HW 09-solutions - cooper(mlc4285 HW 09 florin(55245 This...

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cooper (mlc4285) – HW 09 – florin – (55245) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points The planets in our Solar System have orbits around the Sun that are nearly circular, and v << c . Calculate the period T ( a ”year” - the time required to go around the Sun once) for a planet whose orbit radius is r . This is the relationship discovered by Kepler and explained by Newton. (It can be shown by advanced techniques that this result also ap- plies to elliptical orbits if you replace r by the ”semimajor axis,” which is half the longer, ”major” axis of the ellipse.) Use this ana- lytical solution for circular motion to predict the Earth’s orbital speed, using the data for Sun and Earth on the inside back cover of the textbook. 1. T = (2 π ) radicalbigg R 3 GM s correct 2. T = (4 π ) radicalbigg R 3 GM s 3. T = (4 π ) radicalbigg GM R s 1 4. T = (2 π 2 ) radicalbigg GM R s 5. T = (4 π ) radicalbigg GM R 3 s 6. T = (4 π ) radicalbigg GM R s 7. T = (2 π ) radicalbigg R GM s 1 8. T = (2 π ) radicalbigg R GM s 9. T = (2 π ) R 3 GM s 10. T = (4 π 2 ) radicalbigg R 3 GM s 1 Explanation: Define the system to be Earth. The net force on the Earth is the gravitational force by the Sun on the Earth. Use this to solve for the speed of Earth. | vector F net | = | vector F grav | | vectorp | | vectorv | R = G M m R 2 m | vectorv | 2 R = G M m R 2 | vectorv | 2 = G M R Earth is in a nearly circular orbit, thus its speed is v = 2 π R T . Substitute this to get an expression that relates the period and radius for Earth’s orbit. ( 2 π R T ) 2 = G M R 4 π 2 R 2 T 2 = G M R T 2 = 4 π 2 R 3 G M T = (2 π ) radicalbigg R 3 GM The above expression is known as Kepler’s third law for circular orbits. 002(part2of2)10.0points Calculate the Earth’s orbital speed. [ Use G = 6 . 67 × 10 11 Nm 2 / kg 2 ; M = 1 . 98843 × 10 30 kg; R = 1 . 49619 × 10 11 m ] Correct answer: 29773 . 2 m / s. Explanation: From the solution above, Earth’s orbital speed as determined from Newton’s law of gravitation and The Momentum Principle is: | vectorv | = radicalbigg G M R = radicalBigg (6 . 67 × 10 11 Nm 2 / kg 2 )(1 . 98843 × 10 30 kg) 1 . 49619 × 10 11 m
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cooper (mlc4285) – HW 09 – florin – (55245) 2 = 29773 . 2 m / s Note that the radius of the Earth’s orbit is defined as 1 A.U. ( Astronomical Unit ) which is 1 . 49619 × 10 11 m 003 10.0points A certain weighing scale calculates the mass of an object by first measuring the normal force exerted by the object on the scale and then dividing it by the standard value of g = 9 . 8 m / s 2 . For a given object, the scale reads 35 kg on the surface of the earth at the North Pole, which is at a distance 6360 km from the center of the earth. Calculate the reading on the scale for the same object at the equator, which is at a distance 6378 km from the center of the earth.
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