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PHY303k_-_Spring_2010_-_HW_11

# PHY303k_-_Spring_2010_-_HW_11 - saldana(avs387 Homework 11...

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saldana (avs387) – Homework 11 – florin – (58140) 1 This print-out should have 23 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 9 . 3 × 10 6 m. If the radius of Roton is 4 . 464 × 10 6 m, what is the magnitude of the free-fall acceleration on the surface of Roton? Correct answer: 3 . 19869 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 . If we solve for a , we obtain a = parenleftbigg 4 π 2 T 2 R 2 parenrightbigg r 3 = 3 . 19869 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 Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical orbit of planet B is two times larger than the semi- major axis of the elliptical orbit of planet A. What is the ratio of the orbital period of planet B to that of planet A? 1. T B T A = 2 2. T B T A = 1 8 3. T B T A = radicalbigg 1 8 4. T B T A = 8 5. T B T A = 1 6. T B T A = 1 2 7. T B T A = 1 4 8. T B T A = 8 correct 9. T B T A = 2 10. T B T A = radicalbigg 1 2 Explanation: Basic Concept: Kepler’s Third Law is T 2 = parenleftbigg 4 π 2 G M S parenrightbigg a 3 = K S a 3 ,

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saldana (avs387) – Homework 11 – florin – (58140) 2 where K S = 4 π 2 G M S = 2 . 97 × 10 19 s 2 / m 2 , and where a is the semi-major axis of the elliptical orbit of the planet ( a = r the radius of a planet in a circular orbit). Solution: According to Kepler’s third law, the square of the orbital period is proportional to the cube of the semi-major axis “ a ” of the elliptical orbit. Therefore, T 2 A a 3 A = T 2 B a 3 B . Therefore, T B T A = parenleftbigg a B a A parenrightbigg 3 2 = 2 3 / 2 = 8 . 003 10.0 points Given: G = 6 . 67259 × 10 11 N m 2 / kg 2 Calculate the work required to move a planet’s satellite of mass 1020 kg from a cir- cular orbit of radius 2 R to one of radius 3 R , where 3 . 92 × 10 6 m is the radius of the planet.
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