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THIS PRESENTAION HAS BEEN RATED BY THE CLASSIFICATION AND RATING ADMINISTRATION TG-13 TEACHERS’ GUIDANCE STRONGLY ADVISED Some Material May Be Unintelligible For Students Under 13. Intense Frames of Scientific Instruction, Analysis, Comparing and Contrasting, Description, and for Some Vocabulary. © 1852 All Rights Reserved VOID WHERE PROHIBITED BY LAW
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The authorized reproduction or distribution of this copyrighted work is highly encouraged. Lethargic obtuseness is insubordinate and is discouraged by PBIS, as it may result in little or no monetary gain after secondary education or a fine of $250,000. © 1852 All Rights Reserved VOID WHERE PROHIBITED BY LAW PBIS ANTI-VACUITY
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ASTRONOMY ORBITAL MECHANICS
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OBJECTIVES By the end of this presentation, students will be able to Describe how the orbits of planets are defined using Kepler’s laws. Evaluate the period of an orbit or the relative distance of a planet using Kepler’s Law of Harmony.
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KEPLER’S LAWS OF PLANETARY MOTION 1. Every planet moves around the sun in an elliptical orbit (near circular, ovals) with the sun at one focus. – This describes a simple shape for the orbit. – For a more detailed description, the elements of an orbit need to be defined.
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KEPLER’S LAWS OF PLANETARY MOTION PERIHELION - the point marking the nearest approach of a planet to the sun. APHELION - the point marking the farthest distance of a planet from the sun.
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KEPLER’S LAWS OF PLANETARY MOTION SEMI-MAJOR AXIS (longest radius) - the average distance from the sun. a = ½(aphelion + perihelion) ECENTRICITY - the flatness of the orbit (how oval or circular). e = 1- aphelion = perihelion aa - 1
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The size and shape of the orbit are described by a and e, whereas the orbital orientation is described by the INCLINATION ( i ) - the tilt of the orbital plane to the orbital plane of the earth. KEPLER’S LAWS OF PLANETARY MOTION
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i Plane of Earth’s orbit star PLANET’s ORBIT Planet PERIHELION To Vernal Equinox APHELION r Located a distance r away from the star ANGLE OF INCLINATION ASCENDING NODE DESCENDING NODE
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The distance of the planet from the star can then be determined by measuring the angle between the observed position of the perihelion and the observed position of the planet ( ). PERIOD (T) - the time for one complete revolution around the sun. –Can be used to predict when the planet will pass any expected point. KEPLER’S LAWS OF PLANETARY MOTION
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2. A line drawn from the sun to a planet will sweep out equal areas of space in equal times. = constant = a 2 t T
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KEPLER’S LAWS OF PLANETARY MOTION 2. A line drawn from the sun to a planet will sweep out equal areas of space in equal times. - Indicates that the planet speeds up as it approaches perihelion and slows down as it approaches aphelion. - Also used to calculate and .
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star Planet r A1A1 A2A2 A 1 = A 2 = a 2 t t T
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KEPLER’S LAWS OF PLANETARY MOTION 3. The ratio of the cube of the average orbital radius to the square of its period is a constant. T 2 = K a 3 K = Kepler’s Constant = 4 2 G m o
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KEPLER’S LAWS OF PLANETARY MOTION 3. The ratio of the cube of the average orbital radius to the square of its period is a constant. T 2 = 4 2 a 3 G m o
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KEPLER’S LAWS OF PLANETARY MOTION 3. T 2 = K a 3 Keeping this simple: for objects orbiting the sun, like the earth, the earth is 1.0 A.U. from the sun. It has a period of orbit of 1.0 year. K = 1.0
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According to Bode’s Rule, Saturn orbits the sun at a relative distance of 10.0 A.U. What is its orbital period? T 2 = K a 3 T 2 = 1.0 (10.0) 3 T 2 = 1.0 (1000) T 2 = (1000) T = 31.6 years
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How far away from the sun will an asteroid be if it was to have an orbital period of 2.0 years? T 2 = K a 3 2.0 2 = 1.0 a 3 4.0 = 1.0 a 3 4.0 = a 3 a = 1.26 A.U.
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One of Jupiter’s moons orbits in 1.77 days and is 4.22 units distance from Jupiter. Another moon orbits 6.71 units distance. What is it’s orbital period? T 2 = K a 3 (1.77) 2 = K (4.22) 3 3.1329 = K (75.151) K = (3.1329)/(75.151) K = 0.0417
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One of Jupiter’s moons orbits in 1.77 days and is 4.22 units distance from Jupiter. Another moon orbits 6.71 units distance. What is it’s orbital period? T 2 = K a 3 T 2 = (0.0419) (6.71) 3 T 2 = (0.0419) (302.1) T 2 = (12.66) T = 3.56 days
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A new ‘planet’ has been discovered at 80.0 A.U. from the sun. What is its orbital period? T 2 = K a 3 T 2 = 1.0 (80.0) 3 T 2 = 1.0 (512,000) T 2 = 512,000 T = 716 years
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ASTRONOMY ORBITAL MECHANICS
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