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Astronomy Exam #3 Fall 2012
Name_______________________
Class Meeting Time _________________________
1) The mass of the Sun is closest to which of the values below?
A) 300 Earth masses
D) 300,000 Earth masses
B) 3,000 Earth masses
E) 3,000,000 Earth masses
C) 30,000 Earth masses
2) The Sun is approximately how much larger in radius than the Earth?
D) 100,000 Earth radii
A) 100 Earth radii
B) 1,000 Earth radii
E) 1,000,000 Earth radii
C) 10,000 Earth radii
3) The density of the Sun is closest to the density of ….
A) Iron, 8 g/cm3
B) Rock, 5 g/cm3
C) Water, 1 g/cm3
D) Ice, 0.9 g/cm3
E) Air, 0.001 g/cm3
4) The surface temperature of the Sun is….
A) 32,000 K
B) 15,500 K
C) 7,200 K
D) 5,800 K
E) 1,600 K
5) The core of the Sun has a temperature of…
A) Unknown
B) 15 Million K
C) 50 Million K
D) 100 Million K
E) 1.4 Billion K
6) The Sun can be functionally divided into two layers. Describe the physical characteristics of the two layers
in the space below.
The Core of the Sun
Radius range: bottom 1/5th of the
Sun’s radius
Mass: about 1/3rd’s of the Sun’s mass
The Envelope of the Sun
Radius range: top 4/5th of the Sun’s
radius
Mass: about 1/3rd’s of the Sun’s mass
Temperature: 15 million K
Temperature: about 5 million K
Density: about 100
g/cm3
Average Density: about 1 g/cm3
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7) In a sentence or two, describe the functional difference between the core and envelope. In other words,
what purpose does the core serve in the star’s function and what purpose does the envelope serve in the
star’s function?
The core (1) produces almost all the Sun’s energy and (2) holds up the envelope preventing collapse.
The envelope (1) keeps the core under pressure to allow fusion to continue and (2) thermalizes gamma
rays as they pass through.
8) The picture below portrays a proposed new space mission to establish a Hipparchus-like spacecraft in orbit
around Jupiter for the purpose of measuring stellar parallaxes. In a couple of sentences, explain why a
mission like this would be superior to the original Hipparchus spacecraft in orbit around the Earth during the
early
Proposed orbit of new
1990’s.
spacecraft orbiting
Jupiter
Jupiter
Nearby star
Earth
Sun
A mission like this would be superior to the original Hipparchus spacecraft in orbit around the
Earth during the early 1990’s because, with the larger baseline of Jupiter’s orbit at 5.2 AU, all
stars would have parallax angles about 5 times larger than those measured from Earth. Thus we
would be able to measure the distance to stars with spectroscopic parallax about five times farther
away out to 2500 pc.
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The four simulated images of a star field appear below with the year which they represent. One of the stars is
consistently changing its position relative to the other stars.
9) Identify the star that is changing its position by circling it in each of the simulated images.
10) These simulated images illustrate what type of stellar motion. Choose from the list below.
A) Spaced velocity
D) Proper motion
B) Diurnal motion
E) Annual motion
C) Radial velocity
The table below lists the properties of five stars. Answer the questions following the table based on these
parameters.
Apparent
Absolute
Spectral Type and
Designation
Magnitude, m Magnitude, M Luminosity Class
-0.7
-3.1
F0 II
α
0.0
0.5
A0 V
β
+.9
-4.5
M1 I
γ
0.45
-1.0
B3 V
δ
12.3
14.8
M5 V
ε
11) Which star is the hottest α
12) Which star is the brightest? δ
13) Which star is the least luminous? ε
14) Which star is the largest in radius? γ
15) Which star has a temperature of about 2,500 K? ε or γ
16) Which star is closer star α or star γ? Explain your reasoning. α is closer because (1) α distance
modulus is smaller than γ’s, or because (2) α is less luminous but brighter than γ.
Either reason puts α closer.
17) How many times more luminous is star δ compared to the Sun? δ is 6=5+1 magnitudes more
luminous than the Sun, so it is 100×2.5=250 times more luminous than the Sun.
18) Explain why the star γ is so luminous? γ is so luminous due to its very larvge radius as a
supergiant (luminosity class I).
19) What is the approximate luminosity of star  compared to the Sun? Since ε is 10 magnitudes less
luminous than the Sun, its luminosity is 0.0001 LSun.
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20) What are molecular clouds and how are they related to star formation? Answer in a few sentences? Be
quantitative.
Giant molecular clouds are very large clouds of gas and dust in the inter-stellar medium. These
clouds are composed mostly of H2 and He with trace amounts of other gases and microscopic pieces of
rock and ices that astronomers collectively call dust. These GMC’s can be 300 ly in diameter and
have masses equivalent to 300,000 solar masses. They provide the raw material for stars to form if
their temperatures are below 10 K. One single GMC can form thousands or stars. They are the
birthing ground of stars.
21) At what wavelength is best to observe a protostar when it is still accreting gas.
A) Radio wavelengths
D) Ultraviolet wavelengths
E) X-Ray wavelengths
B) Infrared wavelengths
C) Visual wavelengths
22) What generates the heat in a protostar?
A) Proton-proton chain
B) Core helium fusion
C) Nearby O & B stars
D) Gravitational contraction
E) Radioactive elements
23) What determines the upper mass limit of a star? Answer in a few sentences.
The upper mass limit of a star isa about 100 solar masses. This is the upper limit because any giant
molecular cloud fragment that is more massive than this 100 solar mass limit will break into smaller
fragments due to the large amount of thermal energy it would generate as it rapidly collapses.
24) What determines when a star becomes a main sequence star?
A) Core helium fusion begins.
B) Supernovas ignite the star.
C) Gravitational contraction begins heating the star.
D) Radioactive elements cause the star to glow.
E) Proton-proton chain begins in its core.
25) Define the acronym ZAMS and carefully sketch in the location
of the ZAMS line in the adjacent HR diagram?
ZAMS = Zero Age Main Sequence
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26) What factors determine how long a star stays on the main sequence?
A) Its temperature
D) Its spectral type
B) Its mass
E) All of the above
C) Its luminosity
27) Why do high mass stars have a shorter time on the main sequence than low mass stars? Answer in few
sentences?
High mass stars have higher central pressures that drive faster fusion rates. The faster fusion rates
“burn” the mass of star faster and have tremendously greater luminosities than lower mass stars..
Even though the high mass stars are more massive they burn their mass much much faster than lower
mass stars. Thus the lifetimes of high mass stars are relatively short due to their astronomical
luminosities.
28) How does a star’s luminosity change during star’s main sequence lifetime?
A) It decreases slowly.
D) It varies in a periodic manner.
B) It remains constant.
E) None of the above.
C) It increases slowly.
29) What makes a star move off the main sequence?
A) An iron core forms
B) Neutrinos flood from the core disrupting it.
C) Shell Helium burning begins
D) Core Hydrogen burning ends
E) Gravitational collapse ceases.
30) Connect with arrows the phases of stellar evolution the Sun will go through in the correct chronological
order on the diagram below.
5
6
Planetary Nebula
White Dwarf
3
Yellow Giant
a.k.a. Horizontal Branch
4
2
1
Red Giant
Main Sequence
Mira Variable
a.k.a. Red Super Giant
31) What is a planetary nebula? Answer in a few sentences.
A planetary nebula is the ejected envelope of a low mass star at the end of its lifetime. The nebula is
created by the last episode of shell He-burning in a flash that ejects the envelope and exposes the dead
stellar core of hot carbon that remains after planetary nebula has disappeared from view after about
10,000 years. Most planetary nebula are about 1 light year across.
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32) What is left when a planetary nebula dissipates?
A) A protostar
B) A main sequence star
C) A white dwarf
D) A giant star
E) A brown dwarf
33) What is the energy source for a white dwarf?
A) Proton-proton chain
B) Core helium fusion
C) Nearby O & B stars
D) Gravitational contraction
E) None of the above.
Refer to H-R diagram illustrating the evolutionary track
of a 1 solar mass star to the right to answer the following
questions.
34) Which of the objects listed below would be observed
along the portion of the track marked (d)?
A) White dwarf
B) Planetary Nebula
C) Red Giant
D) Red Super Giant (Mira Variable)
E) None of the above
35) Which of the methods of energy production is at
active along the portion of the track marked (b)?
A) Core H-burning
B) Core He-burning
C) Shell H-burning
e
-10
d
-5
0
M
b
c
+5
a
+10
+15
f
+20
O B A
F G
K
M
D) Shell He-burning
E) Proton-proton chain
36) At the portion of the evolutionary track marked c, what is the source of the star’s energy?
A) Core H-burning
D) Shell He-burning
E) Proton-proton chain
B) Core He-burning
C) Shell H-burning
37) At the stage of evolution marked f, what is the composition of the object (i.e. what is it made of)?
A) Hydrogen
D) Neon
B) Helium
E) Iron
C) Carbon
38) In a few sentences explain how the method of energy production employed along path d creates the object
found along path e.
Along the path d the star is producing energy by shell He-burning in flashs. The last helium flash is
strong enough to lift off the envelope of the star and create a planetary nebula – the object created at
e.
39) Describe the properties of a typical O or M main sequence star as completely as you can in a few sentences
below. See the slide from the PowerPoint presentation Properties of Main Sequence Stars on the next
page. You could describe either an O or an M star.
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Please identify the objects that appear in the images numbered 1 to 5 below
40) The object indicated in image #1 is a(n)
A) Open Cluster
D) OB Association
B) Bok Globule
E) Giant Molecular Cloud
C) Dense Core
F) HII Region
G) Planetary Nebula
H) White Dwarf
41) The object indicated in image #2 is a(n)
A) Open Cluster
D) OB Association
B) Bok Globule
E) Giant Molecular Cloud
C) Dense Core
F) HII Region
G) Planetary Nebula
H) White Dwarf
42) The object indicated in image #3 is a(n)
D) OB Association
A) Open Cluster
B) Bok Globule
E) Giant Molecular Cloud
C) Dense Core
F) HII Region
G) Planetary Nebula
H) White Dwarf
43) The object indicated in image #4 is a(n)
A) Open Cluster
D) OB Association
B) Bok Globule
E) Giant Molecular Cloud
C) Dense Core
F) HII Region
G) Planetary Nebula
H) White Dwarf
44) The object indicated in image #5 is a(n)
A) Open Cluster
D) OB Association
E) Giant Molecular Cloud
B) Bok Globule
C) Dense Core
F) HII Region
G) Planetary Nebula
H) White Dwarf
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45) In a list of bulleted points, describe the character of
the brightest stars in the sky as completely as you
can. Use the HR diagram at the right.




The 91 Brightest Stars
-10
Absolute Magnitude
The HR Diagram of the 91 brightest stars in the sky
appears to the right. The Main Sequence appears as
a dotted line and the Sun appears as the gray
triangle symbol on this HR diagram.
-5
0
5
10
15
The brightest stars in the sky appear to fall
20
O B
A
F
G
K
M
into two basic types; hot main sequence stars
Spectral Type
and cool giant stars.
The hot main sequence stars appear to be mostly B and A spectral type with an absolute
magnitude between +2 and -5. This range in absolute magnitudes corresponds to a range in
luminosity of between 16 and 10,000 solar luminosities. These stars will have a short main
sequence lifetime compared to the Sun’s lifetime. Some as short as a few million years.
The giant stars appear to be mostly G and K spectral types with luminosities over 100 solar
luminosities. These giant stars are at the end of their evolution and will shortly perish.
Although it appears on this HR diagram that the majority of bright stars are hot main sequence
stars, in truth about 75% of the bright stars are giant stars near the end of their evolution.
Use the HR diagram below to answer the following questions. Note: No calculations are needed.
46) The bright star Deneb has a luminosity of 54,000 solar
luminosities (M=-7.0) and a temperature of 8,525 K
(spectral type A2). What is its approximate radius?
HR Diagram
30,000 K
-10
Vega’s approximate radius is 80 solar radii. See
HR diagram.
6,000 K
5,250 K
3,800 K
1,000 R
17.5 M
10 Solar masses
100 R
5.9 M
Absolute Magnitude
Sirius b would have an absolute magnitude of about
-13. This means Sirius b is 8 magnitudes less
luminous than the Sun. Since 8 = 5+1+1+1 then
Sirius is 100×5×5×5=125,000 times less luminous
than the Sun.
7,200 K
Vega
60 M
-5
47) The companion star to the brightest star in the sky is
designated Sirius b. It has a spectral type of A2 and has
a radius of 0.01 that of the Sun. What is its
approximate luminosity?
9,500 K
2.9 M
0
1.8 M
10 R
1.2 M
0.1 R
1.0 M
5
Sun
.67 M
1 R
0.01 R
10
.21 M
Sirius b
0.001 R
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48) What would be the spectral type of a main sequence
star with 10 times the mass of the Sun?
20
The spectral type of a main sequence star with 10
times the mass of the Sun would be a B2 or B3
spectral type.
O
B
A
F
G
K
M
Spectral Type
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49) On this page, write an essay that describes how the Sun produces energy by explaining the net proton-proton
chain reaction, define the symbols in the net reaction, stating the origin of the terms in the net reaction, and
describing the significance of thermalization of solar gamma rays.
Points needed to be made for full credit

The net p-p chain is 411 H 142 He  2  6

The hydrogen ( 11 H ) is primordial created during the Big Bang.

The helium ( 42 He ) is created in the Sun’s core by fusing the four hydrogen atoms.

The neutrinos (ν) are subatomic nearly-massless neutral particles that travel near the speed of
light and only weakly interact with matter passing through most matter without hindrance.
They are created as a byproduct of neutron creation when an electron and proton combine to
form a neutron.

The gamma rays (γ) are high energy photons like X-rays but higher in energy. They are
massless and travel at the speed of light. Gamma rays interact strongly with matter and are
deadly to biologic systems. They are created from a small amount of mass loss during each p-p
chain completion as described by Einstein’s famous equation E=mc2. The Sun is converting
4.25 million metric tons of mass into energy every second to maintain its luminosity.

Deadly gamma rays are transformed by countless collisions within the envelope of the Sun as
they beat their way to the surface losing a bit of energy with each collision. The result is that
the energy lost keeps the envelope very hot (millions of K) even though it produces no energy
and the gamma rays leave the Sun as benign visible and IR photons mostly.
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
Problems: Please answer in the space below each problem.
50) When the Sun expands to be a red giant, its radius may be 150 RSun. What will the average density of the
Sun be at that time?

M
M
1  M Solar
1
M Solar
1





 Solar  3  10 7  Solar
3
3
4 3 4
4
V
R
 150  RSolar 3 150
 RSolar 3 150
3
3
3
The Sun’s density as a red giant will be only 3×10-7 its present density.
51) A star is observed to have a stellar parallax angle of 0.037”. What is the distance to the star? Show your
work.
Dpc 
1
1

 27 pc  80 ly
p 0.037
The distance to the star is about 27 parsecs or 80 light years.
52) The bright star Acrux in the constellation Crux has a
temperature of about 29,000 K. At what wavelength
does it emit light most strongly? Using the
electromagnetic spectrum below comment on
whether humans could live safely around this star?
T
2.9  106 K  nm
 Max 
Max
The Electromagnetic Spectrum
Region
Wavelength Range
Gamma Rays
X-Rays
Ultra-violet Rays
Visible Light
Infrared Radiation
Microwaves
Radio Waves
Less than 0.01 nm
0.01 nm
to 10 nm
10 nm
to 380 nm
380 nm
to 740 nm
740 nm
to 200,000 nm
200,000 nm to 5 x 106 nm
Greater than 5 x 106 nm
2.9  106 K  nm 2.9  106 K  nm

 100 nm
T
9,000K
The star will emit its maximum radiation at 100 nm which is deep into the UV portion of the
electromagnetic spectrum. This would not be a safe star live around because of the very high flux
of UV hard radiation.
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53) The star Vega has a luminosity of 59 L and a temperature of 9,520 K. What is its radius compared to the
Sun?
2
4
This is a Stefan-Boltzmann Law problem; LVega  4RVega
. Taking the ratio of this equation
 TVega
and the same equation applied to the Sun yields
LVega
LSun
LVega
LSun
LVega
LSun

4
2
TVega
4RVega

4
2
TSun
4RSun
4
2
TVega
RVega
 2  4
RSun TSun
R
  Vega
 RSun



2
2
R
59   Vega
 RSun
  9,520 
  

  5,800 
R
59   Vega
 RSun



R
59   Vega
 RSun

  1.64 4

2



T
  Vega
 TSun
 9,520 


 5,800 
4
4
4
2
2
R 
59   Vega   7.26
 RSun 
RVega
59
 2.85

7.26
RSun
Vega’s radius is 2.85 times larger than the Sun’s radius.
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Astronomy Formula and Constants Sheet for Exams
Conversions
Formulas
A
L

2D
360 
Main Sequence Lifetime t 
M
1010 yr
L
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