PHY183-Lecture35 - Final Exam Our Common Comprehensive...

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1 March 29, 2012 Physics for Scientists&Engineers 1 1 Final Exam Our Common Comprehensive Final Exam is on Thursday, May 3 rd from 8:00-10:00PM Location Chemistry (CEM) room 138 Two 8.5 inch by 11 inch sheets of notes (both sides) Calculator Alternate Final Exam University approved excuses only Email me with your reason for taking the exam at the alternate time prior to Friday, April 6 Wednesday, May 2, 7:45 – 9:45 AM (room TBA) March 29, 2012 Physics for Scientists&Engineers 1 2 Physics for Scientists & Engineers 1 Spring Semester 2012 Lecture 35 Kepler’s Second and Third Laws March 29, 2012 Physics for Scientists&Engineers 1 3 Example: Orbital Period of Sedna On November 14, 2003, astronomers discovered a previously unknown object in our solar system in the Kuiper Belt beyond the orbit of Neptune They named this object Sedna, after the Inuit goddess of the sea The average distance of Sedna from the Sun is 17 10 9 km PROBLEM How long does it take Sedna to complete one orbit around the Sun? THINK We can use Kepler’s Laws to relate the distance of Sedna from the Sun to the period of Sedna’s orbit around the Sun March 29, 2012 Physics for Scientists&Engineers 1 4 Example: Orbital Period of Sedna SKETCH A sketch of the distance of Sedna from the Sun compared with the distance of Pluto from the Sun is shown below RESEARCH We can relate the orbit of Sedna to the known orbit of the Earth using Kepler’s Third Law T Earth 2 a Earth 3 = T Sedna 2 a Sedna 3 March 29, 2012 Physics for Scientists&Engineers 1 5 Example: Orbital Period of Sedna SIMPLIFY We can solve for the period of Sedna’s orbit CALCULATE Putting in the numerical values gives us ROUND We report our result to two significant figures T Sedna = T Earth a Sedna a Earth 3/2 T Sedna = 1 year () 17 10 9 km 0.150 10 9 km 3/2 = 1206.53 years T Sedna = 1.2 10 3 years March 29, 2012 Physics for Scientists&Engineers 1 6 Example: Orbital Period of Sedna DOUBLE-CHECK We can compare our result to the tabulated values of the semi-major axes of the orbits of the planets and their orbital periods As our figure shows, our calculated result falls nicely onto the extrapolation of the data from the planets and known dwarf-planets
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2 March 29, 2012 Physics for Scientists&Engineers 1 7 How do we know the mass of the Sun? Kepler’s Third Law: Solve for M : Insert numbers for Earth’s orbit In the same way we can determine the mass of Earth from the Moon’s orbit, or the mass of the galaxy from orbits of stars (more on this later) Or the mass of any object with a satellite Mass of the Sun T 2 r 3 = 4 π 2 GM M = 4 2 r 3 GT 2 M = 4 2 (1.496 10 11 m) 3 (6.67 10 11 m 3 kg -1 s 2 )(3.15 10 7 s) 2 = 1.99 10 30 kg March 29, 2012 Physics for Scientists&Engineers 1 8 Planets’ Data Planet Radius (km) Mass (10 24 kg) g (m/s 2 ) Escape v (km/s) r -orbit (10 6 km) Eccent. T -orbit (years) Mercury 2,440 0.330 3.7 4.3 57.9 0.205 0.241 Venus 6,050 4.87 8.9 10.4 108.2 0.007 0.615 Earth 6,380 5.97 9.8 11.2 149.6 0.017 1 Mars 3,400 0.642 3.7
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This note was uploaded on 04/02/2012 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

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PHY183-Lecture35 - Final Exam Our Common Comprehensive...

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