# Chapter 3 Earth - Sun Relations

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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

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

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

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

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