PHSX 114, Wednesday, September 17, 2003

1. 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

2. 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

3. 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

4. The falling apple • FG=mg • What direction does the apple fall? • Down? (see video clip) • toward the center of the Earth

5. 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

6. 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

7. Is the moon’s acceleration 9.8 m/s2? • No, it’s 0.0027m/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

8. m2 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

9. 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

10. 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

11. 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

12. 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

13. 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)

14. 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

15. 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

16. 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