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PHSX 114, Wednesday, September 17, 2003 • Reading for today: Chapter 5 (5-6 -- 5-8) • Reading for next lecture (Fri.): Chapter 5 (5-8 -- 5-10) • Homework for today's lecture: Chapter 5, question 13; problems 25, 33, 62, 69 Other announcements • Exam #2 is one week from today, covers Chapters 4 and 5 • Review and equation sheets are posted on the web • Monday's class will be review Newton’s Law of Universal Gravitation •perhaps the most important equation in the history of science •Historical context will be discussed Friday •The first great “unification” in physics – motion of a falling apple and motion of the moon about the Earth explained by the same theory The falling apple • FG=mg • What direction does the apple fall? •Down? (see video clip) •toward the center of the Earth An orbiting body is a falling body • the Earth is not flat • curvature is such that the ground drops about 5 m for every 8000 m traveled horizontally • it takes about one second to fall one meter • if your horizontal velocity is about 8000 m/s (18,000 mph), you are always at the same height above the ground Orbiting at the Earth’s surface • r is the Earth’s radius, 6.38 x 106 m • centripetal acceleration is g (9.8 m/s2) • aR= v2/r = (8000 m/s)2/(6.38 x 106 m)= 9.8 m/s2 Is the moon’s acceleration 9.8 m/s2? • No, it’s 0.0027 m/s2 • g (9.8 m/s2) is 3600 times bigger (602) • Moon’s period is 27.3 days=2.36 x 106 s • Earth-moon distance is 3.84 x 108 m • v=2πr/T=2π(3.84 x 108 m)/(2.36 x 106 s) = 1020 m/s • aR=v2/r=(1020 m/s)2/(3.84 x 108 m) = 0.0027 m/s2 • (3.84 x 108 m) /(6.38 x 106 m)= 60 • force of gravity decreases as the distance squared 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 • Valid not just on Earth, but everywhere m2 Example: Find the gravitational force between the Earth and the Moon • F = GmEmM/r2 • mE= 5.97 x 1024 kg; mM= 7.35 x 1022 kg; r=3.84 x 108 m • F = (6.67 x 10-11 N-m2/kg2)(5.97 x 1024 kg)(7.35 x 1022 kg)/(3.84 x 108 m)2 = 1.98 x 1020 N • aM=F/mM = (1.98 x 1020 N)/(7.35 x 1022 kg) =0.0027 m/s2 • Note: Newton's third law applies, aE=F/mE is much smaller Your turn Estimate the force of gravity between you and the person sitting next to you. • Answer: approximately 3 x 10-7 N • F = Gm1m2/r2 • m1= m2= 70 kg (150 lb); r=1 m • F = (6.67 x 10-11 N-m2/kg2)(70kg)(70kg)/(1m)2 = 3 x 10-7 N Cavendish experiment (1798) • In 1665, Newton didn't have the technology to measure G • Henry Cavendish uses a clever "torsion pendulum" apparatus to find G • Video clip • Cavendish is said to have measured the mass of the Earth G and g • two expressions for the force of gravity at the Earth’s surface •FG = GmmE/rE2 • FG = mg • mg = GmmE/rE2 • g = GmE/rE2 • g and rE well known in Cavendish’s time • by measuring G, mass of Earth could now be determined by mE = g rE2/G • example Your turn If g=3.72 m/s2 on the surface of Mars and the radius of Mars is 3400 km, find the mass of Mars. • Answer: 6.4 x 1023 kg (about 1/10 that of Earth) • mM = g rM 2/G=(3.72 m/s2 )(3.4 x 106 m)2/ (6.67 x 10-11 N-m2/kg2) 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 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 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