Chapter 3 Earth - Sun Relations
3.1 Introduction
We saw in the last chapter that the short wave radiation from the sun passes
through the atmosphere and heats the earth, which in turn radiates energy in the
infrared portion of the electromagnetic spectrum. This short wave terrestrial
radiation is absorbed by the atmosphere, thereby heating the atmosphere. The
entire process was called the Greenhouse effect. But we know from experience that
some places on the earth are very hot while others are very cold, and in general, the
temperatures on the earth vary throughout the year. If everyplace on the earth
always received the same amount of radiation, we would expect that every
temperature on the earth would be the same. Since this is not the case, we must
assume that some places on the earth receive more radiation than others, and the
amount of that radiation must vary throughout the year. Hence, we must determine
why some places on the earth receive more radiation than other places.
The amount of radiation received from the sun at a particular place on the
surface of the earth will depend upon:
1. The Solar Constant So, which depends upon:
a) the energy output from the sun,
b) the distance from the sun to the earth.
2. The angle at which the sun’s noon rays strike the earth on the Vernal
Equinox.
3. The angle at which the sun’s noon rays strike the earth on the Summer
and Winter Solstice.
4. The number of hours of daylight in a day.
5. The length of atmosphere that the sun’s rays must pass before reaching
the earth.
3.2 The Solar Constant So
We saw in chapter 2 that the sun radiates energy as a blackbody at a temperature
of 5800 K. The Stefan-Boltzmann law showed us that the energy output of the sun,
the amount of energy radiated per unit time, was Q/t = 3.91  1026 J/s. We further
saw that the amount of that energy impinging upon the upper atmosphere per
second per square meter was called the solar constant and we found that it was
equal to So = Q/At = 1380 J/(s m2). If this amount of radiation truly reached the
entire surface of the earth, then every place on the surface of the earth would
receive the same amount of radiation and the earth would have a uniform
temperature. There would then be no seasons of the year, with the concomitant
large observed changes in temperature. It is of course apparent that this is not the
case. There must be some other aspects of the problem to consider.
We calculated the solar constant So on the assumption that the orbit of the
earth about the sun was a circle. We know of course that this is not correct. The
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Chapter 3 Earth - Sun Relations
orbit of the earth about the sun is elliptical as shown in figure 3.1. Maybe the
elliptical nature of the orbit can explain the seasons, because with an ellipse,
sometimes the earth is very close to the sun and at other times the earth is further
Figure 3.1 The elliptical orbit of the earth about the sun.
away from the sun. When the earth is closest to the sun, a point in the orbit called
the perihelion, the earth is at a distance of 1.47  1011 m. When the earth is
furthest from the sun, a point in the orbit called the aphelion, the earth is at a
distance of 1.53  1011 m. We can reevaluate the solar constant at perihelion, as we
did in example 2.14, in chapter 2, with the value of r = 1.50  1011 m replaced with
the value of r = 1.47  1011 m at perihelion. This is done in Example 3.1.
Example 3.1
The solar constant at perihelion. How much energy from the sun impinges upon the
top of the earth’s atmosphere per unit time per unit area when the earth is closest
to the sun?
Solution
The energy per unit time emitted by the sun is power and was found in example
2.13 to be 3.91  1026 J/s. This total power emitted by the sun does not all fall on
the earth because that power is distributed throughout space, in all directions,
figure 2.13. Hence, only a small portion of it is emitted in the direction of the earth.
To find the amount of that power that reaches the earth at perihelion, we
first find the distribution of that power over a sphere, whose radius is the radius of
the earth’s orbit at perihelion, r = 1.47  1011 m. This will give us the power, or
energy per unit time, falling on a unit area at the distance of the earth from the sun
at perihelion. The area of this sphere is
A = 4πr2 = 4π(1.47  1011 m )2
A = 2.72  1023 m2
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Chapter 3 Earth - Sun Relations
Figure 2.13 Radiation received on the earth from the sun.
The energy per unit area per unit time impinging on the earth at perihelion is
therefore
Q = 3.91  1026 J/s = 1.44  103 W
At
2.72  1023 m2
m2
This value, 1.44  103 W/m2, the energy per unit area per unit time impinging on the
edge of the atmosphere, is called the Solar Constant when the earth is at perihelion,
and will be designated as SoP. Hence,
SoP = 1.44  103 W/ m2
(3.1)
To go to this Interactive Example click on this sentence.
The new value of the solar constant at perihelion so obtained is Sop = 1440 J/(s m2).
A similar analysis for the solar constant at aphelion is shown in Example 3.2.
Example 3.2
The solar constant at aphelion. How much energy from the sun impinges upon the
top of the earth’s atmosphere per unit time per unit area when the earth is at
aphelion, the farthest distance from the sun?
Solution
The energy per unit time emitted by the sun is power and was found in example
2.13 to be 3.91  1026 J/s. This total power emitted by the sun does not all fall on
the earth because that power is distributed throughout space, in all directions,
figure 2.13. Hence, only a small portion of it is emitted in the direction of the earth.
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Chapter 3 Earth - Sun Relations
To find the amount of that power that reaches the earth at aphelion, we first
find the distribution of that power over a sphere, whose radius is the radius of the
earth’s orbit at aphelion, r = 1.53  1011 m. This will give us the power, or energy
per unit time, falling on a unit area at the distance of the earth from the sun at
aphelion. The area of this sphere is
A = 4πr2 = 4π(1.53  1011 m )2
A = 2.94  1023 m2
The energy per unit area per unit time impinging on the earth at aphelion is
therefore
Q = 3.91  1026 J/s = 1.33  103 W
At
2.94  1023 m2
m2
This value, 1.33  103 W/m2, the energy per unit area per unit time impinging on the
edge of the atmosphere, is called the Solar Constant when the earth is at aphelion,
and will be designated as SoA. Hence,
SoA = 1.33  103 W/ m2
(3.2)
To go to this Interactive Example click on this sentence.
Hence the Solar Constant at aphelion is SoA = 1330 J/(s m2).
As can be seen from these two values of the solar constant, more radiation is
received from the sun at perihelion than at aphelion. Therefore it would seem that
since we receive the greatest amount of radiation at perihelion, perihelion should
correspond to our summer time and aphelion should correspond to our winter time.
Alas, this is totally incorrect. The earth is at perihelion on about January 3, our
winter time; and at aphelion on July 4, our summer time. That is, we receive the
maximum amount of radiation from the sun in our winter time, and the minimum
amount of radiation from the sun in our summer time. The conclusion that we must
draw from this, is that the difference in the radiation received on the earth due to
the elliptical nature of the earth’s orbit, and hence the different value of the solar
constant, cannot account for the seasonal variations in the temperature at the
surface of the earth. We must look elsewhere to explain the variations in the
temperature at the surface of our earth that we associate with our seasons. (In
what follows, we will continue to use the mean value of the solar constant So = 1380
J/(s m2) in all our analysis.)
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Chapter 3 Earth - Sun Relations
3.3 The Angle at which the Sun’s Noon Rays Strike the
Earth on the Vernal Equinox
Because of the spherical shape of the earth, the radiation from the sun does
not strike the earth normally (i.e. perpendicularly) everywhere on the surface of the
earth. That is, the angle at which the sun’s rays strikes a point on the surface of the
earth is a function of the latitude of the particular point. This can be seen in figure
3.2, which shows the sun’s rays striking the earth on March 21, this day is called
the Vernal Equinox. On this day at the equator, the sun’s noon rays are
perpendicular to the surface of the earth. That is, the sun is directly overhead at
Figure 3.2 The angle that the sun’s noon rays make at a point on the
surface of the earth (on Mar. 21) is a function of the latitude of the point.
the equator on this date. We can determine the angle that the sun’s noon rays make
at any point on the surface of the earth by drawing a line from the center of the
earth to that point. The line is, of course, just a radius of the earth, and its
extension represents the local vertical direction. As an example, let the point
considered be at 200 N latitude. The radius of the earth when extended through this
point will also make an angle of 200 from the vertical with the sun's rays as seen in
the figure. Since the vertical is at an angle of 900 with the surface, the sun’s noon
rays make an angle of 900 − 200 = 700 with the surface. Notice from the figure that
at 400 N latitude the sun’s noon rays make an angle of 500 with the surface, while at
600 N latitude the sun’s noon rays make an angle of 300 with the surface. These
results from figure 3.2 can be generalized into the formula
θ = 900 − φ
(3.3)
where θ is the angle that the sun’s noon rays make with the horizontal, and φ is the
latitude of the observer. The effect of this angle shows up in the intensity of the
radiation received at the surface of the earth.
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Chapter 3 Earth - Sun Relations
Example 3.3
The angle that the sun’s noon rays make on the Vernal Equinox. Determine the
angle that the sun’s noon rays make at Farmingdale State College, SUNY, which is
at a location of 40.750 N latitude on the surface of the earth.
Solution
The angle that the sun’s noon rays make with the horizontal is found from equation
3.3 as
θ = 900 − φ
θ = 900 − 40.750 = 49.250
To go to this Interactive Example click on this sentence.
Figure 3.3a shows a beam of the sun’s rays hitting the earth normally (900).
The amount of solar energy impinging upon the upper atmosphere per square meter
was called the solar constant and we found that it was equal to So = Q/At = 1380
J/(s m2). Another name for this energy per unit area per unit time is the intensity of
radiation. That is, we can also call the solar constant the solar intensity and we will
label it as Io. Thus Io = So = 1380 J/(s m2). Hence the intensity of the radiation
hitting the surface at the equator, assuming no energy lost through the
Beam of
sun light
Beam of
sun light
φ
θ
L
L0
(a) φ = 0 0 Equator
w
L0
A0 = L0 w
L0
(b) φ = 400
w
L = L0 / sinθ
A= Lw
Figure 3.3 The spreading out of radiation on the surface of the earth caused by the
angle that the sun’s rays make with the surface.
atmosphere, is just Io = So = 1380 J/(s m2). Consider the same beam of light incident
at latitude φ = 400 N, figure 3.3b. As we just saw, at φ = 400 N latitude the sun’s
noon rays make an angle θ = of 500. Notice that the length Lo, the width of the beam
shown in figure 3.3b, when projected onto the horizontal gives the length L. From
the trigonometry of the right triangle we can write
sinθ = Lo / L
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Chapter 3 Earth - Sun Relations
Solving for L we get
L = Lo /sinθ
(3.4)
Since sinθ is generally less than 1, the ratio of (Lo/sinθ) is greater than 1, and hence
L is greater than Lo as can be seen in the figure. The initial area Ao that is
illuminated by the radiation is given by the product of its length Lo times its width
w, hence
(3.5)
Ao = Lo w
When the beam make an angle θ with the horizontal surface, the radiation is
spread out over the area A, as seen in figure 3.3, and is given by
A=Lw
(3.6)
replacing equation 3.4 into equation 3.6 gives
A = Lo w
sinθ
A = Ao
sinθ
(3.7)
Example 3.4
The increased area caused by the angle of the sun for March 21 Find the increased
area that the sun will shine on, caused by the angle of the sun at 400 N latitude for
March 21, which is called the Vernal Equinox.
Solution
As seen in figure 3.2, if the latitude angle is 400 then the sun angle is determined by
equation 3.3 as
θ = 900 − φ
(3.3)
θ = 900 − 400 = 500
Hence the sun’s rays makes an angle θ = 500 with the horizon. If the original area of
the incoming radiation is A0 = 1.00 m2, the new area can now be found from
equation 3.7 as
A = Ao
sinθ
A = 1.00 m2
sin(500)
A = 1.31 m2
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Chapter 3 Earth - Sun Relations
This means that the radiation from the sun, which should be spread across an area
of 1.00 m2, will instead be spread across an area of 1.31 m2. This increased area will
manifest itself as less heating of the earth per square meter and a lower
temperature than would be observed if the sun was directly over head.
To go to this Interactive Example click on this sentence.
The intensity of the radiation from the sun falling on the area A can now be found
as
Q
Q
Q
sin 
=
=
I=
A0t
A0
At
t
sin 
I = Io sinθ
(3.8)
Where I0 is the intensity of the sun’s radiation hitting the earth, also called the
solar constant S0, and θ is the angle that the sun’s rays make with the horizontal.
Equation 3.8 says that the intensity of the radiation received on earth will decrease
from the maximum value I0, when the sun is directly overhead, to zero when the sun
is on the horizon, i.e. θ = 00. The decrease in the intensity of the radiation occurs
because the radiation is now spread out over a greater area and will as a
consequence produce less heating per unit area. Notice the size of the area in figure
3.3a when the sun is directly overhead and in figure 3.3b, when it makes an angle
of 500 with the horizontal.
As an example of the use of equations 3.7 and 3.8, note that when the angle θ
is equal to 900 the area covered by the radiation is Ao and it receives the intensity of
radiation Io. When the angle θ is equal to 450 the area covered by the radiation is
1.41Ao and the intensity of the radiation received is 0.707Io. When the angle θ is
equal to 300 the area covered by the radiation is 2Ao and the intensity of the
radiation received is only 0.5Io.
Example 3.5
The intensity of the sun’s radiation for the Equinoxes. Find the intensity of the sun’s
radiation at 400 N latitude for March 21, the Vernal Equinox.
Solution
As seen in figure 3.2, if the latitude angle is 400 then the sun makes an angle θ
given by equation 3.3 as
θ = 900 − φ
θ = 900 − 400 = 500
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Chapter 3 Earth - Sun Relations
with the horizon. The solar constant is Io = 1380 J/(s m2). The intensity of radiation
can now be found from equation 3.8 as
I = Io sinθ
I = 1380 ( J ) sin(500)
s m2
I = 1060 J/(s m2)
Notice that even though the solar constant is 1380 J/(s m2), the amount of solar
radiation actually hitting the surface of the earth at 400 N latitude is only 1060 J/(s
m2), or 23.2% less than the radiation hitting the surface at the equator. Hence it
will not be as warm at 400 N latitude as it will be at the equator.
To go to this Interactive Example click on this sentence.
We can also express this intensity of radiation in terms of the latitude angle
φ directly by noting that in figure 3.2 that
or
θ + φ = 900
θ = 900 − φ
Hence the intensity of the sun’s radiation hitting the earth, equation 3.8 can also be
written in terms of the latitude angle φ as
I = Io sin(900 − φ )
(3.9)
Example 3.6
The intensity of the sun’s radiation Find the intensity of the sun’s radiation on Mar.
21, called the Vernal Equinox, at the latitude angles of (a) φ = 200, (b) φ = 400, (c) φ =
600, (d) φ = 800.
Solution
The suns radiation at the equator, φ = 00, is just equal to the solar constant and is
Io = 1380 J/(s m2)
(a) For φ = 200, the intensity of the sun’s radiation, found from equation 3.9, is
I = Io sin(900 − φ)
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Chapter 3 Earth - Sun Relations
I = 1380 ( J ) sin(900 − 200)
s m2
I = 1380 ( J ) sin(700)
s m2
I = 1300 J/(s m2)
(b) For φ = 400, the intensity of the sun’s radiation, found from equation 3.9, is
I = Io sin(900 − φ )
I = 1380 ( J ) sin(900 − 400)
s m2
I = 1380 ( J ) sin(500)
s m2
I = 1060 J/(s m2)
(c) For φ = 600, the intensity of the sun’s radiation, found from equation 3.7, is
I = Io sin(900 − 600)
I = 1380 ( J ) sin(900 − 600)
s m2
I = 1380 ( J ) sin(300)
s m2
I = 690 J/(s m2)
(d) For φ = 800, the intensity of the sun’s radiation, found from equation 3.9, is
I = Io sin(900 − 800)
I = 1380 ( J ) sin(900 − 800)
s m2
I = 1380 ( J ) sin(100)
s m2
I = 240 J/(s m2)
Notice how the amount of radiation decreases as we go further away from the
equator, which explains the difference in the temperature distributions across the
surface of the earth from north to south.
To go to this Interactive Example click on this sentence.
The intensity of the sun’s radiation I as a function of the latitude angle φ,
found from equation 3.9, is plotted in figure 3.4. Notice that the maximum intensity
of radiation occurs at the equator, when the sun is directly overhead, and decreases
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Chapter 3 Earth - Sun Relations
as you approach the pole where the angle of the sun’s rays decreases causing the
radiation to be spread out over a larger area, producing less heating per unit area.
As can be seen in figure 3.4 the equator receives the full intensity of 1380 J/s m2
while at 400 latitude the intensity is only a little over 1050 J/s m2, and at 700
latitude, the intensity is a little below 500 J/s m2. Hence, because of the spherical
shape of the earth, the angle at which the sun’s noon rays strike the earth varies
with latitude, and regions near the equator will receive the maximum amount of
radiation and will be the hottest regions, while regions north or south of the equator
will receive less radiation and will be cooler.
Intensity of Sun's Radiation
as a function of lattitude
joules/(s m^2)
Sun's Intensity (Q/tA)
1500
1000
500
0
0
20
40
60
80
100
Lattidude angle phi (deg)
Figure 3.4 The intensity of the sun’s radiation as a function of the latitude angle
on Mar. 21.
3.4 The Angle at which the Sun’s Noon Rays Strike the
Earth on the Summer and Winter Solstice
We have just seen that the intensity of the solar radiation striking the surface of
the earth varies a great deal because of the spherical shape of the earth. In the case
just considered, March 21, the sun was directly overhead at the equator. The
equator received the maximum radiation, and regions north or south of the equator
received less radiation. Hence the equator would be hottest and regions north and
south would be cooler. This picture says that the northern hemisphere and the
southern hemisphere would receive the same amount of radiation, and on that day
considered, March 21, that is correct. However, even though the radiation is the
same in the northern and southern hemisphere, the temperatures will not be the
same. That is, even though the radiation at 400 N latitude and 400 S latitude is the
same, the temperatures are not. When it is hot in the northern hemisphere, it is
cold in the southern hemisphere and vice versa. In the model just considered the
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Chapter 3 Earth - Sun Relations
sun was directly overhead at the equator. Is this always true? The answer is no and
depends on where the earth is in its orbit.
A picture of the earth in its orbit is shown in figure 3.5. Notice that the axis
Figure 3.5 The areas of sunlight upon the surface of the earth as a function of its
position in its orbit about the sun.
of rotation of the earth is not perpendicular to the plane of its orbit about the sun.
Instead, the earth is inclined at an angle of 23.50 to the normal to the plane of the
orbit. The plane of the orbit of the earth about the sun is called the ecliptic. Also
notice that the direction of earth’s axis is fixed in space, pointing toward the North
Star throughout the year. Because of the tilt of the earth’s axis, the radiation from
the sun will cover a different portion of the earth at different times of the year,
which can be seen in figure 3.5. So the amount of radiation received on the earth
will depend on where the earth is in its orbit at a particular time of the year. Figure
3.6 shows that radiation from the sun falling on the earth at three different times of
the year, the summer solstice, the equinoxes, and the winter solstice.
On March 21 and September 21, the Vernal and Autumnal Equinoxes
respectively, the sun’s rays fall evenly on the earth, figure 3.6b. Notice, in figure
3.6, the line called the circle of illumination. The circle of illumination is a circle
around the surface of the earth. It separates the part of the earth that is
illuminated by the sun and hence in daylight, from the part of the earth that does
not receive the sun’s radiation and hence is in darkness. On these dates, the circle
of illumination coincides with the axis of rotation of the earth. Therefore, half the
earth has daylight and half has night. There is an equal amount of day and night.
That is, there are 12 hours of daylight and 12 hours of night. In fact, the word
equinox means equal nights. On either of these days the sun is directly overhead at
the equator. Also note in figure 3.6 the lines around the earth indicating the Arctic
Circle, the Tropic of Cancer, the Equator, the Tropic of Capricorn, and the Antarctic
Circle. Their importance will be mentioned later.
June 21 is called the Summer Solstice. At this time the sun is directly
overhead at the Tropic of Cancer, which is at 23.50 N latitude, figure 3.6a. Recall
that in figure 3.2, which shows the orientation of the earth in space on the vernal
and autumnal equinox that the sun’s noon angle varied with latitude on that
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Chapter 3 Earth - Sun Relations
particular day. The sun’s noon angle also varies with the time of the year. As an
example, let us consider New York City which is approximately at 400 N latitude.
N
0
23.5
pi
Tro
A
Circle of
Illumination
co
f
pr
Ca
rc
nta
ico
C
ti c
rn
tic Cir
North Pole
e
i rc l
Equator
or
ua t
Equator
Eq
er
Arc
f
nc
Ca
ic of Canc
er
co
e
cle
pi
Tro
i rc l
op
Tr
Sun's Rays
Ar
C
c ti c
Circle of
Illumination
S
(a) Summer Solstice
N
Equator
Arctic Circle
p
ro
ic of Canc
Equator
tic Cir
cle
Arc
T
Tropic of Cancer
er
Sun's Rays
Tropic of Capricorn
Circle of
Illumination
North Pole
Antarctic Circle
S
(b) Equinoxes
Circle of
Illumination
N
0
23.5
Tro
p
Arc
An
ic o
tarc
S
fC
ti c
er
or
apr
C ir
ico
rn
cle
r
tic Cir
cle
Tro
p
cle
ce
of C an
Sun's Rays
ua t
anc
C ir
Equator
r ot auq E
Eq
fC
ti c
Tr
op
ic
ic o
Arc
Circle of
Illumination
North Pole
Circle of
Illumination
(c) Winter Solstice
Figure 3.6 Radiation from the sun falling on the earth at different times of the
year.
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Chapter 3 Earth - Sun Relations
On the vernal and autumnal equinox, figure 3.7a, the sun’s rays are
perpendicular to the surface at the equator. A line is drawn from the center of the
earth to NYC which subtends the angle φ = 400. The local zenith is perpendicular to
the surface and as can be seen in figure 3.7a, the sun's rays make an angle of 400
with the zenith or 500 with respect to the horizon.
Figure 3.7 Determining the noon sun angle at 400 N latitude.
On June 21, the summer solstice, the earth’s axis is tilted 23.50 toward the
sun, and the sun is directly overhead at 23.50 N latitude. The radius that is drawn
from the center of the earth to NYC subtends an angle of 400 with the equator, but
as can be seen in figure 3.7b, it makes an angle of 16.50 with the plane of the
ecliptic, and hence the sun’s rays make an angle of 16.50 with respect to the local
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Chapter 3 Earth - Sun Relations
zenith passing through NYC and makes an angle of 900 − 16.50 = 73.50 with the
horizon.
Figure 3.7c shows the orientation of the earth with respect to the plane of the
ecliptic on Dec. 21. Notice that now the radius to NYC subtends an angle of 400 +
23.50 = 63.50 with respect to the ecliptic. Hence the sun’s rays make an angle of
63.50 with respect to the local zenith or 900 − 63.50 = 26.50 with respect to the
horizon. Notice that the sun’s noon rays at NYC make an angle of 500 with respect
to the horizon on Mar. 21, and 73.50 on Jun. 21 and 26.50 on Dec. 21.
We can generalize equation 3.9 for the intensity of radiation, by calling the
angle that the plane of the ecliptic makes with the earth’s axis, β. As we have just
shown, for the summer solstice, the angle that the sun's noon’s rays make with the
horizontal is θ + β and equation 3.3 was θ = 900 − φ, hence the intensity of radiation,
equation 3.8 and 3.9 becomes
I = Io sin(θ + β ) = Io sin(900 − φ + β)
(3.10)
Equation 3.10 gives the intensity of the sun’s radiation on the summer
solstice. In the same way, by observing figure 3.7c for the winter solstice we get for
the intensity of radiation for the winter solstice
I = Io sin(θ − β) = Io sin(900 − φ − β)
(3.11)
Example 3.7
The intensity of the sun’s radiation for the Summer and Winter Solstice Find the
intensity of the sun’s radiation at 400 N latitude for (a) the Summer Solstice and, (b)
the Winter Solstice.
Solution
The solar constant is Io = 1380 J/(s m2).
a. As shown in figure 3.7b for the Summer Solstice, the sun makes an angle of 73.50
with the horizontal. The intensity of radiation can now be found from equation 3.10
as
I = Io sin(900 − φ + β)
I = 1380 ( J ) sin(900 − 400 + 23.50)
s m2
I = 1380 ( J ) sin(73.50)
s m2
I = 1320 J/(s m2)
b. As shown in figure 3.7c for the Winter Solstice, the sun makes an angle of 26.50
with the horizontal. The intensity of radiation can now be found from equation 3.11
as
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Chapter 3 Earth - Sun Relations
I = Io sin(900 − φ − β)
I = 1380 ( J ) sin(900 − 400 − 23.50)
s m2
I = 1380 ( J ) sin(26.50)
s m2
I = 616 J/(s m2)
Notice that because of the difference in the sun’s noon rays at 400 N latitude the
intensity of the sun’s radiation is 1060 J/(s m2) on Mar. 21, the Vernal Equinox
Example 3.3; 1320 J/(s m2) on Jun. 21, the Summer Solstice; 1060 J/(s m2) on Sept.
21, the Autumnal Equinox; and only 616 J/(s m2) on Dec. 21, the Winter Solstice.
This variation in the intensity of the radiation received will account for the
different seasons.
To go to this Interactive Example click on this sentence.
We can now draw a generalization of figure 3.4 here where now, because of
equations 3.10 and 3.11, we can plot the intensity of the sun’s radiation I, as a
function of the latitude angle φ for the Vernal Equinox, from equation 3.9, for the
Summer Solstice, from equation 3.10, and for the Winter Solstice, from equation
3.11. These graphs are plotted in figure 3.8.
Intensity of Radiation
Intensity of Solar Radiation as a
Function of Latitude
1600
1400
1200
Equinoxes
1000
Summer Solstice
800
Winter Solstice
600
400
200
0
0
20
40
60
80
100
Latitude Angle
Figure 3.8 The intensity of the sun’s radiation as a function of the latitude angle
on Mar. 21.
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Chapter 3 Earth - Sun Relations
Notice, that as expected, the greatest amount of radiation occurs for the
summer solstice, and the least for the winter solstice, with the equinoxes between
the two extremes.
Because of the motion of the earth in its orbit, it appears as though the sun
has two motions across the sky, figure 3.9. For the diurnal motion, the sun rises in
Figure 3.9 The diurnal and annual motion of the sun across the sky.
the east and sets in the west, while for the annual motion the sun rises higher and
higher in the sky until, on June 21, it is at its highest position. It then starts to
move down the sky reaching its lowest position on Dec. 21, the Winter Solstice. (See
figure 3.5 and 3.6.) The sun appears to move 23.50 north of the equator on June 21
and 23.50 south of the equator on Dec. 21. The total movement of the sun above the
horizon is therefore 470.
The motions shown in figure 3.9 also have a very interesting historical
significance. The ancient Romans observed not only the daily motion of the sun
across the sky, but they were also aware of its annual motion. As the sun moved
lower and lower in the sky as the Winter Solstice approached, the ancient Romans
feared that one day the sun would keep lowering further and further and one day
might drop below the horizon and would never rise again. This disappearance of the
sun would signify the beginning of the end of the world. So they would observe the
sun lowering on December 21, and would watch it closely every day. By December
25 they were able to observe that the sun was moving back up in the sky. The
lengthening of the days afterward symbolized the rebirth of the gods and the
renewal of life. The Romans made this day a feast, called Saturnalia. Hence,
December 25 was the date for the coming of the sun for the ancient Romans. The
world would not die. Everything would continue to be good.
When Christianity became established in Rome centuries after, the early
Church Fathers really did not know the date for the birth of Jesus Christ. Since,
they always tried to replace “pagan” holidays with “Christian” holidays, they
changed the feast of Saturnalia, to Christmas, because this feast would now be
3-17
Chapter 3 Earth - Sun Relations
considered as the birthday of Jesus Christ, which they felt was the coming of the
son of God. Hence, we celebrate Christmas on December 25, a few days after the
Winter Solstice.
Since the sun is highest in the sky in the northern hemisphere during the
Summer Solstice, the intensity of radiation reaching the earth will be greater
throughout the northern hemisphere, and this will correspond to the beginning of
our summer. Conversely, since the sun is lowest in the sky in the northern
hemisphere during the Winter Solstice, the intensity of radiation reaching the earth
will be least throughout the northern hemisphere, and this will correspond to the
beginning of our winter. We are also aware that the period of daylight is longer in
the summer than in the winter. Hence we must also consider the effect of the longer
period of daylight will have on the heating of the earth.
3.5 The Number of Hours of Daylight in a Day
Have you ever wondered why there are so many more hours of daylight in the
summer time than in the winter? The variation in the length of daylight comes
about because of the spherical shape of the earth, and because its axis of rotation is
not perpendicular to the plane of its orbit about the sun. The duration of daylight
will be greatest when the sun is highest in the sky which occurs on June 21, the
Summer Solstice.
The number of hours of daylight at a particular latitude on June 21 can be
found as follows. The apparent angular velocity ω of the sun as it moves across the
sky is given by
ω=θ/t
(3.12)
where θ is the angle swept out by the sun as it moves across the sky. For a point on
the surface of the earth, such as New York, located at the latitude φ in figure 3.10a,
the point rotates about the earth’s axis in a circle of radius RE cosφ, where RE is the
radius of the earth, figure 3.10b. The sun rises at the point A and sets at the point
Sun's Rays
RE cosφ
β
0
23.5
NY
φ
x
d
RE
B Sunset
RE cosφ
φ
Noon
Sun
ato r
Equ
Circle of
Illumination
2π
−2
γ
γ
γ
x
North Pole
Circle of
Illumination
S
(a) Side View
A
Eq uato r
N
Sunrise
(b) Top View
Figure 3.10 The geometry of the earth on the Summer Solstice to determine the
length of the day.
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Chapter 3 Earth - Sun Relations
B. The angle subtended by the night sky is θ = 2γ, and the daylight sky, θ = 2π − 2γ.
The angular velocity of the sun around the earth is just equal to the angular
velocity of the earth, which is ωE = 0.261799 radians per hour. Therefore the time
for the sun to move across the sky for daylight, tD, is
tD = θ = 2π − 2γ
ω
ω
(3.13)
The angle γ is found from the geometry of figure 3.10b as
x
RE cos φ
(3.14)
x = d sinβ
(3.15)
cosγ =
But from figure 3.10a, we also have
Placing equation 3.15 into equation 3.14 gives
cosγ =
d sinβ
RE cosφ
(3.16)
But from the geometry of figure 3.10a, we have
RE sin φ = d cosβ
Solving for d we get
d = RE sinφ
cosβ
Replacing equation 3.18 into 3.16 gives
Hence, the angle γ is found as
cosγ = RE sinφ sinβ
cosβ RE cosφ
cosγ = tanφ tanβ
γ = cos−1[tanφ tanβ]
(3.17)
(3.18)
(3.19)
(3.20)
(3.21)
The time of daylight is found by substituting equation 3.21 into 3.13 to obtain
tD = 2π − 2cos−1[tanφ tanβ]
ω
(3.22)
Equation 3.22 gives the length of daylight at a location on the surface of the earth
that is situated at the latitude φ. The angle β is the angle between the axis of
rotation of the earth and the circle of illumination. For the Summer Solstice, β =
+23.50, for the Equinoxes β = 00, and for the Winter Solstice, β = −23.50.
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Chapter 3 Earth - Sun Relations
Example 3.8
The length of daylight for the Summer Solstice. Find the length of daylight for the
Summer Solstice, at a location such as New York, which is at a latitude of
approximately 400 N.
Solution
The length of daylight for the Summer Solstice is found from equation 3.22 as
tD = 2π − 2 cos−1[tanφ tanβ]
ω
−1
tD = 2π − 2cos [tan400 tan23.50]
0.261799 rad/hr
tD = 14.85 hr. = 14 hr 51 min. of daylight
Hence, there are 14 hours and 51 minutes of daylight in New York City on June 21,
the Summer Solstice. (Also notice that we used β = +23.50, for the Summer Solstice
in this calculation.)
To go to this Interactive Example click on this sentence.
The number of hours of nighttime is obtained by subtracting the length of the
day tD from 24 hours to obtain
tN = 24 hr − tD
(3.23)
Example 3.9
The length of nighttime for the Summer Solstice. Find the length of nighttime for
Example 3.8.
Solution
Since we found the length of daylight in example 3.8 to be tD = 14.85 hr, the length
of nighttime is found from equation 3.23 as
tN = 24 hr − tD
tN = 24 hr − 14.85 hr = 9.15 hr
tN = 9 hr 9 min
So, as you would expect, the length of the day is relatively long (14 hr 51 min) while
the length of the night is relatively short (9 hr 9 min) for the summer solstice.
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Chapter 3 Earth - Sun Relations
To go to this Interactive Example click on this sentence.
The above calculation was made for a location at 400 north latitude for the
summer solstice. The number of hours of daylight for the Summer Solstice, at any
latitude, equation 3.22, is plotted in figure 3.11 as the blue curve at the top of the
graph. Notice that at the equator the length of the day is 12 hours. As the latitude
Length of Day as a Function of Lattitude for the
Summer Solstice
Length of Day
20.00
15.00
10.00
5.00
0.00
0
10
20
30
40
50
60
70
Lattitude angle
Length of Day
Length of Night
Figure 3.11 The length of daylight as a function of latitude for the Summer
Solstice.
angle increases, the number of hours of daylight increases until at the Arctic Circle,
66.50 N latitude, the length of day is 24 hours. The red curve toward the bottom of
the graph shows the number of hours of nighttime. Again, notice that at the equator
the length of the night is 12 hours. As the latitude increases, the number of hours of
nighttime decreases until at the Arctic Circle, 66.50 N latitude, the length of night
is 0 hours. This means that the sun never sets north of the Arctic Circle on June 21,
and there is no nighttime at all. There is 24 hours of daylight. An interesting
consequence of this long daytime is that northerly locations, such as in the wheat
belts of Canada, which have a very short growing season, are still able to grow
crops because of the very long length of the day. It also doesn’t get very hot at the
Arctic Circle, even though there are 24 hours of daylight, because the sun is still
very low on the horizon, and the solar radiation is spread over a relatively large
area and does not do very much heating as we discussed in section 3.3.
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Chapter 3 Earth - Sun Relations
The number of hours of daylight tD for the Vernal or Autumnal Equinox, at
any latitude, equation 3.22, is now plotted in figure 3.12. Notice that we get a
straight line of 12 hr from the equator all the way north, indicating that the length
Length of Day as a Function of Lattitude for the Equinoxes
Length of Day
20.00
15.00
10.00
5.00
0.00
0
10
20
30
40
50
60
70
Lattitude angle
Length of Day
Length of Night
Figure 3.12 The length of daylight as a function of latitude for the Vernal and
Autumnal Equinox.
of daylight is 12 hours everywhere. This, of course, also means that the duration of
nighttime is also 12 hours everywhere on the earth.
The number of hours of daylight for the Winter Solstice, at any latitude,
equation 3.22, (note that for the Winter Solstice, the angle β in equation 3.22
becomes − 23.50) is plotted in figure 3.13 as the blue curve which is now at the
bottom of the graph. (This is a reversal from figure 3.11 where the blue curve was
on the top of the graph.) Notice that at the equator the length of the day is still 12
hours. As the latitude angle increases, the number of hours of daylight now
decreases until at the Arctic Circle, 66.50 N latitude, the length of daytime is 0
hours. This means the sun never rises in the sky for regions north of the Arctic
Circle for the Winter Solstice. There will be 24 hours of night. The red curve toward
the top of the graph shows the number of hours of nighttime. (Again, this is a
reversal from figure 3.11 where the red curve was on the bottom of the graph.)
Again, notice that at the equator the length of the night is 12 hours. As the latitude
increases, the number of hours of nighttime increases until at the Arctic Circle,
66.50 N latitude, the length of night is 24 hours. This means that the sun never
rises north of the Arctic Circle on December 21, and there is no daytime at all.
There is 24 hours of nighttime.
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Chapter 3 Earth - Sun Relations
The actual time of sunrise and sunset can be obtained from the length of the
day by noting that at 12 noon, the sun is directly overhead at your location. Because
of the symmetry of the spherical earth, the time for the sun to rise to its noon
Length of Day as a Function of Lattitude for the Winter
Solstice
Length of Day
20.00
15.00
10.00
5.00
0.00
0
10
20
30
40
50
60
70
Lattitude angle
Length of Day
Length of Night
Figure 3.13 The length of daylight as a function of latitude for the Winter Solstice.
position, is equal to the time for the sun to set from its noon position. Therefore, the
time from sunrise to noon is half the time of daylight, that is tD/2. Thus the time for
the sun to rise can be found as
tsunrise = 12 noon − tD
(3.24)
_
2
while the time for sunset can be found from
tsunset = 12 noon + tD
_
2
(3.25)
Example 3.10
Time of Sunrise and Sunset. Find the time of sunrise and sunset for example 3.8.
Solution
We found in example 3.8 that the length of daylight was 14.85 hr. The time of
sunrise is found from equation 3.24 as
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Chapter 3 Earth - Sun Relations
tsunrise = 12 noon − tD = 12 − 14.85 = 4.575 hr. AM
2
2
tsunrise = 4:35 AM
The time for sunset is found from equation 3.25 as
tsunset = 12 noon + tD = 12 + 14.85 = 19.425 hr. PM
2
2
tsunset = 7:26 PM
To go to this Interactive Example click on this sentence.
Equation 3.24 is plotted in figure 3.14 as the blue curve toward the bottom of
the graph, and gives the time for sunrise as a function of latitude for the Summer
Time of Sunrise and Sunset as a Function of Lattitude
for the Summer Solstice
Time (hrs)
20.00
15.00
10.00
5.00
0.00
0
10
20
30
40
50
60
70
Lattitude Angle (degrees)
Time of Sunrise
Time of Sunset
Figure 3.14 Time of sunrise and sunset as a function of latitude for the Summer
Solstice.
Solstice. Notice that the time of sunrise decreases as a function of latitude. That is,
the sun rises at 6 AM at the equator, but rises earlier as the latitude increases. At
400 N latitude the time of sunrise, read from the graph, is about 4:30 AM. At 66.50
N latitude (the Arctic Circle) the sun rises at midnight.
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Chapter 3 Earth - Sun Relations
Equation 3.25 is plotted in figure 3.14 as the red curve toward the top of the
graph, and gives the time for sunset as a function of latitude for the Summer
Solstice. Again, notice that the time of sunset increases (gets later) as a function of
latitude. That is, the sun sets at 18:00 hours (on a 2400 hour clock) or at 6:00 PM
(18:00 - 12:00 = 6:00) on the traditional clock, at the equator, but sets later as the
latitude increases. At 400 N latitude the time of sunset, read from the graph, is
about 19.4 hours or about 7.4 hours, or 7:24 PM. At 66.50 N latitude (the Arctic
Circle) the sun sets at midnight. Since the sun sets at midnight and rises at
midnight the sun is always above the horizon north of the Arctic Circle.
The length of day and night, and the time for sunrise and sunset, for any day
between the equinox and the solstices can be approximated by equation 3.22 by
assuming that the earth’s orbit is circular and hence the angle β in equation 3.22
varies linearly from −23.50 at the Winter Solstice to 00 at the Vernal Equinox and to
+23.50 at the Summer Solstice. Equation 3.22 for a position at 400 N latitude is
plotted in figure 3.15 as a function of the number of days before or after the Vernal
Equinox. As an example of the use of figure 3.15, let us find the length of daylight
and the time for sunrise and sunset on May 15. May 15 is 55 days after Mar 21. The
55th day is found on the x axis in figure 3.15, and this intersects the length of
daylight curve at about 13.7 hours. The time for sunrise is found from the graph as
Length of Day and Night and Time of Sunrise and Sunset as a
Function of the number of days before and after the Vernal
Equinox
25.00
20.00
Time
15.00
10.00
5.00
0.00
-100
-80
-60
-40
-20
0
20
40
60
80
100
Days
Length of Daylight
Length of Night
Time of Sunrise
Time of Sunset
Figure 3.15 The length of day and night and the times of sunrise and sunset at 400
N latitude as a function of the number of days before and after the Vernal Equinox.
slightly after 5 AM and sunset slightly before 7 PM. This graph is only good for a
latitude of 400 N. You would need a similar graph for each latitude you are
3-25
Chapter 3 Earth - Sun Relations
interested in. You can see from this analysis why the length of daylight varies with
the time of the year, and the latitude of the observer.
3.6 Summary of Heating of Earth’s Atmosphere
In section 3.1 we said that the amount of radiation received from the sun at a
particular place on the surface of the earth depended upon:
1. The Solar Constant So, which depends upon:
a) the energy output from the sun,
b) the distance from the sun to the earth.
We showed that we receive the maximum amount of radiation from the sun in our
winter time, and the minimum amount of radiation from the sun in our summer
time. The conclusion that we drew from this, is that the difference in the radiation
received on the earth due to the elliptical nature of the earth’s orbit, and hence the
different value of the solar constant, cannot account for the seasonal variations in
the temperatures observed at the surface of the earth.
2. The angle at which the sun’s noon rays strike the earth.
We showed in section 3.1 that because of the spherical shape of the earth, the
radiation from the sun does not strike the earth normally (i.e. perpendicularly)
everywhere on the surface of the earth. That is the angle at which the sun’s rays
strike a point on the surface of the earth is a function of the latitude of the
particular point. Hence, the intensity of the radiation will decrease from when the
sun is directly overhead, at the equator, to when it is at a lower position on the
horizon at some other latitude. The decrease in the intensity of the radiation occurs
because the radiation is now spread out over a greater area and will, as a
consequence, produce less heating per unit area. When the intensity of the sun’s
radiation is plotted as a function of the latitude angle, we see that the maximum
intensity of radiation occurs at the equator, when the sun is directly overhead, and
decreases as you approach the pole where the angle of the sun’s rays decreases
causing the radiation to be spread out over a larger area, producing less heating per
unit area. Hence, because of the spherical shape of the earth, the angle at which the
sun’s noon rays strike the earth varies with latitude, and regions near the equator
will receive the maximum amount of radiation and will be the hottest regions, while
regions north of the equator will receive less radiation and will be cooler.
3. The number of hours of daylight in a day.
As we have just seen in section 3.5 the length of daylight varies with the location on
the surface of the earth and on the time of the year. The results are summarized in
table 3.1 below. The length of daylight is maximum on the Summer Solstice, June
21, and lowest on the Winter Solstice, December 21. The greater the length of
daylight will increase the amount of radiation received on the earth and hence add
to the heating of the earth and contributes to hotter temperatures in the summer.
Conversely, the shorter the length of daylight in the winter will decrease the amount
of radiation received on the earth and contributes to the colder temperatures
observed in the winter.
3-26
Chapter 3 Solar and Terrestrial Radiation
Latitude Angle
φ
0
10
20
30
40
50
60
70
80
90
Table 3.1
Length of Daylight
Summer Solstice
Equinox
Winter Solstice
12 hr
12 hr 35 min
13 hr 13 min
13 hr 56 min
14 hr 51 min
16 hr 10 min
18 hr 31 min
2 mo
4 mo
6 mo
12 hr
11 hr 25 min
10 hr 47 min
10 hr 4 min
9 hr 9 min
7 hr 50 min
5 hr 29 min
0 hr
0 hr
0 hr
12 hr
12 hr
12 hr
12 hr
12 hr
12 hr
12 hr
12 hr
12 hr
12 hr
3. The length of atmosphere that the sun’s rays must pass through
before reaching the earth. When the sun is directly overhead at the equator, it
only has to pass through one atmosphere on its way to the surface of the earth. For
a location north of the equator, the sun’s rays make an angle with the local vertical
and hence have to go through a longer portion of the atmosphere. The further and
further north you go, the greater the length of atmosphere the rays must go
through. The greater the length of the atmosphere, the greater will be the
probability of some of the radiation being reflected or scattered by the atmosphere.
This reflected or scattered radiation will not reach the surface of the earth, and
hence will not heat the earth. Hence the earth’s surface will be cooler than it would
be if all this radiation actually reached the surface. This effect also contributes to
the northern regions being colder than regions near the equator.
Before we leave this section there are two questions that still have to be
answered. If we receive the maximum radiation from the sun at 12 noon on any
day, why does the hottest part of that day occur at about 3 or 4 PM, three or four
hours after the maximum radiation? And similarly, if we receive the maximum
radiation from the sun on the Summer Solstice, June 21, why does the hottest part
of the summer occur about the first two weeks of August, two months after the
maximum radiation from the sun reached the earth? The two questions are closely
related. We will answer the one about the daily maximum temperature first.
If we plot the radiation absorbed from the sun and the radiation lost by the
earth in a 24 hour period, we get the curves shown in figure 3.16. Notice that the
maximum radiation from the sun occurs at 12 noon as expected. It takes time for
the radiation received from the sun to be conducted down into the earth. Hence, the
earth does not attain its largest temperature until a few hours after 12 noon. This
means that the maximum radiation from the earth occurs a few hours after noon.
Even though the absorbed solar radiation is decreasing, the earth's radiation
is increasing. If we add the absorbed solar radiation with the earth’s radiation, the
maximum radiation into the atmosphere occurs at about 3 or 4 PM. Notice in figure
3-27
Chapter 3 Solar and Terrestrial Radiation
Figure 3.16 Lag in the time for the maximum daily temperature.
3.16 that the plot of the temperature of the surface air as a function of time has its
maximum value where the radiant energy absorbed from the sun coincides with the
maximum energy radiated by the earth. Also note in figure 3.16 that during the
daylight hours there is a surplus of energy received on the earth while during the
later hours there is a deficit, as expected.
A similar analysis of the radiation absorbed from the sun and the radiation
lost by the earth in a 12 month period is shown in figure 3.17. Notice that the
Figure 3.17 Lag in the annual maximum temperature.
maximum radiation from the sun occurs on the summer solstice as expected.
However, since it takes time for the radiation received from the sun to be conducted
down into the earth, the earth does not attain its largest temperature until about
two months later. This means that the maximum radiation from the earth occurs a
few months later. Even though the absorbed solar radiation is decreasing, the
earth's radiation is increasing. If we add the absorbed solar radiation with the
earth’s radiation, the maximum radiation into the atmosphere occurs at about the
first or second week in August, which should also be the time for maximum
temperature. Notice in figure 3.17 that the plot of the temperature of the surface air
3-28
Chapter 3 Solar and Terrestrial Radiation
as a function of time has its maximum value where the radiant energy absorbed
from the sun coincides with the maximum energy radiated by the earth. Also note
in figure 3.17 that the amount of radiation from the sun starts increasing after the
winter solstice and continues to increase until it reaches a maximum on the
summer solstice, as expected. The solar radiation then decreases to its minimum at
the winter solstice on December 21. Notice that there is a surplus of energy received
on the earth from late February until early August. After that there is a deficit of
energy going into the fall and winter, as expected.
3.7 Controls of Temperature
Since two particular positions on the surface of the earth at the same latitude
receive the same radiation from the sun you might expect that they should have the
same temperature and weather. We know from our experience that, in general, this
is not the case. Hence, besides the radiation received from the sun, the weather at a
particular location is also influenced by geographic effects. These geographic effects
are called controls of temperature, and are
1. Altitude. In general if you live high up in the mountains it will be colder
than if you live in a valley region, even though you are at the same latitude.
2. Location of land and water surfaces. If you live near the ocean the
ocean will modify your weather, making it usually warmer in the winter and cooler
in the summer than at a place at the same latitude located in the center of a
continent.
3. Mountain barriers. The mountains can act as barriers, blocking much of
the air from reaching a particular location. As an example, on the west coast of the
United States moist air moves off the ocean and tries to move inland, but the
mountains block this motion. On the west side of the mountain there is plenty of
rain and the crops are green and abundant. This moisture from the Pacific Ocean
does not reach the east side of the mountains and the region east of the mountains
is semiarid.
Besides these main controls of temperature, the temperature at a particular
position can also be influenced by the local topography of the area. Prevailing
winds also can play a big factor in the temperatures in your area. As an example, if
you live near the ocean, the cool breezes in the summertime can be very delightful.
An air mass is a large mass of air that has relatively the same temperatures and
humidity throughout it region. So if a cold air mass moves into your region, it will
become cold. On the other hand, if a warm air mass moves into your region, it will
become warm. Hence the advection of these air masses can have a very large affect
on the temperature in a region. Finally there are semipermanent pressure areas
located around the globe. These pressure areas determine the direction of much of
the winds, and cause a great deal of temperature change by the advection of air. We
will speak more about these later.
3-29
Chapter 3 Solar and Terrestrial Radiation
The Language of Meteorology
The Solar Constant So, is the amount of radiation received from the sun at
a particular place on the surface of the earth and depends upon: ( a)
the
energy output from the sun, and ( b) the distance from the sun to the earth. So
= 1380 J/(s m2)
The perihelion The distance from the earth to the sun when the earth is at
its closest distance to the sun. The value of the solar constant here is SoP =
1440 J/(s m2)
The aphelion The distance from the earth to the sun when the earth is at
its farthest distance to the sun . The value of the solar constant here is SoA =
1330 J/(s m2)
As can be seen from these two values of the solar constant, more radiation is
received from the sun at perihelion than at aphelion. The earth is at
perihelion on about January 3, our winter time; and at aphelion on July 4,
our summer time. That is, we receive the maximum amount of radiation from
the sun in our winter time, and the minimum amount of radiation from the
sun in our summer time. The conclusion that we must draw from this, is that
the difference in the radiation received on the earth due to the elliptical
nature of the earth’s orbit, and hence the different value of the solar
constant, cannot account for the seasonal variations in the temperature at
the surface of the earth.
The angle that the sun's noon rays strike the earth. Because of the
spherical shape of the earth, the radiation from the sun does not strike the
earth normally (i.e. perpendicularly) everywhere on the surface of the earth.
That is, the angle θ at which the sun’s rays strikes a point on the surface of
the earth is a function of the latitude angle φ.
The intensity of the sun’s radiation hitting the earth depends upon I0,
also called the solar constant S0, and θ the angle that the sun’s rays make
with the horizontal. The intensity of the radiation received on earth (I = Io
sinθ ) will decrease from the maximum value I0, when the sun is directly
overhead, to zero when the sun is on the horizon, i.e. θ = 00. The decrease in
the intensity of the radiation occurs because the radiation is now spread out
over a greater area and will as a consequence produce less heating per unit
area.
Hottest and Coldest. Because of the spherical shape of the earth, the angle
at which the sun’s noon rays strike the earth varies with latitude, and
regions near the equator will receive the maximum amount of radiation and
3-30
Chapter 3 Solar and Terrestrial Radiation
will be the hottest regions, while regions north or south of the equator will
receive less radiation and will be cooler.
Diurnal and annual motion of the sun across the sky. Because of the
motion of the earth in its orbit, it appears as though the sun has two motions
across the sky. For the diurnal motion, the sun rises in the east and sets in
the west, while for the annual motion the sun rises higher and higher in the
sky until, on June 21, the Summer Solstice, it is at its highest position. It
then starts to move down the sky reaching its lowest position on Dec. 21, the
Winter Solstice. The total movement of the sun above the horizon is 470.
The length of daylight. The length of daylight varies with the location on
the surface of the earth and on the time of the year. The length of daylight is
maximum on the Summer Solstice, June 21, and lowest on the Winter
Solstice, December 21. The greater the length of daylight will increase the
amount of radiation received on the earth and hence add to the heating of the
earth and contributes to hotter temperatures in the summer. Conversely, the
shorter the length of daylight in the winter will decrease the amount of
radiation received on the earth and contributes to the colder temperatures
observed in the winter.
Summary of Important Equations
Solar constant
Solar constant at perihelion
Solar Constant at aphelion
So = Q/At = 1380 J/(s m2)
SoP = 1.44  103 W/ m2
SoA = 1.33  103 W/ m2
(3.1)
(3.2)
θ is the angle that the sun’s noon rays make with the horizontal, and φ is the
latitude of the observer.
(3.3)
θ = 900 − φ
Increased area caused by the angle of the sun for March 21
A = Ao
sinθ
(3.7)
The intensity of the radiation from the sun falling on the area A
I = Io sinθ
(3.8)
The intensity of the sun’s radiation hitting the earth, in terms of the latitude angle
φ
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Chapter 3 Solar and Terrestrial Radiation
I = Io sin(900 − φ )
(3.9)
The intensity of the sun’s radiation on the summer solstice.
I = Io sin(θ + β ) = Io sin(900 − φ + β)
(3.10)
The intensity of radiation for the winter solstice
I = Io sin(θ − β) = Io sin(900 − φ − β)
(3.11)
The length of daylight at the latitude φ.
tD = 2π − 2cos−1[tanφ tanβ]
ω
The number of hours of nighttime
tN = 24 hr − tD
Time for the sun to rise
tsunrise = 12 noon − tD
_
2
Time for sunset
tsunset = 12 noon + tD
_
2
(3.22)
(3.23)
(3.24)
(3.25)
Questions for Chapter 3
1. If the earth is closest to the sun on December 21, why isn't this the
warmest day of the year?
2. If the earth is farthest from the sun on June 21, why isn't this the coldest
day of the year
3. Since the orbit of the earth about the sun is elliptical, does the solar
constant change every day"?
4. What is the difference between perihelion and aphelion?
5. Why should the sun's ray change with the latitude you are located at?
6. Why do you get more heat from the sun when it is overhead than when it
is low on the horizon?
7. If you went to different planets in our solar system, would the Solar
Constant there be different than the Solar Constant here on earth? Why?
8. What does Christmas Day have to do with the Winter Solstice?
Problems for Chapter 3
1. Determine the angle that the sun’s noon rays make if you live at a latitude
of (a) 100, (b) 200, (c) 300, (d) 600, and (e) 800 north latitude.
2. Find the intensity of the sun’s radiation if you live at a latitude of (a) 100,
0
(b) 20 , (c) 300, (d) 600, and (e) 800 north latitude on March 21, the Vernal Equinox.
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Chapter 3 Solar and Terrestrial Radiation
3. For the Summer Solstice, find the intensity of the sun’s radiation (a) 100,
(b)
(c) 300, (d) 600, and (e) 800 north latitude.
4. For the Winter Solstice, find the intensity of the sun’s radiation (a) 100, (b)
0
20 , (c) 300, (d) 600, and (e) 800 north latitude.
5. Find the length of daylight for the Summer Solstice, at the location (a) 100,
0
(b) 20 , (c) 300, (d) 600, and (e) 800 north latitude.
6. Find the length of nightime for the Summer Solstice, at the location (a)
100, (b) 200, (c) 300, (d) 600, and (e) 800 north latitude.
7. Find the time for sunrise on the Summer Solstice, at the location (a) 100,
(b) 200, (c) 300, (d) 600, and (e) 800 north latitude.
8. Find the time for sunrise on the Winter Solstice, at the location (a) 100, (b)
200, (c) 300, (d) 600, and (e) 800 north latitude.
9. Find the time for sunset on the Summer Solstice, at the location (a) 100,
(b) 200, (c) 300, (d) 600, and (e) 800 north latitude.
10. Find the time for sunset on the Winter Solstice, at the location (a) 100, (b)
200, (c) 300, (d) 600, and (e) 800 north latitude.
200,
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