Chapter 3 Earth - Sun Relations - Farmingdale State · PDF fileChapter 3 Earth - Sun Relations 3-2 orbit of the earth about the sun is elliptical as shown in figure 3.1. Maybe the

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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

Chapter 3 Earth - Sun Relations

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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?

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

Solution


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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 At 2.72 1023 m2 m2

= 1.44 103 W

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?

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.

Solution

http://www.farmingdale.edu/faculty/peter-nolan/pdf/meteorology/PhysAtmEx3.01.xls


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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 At 2.94 1023 m2 m2

= 1.33 103 W

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)


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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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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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.

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


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

(a) Equatorφ = 0 0

L0

A = L w 0 0

w(b) φ = 400

A = L w L = L / sin0 θ

w

L0L0

Lφθ

Beam ofsun light

Beam ofsun light

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

Solution



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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

Ao = Lo w (3.5)

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 = L w (3.6)

replacing equation 3.4 into equation 3.6 gives

A = sinθ

Lo w

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.

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 = sin(500)

1.00 m2

A = 1.31 m2

Solution


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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.


The intensity of the radiation from the sun falling on the area A can now be found as

I =

QAt =

QA0

sin t=

QA0t 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.

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

Solution



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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

s m2 ) sin(500)

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.


We can also express this intensity of radiation in terms of the latitude angle

φ directly by noting that in figure 3.2 that

θ + φ = 900 or

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.

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)

Solution



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I = 1380 ( J s m2

) sin(900 − 200)

I = 1380 ( J s m2

) sin(700)

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

s m2 ) sin(900 − 400)

I = 1380 ( J s m2

) sin(500)

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

s m2 ) sin(900 − 600)

I = 1380 ( J s m2

) sin(300)

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

s m2 ) sin(900 − 800)

I = 1380 ( J s m2

) sin(100)

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.


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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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.

0 20 40 60 80 1000

500

1000

1500

joul

es/(s

m^2

)

Lattidude angle phi (deg)

Sun'

s In

tens

ity (Q

/tA)

Intensity of Sun's Radiation as a function of lattitude

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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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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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.

Equator

Tropic of Cancer

Arctic Circle

Tropic of Capricorn

Antarctic Circle

N

S

Circle of Illumination

Sun's Rays


EquatorTropic of CancerArctic Circle

Tropic of Capricorn

Antarctic Circle

N

S

Sun's Rays

23.50

Equator

Tropic of Cancer

Arctic Circle

Tropic of CapricornAntarctic Circle

N

S


Sun's Rays

23.50


Equator

Tropic of Cancer

Arctic Circle

North Pole


Equa

tor

Arctic

CircleTropic of C

ancer

Equ ator

North Pole


Equa

tor

Arctic

Circle

Tropic of Cancer

Equator

North Pole

(a) Summer Solstice

(c) Winter Solstice

(b) Equinoxes

Figure 3.6 Radiation from the sun falling on the earth at different times of the

year.


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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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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.

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

s m2 ) sin(900 − 400 + 23.50)

I = 1380 ( J s m2

) sin(73.50)

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

Solution


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I = Io sin(900 − φ − β) I = 1380 ( J

s m2 ) sin(900 − 400 − 23.50)

I = 1380 ( J s m2

) sin(26.50)

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.


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 Solar Radiation as a Function of Latitude

0

200

400

600

800

1000

1200

1400

1600

0 20 40 60 80 100

Latitude Angle

Inte

nsity

of R

adia

tion

Equinoxes

Summer Solstice

Winter Solstice

Figure 3.8 The intensity of the sun’s radiation as a function of the latitude angle

on Mar. 21.



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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

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

for the ancient Romans. The world would not die. Everything would continue to be good.


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considered as the birthday of Jesus Christ, which they felt was the coming of the son

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.

of God. Hence, we celebrate Christmas on December 25, a few days after the Winter Solstice.

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


Equator

N

S

Sun's Rays

23.50


Equ a

tor

North Pole

NoonSun

SunriseA

B Sunset

γγ

2π − 2γx

(b) Top View

NYxd

β

φ

(a) Side View

cosφRER E

φ

cosφRE

Figure 3.10 The geometry of the earth on the Summer Solstice to determine the

length of the day.


3-19

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

cosγ = x RE cos φ

(3.14)

But from figure 3.10a, we also have x = d sinβ (3.15)

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 (3.17)

Solving for d we get d = RE sinφ

cosβ (3.18)

Replacing equation 3.18 into 3.16 gives

cosγ = RE sinφ sinβ cosβ RE cosφ

(3.19)

cosγ = tanφ tanβ (3.20) Hence, the angle γ is found as

γ = cos−1[tanφ tanβ] (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.


3-20

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.

The length of daylight for the Summer Solstice is found from equation 3.22 as

tD = ω

2π − 2 cos−1[tanφ tanβ]

tD = 2π − 2cos−1[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.)


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.

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.

Solution

Solution

http://www.farmingdale.edu/faculty/peter-nolan/pdf/meteorology/PhysAtmEx3.08-10.xls


3-21


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

0.00

5.00

10.00

15.00

20.00

0 10 20 30 40 50 60 70

Lattitude angle

Leng

th o

f Day

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.



3-22

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

0.00

5.00

10.00

15.00

20.00

0 10 20 30 40 50 60 70

Lattitude angle

Leng

th o

f Day


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.


3-23

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

0.00

5.00

10.00

15.00

20.00

0 10 20 30 40 50 60 70

Lattitude angle

Leng

th o

f Day


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 _ 2

(3.24)

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.

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

Solution


3-24

tsunrise = 12 noon − tD = 12 − 14.85 2 2

= 4.575 hr. AM

tsunrise = 4:35 AM

The time for sunset is found from equation 3.25 as

tsunset = 12 noon + tD = 12 + 14.85 2 2

= 19.425 hr. PM

tsunset = 7:26 PM


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

0.00

5.00

10.00

15.00

20.00

0 10 20 30 40 50 60 70

Lattitude Angle (degrees)

Tim

e (h

rs)

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.



3-25

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

0.00

5.00

10.00

15.00

20.00

25.00

-100 -80 -60 -40 -20 0 20 40 60 80 100

Days

Tim

e

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


3-26

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.

Chapter 3 Solar and Terrestrial Radiation

3-27

Table 3.1

Length of Daylight Latitude Angle

φ Summer Solstice Equinox Winter Solstice

0 10 20 30 40 50 60 70 80 90

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 12 hr 12 hr 12 hr 12 hr 12 hr 12 hr 12 hr 12 hr 12 hr

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

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


3-28

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


3-29

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.


3-30

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


3-31

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 So = Q/At = 1380 J/(s m2) Solar constant at perihelion SoP = 1.44 103 W/ m2 (3.1) Solar Constant at aphelion SoA = 1.33 103 W/ m2 (3.2) θ is the angle that the sun’s noon rays make with the horizontal, and φ is the latitude of the observer.

θ = 900 − φ (3.3) 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


3-32

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 (3.22)

The number of hours of nighttime tN = 24 hr − tD (3.23)

Time for the sun to rise tsunrise = 12 noon − tD _ 2

(3.24)

Time for sunset tsunset = 12 noon + tD

_ 2 (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,

(b) 200, (c) 300, (d) 600, and (e) 800 north latitude on March 21, the Vernal Equinox.


3-33

3. For the Summer Solstice, find the intensity of the sun’s radiation (a) 100, (b) 200, (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) 200, (c) 300, (d) 600, and (e) 800 north latitude.

5. Find the length of daylight for the Summer Solstice, at the location (a) 100, (b) 200, (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.

To go to another chapter, return to the table of contents by clicking on this sentence.

http://www.farmingdale.edu/faculty/peter-nolan/pdf/TOCPhysAtm.pdf

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