2.)
Consider an object, for example, a cloud of gas,
that is quickly contracting due to gravity's inward pull.
a.) Is it in hydrostatic equilibrium?
Answer: No.
b.) Why or why not?
Answer:
In order to have hydrostatic equilibrium, the force inward (due
to gravity + pressure from any overlying material) has to balance
the force outward (due to thermal pressure).
You can assume that if the cloud contracts, there is an
inward acceleration at some time during the contraction.
If there is an inward acceleration,
then the inward directed forces
(gravity + force due to pressure of any overlying material)
is not completely balanced by the outward-directed force
(force due to pressure of the underlying material.)
3.)
a.) If you can see through the chromosphere and the corona, then
they must be: (pick one)
optically thin or optically thick
Answer: Optically thin
b.) If you cannot see through the convection zone or radiation zone, then
they must be: (pick one)
optically thin or optically thick
Answer: Optically thick
4.) How does the convection zone compare with the radiative zone in terms of a.) Temperature Answer: The convection zone is cooler than the radiative zone. b.) Pressure Answer: The convection zone has a lower pressure than the radiative zone. c.) Gas motion Answer: The convection zone is more active -- more of the material is moving faster and further -- than the radiative zone.
5.) The part of the Sun where mass is converted into energy is
the:
a.) Chromosphere
b.) Convection zone
c.) Core
d.) Corona
e.) None of the above
Answer: C, Core
6.)
a.) Where is the Sun's chromosphere?
Answer: The Sun's chromosphere is between the Sun's surface
and the corona. It is part of the Sun's "atmosphere".
b.) Where is the Sun's convection zone?
Answer: The convection zone is between the radiation zone and
the photosphere. Thus, it extends from 2/3 of the radius of
the Sun to almost the visible surface of the Sun.
c.) Where is the Sun's core?
Answer: The core is in the center of the Sun. It extends from
the center to 1/4 of the radius of the Sun.
d.) Where is the Sun's corona?
Answer: The corona and the chromosphere are the Sun's "atmosphere".
The chromosphere is closer to the surface of the Sun and
the corona is further from the surface of the Sun.
The corona extends for 100s of thousands of kilometers.
e.) Where is the Sun's photosphere?
Answer: The photosphere is the visible surface of the Sun.
f.) Where is the Sun's radiation zone?
Answer: The radiation zone is sandwiched between the core and
the convection zone. Thus, it extends from 1/4th of the radius
of the zone to 2/3 of the radius of the Sun.
7.) What are the steps in the nuclear reactions that convert hydrogen into helium + gamma rays + positrons + neutrinos + kinetic energy? Answer: Step A: 2 protons fuse, making a deuterium nucleus (i.e. a nucleus having a proton + a neutron), a positron, and a neutrino. (do this twice to make 2 deuterium nuclei, 2 neutrinos, and 2 positrons) Step B: 1 proton and 1 deuterium fuse to make a 3He nucleus (i.e. a nucleus having 2 protons and 1 neutron) and a gamma ray. (do this twice, making 2 3He nuclei and 2 gamma rays) Step C: 2 3He nuclei fuse, making 1 4He nucleus and 2 protons During the process, mass is converted to energy.
8.)
Every second, 4 x 109 kg of mass is converted into
energy in the Sun.
Use E = m c2 to calculate the amount of energy per second
produced by the Sun as a result.
Answer:
energy produced per second = mass converted per second x c2
energy produced per second = (4 x 109 kg) x (3 x 108 m/s)2
energy produced per second = (4 x 109kg) x (9 x 1016 m2/s2)
energy produced per second = 3.6 x 1026 kg m2/s2)
Note that a kg m2/s2 is the same as a Joule. So,
energy produced per second = 3.6 x 1026 Joules
9.) Use several sentences to describe:
a.) the process of
radiative diffusion in the core and radiation zone and
Answer:
The core and radiation zone are very dense,
so photons cannot travel very far without hitting
material. Most of the time, they hit and bounce off of electrons.
(Sometimes they hit and
bounce off of positive ions. Occasionally,
they are absorbed by an atom that hasn't lost
all its electrons
and later the energy is re-emitted as new photon(s)).
Most of the time,
after the collision, the photons
end up traveling in a different
directions than they started.
So, the photons "bounce" around
within the Sun. After many collisions, most photons
will be further from the center of the Sun
than they started. As time goes by, they tend to
get further and further from the center of the Sun.
As they move outwards, they encounter
cooler material.
Note that the temperature is proportional to the square of the
velocities
of the atoms/ions/electrons and as the temperature
drops, atoms tend to recapture their electrons
(though, at these
temperatures, only heavy atoms can do that). As a result
of colliding with
(or being absorbed and re-emitted by) the cooler (slower) material, the
photons give some
of their energy to the material, leaving
less for themselves. So, the photons slowly transform
from very high energy (gamma rays) to less high energy (X-rays)
as they move outwards
from the Sun's core. Eventually, the photons reach the lower parts
of the convection zone,
where the material absorbs them more easily.
b.) the process of convection in the convection zone.
Answer: The bottom of the convection zone is the part closest to the
Sun's center.
Photons radiate
outwards from the core to the radiation zone and
then from the radiation zone to the bottom
of the convection
zone. The material at the bottom of the convection zone
absorbs the photons
and, as a result, heats up. "Hot spots"
(regions where the temperature is a little hotter
than the surrounding gas)
at the bottom of the convection zone buoyantly rise, reaching
the top of the convection zone, while slightly cool regions near
the top of the convection
zone fall down to the bottom of the convection
zone. This is the process of convection.
10.) (adapted from textbook) Which one of the following choices is sensible and true? a.) A sudden temperature rise in the Sun's core is nothing to worry about because conditions in the core will soon return to normal. b.) If fusion in the solar core ceased today, worldwide panic would break out tomorrow as the Sun began to grow dimmer. c.) Astronomers have recently photographed magnetic fields churning deep beneath the solar photosphere. d.) Neutrinos probably can't harm me, but just to be safe I think I'll wear a lead vest. e.) By observing solar neutrinos, we can learn about motion in the Sun's convection layer. Answer: "a" is the only sensible sentence. "b" doesn't work because of the timescale. The energy from a fusion event in the Sun's core takes a million years to reach the Sun's surface. So, if fusion ceased today, it would take much longer than 1 day for humans to see the effect. "c" doesn't work because we can only see light from material at and above the solar photosphere. We observe magnetic fields by seeing their effects on light, so we can only see the magnetic fields at and above the solar photosphere. "d" is wrong because the neutrinos would go right through both your body and the lead vest. "e" is wrong because neutrinos don't tell us about the Sun's convection layer -- they tell us about fusion rates in the Sun's core.
11.) (adapted from textbook)
During its life, the Sun will burn up the hydrogen in
the core to produce energy, helium, and other particles.
During its lifespan (i.e. starting from birth),
the Sun will burn 1.95 x 1029 kg of hydrogen in its core.
The Sun burns 600 billion kg (that is 600 x 109 kg)
of hydrogen every second. If the burn rate were to be constant,
how long would it take for the Sun to burn through the hydrogen
in its core? Note that this is the total lifetime, starting
from the time of the Sun's birth, not starting from now.
Please provide your final answer in units of years.
Answer:
amount of hydrogen burned = rate x time
time = amount of hydrogen burned / rate
time = 1.95 x 1029 kg of hydrogen / (600 x 109 kg of hydrogen/sec)
time = 3.25 x 1017 seconds
Now, convert that to years:
time = 3.25 x 1017 seconds x (1 min/60 sec) x (1 hr/60 min) x (1 day/24 hr) x (1 yr/365 day)
time = 1.0 x 1010 years
12.) The hydrogen outside the core doesn't burn. Why not?
Answer: Although the pressure in the core is high enough
for hydrogen to fuse, the pressure
outside of the core is not high enough for hydrogen to fuse.
If you want a more
technical
answer, here it is. The hydrogen protons repel
each other (opposite charges repel). So, in order
for the hydrogen protons to get close enough to each other for
the strong force (which is attractive)
to overwhelm
the electrostatic repulsion, the protons have to be approaching
each other at
extremely fast speeds. These fast speeds
imply very high temperatures.
High temperature x high density => high pressure. The core of the sun has high enough
temperature
and density, (thus, enough pressure) for fusion, but the regions
outside of the core do not.
13.) Write a several-sentence long description of the
magnetic fields on the Sun's surface,
including how they are affected by the convection layer,
what they do to electrons and ions, and their role in making sunspots.
Answer:
Magnetic fields are threaded through the convection layer.
Let's think of the magnetic field
like a rope or a thread.
Occasionally, a portion of the field will pop through the
photosphere.
So, a loop of magnetic field will extend
into the chromosphere.
Most of the gas is ionized into electrons and positive ions.
The magnetic field exerts a force
on moving electrons and positive ions, and this force
prevents them from moving away from the field.
So, the magnetic field
"ropes" or "threads" trap electrons and ions, preventing them
from leaving.
Similarly, the magnetic field also prevents other
electrons or ions from entering. While trapped,
the plasma radiates
away some of its energy, thus cools down. So the plasma
in the ropes
becomes cooler than the plasma outside the ropes.
We can see the effect when we see sunspots.
These dark spots are where the magnetic ropes cross the photosphere.
The material here is cool
(because of explanation above) and
so emits less light than the surrounding, hotter gas, and so
looks darker than the surrounding, brighter gas.
14.)
a.) Do neutrinos interact easily with atoms?
Answer: No
b.) How do we detect some of the neutrinos made by the Sun?
Answer: One type of detector is a tank of perchloroethylene
(a dry-cleaning solution that contains chlorine).
A small fraction of the neutrinos that fly into the tank will
interact with the nucleus of one of the chlorine atoms.
The interaction converts the neutrino plus one of the
neutrons in the chlorine atom into a proton and an electron.
Thus, the reaction is:
neutrino + 37Cl (which has 17 protons and 20 neutrons) =
37Ar (which has 18 protons and 19 neutrons) + electron.
Periodically, the scientists working with this experiment
search for the argon atoms. In order for an argon atom to be in
the tank one of the neutrinos must have interacted with a chlorine atom.
So, the existence of an argon atom is a "detection" of a neutrino.
There are other neutrino detectors that use the same principle,
but use a different type of atom. Often that atom is
71Gallium, which converts to 71Germanium.
15.)
What would happen to the core of the Sun if the fusion rate
were to momentarily and suddenly increase by 10%?
(For your answer, consider how the fusion rate affects the production of
energy, how the production of energy affects the temperature of the gas,
how the temperature affects the pressure, how a change in pressure
affects the size of the core, how a change in the
size of the core affects the temperature, and how
the temperature of the core affects the fusion rate)
Answer:
If the fusion rate were to momentarily increase, then the core would
convert more mass into energy and so become hotter.
Thus, the temperature in the core would increase.
Pressure = density x constant x temperature. So, the
pressure would increase.
With a larger pressure, the core would be able to push outwards with
greater strength and thus to expand.
The expansion would cool the core.
Thus, the temperature would drop back to the original temperate
and the fusion rate would drop back to the original fusion rate.
An important consequence is that a slight change in the core doesn't
get out of hand -- it doesn't cause the core to get hotter, and
hotter and hotter, or vice verse.