Download Homework #4 (Ch. 16)

Astronomy 120
HOMEWORK - Chapter 16
The Sun
Use a calculator whenever necessary.
For full credit, always show your work and explain how you got your answer in full, complete
sentences on a separate sheet of paper.
Be careful about units!
Please CIRCLE or put a box around your final answer if it is numerical.
If you wish, you may discuss the questions with friends, but please turn in your own hand-written
solutions, with questions answered in your own way.
1. Chaisson Review and Discussion 16.4
What is luminosity, and how is it measured in the case of the Sun? (3 points)
Luminosity is a measure of the true brightness or total energy output of an object. For the
Sun, it can be measured by experimentally determining how much solar energy is received by
one square meter at the distance of the Earth from the Sun. This is then multiplied by the
surface area of a sphere whose radius is the semi-major axis of the Earth’s orbit.
2. Chaisson Review and Discussion 16.5
What fuels the Sun’s enormous energy output? (4 points)
The Sun’s energy output is fueled by nuclear fusion of hydrogen into helium. In this process
that takes place in the core of the Sun, 4 hydrogen atoms (really just protons) come together
and fuse to form a heavier element, helium. In this process, a small amount of mass is “lost.”
That missing mass has been converted into energy. According to Einstein’s famous equation,
E = mc2, a small amount of mass can become a large amount of energy.
3. Chaisson Review and Discussion 16.6
What are the ingredients and the end result of the proton-proton chain in the Sun? (4 points)
A total of 6 hydrogen atoms go into the proton-proton chain. What comes out is a helium
nucleus, two neutrinos, two positrons (which are quickly annihilated by colliding with
electrons), energy in the form of gamma rays, and two hydrogens. Thus, only 4 hydrogen
atoms are consumed to make the helium.
4. Chaisson Review and Discussion 16.9
What is helioseismology, and what does it tell us about the Sun? (3 points)
Helioseismology is the study of waves that ripple across the surface of the Sun. Some of these
waves travel from deep inside the Sun. Their appearance on the surface provides information
about the interior of the Sun that cannot otherwise be observed, such as temperature, density,
and rotation speed.
5. Chaisson Review and Discussion 16.11
Describe how energy generated in the solar core eventually reaches the Earth. (4 points)
The solar radiation is first produced in the core of the Sun, largely in the form of gamma
rays. Because the gas in the core is totally ionized, it is transparent to radiation and so the
radiation passes through it freely. But as we get closer to the surface, the temperature drops,
and more and more of the gas is not ionized or only partially ionized. Such a gas is opaque to
radiation. At the outer edge of the radiation zone, all of the radiation has been absorbed by
the gas. This heats the gas and it physically rises, while cooler gas from the surface falls.
This is the region of convection. The energy is transported by convection to the photosphere.
Here, the density of the gas is so low that radiation can freely escape into space, and travel
in a straight line to Earth.
6. Chaisson Review and Discussion 16.12
Why does the Sun appear to have a sharp edge? (3 points)
Virtually all the visible radiation we receive from the Sun comes from a thin layer called the
photosphere. It is only 500 km thick; a small fraction of the Sun’s radius. The gas below the
photosphere is too thick for light to escape, and the gas above is too thin to absorb and emit
significant quantities of light. Light can only escape from this narrow region, so the Sun has
a very well-defined edge.
7. Chaisson Review and Discussion 16.14
What is the solar wind? (3 points)
Because the corona of the Sun is hot, some of the gas particles are traveling fast enough to
escape the gravity of the Sun. The gas is mostly composed of the separated components of
ionized hydrogen, protons and electrons. This flow of high-speed particles away from the Sun
is known as the solar wind.
8. Chaisson Review and Discussion 16.15
Why do we say the solar cycle is 22 years long? (3 points)
Activity in the Sun’s magnetic field creates the cycle of sunspots seen on the Sun. The solar
magnetic field reverses itself every 11 years, so it takes 22 years to go through an entire cycle
of magnetic reversals.
9. Chaisson Review and Discussion 16.16
What is the cause of sunspots, flares, and prominences? (4 points)
All of these phenomena are caused by activity in the magnetic field of the Sun. Sunspots are
caused by kinks or loops of magnetic field extending through the lower atmosphere. These
areas of concentrated magnetic field repel hot material trying to rise up from the Sun’s
interior, so the section of the Sun underneath the knot cools off and darkens. Flares, by
contrast, are areas where large amounts of energy are released in a short amount of time.
Their origin is mysterious, but they are somehow connected to instabilities in the magnetic
field. Prominences are caused by material ejected from the Sun’s surface that follows along
huge loops of magnetic field that carry the luminous gas far above the solar surface.
10. Chaisson Review and Discussion 16.18
Why are scientists so interested in solar neutrinos? (4 points)
Neutrinos are produced in the proton-proton chain, which occurs in the core of the Sun. The
neutrinos pass unimpeded through the Sun at nearly the speed of light. So neutrinos, in a
sense, allow astronomers to directly observe the core of the Sun and the processes that occur
there, almost as they happen. For a long time, astronomers were puzzled because the Sun did
not seem to be producing as many neutrinos as predicted. A combination of more
sophisticated measurement techniques and a better understanding of neutrino behavior has
solved this problem, however, and predictions are more in line with observations.
11. How do we know that the photosphere marks the top of the sun’s convective zone? (3 points)
We see photospheric granulation, which marks the top of the convective cells that are
carrying energy upward from the solar interior.
12. Thanks to Copernicus, we know how to find the distances to the planets in terms of the
Earth-Sun distance (1 A.U.). Venus is 0.723 A.U. from the Sun. If we aim radar signals at
Venus and allow them to bounce off its surface and return to Earth, we receive those signals
278.12 seconds later. If light (radar) travels at 2.98  108 m s , how far away is the Sun from
the Earth in meters? How many kilometers? How many miles is that if 1 mile = 1609
.
km?
You will need to make use of the relation: d  vt , where d is distance, v is velocity and t is
time. (6 points)
Sun
0.723 A.U.
Venus
Earth
1 A.U.
the distance from Earth to Venus is:
d = vt = 2.98 x 108 m/s x
278.12 sec
= 4.14 x 1010 m
2
(one-way trip)
the distance from Earth to Venus in A.U.’s is: 1 A.U. – 0.723 A.U. = 0.277 A.U.
then set up a ratio:
4.14 x 1010 m
X
=
0.277 A.U.
1 A.U.
X=
and solve for X.
4.14 x 1010 m
0.277 A.U. x 1 A.U.
X = 1.495 x 1011 m = 1.495 x 108 km
1 mile
1.495 x 108 km x 1.609 km = 9.29 x 107 miles
13. The Sun's angular diameter is 32 arc minutes, and it lies 1 A.U. away from the Earth. If we
convert arc minutes to degrees, and A.U.'s to kilometers, we can figure out the Sun's diameter
in terms of kilometers. If there are 1.496 x 108 km per A.U., what is the Sun's real diameter
in kilometers? in miles? (6 points)
1 A.U.
Sun
Earth
32’
tan  
the opposite side of the 16’ angle is the sun’s radius, so…
opposite = tan x adjacent
first, convert arcmin to degrees….
1o
16’ x 60' = 0.266 o
or the sun’s radius = tan(0.266o) x 1.496 x 108 km = 6.96 x 105 km
1 mile
6.96 x 105 km x 1.609 km = 4.33 x 105 miles
multiply by 2 for the diameter…
opposite
adjacent
Sun’s diameter = 1.392 x 106 km = 8.66 x 105 miles
14. Newton's second law of motion is:
F  ma
where F is the force exerted on an object which has mass m and therefore is accelerated by an
amount a.
Newton's Universal Law of Gravitation is:
F
GMm
r2
where F is the gravitational force between two masses; one large (M) and the other smaller
(m), and r is the distance between their two centers of mass.
if a is the acceleration due to gravity, then ....
F
GMm
 ma
r2
(*)
By setting these two equations equal, we can figure out the centripetal acceleration of the
Earth about the Sun, or conversely, if we know the Earth's acceleration, we can figure out
the Sun's mass.
Centripetal acceleration is a 
v2
, where v is the linear velocity
r
of an object in motion a distance r from the center.
Finally, the Earth's velocity can be found from d  vt . where d is equal to the Earth's
circumference about the Sun, and t is the period (1 year). The formula for circumference is
C  2r . So....
v
d 2r
, then

t
t
v 2 4 2 r 2 4 2 r
a

 2
r
rt 2
t
Thus, if we substitute this value of the acceleration into the above starred (*) equation for F,
we can solve for the Sun's mass (M).
a) Solve for the Sun's mass (6 points)
GMm
= ma
r2
and
a = 42r/t2
upon substitution…..
GMm
= m42r/t2
r2
the m’s cancel and we solve for M (the sun’s mass)
M = 42r3/Gt2
1 year = 60 sec/min x 60 min/hour x 24 hours/day x 365 days/year = 3.15 x 107
seconds
M =
4 x (3.14)2 x (1.496 x 1011m)3
= 2 x 1030 kg
2
Nm
6.673 x 10-11 2 x (3.15 x 107 s)2
kg
mass
; (3 points)
volume
4
and the volume of a sphere is V  r 3, where r is the radius of the sphere. Use your
3
value for the Sun's mass from above and the value for the Sun's radius from problem
#13, to figure out the Sun's average density.
b) Density   
 = 3M / 4r3
=
3 x 2 x 1030 kg
= 1.416 x 103 kg/m3
4 x 3.14 x (6.96 x 108 m)3
15. Chaisson Problem 16.2
Use Wien's law:
 max 
2.9  10 3
meters
T
where max is the wavelength at which the most energy is emitted by an object of
temperature T in Kelvin.
to determine the wavelength corresponding to the peak of the blackbody curve
a) in the core of the Sun, where the temperature is 107 K (3 points)
7
-7
For a core temperature of 10 K, λmax = 2.9 ×10 mm = 0.29 nm. These are X-rays.
b) in the solar convection zone (105 K) (3 points)
5
-5
For the 10 K temperature of the convection zone λmax = 2.9 ×10 mm = 29 nm. This is
far ultraviolet.
c) just below the photosphere (104 K) (3 points)
4
-4
For the 10 K of the lower photosphere λmax = 2.9 ×10 mm = 290 nm. This is the near
ultraviolet, almost in the visual.
d) what form (visible, infrared, X-ray, etc.) does the radiation take in each case?
See above.