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Transcript
Astronomy Assignment #1 Solutions Text Problems: Unit Number 1 2 3 Questions for Review 2, 5, 6 1, 2, 3, 4 None Problems 11,12, 13, 14 10, 11, 14 11, 12, (14 Not included in these solutions – Sorry) Problem Solutions: Unit 1 11. What would be the circumference and diameter of a ball that would represent the Moon if the Earth were a volleyball? What kind of object matches this size? (Note: See Problem #1 for the circumference of a volleyball, 68 cm.) Use a proportion to solve questions like this where different objects appear at the same scale. Required information Circumference of a volleyball = 68 cm (Problem #1) Circumference of the Earth = 2∙R = 2∙6,378 km = 40,074 km (See Appendix Table 5) Circumference of the Moon = 2∙R = 2∙1,738 km = 10,920 km (See Appendix Table 6) Solve the proportion 68cm x 40,074km 10,920km 68cm x 10,920km 18.5cm 40,074km The scaled down Moon would be 18.5 cm in circumference equivalent to 5.90 cm in diameter. This diameter is about 2.3 inches, and this is a bit bigger than a golf ball, but smaller than a tennis ball. A regulation racquetball has a diameter of 2.25 inches and would be a good model for the Moon if the Earth were a volleyball. 12. If the Earth were a volley ball, what would be the diameter of the Sun? Use a proportion to solve questions like this where different objects appear at the same scale. Required information Diameter of a volleyball = 68 cm / () = 21.6 cm Diameter of the Earth = 2 x 6,378 km = 12,756 km Diameter of the Sun = 2 x 696,000 km = 1.39 x 106 km Solve the proportion 21.6cm x 12,756km 1.39 10 6 km 21.6cm x 1.39 10 6 km 2.36 10 3 cm 23.6m 12,756km The scaled down Sun would be 23.6 m (about 75 feet) in Diameter. This is about the width of a pretty large house. The message being that the Sun is really big compared to the Earth, as big as a large house compared to a volley ball. 13. If the Earth were a volley ball how large would an astronomical unit be? Use a proportion to solve questions like this where different objects appear at the same scale. Required information Circumference of a volleyball = 68 cm (Problem #1) Circumference of the Earth = 2R = 26,378 km = 40,074 km (See Appendix Table 5) Length of and astronomical uint = 1 AU = 1.5 x 108 km (See Appendix Table 1) Solve the proportion 68cm x 40,074km 1.5 10 8 km 68cm x 1.5 10 8 km 2.54 10 5 cm 2.54 10 3 m 2.54km 40,074km The scaled down Astronomical Unit would be 2.54 kilometers in length. This length is about 1½ miles. If the Earth were a volleyball then an AU would be about the distance from OCC to the Jreck Sub shop at Valley Drive and Seneca Turnpike in the Valley. Just for reference, on this scale you would be only 25 nanometers tall. A human hair is about 100,000 nanometers. (The moral of this problem is that an AU is really big, and it is the smallest of the special length units astronomers use. 14. During the 1960s and 1970s, The Apollo spacecraft took humans to the Moon in three days. Traveling to Mars requires a trip of around 2 AU in total. How long would this trip take, traveling at the same speed as to the Moon? Use t d to solve problems like this. v Required information Distance of the trip to Mars and back = 2 AU = 3.0 x 108 km (See Appendix Table 1) Speed of trip = Distance from Earth to Moon divided by 3 days 384,000km km (See Appendix Table 6) v 128,000 3days day Solve the problem t d 3.0 10 8 km 2,344days 6.4 yr km v 128,000 day The trip to Mars and back, at Apollo spacecraft speeds, would require over 6 years to complete. The moral of this problem is, again, that space is real big. A six year trip is required to reach one of the nearest planets and return. To reach the nearest stars and return would require a time 136,000 times longer, at this speed, or a total time of 816,000 years! Space is real big. 6. During the 1960s and 1970s, The Apollo spacecraft took humans to the Moon in three days. How long would it take to travel to Pluto, about 40 astronomical units away? Use t d to solve problems like this. v Required information Distance of the trip to Pluto = 40 AU = 60 x 108 km (See Appendix Table 1) Speed of trip = Distance from Earth to Moon divided by 3 days 384,000km km (See Appendix Table 6) v 128,000 3days day Solve the problem d 60.0 108 km t 46,875days 128.3 yr km v 128,000 day The trip to Pluto, at Apollo spacecraft speeds, would require over 128 years to complete. The moral of this problem is, again, that space is real big. For practical reasons, the spacecraft sent to explore the outermost objects of our Solar System must travel very fast. Currently, the New Horizons spacecraft is on its way to Pluto having been launched in January of 2006 and scheduled for Pluto encounter in July 2015. This corresponds (very roughly) to a speed of 1.8 million kilometers per day, over ten times faster than the Apollo astronauts traveled on their way to the Moon. Problem Solutions: Unit 2 10. If the Milky Way were the size of a nickel (about 2 centimeters). a. How big would the Local Group be? b. How big would the Local Supercluster be? c. How big would the visible Universe be? Use a proportion to solve questions like this where different objects appear at the same scale. Scales of the Universe Table 2.1 Object Approximate Radius Earth Moon’s Orbit Sun R = 6,400 km 380,000 km R = 700,000 km = 110 R 1 AU = 150 Million km = 210 R = 23,100 R 40 AU = 8500 R 4.2 ly = 270,000 AU 50,000 ly 1.5 Million ly 40 Million ly 13.2 Billion ly Earth’s Orbit Solar System to Pluto Nearest Star Milky Way Galaxy Local Group Local Super Cluster Visible Universe Required information Radius of a nickel = 2 cm Radius of the Milky Way = 50,000 lyrs (See Table 2.1) Radius of the Local Group = 1.5 million lyrs (See Table 2.1) Solve the following proportion for the diameter of the Local Group 2cm x 50,000ly 1,000,000ly 2cm x 1,000,000ly 60cm 50,000ly The scaled down Local Group of Galaxies would be 60 cm in radius. Solving similar proportions for the Local Super Cluster and Visible Universe yields the following. The Local Super cluster would have a radius of 20 meters and the visible universe would have a radius of about 5.5 km. This last answer is worth discussing briefly. If the Milky Way galaxy were a nickel, then the distance to the edge of the visible universe would be from downtown Syracuse to Liverpool, approximately. One might well ask, what is beyond this visible limit? We don’t know, because the light from that great a distance has not yet reached us. We believe that the universe has a finite age of about 13.7 billion years. Thus, any light emitting object beyond that distance is un-“seeable “ because the light it emits has not yet traveled the distance between it and Earth. The light train, so to speak has not yet reached the Syracuse station. As the universe ages, its visible limit will grow. 11. If we detected radio signals of intelligent origin from the Andromeda Galaxy (M31) and immediately responded with a radio message of our own, how long would we have to wait for a reply? This type of problem is particularly simple if we know the distance to the M31 in light years. Since light travels at the speed of light, the time to travel on e light year is just one year. Proportionally, the time for light to travel 100,000 light years is just 100,000 years. Required information Distance from the Milky Way to M31 = 0.8 Mpc (Mega parsecs) (See Appendix Table 11) Convert 0.8 Mpc to light years 1 10 6 pc 3.26ly 2.61 10 6 ly 0.8Mpc 0.8Mpc 1Mpc 1 pc The one-way travel time for radios wave (radio waves are a form of light) from the Milky Way to the Andromeda Galaxy (M31) is about 2.6 million years. Thus, we would have to wait for 2.6 million years for our radio signal to arrive at M31 and then another 2.6 million years for there reply to reach us. Total communication travel time 5.2 million years – older that the fossil record of the human species. Guess what the moral of the problem is…you got it. Space is real big. 14. The Milky Way is moving toward the larger galaxy M31 at about 100 kilometers per second. M31 is about 2 million light years away. How long will it take before the Milky Way collides with M31 if it continues at this speed? Use t d to solve problems like this. v Required information Distance from the Milky Way to M31 = 2 million lys Speed of approach = 100 km/sec = 1x105 m/s Convert 2 million lyrs to meters 9.46 1015 m 1.89 10 22 m 2.00 10 6 ly 2.00 10 6 ly 1ly Solve the problem d 1.89 10 22 m t 1.89 1017 s 6.0 10 9 yr m v 1. 10 5 s We will collide with the Andromeda Galaxy in about 6 billion years. 15. The most distant galaxies seen by the Hubble Space Telescope are 12 billion light years away. Using the scale in section 2.4 (where the Milky Way is about the size of a large city), haw far would that be in the model? In this model (See Figure 2.4) 1 light-year is equivalent to 1 meter in the model. Galaxies 12 billion light-years away would be at a model distance of 12 billion meters. This distance can also be expressed as 1.2 x 1010 meters or 1.2 x 107 kilometers or 12 million kilometers or 7,460,000 miles. So, if a light year were shrunk to the length of 1 meter, the Earth would be 15 microns from the Sun (the width of a thin human hair the nearest star would be just over 4 meters from the Sun, The typical bright star we see in our sky would be a few hundred meters from the Sun, The Milky Way Galaxy’s diameter would be 100 kilometers (63 miles), and Most distant galaxies would be 1.2 million kilometers away, almost the actual distance between the Earth and Sun. Moral of the Story: Space is real big. Problem Solutions : Unit 3 1. The radius of the Sun is 7x105 kilometers, and that of the Earth is about 6.4x103 kilometers. Show that the Sun’s radius is approximately 100 times the Earth’s radius. 7 10 5 km . 6.4 10 3 km This is a simple ratio problem. Evaluate the ratio 7 10 5 km 7 10 53 1.09 10 2 109 3 6.4 10 km 6.4 Thus, we see that the Sun has a radius about 109 times that of the Earth. Note: The Sun is REALLY big. The fact that it is just over 100 times the radius of the Earth means that the Sun’s volume is over (100)3 times the Earth’s volume. (100)3 equals 1 million. In other words, if the Sun was a gumball machine, it would take over 1 million Earth’s to fill it. 3. Using scientific notation, show that it takes sunlight about 8.5 minutes to reach the Earth. d problem. We are given a distance (1 AU from the Sun to the Earth) and a speed v (speed of light = 3x108 m/s), and asked to find a distance. This is a t t d 1.5 1011 m 500s 8.33 min v 3.0 10 8 m / s Thus, it takes sunlight 8.33 minutes to travel from the Sun to the Earth. 4. How many 1-kiloton bombs would need to be exploded to produce 3.86x1026 Joules, the amount of energy emitted by the Sun in 1 second? This is a simple proportion problem. Evaluate the proportion 1 kiloton x kiloton . 12 4.18 10 Joules 3.86 10 26 Joules First cross multiply and then solve for x 1 kiloton x kiloton 12 4.18 10 Joules 3.86 10 26 Joules 1 kiloton 3.86 10 26 Joules 4.18 1012 Joules x kiloton x kiloton 1 kiloton 3.86 10 26 Joules 4.18 1012 Joules x kiloton 9.23 1013 kiloton Thus, in 1 second the Sun emits an amount of energy equal to 9.23x1013 kilotons of energy. The atomic bombs that were dropped on Hiroshima and Nagasaki in August, 1945 had an energy equivalent of 16 kilotons. The Sun emits 5.8 trillion Hiroshima atomic bombs of energy each second! 4 10 5 10 8 3 7. Using scientific notation, evaluate 4 10 5 10 8 3 6 2 6 2 10 4 3 10 8 52 3 6 2 . 64 10 24 64 10 24 2.56 10 36 25 10 12 25 10 12 The expression evaluates to a whopping 2.56x1036. 3 10 4 2 8. Using scientific notation, evaluate 3 10 4 2 4 10 6 . 4 10 6 4 10 3 2 10 4 2 1 6 2 9 1012 1 2 4 10 1 6 2 9 1012 4.5 1015 2 10 3 The expression evaluates to 4.5x1015. Unit 3 9. The radius of the Sun is 7x105 kilometers, and that of the Earth is about 6.4x103 kilometers. Show that the Sun’s radius is approximately 100 times the Earth’s radius. 7 10 5 km This is a simple ratio problem. Evaluate the ratio . 6.4 10 3 km 7 10 5 km 7 10 53 1.09 10 2 109 3 6.4 10 km 6.4 Thus, we see that the Sun has a radius about 109 times that of the Earth. Note: The Sun is REALLY big. The fact that it is just over 100 times the radius of the Earth means that the Sun’s volume is over (100)3 times the Earth’s volume. (100)3 equals 1 million. In other words, if the Sun was a gumball machine, it would take over 1 million Earth’s to fill it. 11. Using scientific notation, show that it takes sunlight about 8.5 minutes to reach the Earth. d problem. We are given a distance (1 AU from the Sun to the Earth) and a speed v (speed of light = 3x108 m/s), and asked to find a distance. This is a t d 1.5 1011 m t 500s 8.33 min v 3.0 10 8 m / s Thus, it takes sunlight 8.33 minutes to travel from the Sun to the Earth. 12. How many 1-kiloton bombs would need to be exploded to produce 3.86x1026 Joules, the amount of energy emitted by the Sun in 1 second? This is a simple proportion problem. Evaluate the proportion 1 kiloton x kiloton . 12 4.18 10 Joules 3.86 10 26 Joules First cross multiply and then solve for x 1 kiloton x kiloton 12 4.18 10 Joules 3.86 10 26 Joules 1 kiloton 3.86 10 26 Joules 4.18 1012 Joules x kiloton x kiloton 1 kiloton 3.86 10 26 Joules 4.18 1012 Joules x kiloton 9.23 1013 kiloton Thus, in 1 second the Sun emits an amount of energy equal to 9.23x1013 kilotons of energy. The atomic bombs that were dropped on Hiroshima and Nagasaki in August, 1945 had an energy equivalent of 16 kilotons. The Sun emits 5.8 trillion Hiroshima atomic bombs of energy each second! 4 10 Using scientific notation, evaluate 5 10 8 3 7. 6 2 4 10 5 10 8 3 6 2 10 4 3 10 8 52 3 6 2 . 64 10 24 64 10 24 12 2.56 10 36 12 25 10 25 10 The expression evaluates to a whopping 2.56x1036. 3 10 4 2 16. Using scientific notation, evaluate 3 10 4 2 4 10 6 4 10 6 4 10 3 2 10 4 2 1 6 2 The expression evaluates to 4.5x1015. . 9 1012 1 2 4 10 6 1 2 9 1012 3 4.5 1015 2 10