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PHSX 114, Friday, September 19, 2003

PHSX 114, Friday, September 19, 2003. Reading for today: Chapter 5 (5-8 -- 5-10) Reading for next lecture (Mon.): Review Chapters 4 and 5 Homework for today's lecture: Chapter 5, question 18; problems 39, 43, 53, 78. Other announcements. Exam #2 is Wed., covers Chapters 4 and 5

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PHSX 114, Friday, September 19, 2003

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  1. PHSX 114, Friday, September 19, 2003 • Reading for today: Chapter 5 (5-8 -- 5-10) • Reading for next lecture (Mon.): Review Chapters 4 and 5 • Homework for today's lecture: Chapter 5, question 18; problems 39, 43, 53, 78

  2. Other announcements • Exam #2 is Wed., covers Chapters 4 and 5 • Review and equation sheets are posted on the web • Monday's class will be review

  3. Weightlessness and Apparent weight • Astronauts in orbit are “weightless”, yet they are in Earth’s gravity • in orbit, they are falling with downward acceleration g • “apparent weight” is zero

  4. Apparent weight • Scale reads the force w • ΣF=ma= w – FG • Elevator example • FG=mg doesn't change with elevator's acceleration • If a=0 (elevator has constant speed), w = FG • If a>0 (elevator accelerating upward), w > FG • If a<0 (elevator accelerating downward), w < FG • If a=-g (freefall), w=0 (-mg = w – mg) • example

  5. Your turn Find the apparent weight of a 100 kg object if the elevator is accelerating upward with a=3.0 m/s2. • Answer: 1280 N (in kg: 131 kg=1280N / 9.8 m/s2) • ma= w – FG; w=ma+mg=m(a+g)= (100kg)(3.0 + 9.8 m/s2)=1280 N

  6. Gravity, one of the four fundamental forces of nature • 1.gravity • 2. electromagnetism • 3. strong nuclear force • 4. weak nuclear force

  7. Gravity, one of the four fundamental forces of nature • Gravity is the weakest of the four • Gravity is the least understood of the four (at a microscopic level) • Gravity is the most obvious of the four -- all you need is mass

  8. m2 Review: Newton's law of universal gravitation r m1 • F = Gm1m2/r2 • r is the distance separating the two objects • G is the gravitational constant, G= 6.67 x 10-11 N-m2/kg2 • Force is attractive, direction is along line joining the two objects

  9. Universal gravity • Same force that makes an apple fall, keeps the Moon in orbit • Note: the apple falls toward the center of the Earth (r is distance to the center) • 17th century philosophical breakthrough -- terrestrial and celestial objects obey the same rules r

  10. Newton's gravity in historical context • Law of gravity explains planetary motion, final nail in the coffin of the Earth-centered universe • Copernicus (1473-1543) has idea that planets orbit the Sun (radical! blasphemous!) • Tycho Brahe (1546-1601) Danish astronomer gets $$ from the king, builds observatory. 20 years of precise (1/60 of a degree) measurements of planetary positions.

  11. Newton's gravity in historical context • Johannes Kepler (1571-1630) inherits data, deduces three laws of planetary motion • Isaac Newton (1642-1727) explains Kepler's laws with universal gravitation (had to invent calculus to do so)

  12. Kepler's first law of planetary motion • The path of each planet is an ellipse with the Sun at one focus. • Not perfect circles! Radical! • Using calculus, can show you get an elliptical orbit from a 1/r2 central force • An ellipse is the locus of points such that the sum of the distances from the two foci is constant. (When the two foci are in the same spot, you have a circle.)

  13. Kepler's second law of planetary motion • An imaginary line joining any planet to the Sun sweeps out equal areas in equal periods of time. • This means the planet moves faster when it is closer to the Sun. • Explained by angular momentum conservation (Chapter 8)

  14. Kepler's third law of planetary motion • The square of the period of any planet is proportional to the cube of the planet's mean distance from the Sun. • The period (T) is the time an object takes to complete one cycle of its motion. (For a planet, T is the time to complete one orbit.) • T12/ r13 = T22/ r23= constant (see data for our solar system)

  15. Your turn • Imagine two planets orbiting another star. Planet 1 has T1= 1 year and r1= 1.5 x 108 km. If planet 2 has r2= 3.0 x 108 km, find T2. • T12/ r13 = T22/ r23 => T2= T1(r2/ r1)3/2 = (1 year)( 3.0 x 108 km/1.5 x 108 km)3/2 = (1 year)(2) 3/2 = 2.83 years

  16. Kepler's third law and circular orbits • For a circular orbit, straightforward to show T2/ r3= constant and the constant is 4π2/(GM), where M is the mass of the body being orbited (the Sun in Kepler's case) • Circular orbit examples

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