An exposure to Newtonian mechanics: part II

1. An exposure to Newtonian mechanics: part II • Solving 17 problems Motivation Newton’s concept of the Universe was one of crystalline beauty. The future is predictable. The past can be reconstructed. The present can be completely deconstructed. Today, we will explore some examples.

2. Our tools • Key definitions • v= Δx/Δta= Δv/Δt vtotal= v1 + v2 • Newton’s basics • F = ma F1 = -F2 F12 = Gm1m2/R122 • Expressions of energy • K.E.=½mv2 P.E.=mgh P.E.=-GMm/R • Circular force Kinematics • Fc = mvc2/R x= =½a Δt2 R=(v2/g)sin2θ

3. Velocities • 1) A juvenile delinquent skateboarding at • 10 km/hr, throws a bottle forwards at • 20 km/hr relative to him. How fast is the bottle travelling with respect to the Halloween pumpkin that it hits? • Easy! Just use the velocity addition formula. • vtotal = v1 + v2 • vtotal = 10 km/hr + 20 km/hr = 30 km/hr • 2) What if you were travelling at 2/3 the speed of light, and fired a probe at 2/3 the speed of light. How fast does the star P-Umpkin see the probe coming towards it? • vtotal = v1 + v2 • vtotal = (2/3)c + (2/3)c = (4/3)c = 4×108 m/s. NOTE THE USE OF “c”

4. Kinematics • 3) A car drives 150 km in 4 hours. What is its speed? • v= Δx/Δt = 150 km / 4 hr = 37.5 km/hr • 4) A car accelerates from a stop to 100 km/hr in 6 seconds. What is its average acceleration? • First, convert the change in velocity to m/sec… • 100 km/hr × (1000 m/km) × (1hr/3600 s) = 27.8 m/s • a= Δv/Δt = 27.8 m/s / 6 s = 4.6 m/s/s = 4.6 m/s2

5. Kinematics • 5) Thelma and Louise plummet off the edge of the Grand Canyon. How long until they pancake? • Δx= ½at2 → t2=2Δx/g = (2 × 1800 m)/(9.8 m/s2) = 367 s2 • t=19 s • How fast are they going at impact? • v=at = gt = (9.8 m/s2) × 19 s • v=186 m/s = 420 miles/hr

6. Kinematics • 6) At the strike of midnight, a new-year’s reveler shoots his .357 Remington, aimed upwards at a 45° angle. How far away does the bullet land? • vmuzzle = 46 m/s • R=(v2/g)sin2θ = [(46 m/s)2/9.8 m/s] × (sin90°) • R=216 m • At what time does the bullet go through your living room window? • (Magic: vhorizontal = vmuzzle × cosθ° = 32.5 m/s) • t=Δx/v = (216 m)/(32.5 m/s) = 6.6 s • T=12:00:07

7. Newton’s Laws • 7) The space shuttle’s main engines and 2 SRBs provided a total thrust of about 31,000,000 N (force). The shuttle (unloaded) had a mass of about 2×106 kg (2,000,000 kg). • If an empty shuttle, with two SRBs, began travelling in space, what would be its acceleration? • F=ma → a = F/m • a = 31,000,000 N / 2,000,000 kg = 15.5 m/s2

8. Newton’s Laws • 8) Your mass is 80 kg. How hard is the pull of gravity, from an object as massive as the Earth, from a distance of the Earth’s radius? • F = GMEm/RE2 • = (6.67×10-11 N m2/kg2) × (5.97×1024 kg) × (80 kg) • (6,378,000 m)2 • = 783 N. • Since 4.45 N = 1 pound, • F=176 pounds

9. Kepler’s Third Law (1627) • Planets orbiting the Sun follow the law that • P2=Da3 • Where P is the orbital period and a is the distance of the object from the Sun. • 9) Let us look at this for the circular orbit case. • Is it that the orbits are simply larger, and take longer to traverse? • v = Δx/Δt, so Δt = Δx/v • Δx = C = 2πa, • Δt = P =2πa/v, so • P2 = 4π2a2/v2 = (4π2/v2)a2 =Da2 • Close, but not quite Kepler’s Third Law.

10. Newton and Kepler’s Third Law • P2=Da3 • Where P is the orbital period and a is the distance of the object from the Sun. • 10) Now let us try balancing the gravitational force of • attraction with the inward centripetal force needed • to maintain a circular orbit. • F12 = Gm1m2/R2 andFc = m1vc2/R, so • Gm1m2/R2 = m1vc2/R • Gm/a = v2 (I substituted “a” for “R”, as in Kepler’s Third Law) • Since P =2πa/v, therefore v =2πa/P, and so • Gm/a = 4π2a2/P2, so P2 = (4π2/Gm)a3 • Kepler’s Third Law!

11. Energy and free fall • 11) How fast would something be moving if it accelerated at 1g, for a distance equal to the Earth’s diameter? • Easy! Just balance P.E. and K.E. • mgh = ½mv2 • gD = ½v2 • v2 = 2gD, so v = (2gD)½ • v= (2gD)½ =(2×9.8 m/s2×6.4×106 m)½ • v = 11,200 m/s = 25,000 mph! • i.e, Mach 73 (the speed of sound is about 343 m/s). • The Space Shuttle orbited at about 8000 m/s.

12. Energy and free fall • 12) How far would our object in 1g have to fall, to reach the speed of light? • How does this distance compare to interplanetary and interstellar space? • Easy! Just balance K.E. and P.E. • mgh = ½mv2 • gD =½c2 • D = c2/2g • D = (3×108 m/s)2 /(2 × 9.8 m/s2 )=4.6 ×1015 m • Since 1 AU = 1.5 ×1011 m, and 1 LY = 9.46 ×1015 m… • D= 3.1 ×104 AU = 0.49 LY

13. Length measure • 13) You’re sitting at a railroad crossing, and the train—passing at 50 km/hour—takes five minutes to go by. How long is the train? • Easy! Just use the rate equation. • (“The length of the train” is the same as just asking how far does the front of the train travel, in the time it took for the rear of the train to arrive.) • v = Δx/Δt • Δx = vΔt • Δx = (50 km/hr × 1 hr/60 min) × 5 min = 4.2 km = 4200 m

14. More gravity • 14) Professional basketball players can typically jump upwards about 1.10 m. How fast are they jumping? • Easy! Just balance K.E. and P.E. • mgh = ½mv2 • gh = ½v2 • v2 = 2gh, so • v= (2gh)½ =(2 × 9.8 m/s2·1.1m)½ • v = 4.6 m/s = 10.4 mph • 15) How fast is an elevator falling if it plummets three stories? • (Estimate 1 story as 3.7 m; 12 ft.) • v= (2gh)½ =(2 × 9.8 m/s2 × 11.1m)½ • v = 14.7 m/s = 32.9 mph

15. Weight loss • 16) How does your weight compare on a planet with a different mass and radius? • Make a ratio of the law of gravity. • FE = GMEm/RE2 • FP = GMPm/RP2, so • (FP/ FE) = (GMPm/RP2)/(GMEm/RE2) • (FP/ FE) = (MP/RP2)/(ME/RE2) = (MP/ME) × (RE/RP)2 • Mars, M= 0.107 ME, R= 0.53 RE, • FMars = (0.107 ME/ME) × (RE/0.53 RE)2 = 0.38 FE

16. Monkey in a tree • 17) Consider a monkey in a tree. It knows you intend to shoot it with your gun, which is pointed directly at it. The monkey drops from the tree the moment you fire the gun. What happens?

17. Not all physics is easy • Rotations complicate things! • Consider a spinning book!