Lab 6 Near Earth Asteroids ASTRO

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Red Rocks Community College *

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Astronomy

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Oct 30, 2023

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LAB 6: NEAR EARTH ASTEROIDS ANY QUESTIONS WITH ANSWERS COLORED IN RED ARE CORRECTED ANSWERS AND SHOULD BE CONSIDERED OVER NON-COLORED ANSWERES, IF AVAILABLE. 60pts. OBJECTIVES This laboratory is to examine information on Near Earth asteroids and possible collisions. SKILLS/COMPETENCIES  Interpret tables or graphs.  Present data by construction of charts and graphs.  Evaluate the relevancy of data. BACKGROUND ON NEOS AND NEAS Near-Earth Objects (NEOs) and Near Earth Asteroids (NEAs) are a special class of comets and asteroids that have been nudged by the gravitational attraction of nearby planets into orbits that allow them to enter the Earth's neighborhood. Composed mostly of water ice with embedded dust particles, comets originally formed in the cold outer planetary system while most of the rocky asteroids formed in the warmer inner solar system between the orbits of Mars and Jupiter. The scientific interest in comets and asteroids is due largely to their status as the relatively unchanged remnant debris from the solar system formation process some 4.6 billion years ago. The giant outer planets (Jupiter, Saturn, Uranus, and Neptune) formed from an agglomeration of billions of comets and the left over bits and pieces from this formation process are the comets we see today. Likewise, today's asteroids are the bits and pieces left over from the initial agglomeration of the inner planets that include Mercury, Venus, Earth, and Mars. As the primitive, leftover building blocks of the solar system formation process, comets and asteroids offer clues to the chemical mixture from which the planets formed some 4.6 billion years ago. If we wish to know the composition of the primordial mixture from which the planets formed, then we must determine the chemical constituents of the leftover debris from this formation process - the comets and asteroids. NEA Lab 14 pages

NASA's search program designed to discover 90% of the NEO population (1 km in diameter or larger) within 10 years is under way. The chart below shows the cumulative total known near-Earth asteroids versus time. NEA Lab 14 pages

LAB FIGURE 1: Near Earth asteroids up through 2005 The upper curve area shows all known near-Earth asteroids while the lower area shows only large near-Earth asteroids. In this context, "large" is defined as an asteroid having an absolute magnitude (H or brightness) of 18.0 or brighter which roughly corresponds to diameters of 1 km or larger. Programs (and year) that search for NEAs include: Lincoln Near-Earth Asteroid Research, LINEAR (1996) Near Earth Asteroid Tracking, NEAT (2001) Spacewatch (1984) Lowell Observatory Near-Earth Object Search , LONEOS (1993) Catalina Sky Surveys, CSS (2003) Japanese Spaceguard Association, JSGA (2000) Italy’s Asiago DLR Asteroid Survey, ADAS (2001) NEA Lab 14 pages

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TASK 1: INTERPRETING THE NEA GRAPH 1. (2) a. How many total asteroids were discovered by 2000? About 900 b. How many total asteroids were discovered by 2006? About 3800 2. (2) a. How many large asteroids were discovered by 2000? About 400 b. How many large asteroids were discovered by 2006? About 800 3. (2) What was the average asteroid detection rate from 2000 to 2006 (for all NEA’s)? 900 + 1300 + 1700 + 2100 + 2650 + 3200 + 3800 = 15650/7 = About 2236 I had originally misunderstood this question. The average detection rate from 2000 to 2006 is equal to x = (1300 – 900) + (2100 – 1700) + (3200 – 2650), where x = about 417. The average detection rate from 2000 to 2006 is about 417. 900 + 3800 = 4700 AVERAGE = 4700/2 = 2350 4. (2) Assume that the detection rate stays the same. How many total asteroids will be discovered by 2010? 3800 + 417(4) = 5468. If the detection rate stays the same, there will be a total of about 5468 total asteroids by 2010. NEA Lab 14 pages

4700 + 2350(4/3) = 7834 5. (2) a. As more and more of the larger NEAs are discovered, how do you think the shape of the bottom curve will change over to next 10 years? We believe that over the next ten years, the bottom curve will start to flatten out as there are less large NEA’s available to be found. We believe that the curve will grow exponentially and slowly shift towards a completely vertical or 90 degree increase. (2)b. The next 50 years? In the next 50 years the curve will be completely flat because there will be very little large NEA’s left to be found. The curve will continue to grow exponentially, but at this point I believe it will be completely vertical or 90 degrees. On the next page are the figures showing NEA detections as of June 2013. 6. (2) These graphs are presenting slightly different information than the graph of “Known Near Earth Asteroids” that you looked at before. Please explain the difference. The first difference between the two graphs is that the second Near-Earth Asteroid chart is recorded by every ½ year instead of every 1 year. The next difference is that the second Near-Earth Asteroid covers about 7 more years than the first Known Near-Earth Asteroid chart. Another difference between the two graphs is that the second Near-Earth Asteroid chart is not cumulative while the first Known Near-Earth Asteroid chart is cumulative. Overall, the second Near- Earth Asteroid chart is more specific and detailed than the first Near-Earth Asteroid chart because the second chart is recorded by every ½ year, it records NEA Lab 14 pages

more years and it allows a non-cumulative representation of the discovery of Near-Earth Asteroids. The Known Near Earth Asteroids are presented exponentially and the Near-Earth Asteroid Discoveries are presented linearly. 7. (2) What does Figure 1 show about the rate of discovery for all NEA’s from 2006 to 2013? The rate of discovery for all NEA’s from 2006 to 2013 increase slightly following a linear pattern. 8. (2) What does Figure 2 show about the rate of discovery for large NEA’s from 2006 to 2013? The rate of discovery for large NEA’s from 2006 to 2013 decrease slightly following a linear pattern. 9. (1) Do these updated graphs support your conclusion for questions 5a and 5b? These updated graphs do not support my conclusion for questions 5a and 5b, as the discovery of large decrease linearly from 2006 to 2013. NEA Lab 14 pages

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Figure 1: NEA’s discovered every 6 months (01/2012) Figure 2: large (>1 km) NEAS discovered every 6 months (01/2012) TASK 2: UNDERSTANDING IMPACT PROBABILITIES NEA Lab 14 pages

From your textbook, we have Figure 3 that shows the likelihood of space debris impacting the Earth. FIGURE 3: "CURVE FROM TEXTBOOK" 10. (3) NASA is looking for asteroids one km or larger. a. What would the effects be of a one km asteroid striking the Earth? A kilometer would cause tsunamis, widespread devastation and climate change. b. About how large would the crater be? The crater would be about 10 km. NEA Lab 14 pages

c. What is the smallest-sized asteroid that could cause widespread devastation? The smallest sized asteroid that could cause widespread devastation would be about 500m. 11. (2) a. According to Figure 3: "curve from textbook" On average, how often does a 1 km asteroid size strike the Earth? A 1 km asteroid strikes the Earth about once every 1 million years. b. A 100 meter diameter asteroid? A 100 meter diameter asteroid would strike the earth every about 1,000 years. A 100 m asteroid would strike the earth every about 10,000 years. 12. (2) Based on the graph in figure 4 (the curve from the textbook), do you think the NASA NEO programs using a one km size search criteria is the right decision? Why or why not? I think that NASA NEO programs using 1 km size search criteria is not the right decision. I believe this because a 1 km asteroid would strike the Earth every about 1 million years. It is not reasonable for NASA NEO to use a 1 km search criteria because it is not entirely guaranteed that humanity will exist for 1 million years, and by that point if humanity is still alive, they will have the technology to create habitats on other planets in the solar system or even outside of the solar system. 13. (2) Do you think Figure 3 might change as more asteroids are discovered? Why? I think that figure 4 might change as more asteroids are discovered. I believe this because more information about asteroids may present different criteria for how often different sizes of asteroids may strike earth. NEA Lab 14 pages

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TASK 3: INTERPRETING IMPACT DATA 14. (4) How do you think astronomers get their estimates of impact rates? (This is a question about what you think. All thoughtful, well-written answers will receive full credit.) I think that astronomers get their estimates of impact rates by observing craters on earth and dating them. I believe this is true because if astronomers were to find 220 craters on Earth and each one, for example, struck the earth 10 years after the next, then it would be safe for them to assume that asteroids would strike the earth every 10 years. Overall, by using the observations of craters on earth, astronomers will be able to create an estimate about how often asteroids collide with earth. The impact rate would more specifically be understood through the knowledge of the age of the craters. One tool for making predictions about impact rates is to study impact events on Earth through astroblemes. An astrobleme is an impact crater (impact basin) is usually a circular depression on the surface of a body caused by a collision of a smaller body (meteorite, asteroid, comet) with the surface. In the center of craters on Earth a crater lake often accumulates, and a central island or peak caused by rebounding crustal rock after the impact is usually a prominent feature in the lake. 15. (6) Fill in the data for the fourth column in the table below: Name Location Crater Diameter Estimated Impactor Size (crater diameter / 10) Age (years) Vredefort South Africa 300 km 30 km 2 billion+ Sudbury Canada 250 km 25 km 1.85 billion Chicxulub Mexico 170 km 17 km 65 million Popigai Russia 100 km 10 km 35.7 million Manicouagan Canada 100 km 10 km 214 million Acraman Australia 90 km 9 km 590 million Chesapeake Bay US 90 km 9 km 35.5 million Puchezh- Katunki Russia 80 km 8 km 167 million Morokweng South Africa 70 km 7 km 145 million Kara Russia 65 km 6.5 km 70 million Beaverhead US 60 km 6 km 600 million NEA Lab 14 pages

Tookoonooka Australia 55 km 5.5 km 128 million Charlevoix Canada 54 km 5.4 km 342 million Siljan Sweden 52 km 5.2 km 361 million Kara-Kul Tajikistan 52 km 5.2 km 5 million Montagnais Canada 45 km 4.5 km 50 million Araguainha Brazil 40 km 4 km 244 million Woodleigh Australia 40 km 4 km 364 million Mjølnir Norway 40 km 4 km 142 million Saint Martin Canada 40 km 4 km 220 million Carswell Canada 39 km 3.9 km 115 million Clearwater West Canada 36 km 3.6 km 290 million Manson US 35 km 3.5 km 73.8 million Yarrabubba Australia 30 km 3 km 2 billion Slate Islands Canada 30 km 3 km 450 million Shoemaker Australia 30 km 3 km 1.63 billion Keurusselkä Finland 30 km 3 km 1.8 billion Mistastin Canada 28 km 2.8 km 28 million Clearwater East Canada 26 km 2.6 km 290 million Nördlinger Ries Germany 25 km 2.5 km 14.8 million Steinheim crater Germany 3.8 km .38 km 15 million Gatun structure Panama 3.0 km .3 km 20 million Lonar Crater India 1.8 km .18 km 52,000 Meteor Crater US (Arizona) 1.2 km .12 km 49,000 Odessa US 0.2 km .02 km 50,000 16. (6) Now, you’ll use the table of astrobleme data to make your own estimates about the frequency of impacts. a. Do you think that this is a complete record of impacts on Earth? Explain why or why not. I do not believe that this is a complete record of impacts on Earth. I believe this because many asteroids may have landed in the ocean. Also because of the old age of the earth, it is possible for many of the impact craters that landed on land to be covered up. NEA Lab 14 pages

b. Show all your work and/or explain your method for using the information from the table to calculate: how often we should expect to be hit by an object 9-10 km in diameter; how often we should expect to be hit by an object 4.5-5.5 km in diameter; and how often we should expect to be hit by an object 2.8-3.2 km in diameter. By subtracting 35.7 from 214, which equals 178.3, and subtracting 35.5 from 590, which equals 554.5, and collecting the mean of the differences, which is equal to x = (178.3+554.5)/2 where x is equal to 366.5, I am able to conclude that on average an asteroid of 9-10 km diameter will hit the earth every about 367 million years. By subtracting from 35.5 from 214, which equals 178, I am able to conclude that an asteroid of 9-10 km in diameter will hit the earth every about 178 million years. By subtracting 50 from 128, which equals 78, I am able to conclude that an asteroid of 4.5-5.5 km diameter will hit the earth every about 78 million years. By subtracting 1630 from 2000, which equals 370, I am able to conclude that an asteroid of about 2.8-3.2 km diameter will hit the earth every about 370 million years. By subtracting 35.7 from 590, which equals 554.3, and dividing it by 4, I am able to conclude that an asteroid of about 9-10 km diameter will hit the earth every about 138.6 million years. By subtracting 5 from 361, which equals 356, and dividing it by 5, I am able to conclude that an asteroid of about 4.5-5.5 km diameter will hit the earth every about 71.2 million years. By subtracting 28 from 2000, which equals 1972, and dividing it by 5, I am able to conclude that an asteroid of about 2.8-3.2 km diameter will hit the earth every about 394.4 million years. By finding the average of how the age of the 4 specific recorded craters in millions of years, using the equation of (35.7 + 214 + 590 + 35.5)/4, I am able to determine that asteroids in the diameter of 9-10 km will hit the Earth every 218.8 million years. By finding the average of the age of the 5 specific recorded craters in millions of years, using the equation of (128 + 342 + 361 + 5 + 50)/5, I am able to determine that asteroids in the diameter of 4.5-5.5 km will hit the Earth every 177.2 million years. By finding the average of the age of the 5 specific recorded craters in millions of years, using the equation of (2000 + 450 + 1630 + 1800 + NEA Lab 14 pages

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28)/5, I am able to determine that asteroids in the diameter of 2.8-3.2 km will hit the earth ever 1.181.6 billion years. 17.(3) How do your estimates compare to the information from Figure 3: "curve from textbook"? Be specific and, if your estimates a different, explain why you think this might be. Based on figure 4, all of my estimates, besides the estimate for asteroids within the size of 9-10 km, are entirely wrong. Both my estimates and figure 4 assume that an asteroid of 9-10 km of will hit the earth every about 219 million years. Compared to figure 4, my estimates for asteroids of the size of 4.5-5.5 km is off by about 100 million years. The difference in this estimate is large. Finally, my estimates for asteroids of the size of 2.8-3.2 km, in comparison to figure 4, is off by over 1 billion years. This difference is extremely large. I believe that my estimates are off because either the equation that I used to make the estimates are wrong, or there is far more data used in figure 4 than there are in the table given. 18.(3) Now go to https://cneos.jpl.nasa.gov/ Spend some time familiarizing yourself with this website and some of the terminology used for building tables of NEA’s. Check out the NEO Basics, Sentry and Asteroid Watch “Quick Links” on the menu on the left side of the page. Click on the NEO Groups link. What is the orbital criterion for an object receiving an NEO designation? What is the difference between an NEC and an NEA? What are the specific criteria for designating an object as a PHA? NEC NEAR EARTH COMET NEA NEAR EARTH ASTEROID THE SPECIFIC CRITERIA NEA Lab 14 pages

The difference between an NEC and a NEA is that a NEC is a near earth comet and an NEA is a near earth asteroid. Some specific criteria for designating an object as a PHA, or potentially hazardous asteroid, is an NEA whose MOID, or minimum orbit intersection distance, with the Earth is 0.05 AU or less and whose absolute magnitude is 22.0 or brighter. 19.(3) Now click on the Life on Earth link. What event occurred ~65mya? Briefly describe how this event created conditions responsible for the evolution of humans. 65 million years ago the Chicxulub asteroid hit the Yucatan Peninsula in South America. The Yucatan Peninsula is more specifically in Mexico. This asteroid created the conditions responsible for the evolution of humans by eliminating the dinosaurs. Without the dinosaurs, humans were able to become the dominant species on earth. If this impact were not to happen, then dinosaurs would have remained the dominant species instead of humans. 20.(5) Now return to the main menu and click on the Close Approaches link. Then click on the NEO link. Examine the table that’s automatically generated. Find the object with the lowest CA (Closest Approach) LD (Lunar Distance) value. Record its name, CA date and time, LD nominal and minimal distances, its velocity and size. Given the information in Figure 4 above, what would be the result of an impact if this object were to strike downtown Denver? Now find the largest object in the NEO table and record the same info as for the one above. How would the result of this object striking Denver be different? On October 4, 2023, at about 5:42 pm, 2023 TC, which has a LD nominal of 2.215 and a LD minimal of 2.213, will reach within 0.05 AU of the Earth. Its velocity will be about 7.38 km/s and it has a size of 13-30 km in diameter. If this object were to strike downtown Denver, then the Earth would suffer a mass extinction. This is the object with the lowest CA and LD value within 60 days. On October 15, 2023, at about 11:26 pm, 2007 SQ6, which has a LD nominal and minimal of 19.289, will reach within 0.05 AU of the Earth. Its velocity will be about 6.45 km/s and it has a size of 97-220 m in diameter. If this asteroid were to strike downtown Denver, there would be an atmospheric explosion or small impact crater in the middle of the city. In comparison to 2023 TC, 2007 SQ6 is an NEA Lab 14 pages

asteroid that is much safer to land on Earth as it would not cause a mass extinction. 21.(2) Now change the Table settings to Future Only and Nominal Distance <=1LD. Over the next 25 years find the object will have the closest approach to Earth and list the characteristics like you did question 20. What damage would this object likely to produce if it was to impact the downtown Denver area. On April 11, 2090, at 7:57 am, 2015XF261, which has a LD nominal of .993 and a LD minimum of .924, will be about 1 lunar distance, or about 0.00255 AU, away from the Earth. The velocity of this asteroid will be 10.10 km/s and its size will be about 27-59 m in diameter. If this asteroid were to strike downtown Denver, then it would only cause an atmospheric explosion or a small crater in the middle of the city. NEA Lab 14 pages

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