HW07-solutions - baum (lmb2768) HW07 distler (56295) This...

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baum (lmb2768) – HW07 – distler – (56295) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 3) 10.0 points Consider two planets oF mass m and 2 m , respectively, orbiting the same star in circular orbits. The more massive planet is 4 times as Far From the star as the less massive one. What is the ratio F 1 : F 2 oF the gravita- tional Forces exerted on the star by the two planets? Correct answer: 8. Explanation: Let : m 1 = m, m 2 =2 m, and r 2 =4 r 1 . F = Gm 1 m 2 r 2 m 1 m 2 r 2 , so F 1 F 2 = M S m r 1 2 M S (2 m ) (4 r 1 ) 2 = m r 1 2 · (4 r 1 ) 2 2 m = 8 . 002 (part 2 of 3) 10.0 points What is the ratio v 1 : v 2 oF the speeds oF the two planets? Correct answer: 2. Explanation: Gravitational Force supplies the centripetal Force, so F = GM S m r 2 = mv 2 r v = ± GM r ± 1 r v 1 v 2 = ± r 2 r 1 = 4= 2 . 003 (part 3 of 3) 10.0 points What is the ratio T 1 : T 2 oF the orbital periods oF the two planets? Correct answer: 0 . 125. Explanation: T = 2 π r v r V , so T 1 T 2 = r 1 v 1 r 2 v 2 = ² r 1 r 2 ³² v 2 v 1 ³ = 1 4 · 1 2 = 0 . 125 . 004 10.0 points An object is projected upward From the sur- Face oF the earth with an initial speed oF 3km / s. The acceleration oF gravity is 9 . 81 m / s 2 . ±ind the maximum height it reaches. Correct answer: 494 . 312 km. Explanation: Let : R E =6 . 37 × 10 6 m , v i =3km / s=3000m / s , and g =9 . 81 m / s 2 . g = GM E R 2 E and the potential energy at a distance r From the surFace oF the earth is U ( r )= - GM E m r Using conservation oF energy to relate the initial potential energy oF the system to its energy at its maximum height ( K f =0) , K f - K i + U f - U i =0 - K ( R E )+ U ( R E + h ) - U ( R E )=0 - 1 2 mv 2 i + GM E m R E + h - GM E m R E
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baum (lmb2768) – HW07 – distler – (56295) 2 1 2 v 2 i = GM E h R E ( R E + h ) · R E R E v 2 i = 2 GM E R 2 E · hR E R E + h v 2 i = 2 ghR E R E + h ( R E + h ) v i 2 =2 ghR E h = - v 2 i R E v 2 i - 2 gR E Since v 2 i - 2 gR E =(3000m / s) 2 - ± 2(9 . 81 m / s 2 ) × (6 . 37 × 10 6 m) ² = - 1 . 15979 × 10 8 , h = - (3000 m / s) 2 (6 . 37 × 10 6 m) - 1 . 15979 × 10 8 · 1km 1000 m = 494 . 312 km . 005 10.0 points An earth satellite remains in orbit at a dis- tance of 17925 km from the center of the earth. What speed would it have to main- tain? The universal gravitational constant is 6 . 67 × 10 - 11 N · m 2 / kg 2 and the mass of the earth is 5 . 98 × 10 24 kg. Correct answer: 4717 . 2m / s. Explanation: Let; r =17925km=1 . 7925 × 10 7 m , G =6 . 67 × 10 - 11 N · m 2 / kg 2 , and M =5 . 98 × 10 24 kg . The gravitational force supplies the cen- tripetal force, so GMm r 2 = mv 2 r v = ³ GM r = ´ 6 . 67 × 10 - 11 N · m 2 / kg 2 1 . 7925 × 10 7 m × µ 5 . 98 × 10 24 kg = 4717 . 2m / s . 006
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This note was uploaded on 11/30/2011 for the course PHYSICS 302K taught by Professor Irenepolycarpou during the Spring '09 term at University of Texas.

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HW07-solutions - baum (lmb2768) HW07 distler (56295) This...

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