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Satellite communication pratt john willey publications

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Satellite communication pratt john willey publications

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Satellite communication pratt john willey publications xtos Universitarios que necesitas!Li bros y Solucionarios en formato digitalEl complement ideal para estar preparados para los exámenes!Los Solucionar

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Satellite communication pratt john willey publications

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Satellite communication pratt john willey publications centrifugal and centripetal mean with regard to a satellite in orbit around the earth.A satellite is in a circular orbit around the earth. The altitud

e of the satellite’s orbit above the surface of the earth is 1.400 km. (i) What are the centripetal and centrifugal accelerations acting on the satell Satellite communication pratt john willey publications

ite in its orbit? Give your answer in ms2. (ii) What is the velocity of the satellite in this orbit? Give your answer in km's. (iii) What is the orbit

Satellite communication pratt john willey publications

al period of the satellite in this orbit? Give your answer in hours, minutes, and seconds. Note: assume the average radius of the earth is 6.378.137 k

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Satellite communication pratt john willey publications rce on the satellite is a force on the satellite that is directly away from the center of gravity of the earth (Four in Fig.2.1) and the centripetal f

orce is one directly towards the center of gravity of the earth (Fix in Fig. 2.1). The centrifugal force on a satellite will therefore try to fling th Satellite communication pratt john willey publications

e satellite away from the earth while the centripetal force will try to bring the satellite down towaids the earth.(i)From equation (2.1) ceiiripetal

Satellite communication pratt john willey publications

acceleration ứ = where ocis Kepler’s constant. The value of r= 6.378.137 + 1.400 = 7.778.137 km. thus a = 3.986004418 ■ 10? / (7.778.137)2 = 0.0065885

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Satellite communication pratt john willey publications r orbit.. From equation (2.5) V = (o/r)’2 = (3.986004418 • 105 / 7,778.137)’2 = 7.1586494 km s and so 47 = 0.0065885007 km s2 = 6.5885007 ms2. NOTE: s

ince the satellite was in stable orbit, the centrifugal acceleration must be equal to the centripetal acceleration, which we have found to be true her Satellite communication pratt john willey publications

e (but we needed only to calculate one of them).(ii)We have already found out the velocity of the satellite in orbit in part (i) (using equation(2.5))

Satellite communication pratt john willey publications

to be 7.1586494 km s(iii)From equation (2.6), the orbital period T = (2jtr 22) = (2^7.778.137’ 2)'( 3.986004418 • 105)12 = (4.310.158.598)4631.348114

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Satellite communication pratt john willey publications y in radians per second;b.The orbital period in minutes; and

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