Munson - Lab 2
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Kent State University *
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Course
21062
Subject
Astronomy
Date
Feb 20, 2024
Type
Pages
7
Uploaded by MasterBaboon4111
Lab 2: Earth-Sun Relations
Instructions
Watch the lectures on Earth Sun Relations, the analemma, and the altitude of the noon
sun and solar intensity, and watch the video on Seasons and Earth-Sun relations. After
you have reviewed the lectures, notes, and the video, use this document to work
through your lab. Once you have finished answering all questions in this lab, submit
your answers online in the link titled “Lab 2 – Earth/Sun Assessment”. After the due
date when your lab is graded, you will be able to review your lab and answers. When
you first submit your lab, your score may appear low. Keep in mind that your lab
instructor will need to grade your lab, especially the written responses. The computer
can automatically grade and score multiple choice, matching, and true/false questions.
Any written or essay responses will need to be graded by your instructor. Once your lab
instructor grades those questions, your lab grade will be updated to the correct grade.
Goals
o
Identify the positions of the vertical rays on the equinoxes and solstices
o
Identify the dates of the equinoxes and solstices
o
Identify the positions of the vertical rays of the sun throughout the year
o
Identify the positions of the tangent rays of the sun throughout the year
o
Identify the factors that result in the changing seasons
o
Calculate and interpret the sun angle given the coordinates and dates for Kent,
Ohio
o
Identify and explain how the day length changes with latitude and seasons.
Key Terms / Concepts
ANS Solstice
Declination Equinox
Insolation Duration
Solar Radiation Intensity Circle of Illumination
Once you have completed questions 1 through 16 below, fill in your answers in the
assessment link online in module 3 folder titled “ Lab 2 – Earth-Sun Assessment”.
Declination
As the earth rotates and revolves around the sun throughout the year, the suns direct
rays are at a 90° angle to different locations between 23.5° N and 23.5° S. The latitude
at which the sun’s rays are at a 90° angle is known as the sun’s
declination
. An
analemma (Figure 1) is
a
scale-
shaped like the figure eight, showing the declination of
the sun throughout the year.
The sun cannot have a declination greater than 23.5° North
or South. Located on the analemma are the dates at which the sun is at a 90° angle to
that latitude. Each degree of latitude between the Tropic of Cancer and the Tropic of
Capricorn will receive direct rays of the sun on exactly two days for each revolution of
the Earth.
Figure 1: The Analemma Source: Textbook Image
1. For seven days, record the time of sunset in the chart below for your home location.
To find the sunset data, go to the following website:
https://www.timeanddate.com/sun/
Once you have recorded the sunset times for seven days, answer the following
question below.
Day
1
2
3
4
5
6
7
Sunset
Time
5:32
5:33
5:34
5:36
5:37
5:38
5:39
After examining your sunset time observations that you recorded for a week, what does
the change in sunset time tell you? What does it tell you about the declination of the
sun? In your write up, provide the dates and sunset times that you recorded for all
seven days. Two points will be deducted if you do not provide the dates and sunset
times for your location for seven days. (4 points)
This chart represents January 24-30th in Cuyahoga Falls, Oh. In order, they were at
5:32 pm, 5:33 pm, 5:34 pm, 5:36 pm, 5:37 pm, 5:38 pm, and 5:39 pm. The sunset gets
later each night, which means the days are getting longer and we are moving towards
summer.
2. What is the declination of the sun on the June Solstice? (1 point)
23.5°N
3. What is the declination of the sun on the December Solstice? (1 point)
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23.5°S
4. What is the declination of the sun on the March Equinox? (1 point)
0
5. What is the declination of the sun on the September Equinox? (1 point)
0
6. Describe what you see happens to the declination of the sun over a solar year.
(1 point)
Over the course of a year, the sun returns to its ‘original’ position and the cycle
repeats
7. On what day will Kent, Ohio receive the maximum intensity of the sun? (1 point) june
20
8. On what day will Kent, Ohio receive the minimum intensity of the sun? (1 point) dec
21
9. What is happening to the length of daylight in the northern hemisphere when the
declination of the sun is moving south? (2 points)
The days are getting shorter and the sun is setting earlier
10.Between which latitudes (both north and south) will the sun never be directly
overhead? (1 point)
no
Insolation
The
duration
of daylight at a place is directly related to the amount of
insolation
(or
in
coming
so
lar
ra
diation) that a place receives throughout the year. The more daylight
received at a place (portion of a day there is light), the greater the potential for
insolation. Insolation varies on Earth depending on latitude. Two things are important
when considering insolation: intensity and duration. Intensity is measured using the
angle of the Sun respective of latitude and the declination. Using a flashlight as an
example, point the beam from the flashlight directly at a wall. All the light is
concentrated into a small area. Now turn the flashlight to a 45º angle. The area the light
illuminates is larger but overall the intensity of the light is weaker. The same thing
happens with insolation. As the sun angle decreases from 90º, the intensity lessons and
the sunlight is spread out over a greater area. To understand intensity, we will be
looking at the altitude of the noon sun (ANS). ANS will change throughout the year as
the Sun’s declination changes. Then, we will look at the intensity of the sunlight.
To compute ANS at any latitude, use the following equation:
LP
= Latitudinal position
LS
= declination
ANS= 90 – (LP + LS) if LP and LS are in different hemispheres
Or
ANS= 90 –
|
LP – LS
|
if LP and LS are in the same hemispheres
(absolute value)
Example:
The declination of the sun on July 20
th
is 21ºN (LS) (refer to the analemma). If your
latitude is 10ºN (LP), or 11º (21-10) from the declination of the sun, then the sun is 11º
from the straight overhead position. To find this you would use the second equation
above because the declination of the sun and your latitudinal position are in the same
hemisphere. The ANS then is 90-11= 79º. So on July 20
th
at the local noon at 10ºN, the
sun will be at 79º in the sky. The intensity of the sun at the place is less then at places
where the sun is 90º in the sky.
In order to determine the intensity of solar radiation of a place, the ANS for a location is
needed. Continuing with the above example: the declination for 10ºN on July 20
th
is
21ºN. The ANS is then 79º. Using the ANS and Table 1 on the next page, go down to
70º then over to 9º to discover the intensity is 98.1%. Does this make sense? Well, 10º
is within the tropical zone so the area receives a lot of sunlight at nearly-direct insolation
in July. So yes, it makes sense that the answer is 98.1%.
Table 1. Percent of solar radiation intensity
Sun
Angle
0º 1º 2º 3º 4º 5º 6º 7º 8º 9º
0º 0.0 1.8 3.5 5.2 7 8.7 10.5 12.2 13.9 15.6 10º 17.4 19.1 20.8 22.5 24.2 25.9 27.6
29.2 30.9 32.6 20º 34.2 35.8 37.5 39.1 40.7 42.3 43.8 45.4 47 48.5 30º 50 51.5 53
54.5 55.9 57.4 58.8 60.2 61.6 62.9 40º 64.3 65.6 66.9 68.2 69.5 70.7 71.9 73.1
74.3 75.5 50º 76.6 77.7 78.8 79.9 80.9 81.9 82.9 83.9 84.8 85.7 60º 86.6 87.5
88.3 89.1 89.9 90.6 81.4 92.1 92.7 93.4 70º 94 94.6 95.1 95.6 96.1 96.6 97 97.4
97.8 98.1 80º 98.5 98.8 99 99.3 99.5 99.6 99.8 99.9 99.9 99.9
For questions 11 through 16, you will be looking at the altitude of the noon sun (ANS) by
calculating it for given latitude, time of year, and corresponding sun declination. You will
then look at the corresponding solar radiation intensity (%).
Remember: ANS is calculated at any latitude by using one of the following equations:
LP = Latitudinal Position and LS = Declination
• ANS = 90 – (LP + LS) if LP and LS are in different hemispheres
•
ANS = 90 –
|
LP – LS
|
if LP and LS are in the same hemispheres. • To
find the intensity, refer to Table 1 after you have calculated ANS.
11.Using Kent, Ohio’s latitude of 41ºN and the date of February 15, what is the ANS on
this day? (1 point)
ANS = 90 - (41N +13S) = 36
12. Using Table 1, what is the percent of solar radiation intensity for Kent, Ohio on
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February 15? (1 point)
58.8
13. Using Kent, Ohio’s latitude of 41ºN and the date of June 5, what is the ANS on this
day? (1 point)
ANS = 90 - |41-22| = 27
14.Using Table 1, what is the percent of solar radiation intensity for Kent, Ohio on June
5? (1 point)
95.1
15.Using Kent, Ohio’s latitude of 41ºN and the date of November 25, what is the ANS
on this day? (1 point)
ANS = 90 - |41-21| = 70
16.Using Table 1, what is the percent of solar radiation intensity for Kent, Ohio on
November 25? (1 point)
94