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Planetary & Satellite Motion Kepler’s 3 Laws Law of Ellipses The path of the planets about the sun is elliptical in shape, with the center of the sun being located at one focus. Law of Equal Areas An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time Law of Harmonies The ratios of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. Law of Ellipses All planets are orbiting the sun in a path described as an ellipse An ellipse is a special curve in which the sum of the distances from every point on the curve to two other points in constant Law of Equal Areas Describes the speed at which any given planet will move when orbiting the sun A planet moves fastest when it is closest to the sun and slowest when it is furthest from the sun Law of Harmonies Compares the orbital period and radius of orbit of a planet to those of other planets The comparison being made is the ratio of the squares of the periods to the cubes of their average distances from the sun Planet Period (s) Average Dist. (m) Earth 3.156 x 107 1.4957 x 1011 Mars 5.93 x 107 2.278 x 1011 T2/R3 (s2/m3) What did Newton Do??? His comparison of the moons acceleration and the acceleration of objects on earth allowed him to conclude the moons is held in a circular orbit by the force of gravity. The force that is inversely proportional to the distance between the centers He used Kepler’s 3rd law (T2/R3) to find an average ratio of k = 2.97 x 10-19 s2/m3 Example 1 Suppose a planet is discovered that is 14 times as far from the sun as the Earth’s distance is from the sun (1.5 x 1011 m). Use Kepler’s law of harmonies to predict the orbital period of such a planet. GIVEN: T2/R3 = 2.97 x 10-19 m2/s3. 2 2 p e 3 3 e p T T R R Example 2 The average orbital distance of Mars is 1.52 times the average orbital distance of the Earth. Knowing that the Earth orbits the sun in approximately 365 days, use Kepler’s law of harmonies to predict the time for Mars to orbit the sun. 2 mars 3 mars T R 2 Tearth 3 Rearth Satellites Any objects that is orbiting the earth, the sun, or an other massive body A satellite is a projectile The only force governing its motion is the force of gravity How fast do you have to launch it to get it into orbit? Therefore, must be launched at a speed greater than 8000 m/s to orbit earth! Velocity, Acceleration, Force Vectors Velocity Directed tangent to the circle at every point along its path Acceleration Directed toward the center of the circle – toward the central body that it is orbiting Net Force Directed inwards in the same direction as acceleration Elliptical Orbit of Satellites Velocity is tangent to the ellipse Acceleration is directed toward the focus of the ellipse Net Force is directed in the same direction in the same direction as the acceleration Mathematics of Satellite Motion Consider a satellite with mass, Msat orbiting a central body with mass, Mcentral. If it moves in the circular motion, the net centripetal force is Fnet (M sat * v 2 ) R The net centripetal force is a result of the gravitation force: (G * Msat * Mcentral ) Fgrav 2 R Mathematics of Satellite Motion cont. Since Fgrav = Fnet M sat * v 2 (G * M sat * M central ) 2 R R (G * M central ) v R 2 v = velocity of the satellite moving G = 6.673 x 10-11 N*kg2/m2 Mcentral = mass of the central body which the satellite is orbiting R = radius of the orbit of the satellite Mathematics of Satellite Motion cont. Acceleration of gravity G * M central g 2 R Motion of Satellites T2 4 * 2 3 R G * M central Example Problem 1 A satellite wishes to orbit the earth at a height of 100 km (approximately 60 miles) above the surface of the earth. Determine the speed, acceleration and orbital period of the satellite. (Given: Mearth = 5.98 x 1024 kg, Rearth = 6.37 x 106 m) Example Problem 2 The period of the moon is approximately 27.2 days (2.35 x 106 s). Determine the radius of the moon's orbit and the orbital speed of the moon. (Given: Mearth = 5.98 x 1024 kg, Rearth = 6.37 x 106 m) Example Problem 3 A geosynchronous satellite is a satellite that orbits the earth with an orbital period of 24 hours, thus matching the period of the earth's rotational motion. A special class of geosynchronous satellites is a geostationary satellite. A geostationary satellite orbits the earth in 24 hours along an orbital path that is parallel to an imaginary plane drawn through the Earth's equator. Such a satellite appears permanently fixed above the same location on the Earth. If a geostationary satellite wishes to orbit the earth in 24 hours (86400 s), then how high above the earth's surface must it be located? (Given: Mearth = 5.98x1024 kg, Rearth = 6.37 x 106 m)