Chapter 3 Earth - Sun Relations

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1 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 = 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 m 2 ). 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 3-1

2 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 m. When the earth is furthest from the sun, a point in the orbit called the aphelion, the earth is at a distance of m. We can reevaluate the solar constant at perihelion, as we did in example 2.14, in chapter 2, with the value of r = m replaced with the value of r = 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 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 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 = 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 = 4πr 2 = 4π( m ) 2 A = m 2 3-2

3 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 = J/s = W At m 2 m 2 This value, W/m 2, 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 = W/ m 2 (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 m 2 ). 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 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 Hence, only a small portion of it is emitted in the direction of the earth. 3-3

4 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 = 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 = 4πr 2 = 4π( m ) 2 A = m 2 The energy per unit area per unit time impinging on the earth at aphelion is therefore Q = J/s = W At m 2 m 2 This value, W/m 2, 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 = W/ m 2 (3.2) To go to this Interactive Example click on this sentence. Hence the Solar Constant at aphelion is SoA = 1330 J/(s m 2 ). 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 m 2 ) in all our analysis.) 3-4

5 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 20 0 N latitude. The radius of the earth when extended through this point will also make an angle of 20 0 from the vertical with the sun's rays as seen in the figure. Since the vertical is at an angle of 90 0 with the surface, the sun s noon rays make an angle of = 70 0 with the surface. Notice from the figure that at 40 0 N latitude the sun s noon rays make an angle of 50 0 with the surface, while at 60 0 N latitude the sun s noon rays make an angle of 30 0 with the surface. These results from figure 3.2 can be generalized into the formula θ = 90 0 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. 3-5

6 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 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 = 90 0 φ θ = = 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 (90 0 ). 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 m 2 ). 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 m 2 ). Hence the intensity of the radiation hitting the surface at the equator, assuming no energy lost through the Beam of sun light L0 φ L0 θ L Beam of sun light (a) φ=0 0 Equator w L0 A 0= L 0 w (b) φ = 40 0 w L = L 0 / sinθ A = L w 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 m 2 ). Consider the same beam of light incident at latitude φ = 40 0 N, figure 3.3b. As we just saw, at φ = 40 0 N latitude the sun s noon rays make an angle θ = of 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 = L o / L 3-6

7 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 replacing equation 3.4 into equation 3.6 gives A = L w (3.6) 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 40 0 N latitude for March 21, which is called the Vernal Equinox. Solution As seen in figure 3.2, if the latitude angle is 40 0 then the sun angle is determined by equation 3.3 as θ = 90 0 φ (3.3) θ = = 50 0 Hence the sun s rays makes an angle θ = 50 0 with the horizon. If the original area of the incoming radiation is A 0 = 1.00 m 2, the new area can now be found from equation 3.7 as A = Ao sinθ A = 1.00 m 2 sin(50 0 ) A = 1.31 m 2 3-7

8 This means that the radiation from the sun, which should be spread across an area of 1.00 m 2, will instead be spread across an area of 1.31 m 2. 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 I = Q Q Q = = At A 0 sin t A 0 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 = 0 0. 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 50 0 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 90 0 the area covered by the radiation is Ao and it receives the intensity of radiation Io. When the angle θ is equal to 45 0 the area covered by the radiation is 1.41A o and the intensity of the radiation received is 0.707Io. When the angle θ is equal to 30 0 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 40 0 N latitude for March 21, the Vernal Equinox. Solution As seen in figure 3.2, if the latitude angle is 40 0 then the sun makes an angle θ given by equation 3.3 as θ = 90 0 φ θ = =

9 with the horizon. The solar constant is Io = 1380 J/(s m 2 ). The intensity of radiation can now be found from equation 3.8 as I = Io sinθ I = 1380 ( J ) sin(50 0 ) s m 2 I = 1060 J/(s m 2 ) Notice that even though the solar constant is 1380 J/(s m 2 ), the amount of solar radiation actually hitting the surface of the earth at 40 0 N latitude is only 1060 J/(s m 2 ), or 23.2% less than the radiation hitting the surface at the equator. Hence it will not be as warm at 40 0 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 = 90 0 θ = 90 0 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(90 0 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 = 20 0, (b) φ = 40 0, (c) φ = 60 0, (d) φ = Solution The suns radiation at the equator, φ = 0 0, is just equal to the solar constant and is Io = 1380 J/(s m 2 ) (a) For φ = 20 0, the intensity of the sun s radiation, found from equation 3.9, is I = Io sin(90 0 φ) 3-9

10 I = 1380 ( J ) sin( ) s m 2 I = 1380 ( J ) sin(70 0 ) s m 2 I = 1300 J/(s m 2 ) (b) For φ = 40 0, the intensity of the sun s radiation, found from equation 3.9, is I = Io sin(90 0 φ ) I = 1380 ( J ) sin( ) s m 2 I = 1380 ( J ) sin(50 0 ) s m 2 I = 1060 J/(s m 2 ) (c) For φ = 60 0, the intensity of the sun s radiation, found from equation 3.7, is I = Io sin( ) I = 1380 ( J ) sin( ) s m 2 I = 1380 ( J ) sin(30 0 ) s m 2 I = 690 J/(s m 2 ) (d) For φ = 80 0, the intensity of the sun s radiation, found from equation 3.9, is I = Io sin( ) I = 1380 ( J ) sin( ) s m 2 I = 1380 ( J ) sin(10 0 ) s m 2 I = 240 J/(s m 2 ) 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 3-10

11 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 m 2 while at 40 0 latitude the intensity is only a little over 1050 J/s m 2, and at 70 0 latitude, the intensity is a little below 500 J/s m 2. 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 Sun's Intensity (Q/tA) joules/(s m^2) Lattidude angle phi (deg) Figure 3.4 The intensity of the sun s radiation as a function of the latitude angle on Mar 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 40 0 N latitude and 40 0 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 3-11

12 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 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 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 3-12

13 Equator 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 40 0 N latitude. Sun's Rays Sun's Rays Circle of Illumination N Circle of Illumination Arctic Circle Tropic of Cancer Equator Tropic of Capricorn Antarctic Circle N Arctic Circle Tropic of Cancer Equator Tropic of Capricorn Antarctic Circle S (a) Summer Solstice North Pole Arctic Circle Circle of Illumination Equator Tropic of Cancer Arctic Circle Tropic of Cancer Equator Equator North Pole S (b) Equinoxes Circle of Illumination Circle of Illumination N Sun's Rays Tropic of Cancer Equator Tropic of Capricorn Antarctic Circle Arctic Circle r toauq E Tropic of Cancer Arctic Circle North Pole S Circle of Illumination (c) Winter Solstice Figure 3.6 Radiation from the sun falling on the earth at different times of the year. 3-13

14 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 = 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 40 0 with the zenith or 50 0 with respect to the horizon. Figure 3.7 Determining the noon sun angle at 40 0 N latitude. On June 21, the summer solstice, the earth s axis is tilted toward the sun, and the sun is directly overhead at N latitude. The radius that is drawn from the center of the earth to NYC subtends an angle of 40 0 with the equator, but as can be seen in figure 3.7b, it makes an angle of with the plane of the ecliptic, and hence the sun s rays make an angle of with respect to the local 3-14

15 zenith passing through NYC and makes an angle of = 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 = with respect to the ecliptic. Hence the sun s rays make an angle of with respect to the local zenith or = with respect to the horizon. Notice that the sun s noon rays at NYC make an angle of 50 0 with respect to the horizon on Mar. 21, and on Jun. 21 and 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 = 90 0 φ, hence the intensity of radiation, equation 3.8 and 3.9 becomes I = Io sin(θ + β ) = Io sin(90 0 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(90 0 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 40 0 N latitude for (a) the Summer Solstice and, (b) the Winter Solstice. Solution The solar constant is Io = 1380 J/(s m 2 ). a. As shown in figure 3.7b for the Summer Solstice, the sun makes an angle of with the horizontal. The intensity of radiation can now be found from equation 3.10 as I = Io sin(90 0 φ + β) I = 1380 ( J ) sin( ) s m 2 I = 1380 ( J ) sin( ) s m 2 I = 1320 J/(s m 2 ) b. As shown in figure 3.7c for the Winter Solstice, the sun makes an angle of with the horizontal. The intensity of radiation can now be found from equation 3.11 as 3-15

16 I = Io sin(90 0 φ β) I = 1380 ( J ) sin( ) s m 2 I = 1380 ( J ) sin( ) s m 2 I = 616 J/(s m 2 ) Notice that because of the difference in the sun s noon rays at 40 0 N latitude the intensity of the sun s radiation is 1060 J/(s m 2 ) on Mar. 21, the Vernal Equinox Example 3.3; 1320 J/(s m 2 ) on Jun. 21, the Summer Solstice; 1060 J/(s m 2 ) on Sept. 21, the Autumnal Equinox; and only 616 J/(s m 2 ) 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 These graphs are plotted in figure 3.8. Intensity of Solar Radiation as a Function of Latitude 1600 Intensity of Radiation Equinoxes Summer Solstice Winter Solstice Latitude Angle Figure 3.8 The intensity of the sun s radiation as a function of the latitude angle on Mar

17 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 north of the equator on June 21 and south of the equator on Dec. 21. The total movement of the sun above the horizon is therefore 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

18 Equator 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 R E 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 N β Equator x d R E cosφ NY φ φ R E Noon Sun R E cosφ 2π 2γ North Pole B Sunset γ γ x Circle of Illumination S (a) Side View Circle of Illumination A Sunrise (b) Top View Figure 3.10 The geometry of the earth on the Summer Solstice to determine the length of the day. 3-18

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 = radians per hour. Therefore the time for the sun to move across the sky for daylight, t D, is td = θ = 2π 2γ ω ω (3.13) The angle γ is found from the geometry of figure 3.10b as cosγ = But from figure 3.10a, we also have x RE cos φ Placing equation 3.15 into equation 3.14 gives (3.14) x = d sinβ (3.15) cosγ = d sinβ RE cosφ (3.16) But from the geometry of figure 3.10a, we have R E sin φ = d cosβ (3.17) Solving for d we get d = RE sinφ (3.18) cosβ Replacing equation 3.18 into 3.16 gives Hence, the angle γ is found as cosγ = RE sinφ sinβ (3.19) cosβ RE cosφ cosγ = tanφ tanβ (3.20) 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 = , for the Equinoxes β = 0 0, and for the Winter Solstice, β =

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 40 0 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 td = 2π 2cos 1 [tan40 0 tan ] rad/hr td = 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 = , 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 t D = hr, the length of nighttime is found from equation 3.23 as t N = 24 hr td t N = 24 hr hr = 9.15 hr t N = 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. 3-20

21 To go to this Interactive Example click on this sentence. The above calculation was made for a location at 40 0 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 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, 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, 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

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 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 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 ) 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, 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, 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. 3-22

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 Length of Day 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 t D /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 hr. The time of sunrise is found from equation 3.24 as 3-23

24 tsunrise = 12 noon td = = hr. AM 2 2 tsunrise = 4:35 AM The time for sunset is found from equation 3.25 as t sunset = 12 noon + td = = 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) 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 40 0 N latitude the time of sunrise, read from the graph, is about 4:30 AM. At N latitude (the Arctic Circle) the sun rises at midnight. 3-24

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 40 0 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 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 at the Winter Solstice to 0 0 at the Vernal Equinox and to at the Summer Solstice. Equation 3.22 for a position at 40 0 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 Time 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 40 0 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 40 0 N. You would need a similar graph for each latitude you are 3-25

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 S o, 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

27 Chapter 3 Solar and Terrestrial Radiation Latitude Angle φ 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 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 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

28 Chapter 3 Solar and Terrestrial Radiation Figure 3.16 Lag in the time for the maximum daily temperature 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 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

29 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

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