Download Astronomy Assignment #1

Astronomy Assignment #1 Solutions
Text Problems:
Unit Number
1
2
3
Questions for Review
2, 5, 6
1, 2, 3, 4
None
Problems
11,12, 13, 14
10, 11, 14
11, 12, (14 Not
included in these solutions
– Sorry)
Problem Solutions: Unit 1
11. What would be the circumference and diameter of a ball that would represent the Moon if the
Earth were a volleyball? What kind of object matches this size? (Note: See Problem #1 for the
circumference of a volleyball, 68 cm.)
Use a proportion to solve questions like this where different objects appear at the same scale.
Required information
Circumference of a volleyball = 68 cm (Problem #1)
Circumference of the Earth = 2∙R = 2∙6,378 km = 40,074 km (See Appendix Table 5)
Circumference of the Moon = 2∙R = 2∙1,738 km = 10,920 km (See Appendix Table 6)
Solve the proportion
68cm
x

40,074km 10,920km
68cm
x
 10,920km  18.5cm
40,074km
The scaled down Moon would be 18.5 cm in circumference equivalent to 5.90 cm in
diameter. This diameter is about 2.3 inches, and this is a bit bigger than a golf ball, but
smaller than a tennis ball. A regulation racquetball has a diameter of 2.25 inches and
would be a good model for the Moon if the Earth were a volleyball.
12. If the Earth were a volley ball, what would be the diameter of the Sun?
Use a proportion to solve questions like this where different objects appear at the same scale.
Required information
Diameter of a volleyball = 68 cm / () = 21.6 cm
Diameter of the Earth = 2 x 6,378 km = 12,756 km
Diameter of the Sun = 2 x 696,000 km = 1.39 x 106 km
Solve the proportion
21.6cm
x

12,756km 1.39  10 6 km
21.6cm
x
 1.39  10 6 km  2.36  10 3 cm  23.6m
12,756km
The scaled down Sun would be 23.6 m (about 75 feet) in Diameter. This is about the width
of a pretty large house. The message being that the Sun is really big compared to the
Earth, as big as a large house compared to a volley ball.


13. If the Earth were a volley ball how large would an astronomical unit be?
Use a proportion to solve questions like this where different objects appear at the same scale.
Required information
Circumference of a volleyball = 68 cm (Problem #1)
Circumference of the Earth = 2R = 26,378 km = 40,074 km (See Appendix Table 5)
Length of and astronomical uint = 1 AU = 1.5 x 108 km (See Appendix Table 1)
Solve the proportion
68cm
x

40,074km 1.5  10 8 km
68cm
x
 1.5  10 8 km  2.54  10 5 cm  2.54  10 3 m  2.54km
40,074km
The scaled down Astronomical Unit would be 2.54 kilometers in length. This length is
about 1½ miles. If the Earth were a volleyball then an AU would be about the distance
from OCC to the Jreck Sub shop at Valley Drive and Seneca Turnpike in the Valley. Just
for reference, on this scale you would be only 25 nanometers tall. A human hair is about
100,000 nanometers. (The moral of this problem is that an AU is really big, and it is the
smallest of the special length units astronomers use.
14. During the 1960s and 1970s, The Apollo spacecraft took humans to the Moon in three days.
Traveling to Mars requires a trip of around 2 AU in total. How long would this trip take,
traveling at the same speed as to the Moon?
Use t 
d
to solve problems like this.
v
Required information
Distance of the trip to Mars and back = 2 AU = 3.0 x 108 km (See Appendix Table 1)
Speed of trip = Distance from Earth to Moon divided by 3 days
384,000km
km
(See Appendix Table 6)
v
 128,000
3days
day
Solve the problem
t
d 3.0  10 8 km

 2,344days  6.4 yr
km
v
128,000
day
The trip to Mars and back, at Apollo spacecraft speeds, would require over 6 years to
complete. The moral of this problem is, again, that space is real big. A six year trip is
required to reach one of the nearest planets and return. To reach the nearest stars and
return would require a time 136,000 times longer, at this speed, or a total time of 816,000
years! Space is real big.
6. During the 1960s and 1970s, The Apollo spacecraft took humans to the Moon in three days.
How long would it take to travel to Pluto, about 40 astronomical units away?
Use t 
d
to solve problems like this.
v
Required information
Distance of the trip to Pluto = 40 AU = 60 x 108 km (See Appendix Table 1)
Speed of trip = Distance from Earth to Moon divided by 3 days
384,000km
km
(See Appendix Table 6)
v
 128,000
3days
day
Solve the problem
d 60.0  108 km
t 
 46,875days  128.3 yr
km
v
128,000
day
The trip to Pluto, at Apollo spacecraft speeds, would require over 128 years to complete.
The moral of this problem is, again, that space is real big. For practical reasons, the
spacecraft sent to explore the outermost objects of our Solar System must travel very fast.
Currently, the New Horizons spacecraft is on its way to Pluto having been launched in
January of 2006 and scheduled for Pluto encounter in July 2015. This corresponds (very
roughly) to a speed of 1.8 million kilometers per day, over ten times faster than the Apollo
astronauts traveled on their way to the Moon.
Problem Solutions: Unit 2
10. If the Milky Way were the size of a nickel (about 2
centimeters).
a. How big would the Local Group be?
b. How big would the Local Supercluster be?
c. How big would the visible Universe be?
Use a proportion to solve questions like this where different
objects appear at the same scale.
Scales of the Universe
Table 2.1
Object
Approximate Radius
Earth
Moon’s Orbit
Sun
R = 6,400 km
380,000 km
R = 700,000 km
= 110  R
1 AU = 150 Million km
= 210  R
= 23,100  R
40 AU = 8500  R
4.2 ly = 270,000 AU
50,000 ly
1.5 Million ly
40 Million ly
13.2 Billion ly
Earth’s Orbit
Solar System to Pluto
Nearest Star
Milky Way Galaxy
Local Group
Local Super Cluster
Visible Universe
Required information
Radius of a nickel = 2 cm
Radius of the Milky Way = 50,000 lyrs (See Table 2.1)
Radius of the Local Group = 1.5 million lyrs (See Table 2.1)
Solve the following proportion for the diameter of the Local Group
2cm
x

50,000ly 1,000,000ly
2cm
x
 1,000,000ly  60cm
50,000ly
The scaled down Local Group of Galaxies would be 60 cm in radius. Solving similar
proportions for the Local Super Cluster and Visible Universe yields the following. The
Local Super cluster would have a radius of 20 meters and the visible universe would have a
radius of about 5.5 km. This last answer is worth discussing briefly. If the Milky Way
galaxy were a nickel, then the distance to the edge of the visible universe would be from
downtown Syracuse to Liverpool, approximately. One might well ask, what is beyond this
visible limit? We don’t know, because the light from that great a distance has not yet
reached us. We believe that the universe has a finite age of about 13.7 billion years. Thus,
any light emitting object beyond that distance is un-“seeable “ because the light it emits has
not yet traveled the distance between it and Earth. The light train, so to speak has not yet
reached the Syracuse station. As the universe ages, its visible limit will grow.
11. If we detected radio signals of intelligent origin from the Andromeda Galaxy (M31) and
immediately responded with a radio message of our own, how long would we have to wait for a
reply?
This type of problem is particularly simple if we know the distance to the M31 in light years.
Since light travels at the speed of light, the time to travel on e light year is just one year. Proportionally,
the time for light to travel 100,000 light years is just 100,000 years.
Required information
Distance from the Milky Way to M31 = 0.8 Mpc (Mega parsecs) (See Appendix Table 11)
Convert 0.8 Mpc to light years
 1  10 6 pc   3.26ly 
  
  2.61  10 6 ly
0.8Mpc  0.8Mpc  
 1Mpc   1 pc 
The one-way travel time for radios wave (radio waves are a form of light) from the Milky
Way to the Andromeda Galaxy (M31) is about 2.6 million years. Thus, we would have to
wait for 2.6 million years for our radio signal to arrive at M31 and then another 2.6 million
years for there reply to reach us. Total communication travel time 5.2 million years – older
that the fossil record of the human species. Guess what the moral of the problem is…you
got it. Space is real big.
14. The Milky Way is moving toward the larger galaxy M31 at about 100 kilometers per second.
M31 is about 2 million light years away. How long will it take before the Milky Way collides with
M31 if it continues at this speed?
Use t 
d
to solve problems like this.
v
Required information
Distance from the Milky Way to M31 = 2 million lys
Speed of approach = 100 km/sec = 1x105 m/s
Convert 2 million lyrs to meters
 9.46  1015 m 
  1.89  10 22 m
2.00  10 6 ly  2.00  10 6 ly  
1ly


Solve the problem
d 1.89  10 22 m
t 
 1.89  1017 s  6.0  10 9 yr
m
v
1.  10 5
s
We will collide with the Andromeda Galaxy in about 6 billion years.
15. The most distant galaxies seen by the Hubble Space Telescope are 12 billion light years away.
Using the scale in section 2.4 (where the Milky Way is about the size of a large city), haw far would
that be in the model?
In this model (See Figure 2.4) 1 light-year is equivalent to 1 meter in the model.
Galaxies 12 billion light-years away would be at a model distance of 12 billion meters. This
distance can also be expressed as 1.2 x 1010 meters or 1.2 x 107 kilometers or 12 million
kilometers or 7,460,000 miles. So, if a light year were shrunk to the length of 1 meter,
 the Earth would be 15 microns from the Sun (the width of a thin human hair
 the nearest star would be just over 4 meters from the Sun,
 The typical bright star we see in our sky would be a few hundred meters from the Sun,
 The Milky Way Galaxy’s diameter would be 100 kilometers (63 miles), and
 Most distant galaxies would be 1.2 million kilometers away, almost the actual distance
between the Earth and Sun.
Moral of the Story: Space is real big.
Problem Solutions : Unit 3
1. The radius of the Sun is 7x105 kilometers, and that of the Earth is about 6.4x103 kilometers.
Show that the Sun’s radius is approximately 100 times the Earth’s radius.
7  10 5 km
.
6.4  10 3 km
This is a simple ratio problem. Evaluate the ratio
7  10 5 km
7

 10 53  1.09  10 2  109
3
6.4  10 km 6.4
Thus, we see that the Sun has a radius about 109 times that of the Earth.
Note: The Sun is REALLY big. The fact that it is just over 100 times the radius of the Earth
means that the Sun’s volume is over (100)3 times the Earth’s volume. (100)3 equals 1 million.
In other words, if the Sun was a gumball machine, it would take over 1 million Earth’s to fill it.
3. Using scientific notation, show that it takes sunlight about 8.5 minutes to reach the Earth.
d
problem. We are given a distance (1 AU from the Sun to the Earth) and a speed
v
(speed of light = 3x108 m/s), and asked to find a distance.
This is a t 
t
d
1.5  1011 m

 500s  8.33 min
v 3.0  10 8 m / s
Thus, it takes sunlight 8.33 minutes to travel from the Sun to the Earth.
4. How many 1-kiloton bombs would need to be exploded to produce 3.86x1026 Joules, the amount
of energy emitted by the Sun in 1 second?
This is a simple proportion problem. Evaluate the proportion
1  kiloton
x  kiloton

.
12
4.18  10 Joules 3.86  10 26 Joules
First cross multiply and then solve for x
1  kiloton
x  kiloton

12
4.18  10 Joules 3.86  10 26 Joules
1  kiloton  3.86  10 26 Joules  4.18  1012 Joules  x  kiloton

 
x  kiloton 

1  kiloton  3.86  10 26 Joules 
4.18  1012 Joules
x  kiloton  9.23  1013  kiloton
Thus, in 1 second the Sun emits an amount of energy equal to 9.23x1013 kilotons of energy. The
atomic bombs that were dropped on Hiroshima and Nagasaki in August, 1945 had an energy
equivalent of 16 kilotons. The Sun emits 5.8 trillion Hiroshima atomic bombs of energy each
second!
4  10 
5  10 
8 3
7. Using scientific notation, evaluate
4 10 
5 10 
8 3

6 2
6 2
 
 10 
4 3  10 8
52
3
6 2

.
64  10 24 64 10 24


 2.56  10 36
25  10 12 25 10 12
The expression evaluates to a whopping 2.56x1036.
3 10 
4 2
8. Using scientific notation, evaluate
3  10 
4 2
4  10 6
.
4  10 6
 

4  10 
3 2  10 4
2
1
6 2

9  1012
1
2

4  10

1
6 2

9 1012

 4.5  1015
2 10 3
The expression evaluates to 4.5x1015.
Unit 3
9. The radius of the Sun is 7x105 kilometers, and that of the Earth is about 6.4x103 kilometers.
Show that the Sun’s radius is approximately 100 times the Earth’s radius.
7  10 5 km
This is a simple ratio problem. Evaluate the ratio
.
6.4  10 3 km
7  10 5 km
7

 10 53  1.09  10 2  109
3
6.4  10 km 6.4
Thus, we see that the Sun has a radius about 109 times that of the Earth.
Note: The Sun is REALLY big. The fact that it is just over 100 times the radius of the Earth
means that the Sun’s volume is over (100)3 times the Earth’s volume. (100)3 equals 1 million.
In other words, if the Sun was a gumball machine, it would take over 1 million Earth’s to fill it.
11. Using scientific notation, show that it takes sunlight about 8.5 minutes to reach the Earth.
d
problem. We are given a distance (1 AU from the Sun to the Earth) and a speed
v
(speed of light = 3x108 m/s), and asked to find a distance.
This is a t 
d
1.5  1011 m
t 
 500s  8.33 min
v 3.0  10 8 m / s
Thus, it takes sunlight 8.33 minutes to travel from the Sun to the Earth.
12. How many 1-kiloton bombs would need to be exploded to produce 3.86x1026 Joules, the
amount of energy emitted by the Sun in 1 second?
This is a simple proportion problem. Evaluate the proportion
1  kiloton
x  kiloton

.
12
4.18  10 Joules 3.86  10 26 Joules
First cross multiply and then solve for x
1  kiloton
x  kiloton

12
4.18  10 Joules 3.86  10 26 Joules
1  kiloton  3.86  10 26 Joules  4.18  1012 Joules  x  kiloton

 
x  kiloton 

1  kiloton  3.86  10 26 Joules 
4.18  1012 Joules
x  kiloton  9.23  1013  kiloton
Thus, in 1 second the Sun emits an amount of energy equal to 9.23x1013 kilotons of energy. The
atomic bombs that were dropped on Hiroshima and Nagasaki in August, 1945 had an energy
equivalent of 16 kilotons. The Sun emits 5.8 trillion Hiroshima atomic bombs of energy each
second!
4  10 
Using scientific notation, evaluate
5  10 
8 3
7.
6 2
4 10 
5 10 
8 3
6 2

 
 10 
4 3  10 8
52
3
6 2

.
64  10 24 64 10 24

 12  2.56  10 36
12
25 10
25  10
The expression evaluates to a whopping 2.56x1036.
3 10 
4 2
16. Using scientific notation, evaluate
3  10 
4 2
4  10 6

4  10 6
 
4  10 
3 2  10 4
2
1
6 2
The expression evaluates to 4.5x1015.

.
9  1012
1
2

4  10 6

1
2

9 1012
 3  4.5  1015
2 10