Week 6 - Kepler's Laws and Retrograde Motion

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Dec 6, 2023

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Name: Key Kepler’s Laws and Retrograde Motion 1 st Law 1. Consider the 4 ellipses above. Just by looking at them, which of the ellipses has the lowest eccentricity? Explain how you know. Ellipse A because it is the least squashed (closest to a circle). Just by looking at them, which of the ellipses has the highest eccentricity? Explain how you know. Ellipse D because it is the most squashed/stretched out. 2. Now we will calculate the eccentricity of the 4 ellipses above. For each ellipse, use a ruler to measure the length of the major axis and the minor axis. The major axis is the longer direction and the minor axis is the shorter direction. Make your measurements in centimeters and record in the table on the next page. Once you have measured the length of the major and minor axis, calculate the eccentricity for each ellipse using the following equation: Eccentricity = 1 Minor Axis 2 Major Axis 2 A B D C
Ellipse Major Axis (in cm) Minor Axis (in cm) Eccentricity A 4 4 0.00 B 4.2 4 0.30 C 5 4 0.60 D 10 4 0.92 3. a) Using the eccentricities you just calculated for the four ellipses, which ellipse has the lowest eccentricity? Ellipse A b) Which ellipse has the highest eccentricity? Ellipse D c) Do your answers match your answers in Question 1? Yes 4. a) The Earth has an eccentricity of 0.016. Which of the four ellipses on the first page best matches the orbit of the Earth around the Sun? Ellipse A b) Halley’s Comet has an eccentricity of 0.967. Draw an ellipse that looks like the orbit of Halley’s Comet. The ellipse will look a lot like Ellipse D which has an eccentricity of 0.92, but just a little more squished.
5. Since planets do not orbit the Sun in perfect circles, their distance from the Sun is constantly changing. We can calculate the closest and farthest distances that an object gets from the Sun using the following equations: Closest Distance = ( 1 eccentricity ) × ( Semimajor axis ) Farthest Distance = ( 1 + eccentricity ) × ( Semimajor axis ) The semi-major axis is just the average distance of the planet from the Sun. For instance, the semi-major axis of the Earth is 1 AU. Calculate the closest and farthest distance that each of these objects gets from the Sun: Object Average Distance (in AU) Eccentricity Closest Distance (in AU) Farthest Distance (in AU) Earth 1.00 AU 0.016 0.984 1.016 Mars 1.52 AU 0.093 1.379 1.661 Neptune 30.1 AU 0.008 29.859 30.341 Pluto 39.5 AU 0.248 29.704 49.296 Halley’s Comet 17.8 AU 0.967 0.587 35.013 Comet Hyakutake 1700 AU 0.9999 0.170 3399.830 a) Your friend claims that there are times in its orbit that Mars gets just as close to the Earth as the Moon does. The Moon never gets farther away from the Earth than 0.003 AU. Using the table above, explain whether you agree or disagree with your friend. I disagree. The distance from Earth to Mars will be the smallest when Mars is the closest it gets to the Sun and the Earth is the farthest it gets from the Sun. The closest Mars ever gets to the Sun is 1.379 AU. The farthest the Earth gets from the Sun is 1.016 AU. They are still 0.363 AU apart, which is more than 100x bigger than the distance from the Earth to the Moon. b) A different friend claims that there are times when Pluto (the old 9 th planet) is actually closer to the Sun than Neptune (the 8 th planet). Based on the table above, could this be a possibility? Explain. Yes it could. The closest Pluto gets to the Sun is 29.704 AU which is closer than Neptune ever gets to the Sun. So Pluto will be closer to the Sun than Neptune for part of its orbit (it turns out that Pluto is closer than Neptune for about 20 years in its 240 year orbit).
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2 nd Law 6. The figure shows the orbit of a comet around the Sun. Put a circle around the point where the comet is moving the fastest. Put a square around the point where it is moving the slowest. Closer it is to the Sun the faster it moves. 7. Consider the 3 ellipses above. Which of the 3 ellipses would have the largest change in speed during its orbit? Explain why. Ellipse C because the distance from the Sun would change the most. Which of the 3 ellipses would have the least change in speed during its orbit? Explain why. Ellipse A because the distance from the Sun would change the least. 3 rd Law 8. Consider two planets that are orbiting a star like the Sun. Micol is a very heavy planet (300x heavier than the Earth) and is orbiting very close to the star. Christopher is a light planet (about the same weight as the Earth) and is orbiting far away from the star. a) Which of the two planets (Micol or Christopher) will orbit the star in the least amount of time? Explain your answer. Planet Micol because it is closest to the Sun. Distance is all that matters – the closer you are the less amount of time you need to orbit. b) If Micol and Christopher switched positions (so that Christopher was now very close to the star), which planet would orbit in the least amount of time? Explain your answer. Now Planet Christopher would orbit in less time because it is closer to the star.
c) Imagine that the weight of Christopher increased by 100x but it stayed the same distance from the star. Would Christopher orbit in less time, more time, or the same amount of time as before? Explain your answer. Same amount of time as before – only distance matters in figuring out time to orbit. Weight does not have an effect. d) Finally, imagine that Micol (300x the weight of Earth) and Christopher (same weight as Earth) were the same distance from the star. Which would move around the star in the least amount of time? Explain your answer. They would orbit in the same time because they are the same distance away. 9. Use the graph above to help you answer the following questions. a) Does the orbital period of planets increase, decrease, or stay the same as the distance is increased? As distance gets bigger, the orbital period also increases. b) How long would it take a planet to orbit if it is 1 AU from the star? On the graph, 1 AU lines up with 1 year. c) How long would it take a planet to orbit if it is 2 AU from the star? On the graph, 2 AU lines up with 2.8 years. d) Based on your answers to b and c, when the distance to the star is doubled, will the period of the star: DECREASE BY HALF, NOT CHANGE, DOUBLE, or MORE THAN DOUBLE? Explain your answer. A planet at a distance of 2 AU takes more than 2 years to orbit, so the period MORE THAN DOUBLES when the distance doubles.
Sun Earth Mars’s Orbit X 10. Kepler’s 3 rd law says that the time it takes a planet to orbit the Sun (the period) in years can be calculated using the following equation: Average Distance ( Period years ) 2 =( ¿ Sun AU ) 3 Using the data in the table on Question 5, calculate the time that it takes Halley’s Comet and Comet Hyakutake to each orbit the Sun. Halley’s Comet: 75.1 years Comet Hyakutake: 70,093 years Retrograde Motion 1. What is retrograde motion? The backwards motion of a planet or other object compared to background stars. 2. Use the image below to explain the conditions that create retrograde motion. In other words, according to the heliocentric model, when does retrograde motion occur? Retrograde motion occurs when one object passes by another object. On Earth, this happens either when the Earth passes by a slower object or a faster object passes by the Earth. 3. On the diagram to the right, you see the orbit of the Earth and Mars around the Sun. Put an “X” where Mars would be located when Mars is going through retrograde motion. Explain your answer. When Mars is at the X, Earth is passing by it. from NASA SP-4212
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4. T/F: All planets (and in fact all objects that directly orbit the Sun) should go through retrograde motion as seen from the Earth. Explain your answer. True --- Earth will pass by any object that orbits the Sun at a distance > 1 AU, creating retrograde motion. Any object with an orbit around the Sun at a distance of < 1AU will pass by the Earth, also creating retrograde motion. 5. On the next page, you will find a plot of the location of Mars compared to the background stars over the course of 6 months. Use the chart to help answer the questions below: a) What dates did retrograde motion begin and end for Mars? Explain how you decided upon the dates. Beginning: 1/21/2012; End: 4/20/2012 --- these were the dates when Mars’s motion will seem to change directions. For example – before 1/21/2012, the RA is increasing, but after 1/21/2012 the RA is decreasing. Similarly, before 4/20/2012, the RA is decreasing, but after 4/20/2012, the RA is increasing. b) What date is closest to the middle of the retrograde motion period for Mars? What is the right ascension for Mars on that date? Based on the beginning and end dates, the middle of the retrograde motion period is 3/6/2012 – this is 45 days from the beginning date and 45 days from the end date. The RA for Mars on 3/6/2012 is around 11h 2m. c) Using your answers to the previous question, and the Equatorial Coordinates star chart from the Coordinates Lab, calculate what time Mars will transit on the date that is closest to the middle of the retrograde motion period. According to the star chart, on March 6 th , RA = 7hr transits at 8pm. Mars is at a RA of about 11 hr, which is 4 hr bigger. This means that on March 6 th , Mars will transit around (8pm + 4 hours) = Midnight. d) You should have calculated a time near midnight in the question above. Explain why this answer makes sense given the position of the Earth, Sun, and Mars when Mars is in retrograde motion. What would be different about Venus or a planet closer to the Sun than the Earth? Mars is in retrograde motion when the Earth passes by. At this moment, Mars is on the opposite side of the Earth from the Sun, so when Mars is high in the sky, the Sun will be as far below the horizon as it gets -> in other words MIDNIGHT! For a planet closer to the Sun that the Earth, it will pass by Earth when it is in the same direction as the Sun as viewed from Earth. This means the planet will be transiting around noon (since it is the same direction as the Sun) when it is in retrograde motion.