Summer Astronomy Assignment #2
Text Problems:
Unit Number
6
7
Review Questions
5, 8, 10, 11
4, 5
Quantitative Problems
15See note below, 16
12, 14
Regarding Problem 15 in Unit 6: To answer this question you must complete the 2-D Local Horizon
Map of the Sun’s maximum altitude when it is at the winter solstice (solar declination =-23.5),
equinoxes (solar declination =0), and the summer solstice (solar declination =+23.5) in order to
determine the Sun’s maximum altitude as it crosses the meridian. Additionally, you should address
where around the horizon the sun rises and sets and how long it will be above the horizon.
Unit 6 Problems
15. Describe the motion you would see on the solstices and the equinoxes if you were observing the
Sun from the Arctic Circle at a latitude of 66.5°N.
This problem is really three problems; the apparent motion of the Sun on the Summer Solstice, Winter
Solstice and on the Equinoxes. The declination of the Sun on those three times is +23.5°dec, -23.5°dec
and 0°dec, respectively. Let’s examine one at a time.
Before looking at the apparent motion of the Sun, let’s first get to know the sky at 66.5°N latitude.
Examine the 2-D Local Horizon Map below drawn for 66.5°N latitude.


You can see that the
boundary of the
circumpolar region is at
23.5° declination. Thus
any star or other celestial
object that has a
declination of23.5° or
greater will be above the
horizon all the time at
this latitude.
The southernmost visible
star has a declination of 23.5°. Thus, any star or
other celestial object that
has a declination of 23.5° or less (meaning
farther south) will never
appear above the horizon
at this latitude.
Now for the Sun.
NCP (dec = 90)
Sun on the Summer
Solstice. +23.5°dec

23.5
Celestial Equator
(dec = 0)
(dec = -23.5) S
The Southernmost
Visible Star
66.5
66.5
Horizon Line
Horizon Line
23.5
66.5
66.5
23.5
SCP (dec = -90)
N (dec = 23.5)
The Boundary of the
Circumpolar Region
On the Summer Solstice the Sun will have a declination of +23.5°. Thus the Sun will be circumpolar at
the arctic circle on that day. The Sun will not set that day. We can figure out the maximum altitude of
the Sun by plotting the Sun on the above diagram and adding some angles together to get 47° maximum
altitude at transit.
On the Winter Solstice the Sun will have a declination of -23.5°. Thus the Sun will be never rise at the
arctic circle on that day.
On the Equinoxes the Sun will have a declination of 0°. The Sun will be on the celestial equator those
two days. It will rise due east and set due west and be above the horizon for 12 hours. We can figure
out the maximum altitude of the Sun by determining the maximum altitude of the celestial equator
which you can see from the diagram is 23.5°.
16. If you wished to observe a star with a right ascension of 12h, what would be the best time of the
year to observe it? What would be the best time of the year to observe a star with a right
ascension of 6h?
This question is best answered by first sketching the Whole Sky Map as shown below
The Whole Sky Map
Star at 6h RA
Star at 12h RA
0°dec
Sun 12h RA
from Star
Ecliptic

Celestial Equator
Sun 12h RA
from Star
12h
6h
Celestial Equator

0h
18h
Ecliptic
12h
The principle is this: Stars are best seen when they are farthest from the Sun – in opposition to the Sun.
That is a star is easiest to observe when the Sun is on the opposite side of the celestial sphere, at least
opposite in RA. Thus, the best observing of a star occurs when the Sun is 12h of RA away from the Sun.
So for a star at 12h RA, the Sun must be at 0h RA or at the Spring Equinox on Mar 22.
So for a star at 6h RA, the Sun must be at 18h RA or at the Winter Solstice on Dec 22.
Instructor Assigned Topic. You have been hired by the City of Syracuse to make a scale model of the
Milky Way Galaxy such that the Sun would be placed on a stone marker at the center of Clinton Square
in downtown Syracuse and the center of the Milky Way would be marked on a similar stone marker at
the Regional Market on Park Street in the north side of Syracuse. The concept being that visitors would
start at Clinton Square and walk down North Salina Street passing similar stone markers that denote the
position of significant objects as they walk to the center of the Milky Way Galaxy. The distance of the
walk would be 3.3 km (about 2 miles). See map below. You will need to look up the distance the Sun
of from the center of the Milky Way Galaxy.
Using proportions answer the following questions:
1. If the first stone marker in Clinton Square were to contain a pictorial representation of the Sun,
what would be its diameter in cm?
2. How far from that first marker would a marker indicating the edge of the Solar System (actual
distance 40 AU from the
Sun) be?
3. How far away from that
first marker would you
place a third marker that
represents the distance
from the Sun to the
nearest star, Proxima
Centauri (actual distance
from the Sun 4.3 ly)?
4. How far away would
you place a fourth
marker that would
represent the distance to
Polaris (actual distance
to Polaris is 434 ly)?
5. The bright star Deneb is
one of the most distant
bright stars in the
summer sky in the
constellation of Cygnus.
Deneb is 2,600 ly away
from the Sun. How far
away would you place a
fifth marker that would
represent the distance to
Deneb? (Aside: You can
see the first magnitude
star Deneb at the zenith
at 4:00 a.m. all this
week)
6. The most distance
visible star to the nakedeye may be Rho
Cassiopeiae at a distance
of 9,000 ly. How far away would you place a sixth marker that would represent the distance to
Rho Cassiopeiae?
7. What would be the distance on the walk from this sixth marker of the most distant visible star
from the Earth to the center of the galaxy? (Note this is not a proportion problem.)
Extra Credit: Mark the position of the stone
bymap
number
on0.167
the actual
abovekilometers
map of the
The markers
scale of this
is about
per walk.
1 cm on the map.