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HW 12 solution_pdf

# HW 12 solution_pdf - ghei(mg42382 HW 12 opyrchal(11130 This...

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ghei (mg42382) – HW 12 – opyrchal – (11130) 1 This print-out should have 7 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An object is dropped from a height of 6 . 9 × 10 6 m above the surface of the Earth. What is its initial acceleration? The accel- eration of gravity is 9 . 81 m / s 2 and the radius of the earth is 6 . 37 × 10 6 m . Correct answer: 2 . 26051 m / s 2 . Explanation: Let : R E = 6 . 37 × 10 6 m , h = 6 . 9 × 10 6 m , and g = 9 . 81 m / s 2 . R E + h = 6 . 37 × 10 6 m + 6 . 9 × 10 6 m = 1 . 327 × 10 7 m . Applying Newton’s second law and the law of gravity, summationdisplay F radial = G m M E ( R E + h ) 2 = m a a = G M E ( R E + h ) 2 . (1) When it is at the surface of the earth, summationdisplay F radial = G m M E R E 2 = m g g = G M E R E 2 . (2) Dividing equation (1) by equation (2), a g = R E 2 ( R E + h ) 2 a = g R E 2 ( R E + h ) 2 = (9 . 81 m / s 2 ) parenleftbigg 6 . 37 × 10 6 m 1 . 327 × 10 7 m parenrightbigg 2 = 2 . 26051 m / s 2 . 002 10.0 points A satellite is in an elliptical orbit around a planet as shown, with r 1 and r 2 being its closest and farthest distances, respectively, from the center of the planet. r 1 r 2 vectorv 1 vectorv 2 1 2 Planet Satellite If the satellite has a speed v 1 ≡ bardbl vectorv 1 bardbl at its closest distance, what is its speed at its farthest distance? 1. bardbl vectorv 2 bardbl = r 2 r 1 v 1 2. bardbl vectorv 2 bardbl = r 2 + r 1 r 1 r 2 v 1 3. bardbl vectorv 2 bardbl = ( r 2 r 1 ) v 1 4. bardbl vectorv 2 bardbl = r 1 r 2 v 1 correct 5. bardbl vectorv 2 bardbl = r 2 2 r 2 1 v 1 6. bardbl vectorv 2 bardbl = r 2 r 1 r 1 + r 2 v 1 7.

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