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hmwk04 - Su(ycs73 HW04 Tsoi(58020 This print-out should...

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Su (ycs73) – HW04 – Tsoi – (58020) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Two satellites A and B orbit the Earth in the same plane. Their masses are m and 6 m , respectively, and their radii r and 3 r , respectively. 3 r r 6 m B m A What is the ratio of the orbital speeds? 1. v B v A = 3 2. v B v A = 3 3. v B v A = 1 2 4. v B v A = 1 3 5. v B v A = 1 3 correct 6. v B v A = 1 9 7. v B v A = 2 8. v B v A = 1 2 9. v B v A = 2 10. v B v A = 9 Explanation: The force of gravity is responsible for hold- ing a satellite in its orbit, so the orbital cen- tripetal force is equal to the force of gravity: F r = m v 2 r = G M E m r 2 , where M E is the mass of the Earth, m is the mass of the satellite, and r is the radius of the orbit (from the Earth’s center). Thus the tangential speed v of an orbit at distance r is v = radicalbigg G M r radicalbigg 1 r . Since r A r B = 1 3 , the ratio v B v A = radicalbigg r A r B = 1 3 . 002 (part 2 of 2) 10.0 points Let 10 R E be the distance of the satellite A from the center of the Earth, where R E is the radius of the Earth. What is the gravitational acceleration due to the Earth at satellite A ? g is the gravita- tional acceleration at the surface of the Earth. 1. g A = g 11 2. g A = g 100 correct 3. g A = g 81 4. g A = g 11 5. g A = g 3 6. g A = g 10 7. g A = g 9 8. g A = g 121 9. g A = g 10 10. g A = g Explanation: g A = G M E r 2 A = G M E (10 R E ) 2 = 1 100 g .
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Su (ycs73) – HW04 – Tsoi – (58020) 2 003 (part 1 of 2) 10.0 points A small metal ball is suspended from the ceil- ing by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread’s trajectory describes a cone. The acceleration of gravity is 9 . 8 m / s 2 . v r g m θ What is the speed of the ball when it is in circular motion? Answer in terms of g , and θ . 1. v = g ℓ tan θ 2. v = radicalbig g ℓ sin θ 3. v = g tan θ 4. v = radicalbig g ℓ cos θ 5. v = tan θ 6. v = radicalbig g ℓ cos θ 7. v = radicalbig g ℓ sin θ 8. v = g ℓ sin θ 9. v = radicalbig g ℓ tan θ 10. v = radicalbig g ℓ tan θ sin θ correct Explanation: Use the free body diagram below. T m g θ The tension on the string can be decom- posed into a vertical component which bal- ances the weight of the ball and a horizontal component which causes the centripetal ac- celeration, a centrip that keeps the ball on its horizontal circular path at radius r = sin θ . If T is the magnitude of the tension in the string, then T vertical = T cos θ = m g (1) and T horiz = m a centrip or T sin θ = m v 2 ball sin θ . (2) Solving (1) for T yields T = m g cos θ (3) and substituting (3) into (2) gives m g tan θ = m v 2 ball sin θ . Solving for v yields v = radicalbig g ℓ tan θ sin θ .
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