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Unformatted text preview: CHAPTER 11 Gravity 1* Â· True or false: ( a ) Keplerâ€™s law of equal areas implies that gravity varies inversely with the square of the distance. ( b ) The planet closest to the sun, on the average, has the shortest period of revolution about the sun. ( a ) False ( b ) True 2 Â· If the mass of a satellite is doubled, the radius of its orbit can remain constant if the speed of the satellite ( a ) increases by a factor of 8. ( b ) increases by a factor of 2. ( c ) does not change. ( d ) is reduced by a factor of 8. ( e ) is reduced by a factor of 2. ( c ) 3 Â· One night, Lucy picked up a strange message on her ham radio. â€śHelp! We ran away from Earth to live in peace and serenity, and we got disoriented. All we know is that we are orbiting the sun with a period of 5 years. Where are we?â€ť Lucy did some calculations and told the travelers their mean distance from the sun. What is it? Use Equ. 112; R = R ES (5/1) 2/3 ; R ES = 1.5 Ă— 10 11 m R = (1.5 Ă— 10 11 )(5) 2/3 m = 4.39 Ă— 10 11 m 4 Â· Halleyâ€™s comet has a period of about 76 y. What is its mean distance from the sun? R mean = (1 AU)(76) 2/3 (see Problem 3) R mean = 1.5 Ă— 10 11 Ă— 76 2/3 m = 26.9 Ă— 10 11 m 5* Â· A comet has a period estimated to be about 4210 y. What is its mean distance from the sun? (4210 y was the esti mated period of the comet Haleâ€“Bopp, which was seen in the Northern Hemisphere in early 1997. Gravitational interactions with the major planets that occurred during this apparition of the comet greatly changed its period, which is now expected to be about 2380 y.) R mean = (1 AU)(4210) 2/3 (see Problem 4) R mean = 1.5 Ă— 10 11 Ă— 4210 2/3 m = 3.91 Ă— 10 13 m 6 Â· The radius of the earthâ€™s orbit is 1.496 Ă— 10 11 m and that of Uranus is 2.87 Ă— 10 12 m. What is the period of Uranus? Use Equ. 112; T U = (1 y)( R U /1 AU) 3/2 T U = (2.87 Ă— 10 12 /1.5 Ă— 10 11 ) 3/2 y = 83.7 y 7 Â· The asteroid Hektor, discovered in 1907, is in a nearly circular orbit of radius 5.16 AU about the sun. Determine the period of this asteroid. Chapter 11 Gravity T H = (1 y)(5.16/1) 3/2 (see Problem 6) T H = 11.7 y 8 Â·Â· The asteroid Icarus, discovered in 1949, was so named because its highly eccentric elliptical orbit brings it close to the sun at perihelion. The eccentricity e of an ellipse is defined by the relation d p = a (1  e ), where d p is the perihelion distance and a is the semimajor axis. Icarus has an eccentricity of 0.83. The period of Icarus is 1.1 years. ( a ) Determine the semimajor axis of the orbit of Icarus. ( b ) Find the perihelion and aphelion distances of the orbit of Icarus. ( a ) Use Keplerâ€™s third law; a = (1 AU)( T I ) 2/3 ( b ) d p = a (1  e ); d a + d p = 2 a ; d a = 2 a d p a = 1.5 Ă— 10 11 Ă— 1.1 2/3 m = 1.6 Ă— 10 11 m d p = (1.6 Ă— 10 11 Ă— 0.17) m = 2.72 Ă— 10 10 m; d a = 2.93 Ă— 10 11 m 9* Â· Why donâ€™t you feel the gravitational attraction of a large building when you walk near it?...
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 Spring '09
 Eduardo
 Physics, Gravity, Mass, Potential Energy, General Relativity, Gravitational constant

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