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A satellite moves in a circular earth orbit that has a radius of 7.67 x 10 6 m. A model airplane is flying on a 14.7-m guideline in a horizontal circle. The guideline is nearly parallel to the ground. Find the speed of the plane such that the plane and the satellite have the same centripetal acceleration. Solution We know that F = G *m *M/R 2 ----------- Equation 1 where M - Mass of Earth, m - Mass of Satellite, F - Gravitational Force, R - Distance measured from center of Earth. We also know that F = m*g(R) ------- Equation 2 Equating 1 and 2, we get g(R) = G*m / R 2 --------- Equation 3 This is equivalent to satellite centripetel acceleration. a sat = G*m / R 2 For Plane, a plane = V 2 / r Since two accelerations are equal, Thus G*m / R 2 = V 2 / r ------------ Equation 4 We know that For G = 6.67 Ã— 10 -11 m 3 / kg. s 2 M = 5.97 Ã— 10 24 kg. R = 7.67 Ã— 10 6 m Substituting all values in Equation 4, we get V = sqrt (99.50) V = 9.97 m/s .

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A satellite moves in a circular earth orbit that has a radius of 7.67 x 10 6 m. A model airplane is flying on a 14.7-m guideline in a horizontal circle. The guideline is nearly parallel to the ground. Find the speed of the plane such that the plane and the satellite have the same centripetal acceleration. Solution We know that F = G *m *M/R 2 ----------- Equation 1 where M - Mass of Earth, m - Mass of Satellite, F - Gravitational Force, R - Distance measured from center of Earth. We also know that F = m*g(R) ------- Equation 2 Equating 1 and 2, we get g(R) = G*m / R 2 --------- Equation 3 This is equivalent to satellite centripetel acceleration. a sat = G*m / R 2 For Plane, a plane = V 2 / r Since two accelerations are equal, Thus G*m / R 2 = V 2 / r ------------ Equation 4 We know that For G = 6.67 Ã— 10 -11 m 3 / kg. s 2 M = 5.97 Ã— 10 24 kg. R = 7.67 Ã— 10 6 m Substituting all values in Equation 4, we get V = sqrt (99.50) V = 9.97 m/s .

- 1. A satellite moves in a circular earth orbit that has a radius of 7.67 x 10 6 m. A model airplane is flying on a 14.7-m guideline in a horizontal circle. The guideline is nearly parallel to the ground. Find the speed of the plane such that the plane and the satellite have the same centripetal acceleration. Solution We know that F = G *m *M/R 2 ----------- Equation 1 where M - Mass of Earth, m - Mass of Satellite, F - Gravitational Force, R - Distance measured from center of Earth. We also know that F = m*g(R) ------- Equation 2 Equating 1 and 2, we get g(R) = G*m / R 2 --------- Equation 3 This is equivalent to satellite centripetel acceleration. a sat = G*m / R 2 For Plane, a plane = V 2 / r Since two accelerations are equal, Thus G*m / R 2 = V 2 / r ------------ Equation 4 We know that For G = 6.67 Ã— 10 -11 m 3 / kg. s 2 M = 5.97 Ã— 10 24 kg. R = 7.67 Ã— 10 6 m
- 2. Substituting all values in Equation 4, we get V = sqrt (99.50) V = 9.97 m/s