F = G m 1m 2 r 2. F = G 4πmρ 3

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2 apple A uniform spherical shell of matter attracts a particle that is outside the shell as if all the shell s mass were concentrated at its center. F = G m 1m 2 r 2 apple A uniform shell of matter exerts no net gravitational force on a particle located inside it. F = G 4πmρ 3 r

3 Gravitational Potential Energy U U(r)= W = GmM 0 1 R U(r)= GmM r Escape speed v escape = 2GM R E = KE KE = 1 2 U Mechanical Energy E mech = U + KE = GMm 2r KE = GMm 2r U = GMm r

4 Solar System Planets and Satellites: Kepler s Laws Kepler Brahe Newton ) Law of Orbits - all planets move in elliptical orbits with the Sun at one focus Ellipse criterion: Fm+F m=const Assumes M Sun >> M E a = semimajor axis e = eccentricity sun-earth farthest point R a = aphelion closest point R p = perihelion earth-moon = apogee = perigee R a = a(1+ e) R p = a(1 e)

5 Planets and Satellites: Kepler s 2 nd Law 2) Law of equal areas - A line that connects a planet to the Sun sweeps out equal areas in equal time. - Planets move slowest when furthest away from Sun (at R a ) - Planets move fastest when closest away from Sun (at R p ) ANGULAR MOMENTUM IS CONSERVED Angular momentum L = r p area of triangle da = 1 2 ( rdθ)r da dt = 1 2 r 2 dθ dt = 1 2 r 2 ω L = rp = r( mv) = r( m(ωr) ) = mr 2 ω L m = r 2 ω da dt = L 2m = constant

6 Planets and Satellites: Kepler s 3 rd Law 3) The Law of Periods : The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit Gm M sat E = m 2 sat r sat 2πr sat T r sat 2 3 r sat = GM 4π 2 T 2 = 4π 2 GM T 2 3 r sat r 3 T 2 Circular Orbit Elliptical Orbit When used in the equations, r (radius) and a (semimajor axis) are synonymous

7 Kepler s 3 rd Law: Applica:on A planet travels in an elliptical orbit about a star X as shown Q: The magnitude of the acceleration of the planet is (a) greatest at point U (b) greatest at point S (c) greatest at point W (d) the same at all points Q: At what pair of points is the speed of the planet the same? (a) P and R (b) W and S (c) Q and U (d) U and R Q: At what point is the speed of the planet minimum? (a) Point W (b) Point P (c) Point S (d) point Q

8 1) Law of Orbits - all planets move in elliptical orbits with the Sun at one focus 2) Law of equal areas - ANGULAR MOMENTUM IS CONSERVED A line that connects a planet to the Sun sweeps out equal areas in equal time. 3) The Law of Periods : The square of the period of any planet is proportional to the cube of the semimajor axis of its orbit 3 r sat = GM T 2 4π 2 T 2 = 4π 2 3 r sat GM

9 Chapter 14: Fluids In this chapter we will explore the behavior of fluids. In particular we will study the following: Static fluids: Pressure exerted by a static fluid Methods of measuring pressure Pascal s principle Archimedes principle, buoyancy

10 ΔV Δm Fluids As the name implies, a fluid is defined as a substance that can flow. Fluids conform to the boundaries of any container in which they are placed. A fluid cannot exert a force tangential to its surface. It can only exert a force perpendicular to its surface. Liquids and gases are classified together as fluids to contrast them with solids. In crystalline solids the constituent atoms are organized in a rigid three-dimensional regular array known as the "lattice." Density : Consider the fluid shown in the figure. It has a mass Δm and volume ΔV. The density (symbol ρ ) is defined as the ratio of the mass over the volume: ρ = Δm ΔV. SI unit: kg/m 3 If the fluid is homogeneous, the above equation has the form ρ = m V.

11 Density density =ρ = m V Note: density solid > liquid > gas air = 1.21 kg/m 3 wood = 550 kg/m 3 water = 1000 kg/m 3 Al = 2700 kg/m 3 = 2.7 g/cm 3 Cu = 8960 kg/m 3 = 8.9 g/cm 3 Vacuum What is pressure?

12 Pressure P = F A collisional force Units: pascal = Pa = N/m 2 1 atmosphere = 1 atm = Pa = 760 torr = 14.7 lb/in 2 The collisions of gas molecules on the wall of the tire keep it inflated Uniform force on flat area Hydrostatic Pressure: Water applies a force perpendicular to all of the surfaces in the pool, including the swimmer, the walls of the pool etc.

13 Fluids at Rest F = 0 Static Equilibrium F = 0 F bottom F top mg = 0 a = 0 p bottom A = p top A + mg p bottom = p top + mg A = p top + mg A h h p bottom = p top + gh m V p bottom = p top + ρgh Pressure depends on depth NOT horizontal dimensions p at h = p atm + ρgh p at h > p atm for h down p at h < p atm for h up

14 Example 1. What is the pressure head at the bottom of a 98 ft (30 m) water tower?

15 2. What is the net force on Grand Coulee dam (width 1200 m - height 150 m)? p at h = p atm + ρgh df = ρgy ( )da F = D 0 ( ρgy)w dy = 1 2 ρgwd2 = N F = lb

16 3. At what depth is the pressure two-times that of atmosphere? p at h = p atm + ρgh 2 p atm = p atm + ρgh h = p atm ρg = Pa ( 1000 kg / m 3 ) 9.8 m / s 2 = 10.3 m = 33.8 ft ( )

17 4. What is the maximum height you can suck water up a straw? p = p atm + ρgh 0 = p atm + ρgh h = p atm ρg = 10.3 m = 33.8 ft

18 Blood Pressure Blood pressure of 120/80 is considered normal what are these units? How much pressure is this? WHY? ρ Hg gh = (13,600 kg 3)(9.8 m m )(0.12m) s 2 120mmHg = Pa 80mmHg = Pa Difference = Pa Systolic Dystolic What is the pressure difference between your heart and your feet? (Density of blood is 1060 kg/m 3 ) P 2 P 1 = ρgh = (1060 kg 3)(9.8 m m )(1.35m) s 2 = Pa

19 Pressure vs height: gasses Remember: if h is down, pressure goes up; if h is up, pressure goes down Δp = ρgh What is the air pressure at 18,000 ft (5,500 m)? (elevation affects pressure- how?) Assume that the density of air is proportional to the pressure (compressible fluid) ρ h ρ 0 = p h ρ ρ h = p 0 h p 0, where at 0 º C & sea level ρ 0 = 1.29 kg/m 3 & p 0 =1 atm = Pa p 0 ρ dp = p 0 h gdy dp = ρ 0 p h gdy p0 dp p h p 0 p H = ρ 0 Negative because pressure is decreasing as you go up. p 0 p 0 Δp = ρgh dp = ρg dy H g dy 0 ln p H = ρ 0 g H 0 p 0 p H p 0 = p 0 e ρ 0 p 0 ( ) gh p 18,000 ft = 1 2 p atm

20 Measuring Pressure Torricelli ( ) Closed-end Manometer (Hg Barometer) 1mm of Mercury = 1 torr Open-end Manometer p at h = 0 + ρgh p at h = ρgh p at h = p atm + ρgh Δp = ρgh Absolute Pressure Gauge Pressure = p g = Δp h Hg = p atm ρg = N / m 2 13,550 kg / m 3 = 760 mm Hg ( )( 9.8 m / s 2 )

21 Question Is the gauge pressure at the bo.om of a 1 m high tube of water on the earth the same as is on the moon? 1. Yes 2. No 3. Some=mes ΔP moon = ρg earth h ΔP earth = ρg moon h g moon < g earth, so that ΔP moon < ΔP earth

22 Pascal s Principle Blaise Pascal ( ) Pressure applied to a confined fluid increases the pressure throughout by the same amount. What is the pressure 100m below sea level? Transmitted throughout the whole 100m Hydraulic Lever p 100m = p atm + ρgh p 100m = 1 atm atm = 10.7 atm Mechanical Advantage p out = p in F out A out = F in A in F out = F in A out A in Incompressible Fluid: V = d in A in = d out A out d out = d in A in Work: W out = F d out = out F in A out = F in d in = W in With a hydraulic lever, a given force applied over a given distance can be transformed to a greater force over a smaller distance A out A in d in A in A out

23 Examples of Pascal s Principle Piston B A person m=75kg stands on circular piston A (diameter=0.40m) of a hydraulic pump. If you want to lift an elephant weighing 1500kg, what is the minimum diameter of circular piston B? P = F A A A = F B A B π d B 2 A B = A A 2 F B F A = π d A 2 F B 2 F A d B = 2 d A 2 = 1.8m F B F A