lab1

2. Hands-on Solar Zenith Angle and Obliquity Exercises Required Materials (provided by the lab):  Flashlight  Globe beach ball fully inflated Note: If the beach ball is not fully inflated (e.g., wrinkles are visible) ask a TA to pump air into it. Split into groups of approximately 3-4 and write the names of the other group members below. Other group members: Punit Reyat, Pasquale Fioccola, Ngoc Thach Pham (a) Ask one group member to hold a flashlight while another classmate holds a globe. Centre the flashlight beam over Botswana, and tilt the beach ball relative to the flashlight to resemble noontime summer conditions in Botswana with Earth’s current obliquity. The flashlight beam represents one portion of the solar radiation traveling toward Earth, and the flashlight should be parallel to the path taken by rays of the Sun. You can half-push the flashlight button to find the setting with the brightest steady beam, and you can extend the flashlight head to make the beam more focused. The orientation of the globe and flashlight should resemble one of the diagrams below. Where indicated below, please circle A or B to indicate which diagram closely resembles the way you have oriented the globe and flashlight. Also circle JULY or DECEMBER below to indicate whether the diagram you have chosen more closely resembles conditions during July or December. No further justifications are required. (2 points.) Circle one: A or B

(a) In this location, how can summer be warmer than winter if Earth is farther away from the Sun at that time? Use physical concepts and mathematical expressions to explain your answer. (7 points.) The variation in temperature throughout the year is primarily influenced by the tilt of the Earth's axis and its orbit around the Sun. The axis of the Earth is tilted relative to its orbital plane, and this tilt causes different parts of the Earth to receive varying amounts of solar radiation at different times of the year. During the summer solstice, the Northern Hemisphere is tilted towards the Sun, leading to more direct sunlight and longer daylight hours. When the Sun is directly overhead, the solar zenith angle is 0 degrees, and as it moves towards the horizon, the solar zenith angle increases. At Nodlysstasjonen, the Sun can remain above the horizon for an extended period during the summer, even at midnight, due to the tilt of the Earth's axis. This results in a more direct angle of sunlight, and thus, more solar energy reaching the surface. solarenergy = solarconstant × d p 2 d 2 × cos ( solar zenithangle ) Using the mathematical formula for calculating solar energy, and from our understanding that the solar zenith angle is 0, gives the maximum solar energy since cos ( 0 ) is equal to 1. (b) During approximately which date ranges is there a complete lack of daylight at Nodlysstasjonen? (4 points.) Approximately, the first and last 50 days of the year, there is a complete lack of sunlight throughout its 24 hours. This can be translated into mid-October to mid- February as extrapolated from the above diagram. (c) On May 10, the Earth-Sun distance is 1.00975 AU. Use this information and information from the above figure to approximate the downward solar radiation at Nodlysstasjonen on May 10. Express your answer to 4 significant digits. (5 points.) solarenergy = solarconstant × d p 2 d 2 × cos ( solar zenithangle ) solarenergy = ( 1362 Wm − 2 ) × ( 1 m ) 2 ( 1.00975 m ) 2 × cos ( 1 3 π ) approximate days ¿ May 10 th = 131 sotake 130 ∧ usingthis ,weget 60 degreesduring solarnoon change 60 degrees ¿ radians→ 60 × π 180 = 1 3 π ∴ solar energy = 667.9122183 ≈ 667.9 Wm − 2 (d) Compared to your answer in part (c), would you expect your answer to increase or

 Globe beach ball  Measuring tape Note: make sure that the globe is fully inflated (no visible wrinkles). If it requires more air, please ask a TA for assistance to pump additional air into the beach ball. Split into groups of approximately 3-4 and write the names of the other group members below. Other group members: Punit Reyat, Pasquale Fioccola, Ngoc Thach Pham (a) Locate the legend on the globe beach ball. Measure in mm the length corresponding to 1000 km. Do your best to estimate to one decimal place. Write this value below. and use this conversion factor for subsequent problems. No additional explanation is required. (2 points.) 21.5 mm For the following questions, assume that the Earth is a sphere of radius r e = 6378.1 km. Calculate the distance along the surface between the points following the shortest possible path that adheres to the specified requirements. For distance, only the absolute value matters, and none of your answers should have a negative sign. First measure the distance on the globe beach ball, then calculate it theoretically assuming that the Earth is a perfect sphere. Then calculate the percent difference and discuss possible reasons for differences. Note that the solid latitude and longitude lines on the globe are printed in increments of 10 o . (b) Consider two points on the equator at coordinates (latitude, longitude) of (0 0 N, 20 0 E) and (0 0 N, 20 0 W). (7 points.) Measured distance on globe (in mm to one decimal place, no need to show work): 96.5 mm Measured distance converted (to nearest km): 21.5:1000 96.5: x x = 4488.3720930232 ≈ 4488 km Theoretical calculation (to nearest km): 2 π 9 × cos ( 0 ) × 6378.1 = 4452.753801 ≈ 4453 km Percent difference between measured and theoretical (to one decimal place):

2372 − 2226 2226 × 100 = 6.6037735849056 ≈ 6.6 Discuss any difference between measured and theoretical values: The difference is small because of the very small percentage difference hence it is accurate. (g) Compare your answers for parts (e) and (f). Do you expect the answers to be approximately the same or different? Why? If you expected the answers to be approximately the same, discuss any difference. (4 points.) Same, they are approximately the same because their percentage difference is close. 5. The Density of the Earth (a) Assume that Earth is spherical with radius of 6378.1 km. What is the volume of Earth? Express your answer to 4 significant digits. (4 points.) Volumeof sphere,V = 4 3 πr 3 V = 4 3 ( π ) ( 6.3781 × 10 8 cm ) 3 change km ¿ cm ,becausethat canbe seen ∈ part ( b ) 6378.1 km→ 6.3781 × 10 8 cm ∴ volumeof the Earth = 1.0886832412 × 10 8 cm 3 ≈ 1.089 × 10 8 cm 3 (b) If the mass of Earth is 5.972 x 10 24 kg, use your answer to part (a) to compute the average density of Earth. Express your answer to 4 significant digits in units of g cm -3 . (4 points.) De nsity ,ρ = m V ρ = 5.972 × 10 27 g 1.0886832412 × 10 8 cm 3 change 5.972 × 10 24 kg ¿ g→ 5.972 × 10 27 g ∴ density of the Earth = 5.49486741 gcm − 3 ≈ 5.495 g cm − 3 (c) How does this mean density for the whole Earth compare with the density of minerals