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HW18_Solutions

# HW18_Solutions - hiarker(srh959 homework 18 Turner(58220...

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hiarker (srh959) – homework 18 – Turner – (58220) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points On the way from a planet to a moon, as- tronauts reach a point where that moon’s gravitational pull is stronger than that of the planet. The masses of the planet and the moon are, respectively, 5 . 65 × 10 24 kg and 7 . 36 × 10 22 kg. The distance from the cen- ter of the planet to the center of the moon is 3 . 46 × 10 8 m. Determine the distance of this point from the center of the planet. The value of the universal gravitational constant is 6 . 67259 × 10 - 11 N · m 2 /kg 2 . Correct answer: 3 . 10555 × 10 8 m. Explanation: Let : M p = 5 . 65 × 10 24 kg , M m = 7 . 36 × 10 22 kg , and R = 3 . 46 × 10 8 m . Consider the point where the two forces are equal. If r p is the distance from this point to the center of the planet and r m is the distance from this point to the center of the moon, then G m M p r 2 p = G m M m r 2 m r 2 m r 2 p = M m M p r m = radicalBigg M m M p r p . It is also true that R = r p + r m R = r p + radicalBigg M m M p r p r p = R 1 + radicalBigg M m M p = 3 . 46 × 10 8 m 1 + radicalBigg 7 . 36 × 10 22 kg 5 . 65 × 10 24 kg = 3 . 10555 × 10 8 m . 002 10.0 points How high does a rocket have to go above the Earth’s surface until its weight is 0 . 8 times its weight on the Earth’s surface? The radius of the earth is 6 . 37 × 10 6 m and the acceleration of gravity is 9.8 m/s 2 . Correct answer: 751 . 877 km. Explanation: Let : R e = 6 . 37 × 10 6 m . By Newton’s universal law of gravitation, W 1 r 2 . The rocket will be at a distance of h + R e from the center of the Earth when it weighs n W , so n W W = 1 ( h + R e ) 2 1 R 2 e = R 2 e ( h + R e ) 2 n ( h + R e ) 2 = R 2 e ( h + R e ) 2 = R 2 e n h + R e = R e n h = R e n - R e = R e parenleftbigg 1 n - 1 . 0 parenrightbigg = (6 . 37 × 10 6 m) parenleftbigg 1 0 . 8 - 1 . 0 parenrightbigg = 751 . 877 km .

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hiarker (srh959) – homework 18 – Turner – (58220) 2 003 10.0 points A satellite circles planet Roton every 6 h in an orbit having a radius of 1 . 4 × 10 7 m. If the radius of Roton is 1 . 064 × 10 7 m, what is the magnitude of the free-fall acceleration on the surface of Roton? Correct answer: 2 . 05094 m / s 2 . Explanation: Basic Concepts: Newton’s law of gravi- tation F g = G m 1 m 2 r 2 . Kepler’s third law T 2 = parenleftbigg 4 π 2 G M parenrightbigg r 3 . The free-fall acceleration a on the surface of the planet is the acceleration which a body in free fall will feel due to gravity F g = G M m R 2 = m a , where M is the mass of planet Roton. This acceleration a is a = G M R 2 , (1) the number which is g on Earth. Here, how- ever, the mass M is unknown, so we try to find this from the information given about the satellite. Use Kepler’s third law for the period of the orbit T 2 = parenleftbigg 4 π 2 G M parenrightbigg r 3 . (2) By multiplying both sides with R 2 and com- paring to equation (1), we can identify our a in the right hand side T 2 R 2 = parenleftbigg 4 π 2 a parenrightbigg r 3 .
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