lab1

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York University *

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Apr 3, 2024

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LE/ESSE 1012 3.0 The Earth Environment Winter 2024 Lab. Section: 3 Name: Tahman Ahmed Student Number: 220825071 Lab 1: EARTH IN SPACE DUE: JANUARY 26, 2024, 10:00 PM ET IMPORTANT: If you have not already done so, you must complete the course policy quiz on eClass and get all answers correct (multiple tries are okay) before submitting this lab, otherwise your lab will not be accepted . Unless otherwise indicated, show your work for all problems. You can either enter your answers into this document electronically using a computer or tablet, or you can print this document, handwrite your answers in the spaces provided, and scan the pages. If you need additional space, you can insert additional pages or you can add additional space within the Word document. For all numerical answers, the units should be indicated. Students can discuss this lab with each other, but copying from each other or copying from other sources is cheating and is not permitted. You should not share your answer sheets with other students or look at the answer sheets of other students. You should understand the concepts well enough to explain your answers in your own words. Your answers for hands-on portions of the lab should be based on work that you yourself performed in the lab location. If the lab procedure indicates that you can form groups to complete particular tasks, then you should still be physically present in the lab location contributing to the completion of those tasks, and you should write the names of other group members on your answer sheets. If your work relies on information that is obtained from a legitimate source other than ESSE 1012 course materials, please indicate the source of that information with enough detail so that someone else can locate the source. Please see the course outline for detailed policies.

1. Understanding Plagiarism The Wikipedia entry for the origin of water on Earth contains the following text: In response to a question about the origin of water on Earth, a student submits the following response: “ It was long believed that Earth’s water didn ’ t originate from the planet’s region of the protoplanetary disk. Rather, the predominant view has been that asteroids delivered water to Earth and those asteroids either contained water in ice form or hydrogen and oxygen that reacted with each other after impact. Other research has also suggested that some of the hydrogen and oxygen needed to form water might have come from either Earth itself or from gases from the solar nebula in the early formation of the solar system. ” No quotation marks are included in the student’s response. a. On its own, based on just the evidence presented, w ould the first sentence of the student’s response be considered plagiarism? Why or why not? (3 points.) Yes, it would be because there was little to no effort to either paraphrase the sentence or even cite where the information was taken from. There are some synonyms used to replace other words but no real evidence of the student understanding and conveying the information in his/her words. b. On its own, based on just the evidence presented, would the second sentence of the student’s response be considered plagiarism? Why or why not? (3 points.) It still would be even though in this sentence, it was altered using paraphrasing techniques, synonyms and a completely different writing style, just because it was not cited. c. Based on the entire response submitted by the student and just the evidence presented, has the student committed plagiarism? Why or why not? (3 points.) Yes, he has because even if one sentence of entire paragraph is not cited properly, it still is plagiarism.

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 Circle one: JULY or DECEMBER

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(b) Keeping the flashlight beam centred over Botswana, tilt the beach ball relative to the flashlight to resemble winter noontime conditions in Botswana with Earth’s current obliquity. Which diagram above (A or B) would more closely resemble winter noontime conditions in Botswana? Does the diagram you have chosen more closely resemble July or December conditions? Does the flashlight spot look brighter or dimmer than in part (a)? What does that tell you about the solar zenith angle in winter compared to summer over Botswana? What does that tell you about average temperature in Botswana during winter compared to summer? (5 points.) Diagram A December Dimmer Solar zenith angle in winter is more than summer. The average temperature is going to be lower in winter than summer. (c) Keep the globe and flashlight in the same positions as in part (b), except tilt the globe so as to increase Earth’s obliquity. Does the flashlight spot over Botswana become brighter or dimmer than before? What does this tell you about the effect of increasing obliquity on winter solar zenith angle and temperature? Conversely, what effect does decreasing obliquity have on winter solar zenith angle and temperature? (5 points.) Dimmer Increasing winter solar zenith angle, it decreases the average temperature. Decreasing winter solar zenith angle, it increases the average temperature.

3. Solar Zenith Angle and Obliquity Calculations The figure below shows a plot of solar zenith angle over the Nodlysstasjonen laboratory (78.2 o N latitude) located on the Norwegian island of Svalbard. (Source: Robertson et al., Annales Geophysicae , 2006, doi:10.5194/angeo-24-2543-2006.) The plot includes values at midnight and at noon for each day of the year, assuming a non-leap year. For this problem, assume a non-leap year and assume that the insolation is 1362 W m -2 . Note that the y axis decreases upward, and when the solar zenith angle is greater than 90 o , the Sun is below the horizon, and it is considered evening or nighttime. Also, the x axis indicates the day of the year , ranging from 0 to 365. You can use the table at the following link to convert between day of the year and date: https://nsidc.org/support/faq/day-year-doy-calendar . If a question below asks for a date, you should give a month and a day of the month as a final answer, not the day of year. Evening/Nighttime

(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. ????? ?????? = ????? ???????? × ? 𝑝 2 ? 2 × cos (????? ???𝑖?ℎ ?????) 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 ???(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.) ????? ?????? = ????? ???????? × ? 𝑝 2 ? 2 × cos(????? ???𝑖?ℎ ?????) ????? ?????? = (1362 ?? −2 ) × (1 ?) 2 (1.00975 ?) 2 × cos ( 1 3 ?) ??????𝑖???? ???? ?? ??? 10?ℎ = 131 ?? ???? 130 ??? ??𝑖?? ?ℎ𝑖?, ?? ??? 60 ??????? ???𝑖?? ????? ???? ?ℎ???? 60 ??????? ?? ???𝑖??? → 60 × ? 180 = 1 3 ? ∴ ????? ?????? = 667. 9122183 ≈ 667.9 ?? −2

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(d) Compared to your answer in part (c), would you expect your answer to increase or decrease if the location were Ottawa instead of Nodlysstasjonen? Why? (4 points.) If we consider the location to be Ottawa instead, we would generally expect the solar energy to increase. Ottawa (45.4 °N), being at a lower latitude than Nodlysstasjonen (78.2°N), receives more direct sunlight throughout the year. Higher latitudes, like Nodlysstasjonen, tend to experience lower solar energy levels due to the lower angle of the Sun in the sky even during the daytime. Refer to the figure below from the lecture slides, which shows the downward component of the incoming solar radiation ( Q ) as a function of latitude averaged over all years (green), the day of the northern summer solstice (red) and the day of the northern winter solstice (blue). Confusingly, this is sometimes referred to as “insolation” (as in the figure), even though strictly speaking, it is the downward component of the insolation. (e) Assuming no absorption or reflection of solar radiation by the atmosphere, and assuming an efficiency of 60%, how much solar power would a 10 m by 10 m solar panel centred over Nodlysstasjonen generate on average over the course of the year? Assume that the solar panel is oriented parallel to the ground. Express your answer to three significant digits. (5 points.) ?????? ??????? 𝑖??????𝑖?? ?? 78.2°?: ??????? 𝑖??????𝑖?? = (180 ?? −2 ) × (100 ? 2 ) × (. 60) ??𝑖?? ??????𝑖?????? 80°?, ?ℎ? ??????? ??𝑖?? 𝑖??????𝑖?? 𝑖? ????? 180 ?? −2 ∴ ????? ????? ????????? ???? ? ?????? ?? ?ℎ? ???? = 10800 ?

(f) If you wished to generate more power (on average) than in part (e), would you tilt the solar panel to face more toward the south or the north? Why? (4 points.) Tilt the solar panels south because it is related to the tilt of the Earth's axis and the path of the Sun in the sky. In high latitudes, the Sun's path across the sky can be at a lower angle compared to equatorial regions. Tilting the solar panels toward the south optimizes their exposure to sunlight throughout the day. When panels face south, they can capture more sunlight during the Sun's journey across the southern part of the sky, maximizing the solar energy harvested.

4. Geographic Coordinates Required materials (provided by the lab): • 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 ∶ ? ? = 4488.3720930232 ≈ 4488 ??

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Theoretical calculation (to nearest km): 2? 9 × cos(0) × 6378.1 = 4452.753801 ≈ 4453 ?? Percent difference between measured and theoretical (to one decimal place): 4488 − 4453 4453 × 100 = 0.7859869751 ≈ 0.8 Discuss any difference between measured and theoretical values: The difference is very small because of the very small percentage difference hence it is accurate. (c) Consider two points with the same longitudes as in part (b) but at latitude 40 o N. (7 points.) Measured distance on globe (in mm to one decimal place, no need to show work): 87.5 mm Measured distance converted (to nearest km): 21.5 ∶ 1000 87.5 ∶ ? ? = 4069.7674418604 ≈ 4070 ?? Theoretical calculation (to nearest km): 2? 9 × cos(40) × 6378.1 = 3411.007306 ≈ 3411 ?? Percent difference between measured and theoretical (to one decimal place): 4070 − 3411 3411 × 100 = 19.31984755 ≈ 19.3 Discuss any difference between measured and theoretical values: The difference is big because of the large percentage difference hence it is not as accurate.

(d) Compare your answers to parts (b) and (c). Did you expect the answers to be approximately the same or different? Why? If you expected the answers to be the same, explain any difference. (4 points.) Same, because the measurements were done repeatedly hence accurate. The answer in (b) is accurate while the answer in (c) is not. This may be because the beach ball is not a perfect sphere. And we assumed that it is. (e) Now consider two points on the same meridian of longitude, but with different latitudes, specifically (20 o N, 120 o W) and (40 o N, 120 o W). (7 points.) Measured distance on globe (in mm to one decimal place, no need to show work): 49.5 mm Measured distance converted (to nearest km): 21.5 ∶ 1000 49.5 ∶ ? ? = 2302.3255813953 ≈ 2302 ?? Theoretical calculation (to nearest km): ? 9 × 6378.1 = 2226.3769 ≈ 2226 ?? Percent difference between measured and theoretical (to one decimal place): 2303 − 2226 2226 × 100 = 3.4591194968553 ≈ 3.5 Discuss any difference between measured and theoretical values: The difference is small because of the very small percentage difference hence it is accurate.

(f) Repeat for (20 o N, 20 o W) and (40 o N, 20 o W). (7 points.) Measured distance on globe (in mm to one decimal place, no need to show work): 51.0 mm Measured distance converted (to nearest km): 21.5 ∶ 1000 51.0 ∶ ? ? = 2372.0930232558 ≈ 2372 ?? Theoretical calculation (to nearest km): ? 9 × 6378.1 = 2226.3769 ≈ 2226 ?? 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.

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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.) ?????? ?? ??ℎ???, ? = 4 3 ?? 3 ? = 4 3 (?)(6.3781 × 10 8 ??) 3 ?ℎ???? ?? ?? ??, ??????? ?ℎ?? ??? ?? ???? 𝑖? ???? (?) 6378.1 ?? → 6.3781 × 10 8 ?? ∴ ?????? ?? ?ℎ? ????ℎ = 1.0886832412 × 10 8 ?? 3 ≈ 1.089 × 10 8 ?? 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.) ????𝑖??, ? = ? ? ? = 5.972 × 10 27 ? 1.0886832412 × 10 8 ?? 3 ?ℎ???? 5.972 × 10 24 ?? ?? ? → 5.972 × 10 27 ? ∴ ????𝑖?? ?? ?ℎ? ????ℎ = 5.49486741 ? ?? −3 ≈ 5.495 ? ?? −3 (c) How does this mean density for the whole Earth compare with the density of minerals found in rock samples from the upper mantle (olivine, pyroxene, garnet)? Reference density values are provided below. Does the comparison surprise you or not? Why or why not? (4 points.) Densities of upper mantle minerals (g cm -3 ) Olivine Pyroxene Garnet Typical Mix (40% Olivine, 50% Pyoxene, 10% Garnet) 3.31 3.34 4.16 3.39 The mean density of the whole Earth is higher compared to the density of minerals found in rock samples from the upper mantle. It does not surprise me because it makes sense, the Earth's structure consists of different layers with varying compositions and densities. The dense metallic core, which is composed mainly of iron and nickel, significantly influences the mean density of the entire planet. The crust and upper mantle, where minerals like olivine, pyroxene, and garnet are found, have lower densities in comparison.

lab1

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