Download Astronomy Assignment #1

Astronomy Assignment #2 Solutions
Text Problems:
Unit Number
1
2
5
Questions for Review
5, 7, 9
5, 6, 7, 8
4, 5, 7, 8
Problems
10c
10, 11, 14
Problem Solutions: Unit 2
10. If the Milky Way were the size of a nickel (about 2 centimeters).
c.
How big would the visible Universe be?
Use a proportion to solve questions like this where different objects appear
at the same scale.
Required information
Radius of a nickel = 2 cm
Radius of the Milky Way = 50,000 lyrs (See Table 2.1)
Radius of the Visible Universe = 13.7 billion lyrs (See Table 2.1)
Solve the following proportion
50,000 ly
2 cm

9
13.7  10 ly
x

50,000 ly  x  2 cm  13.7  109 ly
x


Scales of the Universe
Table 2.1
Object
Approximate Radius
Earth
R = 6,400 km
Moon’s Orbit
380,000 km
Sun
R = 700,000 km
= 110  R
Earth’s Orbit
1 AU = 150 Million km
= 210  R
= 23,100  R
Solar System to Pluto 40 AU = 8500  R
Nearest Star
4.2 ly = 270,000 AU
Milky Way Galaxy
50,000 ly
Local Group
1.5 Million ly
Local Super Cluster
40 Million ly
Visible Universe
13.7 Billion ly

2 cm  13.7  109 ly
 548,000 cm  5,480 m  5.48 km
50,000 ly
The scaled down Visible Universe would be 5.48 km in radius. If the Milky Way galaxy were a nickel, then the
distance to the edge of the visible universe would be from downtown Syracuse to the NYS Fairgrounds,
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.
Problem Solutions: Unit 5
10. What are your latitude and longitude? What are the latitude and longitude at the point on the Earth exactly opposite
you? What geographic features are located there?
A Google search of “latitude and longitude of Syracuse, NY” yields our latitude is about 43
about 76 degrees W.
degrees N and longitude is
Just opposite us on the globe implies that we (1) take the same latitude but opposite hemisphere and (2) add 180 degrees to
the longitude (and then play with the number so that it is between zero and 180 degrees either E or W. So…the point
opposite Syracuse, NY on the Earth has a latitude of 43 degrees S and the longitude is (76 W + 180 = 256 W = ) 104 degrees
E longitude.
On the map below the position of Syracuse, NY and its opposite point are marked with Xs. You can see that our opposite
point is in the middle of the ocean off the southwest corner of Australia.
X
X
11. Your observatory lies at a latitude of 45N and a longitude of 90W. You wish to observe three celestial objects that have
declinations of +87, -40, and -67. Which of these objects can you observe?
This question requires a fairly complete understanding of the celestial coordinate system and the celestial sphere model.
According to the last sentence on page 43 “if a star’s declination matches your latitude it will pass over through zenith”.
This means that the declination of your zenith is just your latitude. Observing from 45 N will mean that 45 declination will
be directly over your head. Following Figure 5.4A on page 42, you can see that the horizon is 90 from the zenith. Thus, any
star that is within 90s from the declination of your zenith will be visible. As seen from 45N, all stars between +90
declination and -45 declination will eventually be visible over the horizon. So, you will be able to see the objects at +87
and -40 declination, but not the object at -67 declination..
Pictorially, we can draw the simplified celestial sphere as shown below. The three stars in the question are plotted in red.
Clearly the stars at 87 degrees dec and -40 degrees dec are visible from this location because they appear above the horizon.
The star at -67 degrees dec appears below the horizon and thus is not visible from that latitude.
Zenith
(dec= 45)
Celestial Equator
(dec = 0)
dec = 87
NCP
(dec = 90)
45
dec = -40
(dec = -45) S
Horizon Line
45
45
The Southernmost
Visible Star
45
45 Horizon Line
N (dec = 45)
The Boundary of the
Circumpolar Region
dec = -67
SCP
(dec = -90)
Like 13. Boston and Rome are at the same latitude, but Boston is at a longitude of 71 W, while Rome is at 12E. If M31
passes through the zenith at a particular time in Rome, how many hours later will it pass through the zenith in Boston? This
question was not assigned, but questions like it may appear on the exam, as stated in the study guide.
This question requires knowledge of the relationship between the celestial coordinate system and the rotation of the Earth.
As stated on page 43, “Each hour of RA equals 15;…if a star at RA =2h is overhead now, a star at RA =5h will be overhead
three hours from now.” Thus, if were to calculate the number of degrees longitude between Boston and Rome, then divide
that difference by 15 per hour, we would have the time in hours between M31 passing over both cities.
Number of degrees longitude between Boston and Rome is 83. This divided by 15 per hour gives a time difference
between M31 over Rome than Boston as 5.53 hours or 5 hours and 32 minutes.
14. How far is the celestial equator from your zenith if your latitude is 42N? 34S?
This question requires a fairly complete understanding of the celestial coordinate system and the celestial sphere model.
According to the last sentence on page 43 “if a star’s declination matches your latitude it will pass over through zenith”.
This means that the declination of your zenith is just your latitude. Observing from 42N will mean that 42 declination will
be directly over your head. Since 42 declination is 42 away from the celestial equator, the celestial equator must be 42
away from your zenith at this location. Observing from 34S latitude, will mean that the celestial equator is 34 away from
your zenith at that location.
The picture you could draw for each location appears below. Be certain that you understand how to construct these pictures
to answer similar questions.
If you were observing at 42N latitude the picture would be as follows: As you can see the celestial equator is 42 south of
the zenith.
Zenith (dec = 42)
Celestial Equator
(dec = 0)
NCP
(dec = 90)
42
(dec = -47) S
47
42 Horizon Line
47
Horizon Line
N (dec = 47)
42
The Southernmost
Visible Star
The Boundary of the
Circumpolar Region
SCP
(dec = -90)
If you were observing at 34S latitude the picture would be as follows: As you can see the celestial equator is 34 north of the
zenith. Notice that the circumpolar region is in the southern sky, not the northern sky, and that there is a portion of the
northern hemisphere that is never visible.
Zenith (dec = -34)
Celestial Equator
(dec = 0)
SCP
(dec = -90)
56
(dec = 47) S
The Boundary of the
Circumpolar Region
Horizon Line
34
34
56
Horizon Line
-34
N (dec = -56)
The Northernmost
Visible Star
NCP
(dec = 90)