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ASTRONOMY 120: GALAXIES AND THE UNIVERSE
HOMEWORK #1 SOLUTIONS
FALL 2016
1. Scientific Theory (10 points)
A scientific theory is fundamentally different than the everyday use of the
word “theory”. Briefly explain the difference.
In everyday language, the word ‘theory’ is often used to mean an idea that looks good on
paper and may or may not have anything to do with reality. A scientific theory is a body
of related hypotheses pieced together into a self-consistent description of nature and must
be falsifiable. An important part of a scientific theory is its ability to make predictions
that can be tested by other scientists.
2. Reason for the Seasons (20 points)
Why do we have seasons? Briefly explain the 2 main reasons. Use diagrams.
You must talk about energy, and say more than “tilt of axis”.
The cycle of the seasons occurs because Earth’s axis of rotation is inclined 23.5◦ from
the perpendicular to its orbital plane. Although precession moves the axis very slowly, the
axis is essentially fixed in space. As Earth orbits the Sun, its axis remains pointing in the
same direction in space. Seasonal temperature depends on the amount of heat (energy)
we receive from the Sun. Two factors are most responsible for the change in this amount.
Firstly, and most importantly, the Sun stands higher in the sky at noon on a summer
day and shines almost straight down, providing more energy per unit area. On a winter
day, however, the noon sun is lower and hence it spreads out and we receive less energy per
unit area (Figure 1). In other words, the amount of total energy coming towards the Earth
from the Sun being almost the same throughout the year, during summer it is intercepted
by a smaller area on the ground (Sun shining at a more direct angle), thereby increasing
the energy per unit area. During winter, the same amount of energy is spread out over a
larger area (Sun shining at a more inclined angle), and therefore the energy per unit area
is decreased.
Secondly, the summer Sun is above the horizon for more hours each day than the winter
Sun: summer days are longer, and winter days are shorter. Because the Sun is above our
horizon longer in summer, we receive more energy each day. The earth, especially the
oceans and the atmosphere, retains this heat somewhat even during the night, thereby
creating a generally warmer temperature (otherwise, the nights would be equally cold
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ion
at
Rot
Less area,
More energy / area
Axi
s
23.5
ator
Equ
SUN
More area
Less energy / area
Figure 1. Sunlight strikes a given area of the Earth at different angles as
we progress through the seasons since the direction of Earth’s rotation axis
is essentially fixed in space.
throughout the year). In fact, this storage of the energy also causes the slight delay in the
seasons — in the northern hemisphere, July/August are warmer than 21 June (the summer
solstice), and January/February are colder than 21 December (the winter solstice).
3. Tennis Ball Sun (10 points)
Suppose you are building a scale model of the nearby stars in our Galaxy.
The Sun, which has a radius of 696,000 km, is represented by a tennis ball with
a radius of 6 cm. On this scale, how far away would the nearest star, Proxima
Centauri, be? Proxima Centauri, which is probably part of the Alpha Centauri
star system, is located 4.2 light years away. Express your answer in km.
To solve this problem, we must first convert the numbers given in the problem into a
consistent set of units. Let’s choose an SI unit: meters (but we must remember to quote
our final answer in km). So we need to kow how to convert light years into meters.
1 ly = 9.461 × 1015 m
4.2 ly = 3.974 × 1016 m
(1pt)
We must also convert the actual radius of the Sun into meters as well as the radius of
the Sun in our model from km to meters:
696, 000 km = 6.96 × 108 m
−2
6 cm = 6 × 10
m
(1pt)
(1pt)
ASTRONOMY 120: GALAXIES AND THE UNIVERSE
HOMEWORK #1 SOLUTIONS
3
We solve the problem by equating the ratio between the actual solar radius (R) and the
model solar radius (r) to the ratio between the actual distance of Proxima Centauri (D)
to the model distance of Proxima Centauri (d):
R
r
=
D
d
(4pts)
Then it follows that:
d = D×
r
R
(1pt)
6 × 10−2 m
6.96 × 108 m
6
= 3.43 × 10 m
(1pt)
= 3430 km
(1pt)
= 3.974 × 1016 m ×
4. Some Sirius Questions (24 points)
The brightest star in the sky is Sirius, which is located at a distance of 2.6
pc.
a. What is the distance to Sirius in units of light-years?
b. How long does it take light to reach the Earth from Sirius?
c. If a planet were orbiting Sirius at the same distance that the Earth is
from the Sun, what would be the angular separation on the sky between
Sirius and this hypothesized planet?
4.a. There are two ways for answering this question.
In the first way we use that a light year is the distance that light travels in a year. Then,
for instead, if light takes 3.1 years to reach a point it means that the point is 3.1 light years
away. Having this in mind we can use the relation between distance d travelled in time t
by object moving at speed c:
d=c×t
(5pts)
(1)
where c = 3 × 108 m is speed of light.
Convert the distance of Sirius to meters:
d = 2.6 pc
3.09 × 1016 m
= 8.03 × 1016 m
1 pc
(1pt)
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FALL 2016
Using relation 1:
d
(1pt)
c
8.03 × 1016 m
3 × 108 m/s
t =
=
= 2.68 × 108 s
(1pt)
To convert travel time to Sirius in units of years, we calculate number of seconds in a
year:
N
= 365.25 days/yr × 24 hr/day × 60 min/hr × 60 s/min
= 3.16 × 107 s/yr
Finally, the amount of time it takes light to get from Earth to Sirius:
2.68 × 108 s
2.68 × 108 s
=
N
3.16 × 107 s/yr
= 8.48 yr
(1pt)
t =
Then, since light takes 8.48 years to travel to Sirius we have that Sirius is 8.48 light
years away. (1pt)
The second way would be doing a conversion of units from parsecs to light years. Thus,
d = 2.6 pc
3.26 ly
= 8.48 ly
1 pc
(10pt)
4.b. A light-year is the distance that light travels in one year. Since Sirius is at a distance
of 8.48 ly (see 4.a) it takes 8.48 years for light to travel from Sirius .(3pts)
4.c. Since Sirius is far away from the Earth (at distance d), we suppose that the hypothesized planet would be observed very close to Sirius, subtending small angle α. In this
regime we can use the small angle formula:
D=
αd
206, 265
(6pts)
(2)
where α is angular separation of Sirius and its planet (measured in arc seconds), D is linear
separation, ie distance from planet to Sirius, in this case D = 1 AU, and d is the distance
to Sirius. Note that D and d must be in the same units.
Since
1 AU = 1.496 × 1011 m
(2pt)
ASTRONOMY 120: GALAXIES AND THE UNIVERSE
HOMEWORK #1 SOLUTIONS
5
and distance to Sirius d was already calculated in part 4.a, we have for α:
D
(1pt)
d
(206, 265)(1.496 × 1011 m)
=
8.03 × 1016 m
= 0.38 arc sec
(2pt)
α = (206, 265)
5. Better Days (30 points)
One earth year is 365.244 (solar) days. The professor of this course would
be much happier if there were an integral number of days in a year. Using
Newton’s form of Kepler’s Third Law, figure out 3 ways to make the length of
earth’s year exactly 354.000 (solar) days. (This would also make the months
be more coordinated with the phases of the moon! We could have 6 months
of 29 days and 6 months of 30 days. Since the orbital period of the moon is
about 29.5 days, and 12 months x 29.5 days/month = 354 days, I think this
would make a great calendar!)
(HINT: Check your answers experimentally by actually changing the mass of the Sun
and earth, and the orbital radius of the earth.)
(REAL HINT: Take care with significant figures for this problem. See help on the website.)
Please review the rules on significant figures at
http://www.astro.yale.edu/astro120/SigFig.pdf
5.a. What would the mass of the sun have to be changed to, if nothing else
were changed?
Considering Newton’s Form of Kepler’s Third Law :
#
"
2
4π
a3
(3)
P2 =
G(m1 + m2 )
m1 = M⊕ = 5.974 × 1024 kg
m2 = M = 1.989 × 1030 kg.
a = 1AU = 1.496 × 1011 m
m3
G = 6.67 × 10−11
kg s2
hours
minutes
seconds
P = 354.000days × 24
× 60
× 60
day
hour
minute
= 30585600sec
= 3.05856 × 107 sec
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FALL 2016
"
#
4π 2 3
P ∗ (M⊕ +
=
a
G
#
"
4π 2 3
new
a
(M⊕ + M ) =
P 2G
"
#
2
4π
Mnew =
a3 − M ⊕
P 2G
"
2
Mnew )
Mnew =
(4)
#
4π 2
(3.05856 × 107 sec)2 · 6.67 × 10−11 kgm
Mnew = 2.11833 × 1030 kg ×
(1.496 × 1011 m)3 − 5.974 × 1024 kg
3
s2
1M
1.98892 × 1030 kg
Mnew = 1.06507M
There are 6 significant figures in the Period of 354.000, but since the masses and distances
are given to 4 significant figures we can round down to 4.
Mnew = 2.118 × 1030 kg
or
new
M
= 1.065M
5.b. What would the mass of the earth have to be changed to, if nothing else
were changed?
Solve equation (4) for M⊕new :
"
M⊕new =
#
4π 2 3
a − M
P 2G
M⊕new = 1.293 × 1029 kg ×
1M⊕
5.9742 × 1024 kg
M⊕new = 2.164 × 104 M⊕
5.c. What would the orbital radius of the earth have to be changed to, if
nothing else were changed?
Solve equation (4) for anew :
ASTRONOMY 120: GALAXIES AND THE UNIVERSE
"
G(M⊕ + M )P 2
4π 2
#
"
G(M⊕ + M )P 2
4π 2
#1/3
(anew )3 =
anew =
HOMEWORK #1 SOLUTIONS
anew = 1.465 × 1011 m ×
7
1AU
1.49598 × 1011 m
anew = 0.979AU
NOTE: This problem can be done in a much simpler way if you use the ratio method.
e.g., for part (a) when we are solving for Mnew :
"
P12
P22
P1
P2
!2
#
4π 2
=
a3
G(M⊕ + M ) 1
"
#
4π 2
=
a3
G(M⊕ + Mnew ) 2
"
#
!3
4π 2
a1
G(M⊕ +M )
=
4π 2
a2
new
G(M⊕ +M
(5)
)
Notice that all the factors of (4π 2 /G) cancel and you don’t have to worry about their
units anymore. Also, since (P1 /P2 ) and (a1 /a2 ) are in a ratio, their units cancel and you
don’t have to convert from days to seconds.
P1
P2
!2
"
=
(M⊕ + Mnew )
M ⊕ + M
#
a1
a2
!3
(M⊕ + Mnew )
since a1 = a2
M ⊕ + M
(M⊕ + Mnew )
M ⊕ + M
(M⊕ + Mnew )
M ⊕ + M
(M⊕ + Mnew )
M ⊕ + M
=
(
365.244days 2
) =
354.000days
1.031762 =
1.06453 =
Then just a little algebra allows you to solve for Mnew in terms of M and M⊕ , which
you substitute at the end.
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FALL 2016
6. Radio Stations (20 Points)
Compare these properties of the radio stations WFOX (95.9 FM) and WINE
(940 AM).
a.) What is the wavelength of the radio waves transmitted by each station?
b.) What is the frequency of the radio waves transmitted by each station?
c.) What is the speed of the radio waves transmitted by each station?
d.) What is the energy of the radio photons transmitted by each station?
What is the wavelength of the radio waves transmitted by each station?
The relationship between the frequency and wavelength of an electromagnetic wave is a
simple one. Because light moves at a constant speed, c = 2.9979×108 m/s in a vacuum (see
Appendix 7), if the wavelength (distance from one crest to the next) is made shorter, the
frequency must increase (more closely spaced crests pass you each second), the frequency
ν of light is related to its wavelength λ by,
ν = λc
where, ν = frequency of an electromagnetic wave (in Hz)
c = speed of light = 2.9979 × 108 m/s.
λ = wavelength of the wave (in m).
So if we want to get the wavelength from the information on part a) we must use, λ = νc
and so for each station we have,
Station frequency in Hz wavelength in m
WFOX
9.59 × 107
3.12
WINE
9.40 × 105
318.9
What is the frequency of the radio waves transmitted by each station?
AM radio stations broadcast at frequencies between 535 and 1605 kHz (kilohertz), while
FM radio stations broadcast at frequencies in the range from 88 to 108 M Hz (megahertz).
So, using this information, we have that our WFOX FM radio station must be transmitting
at 95.9 M Hz and our WINE AM radio station must be transmitting at 940 kHz. So
converting to comparable units we have to remember that mega stands for 106 and kilo
for 103 . Remember 1 Hz = 1 s−1 .
Station frequency in Hz
WFOX
9.59 × 107
WINE
9.40 × 105
What is the speed of the radio waves transmitted by each station?
As we said above, radio waves are part of the electromagnetic spectrum and therefore
travel at the speed of light, c = 2.9979 × 108 m/s.
What is the energy of the radio photons transmitted by each station?
We know that the energy of a photon in terms of the frequency is defined by Planck’s law,
E=hν
where,
E = energy of the photon (in J)
ASTRONOMY 120: GALAXIES AND THE UNIVERSE
HOMEWORK #1 SOLUTIONS
9
h = Planck’s constant = 6.6261 × 10−34 J s (see Appendix 7) here J stands for Joule,
the unit of energy in the MKS system and is equal to 1 J = 1 kg m2 /s2 .
ν = frequency of light (in Hz).
so we will have,
Station frequency in Hz wavelength in m energy in J
WFOX
9.59 × 107
3.12
6.34 × 10−26
5
WINE
9.40 × 10
318.9
6.23 × 10−28