Download astronomy 31 - UNC Physics

ASTRONOMY 101
SECTION 2
EXAM 1
1. How many arcseconds are in a degree?
A.
B.
C.
D.
0.00028
0.017
60
3,600
D. 1 degree × (60 arcminutes / 1 degree) × (60 arcseconds / 1 arcminute) = 3,600 arcseconds
2. The sun is approximately how many Earth’s across?
A.
B.
C.
D.
1
10
100
1,000
C. From class notes on size and scale.
3. You watch a bright star rotate 90 degrees around the south celestial pole. How long have you been watching
this star?
A.
B.
C.
D.
3 hours
6 hours
9 hours
12 hours
B. 90 degrees × (24 hours / 360 degrees) = 6 hours
4. All stars are circumpolar:
A.
B.
C.
D.
E.
On the equator
At the north pole
At the south pole
B&C
A&B&C
D. Because Earth rotates, the stars appear to move around the north celestial pole in the northern hemisphere
and the south celestial pole in the southern hemisphere. When standing at the north pole, the north celestial
pole is directly overhead and when standing at the south pole the south celestial pole is directly overhead.
Hence, at the north and south poles all of the stars appear to move around the point directly overhead, meaning
that they do not rise or set, meaning that they are circumpolar. When standing on the equator, the north celestial
pole is on the northern horizon and the south celestial pole is on the southern horizon. Hence, on the equator all
of the stars appear to move around these two points on opposite horizons, meaning that all of the stars rise in the
eastern sky and all of the stars set in the western sky, meaning that no star is circumpolar.
5. After how many months are constellations that were setting at midnight now rising at midnight?
A.
B.
C.
D.
3
6
9
12
B. Constellations that were setting but are now rising at any fixed time have shifted 180 degrees in the sky:
180 degrees × (12 months / 360 degrees) = 6 months
6. It is hotter in summer and colder in winter because:
A.
B.
C.
D.
E.
Earth is closer to the sun in summer.
Earth’s orbit around the sun is elliptical.
Your hemisphere of Earth is tipped toward the sun in summer.
A&B
A&B&C
C. The seasons are the result of how directly the sun’s light strikes Earth: It strikes most directly the
hemisphere that is tipped toward the sun, causing summer there, and least directly the hemisphere that is tipped
away from the sun, causing winter there.
7. When it is winter in the northern hemisphere, it is what in the southern hemisphere?
A.
B.
C.
D.
Winter
Spring
Summer
Fall
C. If it is winter in the northern hemisphere, the northern hemisphere must be tipped away from the sun,
meaning that the southern hemisphere must be tipped toward the sun, meaning that it must be summer in the
southern hemisphere.
8. The day is now:
A.
B.
C.
D.
E.
≈6.5 hours long
≈9 hours long
≈11.5 hours long
≈14 hours long
≈16.5 hours long
C. You are supposed to be keeping track of this, at least approximately.
9. The moon revolves around Earth once every 27.3 days. The moon rotates once every:
A.
B.
C.
D.
27.3 days
29.5 days
365.2 days
The moon does not rotate.
A. The moon rotates and revolves at the same rate, due to tidal locking. Otherwise, the side of the moon that
faces Earth would change with time.
10. Which of the following statements are true?
A.
B.
C.
D.
When lunar eclipses occur, the moon is new.
When solar eclipses occur, the moon is new.
When the moon is new, lunar eclipses occur.
When the moon is new, solar eclipses occur.
B. When solar eclipses occur, the moon is directly between the sun and Earth and consequently its phase must
be new. However, solar eclipses do not occur every time the moon is new because the Earth-Moon plane is
tipped, by about 5 degrees, with respect to the Sun-Earth plane, meaning that the moon, although new, is not
usually directly in front of the sun, but slightly above or below it.
11. After how many eclipse years does the sequence of eclipses repeat?
A.
B.
C.
D.
0.5
1
18
19
D. 19 eclipse years is very close to an integer number of lunar months, meaning that the sequence of eclipses
will repeat every 19 eclipses years. (Since the line of nodes regresses, an eclipse year is shorter than a regular
year.)
12. Perfected the heliocentric model of the universe:
A.
B.
C.
D.
E.
Aristarchus
Copernicus
Galileo
Kepler
Ptolemy
D. Kepler perfected the heliocentric model with his three empirical laws, which were later explained on
physical grounds by Newton.
13. Not discovered by Galileo:
A.
B.
C.
D.
E.
Lunar features
Sunspots
Jovian moons
Venusian phases
Stellar parallaxes
E. Stellar parallaxes, which prove that Earth is moving around the sun, were not measured until the mid-19th
century.
14. Every year, a star is observed to move 0.1 arcseconds with respect to background stars for six months,
followed by a return to its original position over the course of the next six months. How far away is this star?
A.
B.
C.
D.
1,100 AU
69,000 AU
2,100,000 AU
4,100,000 AU
D. distance = (baseline / 2× (360 degrees / angular shift. baseline = 2 AU. angular shift = 0.1 arcseconds ×
(1 arcminute / 60 arcseconds) × (1 degree / 60 arcminutes) = 2.8 × 10-5 degrees. distance = (2 AU / 2× 
degrees2.8 × 10-5 degrees = 4.1 × 106 AU
15. A new planet is observed to orbit the sun once every 1,000 years. What is the semi-major axis of its orbit?
A.
B.
C.
D.
100 AU
680 AU
4,600 AU
32,000 AU
A. a = (P / 1 year)2/3 AU = (1,000 years / 1 year)2/3 AU = 100 AU
16. We put a satellite into a circular orbit around Earth at half the distance to the moon. What is the period of
its orbit?
A.
B.
C.
D.
0.35 lunar months
0.63 lunar months
1.6 lunar months
2.8 lunar months
A. P = (a / 1 Earth-Moon distances)3/2 lunar months = (0.5 Earth-Moon distances / 1 Earth-moon distance)3/2
lunar months = 0.35 lunar months
17. The frequency of 1-m radio waves is:
A.
B.
C.
D.
3 × 105 Hz
3 × 108 Hz
3 × 1011 Hz
3 × 1014 Hz
B.  = c /  = (3 × 108 m/s) / (1 m) = 3 × 108 s-1 = 3 × 108 Hz
18. The peak wavelength of a 10,000 K star is:
A.
B.
C.
D.
0.29 nm
290 nm
2.9 m
290 m
peak = 2,900 nm / (T / 1,000 K) = 2,900 nm / (10,000 K / 1,000 K) = 290 nm
19. With the human eye, a 10,000 K star is:
A.
B.
C.
D.
Too red to see
Red
Blue
Too blue to see
C. Although the peak wavelength is too blue to see with the human eye, the spectrum is broad and extends well
past the visible band. Consequently, the star is visible to the naked eye and since more blue light is emitted than
red, the star is blue in color.
20. Oven A is twice as hot as oven B, yet otherwise identical. How many times more energy does oven A emit
than oven B?
A.
B.
C.
D.
2
4
8
16
D. Since energy flux is proportional to T4, energy fluxA / energy fluxB = (TA / T B)4 = 24 = 16
21. Light from a star passes though a cold hydrogen cloud on its way to Earth. We receive a:
A.
B.
C.
D.
Lyman series absorption spectrum
Balmer series absorption spectrum
Lyman series emission spectrum
Balmer series emission spectrum
A. By Kirchhoff’s Third Law, we receive an absorption spectrum. Since the cloud is cold, the hydrogen atoms
begin in the ground state, resulting in a Lyman series absorption spectrum.
22. Aliens observe at this cloud from a completely different direction. They receive a:
A.
B.
C.
D.
Lyman series absorption spectrum
Balmer series absorption spectrum
Lyman series emission spectrum
Balmer series emission spectrum
C. By Kirchhoff’s Second Law, the absorbed light is then re-emitted, but in all directions, resulting in a Lyman
series emission spectrum in every other direction.
23. A hydrogen atom is in an excited state! Its electron decays from the fourth excited state to the second
excited state. What type of photon is released?
A.
B.
C.
D.
E.
Lyman alpha
Balmer beta
Balmer delta
Paschen beta
Paschen delta
D. All transitions down to the second excited state result in Paschen series photons. The first transition in the
series (third excited state to second excited state) is called Paschen alpha. The second transition in the series
(fourth excited state to second excited state) is called Paschen beta.
24. Lyman-alpha light has a wavelength of 91 nm. A galaxy, moving away from us at a speed of 300 km/s,
emits Lyman-alpha light. We detect this Lyman-alpha light at wavelengths that are:
A.
B.
C.
D.
Shorter than 91 nm
Equal to 91 nm
Longer than 91 nm
None of the above
C. Since the galaxy is moving away from us, its light is redshifted, meaning that we detect the emitted light at
longer wavelengths.
25. What is the change in wavelength of this light?
A.
B.
C.
D.
0 nm
0.000091 nm
0.091 nm
91 nm
C.  = emitted × (v / c) = 91 nm × (300 km/s / 3 × 105 km/s) = 0.091 nm