h18 - hernandez(ejh742 – homework 18 – Turner –(58120...

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Unformatted text preview: hernandez (ejh742) – homework 18 – Turner – (58120) 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 . 86 × 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 . 4 × 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 . 05736 × 10 8 m. Explanation: Let : M p = 5 . 86 × 10 24 kg , M m = 7 . 36 × 10 22 kg , and R = 3 . 4 × 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 GmM p r 2 p = GmM 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 . 4 × 10 8 m 1 + radicalBigg 7 . 36 × 10 22 kg 5 . 86 × 10 24 kg = 3 . 05736 × 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 . 6 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: 1853 . 63 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 nW , so nW 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 . parenrightbigg = (6 . 37 × 10 6 m) parenleftbigg 1 √ . 6- 1 . parenrightbigg = 1853 . 63 km . hernandez (ejh742) – homework 18 – Turner – (58120) 2 003 10.0 points A satellite circles planet Roton every 7 . 3 h in an orbit having a radius of 5 . 8 × 10 6 m. If the radius of Roton is 3 . 538 × 10 6 m, what is the magnitude of the free-fall acceleration on the surface of Roton? Correct answer: 0 . 890999 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 GM 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 = ma, 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 GM parenrightbigg...
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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h18 - hernandez(ejh742 – homework 18 – Turner –(58120...

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