comples process
is followed, rather than a one-step reaction, because the probability of four hydrogen atoms colliding
simultaneously is very rare and the velocity needed to get four nuclei close together is much larger (about
three times larger requiring temperatures that are nine times higher) that tose velocities to collide just
two hydrogen (or two deuterium atoms –heavy hydrogen). The “bar is lower” for this more round about
three step process that for a single one step process.
8. Where are neutrinos produced? What information can they tell you about interior conditions in the Sun?
Neutrinos are produced in the core of the Sun as a by-product of the p-p chain. Astronomers and
physicists can measure the energy of the neutrinos from the Sun and deduce rate of nucear reations and
the temperature of the Sun’s core. This is the primary experimental method for measuring the Sun’s core
temperature. The other method of determining the Sun’s core temperature is via stellar models.
9. What was the solar neutrino problem? How was the problem solved and what are the implications of that
solution?
The Solar neutrino problem was that the number of neutrinos measured from the Sun was about 1/3rd
those expected to be measured. That is, the number of solar neutrinos detected, by a variety of
experiments, was significantly less that the number predicted from theory. In short, there were too few
neutrinos from the Sun – far too few.
The problem of too few solar neutrinos was solved when theorists predicted that neutrinos come in three
different varieties (electron neutrinos, tau neutrinos and muon neutrinos) and that as a neutrino interacts
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(albeit very weakly) with electrons in the Sun, the neutrinos oscillate between the three different varieties.
That is the P-P chain produces only electron neutrinos, but 2/3rd’s of the electron neutrinos change into
tau and muon neutrinos as they exit the Sun. Theorists state that these “neutrino oscillations” are
predicted from the electromagnetic weak interaction, which I know nothing about, but I believe them.
So the “too few neutrino problem” was due to detecting only the electron neutrinos and not the tau and
muon neutrinos. Once the theory was better understood, the problem was no longer a problem.
10. How can you determine what the interiors of stars are like?
The interior of stars are determined principally by stellar models – that is computer programs into which
the physics of the star is encoded. One such model is The Standard Solar Model that predicts the interior
pressure, temperature, density and other parameters of the Sun. These stellar models are continually
improved as computers, computing techniques, and our physical understanding of stars improves.
In the case of the Sun, since it is close enough to study its surface in detail, solar seismology – the study
of pulsations and oscillations on the Sun’s surface –can reveal conditions within the solar interior.
11. What is being equilibrated in hydrostatic equilibrium? How does hydrostatic equilibrium explain why the
temperature and density increases inward toward the core of a star?
Hydrostatic equilibrium is a balance between outward thermal forces, controlled by the rate of nuclear
reactions, and inward gravitational forces, from the weight of the overlying layers of the star.
Hydrostatic equilibrium requires that temperature and density increases inward toward the core of a star
because at deeper levels there is more mass above and, therefore, more gravitational pressure on the
deeper levels. The increased pressure compresses the deeper levels and whenever a gas is compressed its
density and temperature increase. Thus with more weight on the deeper levels of a star, the deeper levels
become denser and hotter.
12. How does hydrostatic equilibrium control the fusion rate in the Sun?
Hydrostatic equilibrium controls the fusion rate in the Sun by the following sequence of processes.
Imagine that the fusion rate in the core of the Sun increases slightly. The increased fusion rate produces
more energy and the core expands slightly against the weight of the envelope. The expansion of the core,
however, causes the density of the core to slightly decrease, thus reducing the probability of collisions in
the core and the fusion rate drops slightly. The result is that the core returns to its initial equilibrium
state.
Now, imagine that the fusion rate in the core of the Sun decreases slightly. The decreased fusion rate
produces less energy and the core compresses slightly against the weight of the envelope. The
compression of the core, however, causes the density of the core to slightly increase, thus increasing the
probability of collisions in the core and the fusion rate rises slightly. The result is that the core returns to
its initial equilibrium state.
Thus, as a consequence of being in balance with the weight of the envelope, the core has a built in
thermostat that returns it to the initial state under small perturbations. As a result, the luminosity of the
Sun is quite uniform over long time periods and does not fluctuate wildly.
13. What would happen to the size of a star if its core steadily produced more energy than it did at some earlier time
(e.g., when a main sequence star becomes a red giant)?
If a star’s core steadily produced more energy than it did at some earlier time, then thermal forces from
the core would increase. Since the mass of the envelope does not change, the larger thermal forces from
the core lift the envelope increasing the overall diameter of the star and the star becomes a giant star.
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14. What would happen to the size of a star if its core steadily produced less energy than it did at some earlier time
(e.g., when a star stops fusing nuclei in its core)?
If a star’s core steadily produced less energy than it did at some earlier time, then thermal forces from
the core would decrease. Since the mass of the envelope does not change, the smaller thermal forces
cannot counter the gravitational forces of the weight of the envelope and the envelope crushes the core
decreasing the overall diameter of the star.
15. Do photons produced in the core zip right out from the Sun or does it take longer? Explain why.
Photons (i.e. gamma rays) produced in the core do not zip right out from the Sun. Instead, because
gamma rays interact so strongly with matter, they scatter (bounce) off every atom they encounter and in
the core of the Sun there are a lot of atoms pressed close together. The result is that the photons bounce
around and slowly work their way down the pressure gradient towards the Sun’s “surface” (i.e. the
photosphere). This process has two effects: (1) the photons lose a bit of energy with each bounce and exit
the Sun as lower energy visible and infrared photons, not the very high energy gammas rays that they
started as, (2) the bit of energy the photons lose on each “bounce” heats up the core and envelope to
millions of Kelvins. On average it takes several tens of thousands of years for a photon to bounce its way
out of the Sun, much like a ball in a pin-ball machine can take a long time to reach the bottom of the
board (Unless I played the ball when it invariably seems to miss every obstacle and zip straight to the
bottom.)
Instructor Assigned Topic:
1.
Answer the question below following our class lecture on how the Sun produces energy. This identical question will be on
the next exam – so do it correctly now.
Describe how the Sun produces energy by describing the net proton-proton chain, as
presented in class, including the origin of the particles in the net p-p chain reaction.
Include the effect of thermalization on the gamma rays and a summary of the twolayer model of the Sun’s interior.
Grading Rubric:Full credit awarded to responses that address the following points:
1.
The Sun produce energy by nuclear fusion through the Proto-Proton Chain
2.
Details of the reactions as covered in class: 41 H 12 He  2  6 , 4 points
i. Four hydrogen atoms, that were created during the Big Bang and the beginning of the Universe, combine
through a series of steps into one helium atom that is created by fusing the four hydrogens and produces
two neutrinos as a by-product of neutron creation and six gamma rays created from mass transformed into
energy.
b. Brief discussion of neutrinos, 2 points
i. Neutrinos are subatomic particles that are neutral in charge, travel near the speed of light, interact VERY
weakly with matter and have a VERY LOW mass, 100,000 times less massive than the electron. Neutrinos
are created when a proton and an electron combine to form a neutron and a neutrino as part of the p-p
chain: p+ + e- →n0+ ν.
c. Brief discussion of gamma rays including thermalization, 2 points
i. Gamma rays are high energy photons like X-rays but much higher energy. They travel at the speed of light
(because they are a form of light) and have no mass. Gamma rays STRONGLY interact with matter and
are deadly to biological systems. They are the original form of the Sun’s energy. A collisional process
called thermalization transforms the deadly gamma rays into benign visible and infrared photons as they
undergo countless collisions on their way out of the Sun. The origin of the gamma rays is from the
conversion of a small amount of mass that is lost during the p-p chain. This lost mass is converted into
energy via Einstein’s famous E = mc2 equation.
Mass Loss is associated with energy production, 2 points
a.
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The Sun produces its energy by converting its own mass into energy following Einstein’s famous equation E = mc2.
About 4.3 million metric tons of mass are converted into energy every second in the Sun’s core to maintain the
Sun’s luminosity.
The Two-Layer Model of the Sun, 4 points
a. The Sun can be divided functionally into two layers. The core is the inner layer that produces all the energy and the
envelope is the outer layer that, through hydrostatic equilibrium, holds the core together so it can continue to
produce energy. The Envelope and Core both participate in the thermalization of gamma rays.
b. The core is the inner 1/5th of the Sun’s radius, contains 1/3 the Sun’s mass, has a temperature of 15 million K and a
density of 100 g/cm3.
c. The envelope is the outerer 4/5th’s of the Sun’s radius, contains 2/3 the Sun’s mass, has a temperature of several
million K and a density of 1 g/cm3.
d.
a.
3.
Full credit for
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Define what H is and state its origin.
Define what He is and state its origin (including the origin of the neutrons).
Define and describe what ν’s are and state their origin.
Define and describe what γ’s are and state their origin.
Describe the collisional process by which gamma rays lose energy as they scatter their way out of the envelope transforming
from deadly high energy photons to benign visible and infrared (mostly) photons (i.e. thermalization).
State the Sun’s luminosity comes from the conversion of mass into energy through Einstein’s E = mc2 equation. State that the
Sun converts 4.3 million metric tons of mass into energy each second to maintain its luminosity.
Summarize the Two-Layer Model of the Sun properties of the core and envelope (functionality, radius, mass, density and
temperature).
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