HW 13-solutions

HW 13-solutions - kuruvila(lk5992 – HW 13 – opyrchal...

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Unformatted text preview: kuruvila (lk5992) – HW 13 – opyrchal – (11113) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A satellite circles planet Roton every 6 . 2 h in an orbit having a radius of 1 . 7 × 10 7 m. If the radius of Roton is 1 . 088 × 10 7 m, what is the magnitude of the free-fall acceleration on the surface of Roton? Correct answer: 3 . 28897 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 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 . If we solve for a , we obtain a = parenleftbigg 4 π 2 T 2 R 2 parenrightbigg r 3 = 3 . 28897 m / s 2 which is our answer. Although identifying a in this way is a “quick” way of solving the problem, we could just as well have calculated the planet mass M explicitly from equation (2) and inserted into equation (1). 002 10.0 points An object is projected upward from the sur- face of the earth with an initial speed of 5 km / s....
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HW 13-solutions - kuruvila(lk5992 – HW 13 – opyrchal...

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