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Chap. 13: Gravitation Newton’s Law of Universal Gravity – Attractive Force (Vector) – Inverse Square Law – Universality Kepler’s Laws Taken from Fig. 2.16, “the Universe in a Nutshell” by Stephen Hawking 1 Goals for Chapter 13 • To calculate the gravitational forces that bodies exert on each other • To relate weight to the gravitational force • To use the generalized expression for gravitational potential energy • To study the characteristics of circular orbits • To investigate the laws governing planetary motion • To look at the characteristics of black holes Introduction • What can we say about the motion of the particles that make up Saturn’s rings? • Why doesn’t the moon fall to earth, or the earth into the sun? • By studying gravitation and celestial mechanics, we will be able to answer these and other questions. Newton’s Law of Gravitation • Law of universal gravitation: Every particle of matter attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. • The gravitational force can be expressed mathematically as: m1m2 Fg G 2 r̂ r where G is the gravitational constant. Newton’s Law of Universal Gravitation Newoton's Law of Universal Gravitation 5 Newton’s Law of Universal Gravitation m1m2 F G 2 r̂ r Attractive Force m1 = MM r Magnitude of the force between any 2 masses 6.67x1011 Nm2/kg2 m2 = ME Newoton's Law of Universal Gravitation 6 Parameters • • • • • • • • G = 6.67 x 1011 Nm2/kg2 ME = 5.98 x 1024 kg RE = 6.38 x 103 km MS = 1.99 x 1030 kg RS = 6.96 x 105 km rE-S = 150 x 106 km Mmoon = 7.35 x 1022 kg Rmoon = 1.74 x 103 km Newoton's Law of Universal Gravitation 7 Spherically Symmetric Bodies • The gravitational force between bodies with spherically symmetric mass distributions is the same as if all their mass were concentrated at their centers. • See Figure 13.2 at the right. Problem 1: Find FME (Force on Moon due to Earth) m1m2 F G 2 r̂ r • • • • • • 3. Give the vector an appropriate label. G = 6.67 x 1011 Nm2/kg2 ME = 5.98 x 1024 kg RE = 6.38 x 103 km MS = 1.99 x 1030 kg Mmoon = 7.35 x 1022 kg Rmoon = 1.74 x 103 km Earth 1. Represent the Moon as a particle and Place the tail of the force vector on the Moon’s center. FME r r = 3.80 x 108 m Answer: FME = 2.03 x 1020 N Moon 2. Draw the vector force as an arrow pointing in the Earth’s center. Newoton's Law of Universal Gravitation 10 Gravitational Forces: Vector FMnet = FME + FMS MM ME MM ME FME = G rME2 Univeral Gravitational Constant vector MM MS FMS = G MS rMS2 Newoton's Law of Universal Gravitation 11 Problem 2: Find the Net Force on Sun. Assume that the masses and the coordinates are known. y Mars (x3, y3) m3 NOTE The gravitational force is: • … m1 (x2, 0) x m2 Sun Earth Newoton's Law of Universal Gravitation 12 More Example The starship Enterprise (mass m) is investigating a new star of mass M1, and is holding position a distance Y from the star. There is another star of mass M2 a distance Y from the first along a line perpendicular to that connecting the Enterprise with the first star. Suppose the Enterprise experiences total engine failure. Express the x and y components of the net gravitational force on the Enterprise. Use the coordinate system provided. Don’t forget that force is a vector. M2 y Y x M1 Starship Enterprise (not to scale) X Newoton's Law of Universal Gravitation 14 Satellite Orbital Period : T = (2pr)/v Try to relate between F and v. v FG Newoton's Law of Universal Gravitation 16 Satellite Orbital Newton’s Law of Universal GravitationPeriod : T = (2pr)/v Try to relate between F and v. m =M G FFGG == G m11 m m22 m 1 M rr 22 Gravity = Center-seeking Force Attractive Force r 6.67x1011 Nm2/kg2 mS , ME , r = 2rE , G v Magnitude of the force between any 2 masses m2 = ME Newoton's Law of Universal Gravitation F Dynamics G of FG Uniform Circular Motion Center-seeking Acceleration and Force Frad = m arad FG = mS v 2 / (2 rE) Newoton's Law of Universal Gravitation Circular Motion 17 Problem 3: Find T v FME Earth r Moon r = 3.80 x 108 m FME = 2.03 x 1020 N Newoton's Law of Universal Gravitation 18 Problem 4: Geosynchronous Satellite A geosynchronous satellite, which is one that stays above the same point on the Earth, is used for such purpose as cable TV transmission, for weather forecasting, and as communication relays. a) Determine the height above the Earth’s surface. b) Determine the speed of the satellite. Newoton's Law of Universal Gravitation 20 Gravitational Acceleration: Constant? FG = m g (Newton’s 2nd Law) m ∴ g = FG / m FG r where FG = G m M / r 2 = m (G M / r 2) M R Newoton's Law of Universal Gravitation 23 Gravity: Near the Earth’s Surface m g(h) = G M / r 2 = G M / (R+h) h 2 FG r M R Newoton's Law of Universal Gravitation 25 Gravity: Near the Earth’s Surface h = altitude above the surface MEarth g(h) = G ----------------(REarth + h) 2 e.g., h = 0 m g = 9.80 m/s2 Newoton's Law of Universal Gravitation 26 Gravity: Near the Earth’s Surface 9.9 9.80 9.77 9.8 m/s2 9.7 9.6 9.5 7.33 g 9.4 9.3 9.2 9.1 9 New York h=0m Mt. Everest h=1000km h = 8848 m Newoton's Law of Universal Gravitation 27 Gravity: Near the Moon’s Surface h = altitude above the surface MMoon g(h) = G ----------------(RMoon + h) 2 e.g., h = 0 m g = 1.62 m/s2 Newoton's Law of Universal Gravitation 28 Gravimeters 1. Accuracy: Dg/g ~ 1 part in 109 2. Applications: Mineral deposits r > rave g > gave Salt domes (i.e., petroleum) r < rave g < gave Newoton's Law of Universal Gravitation 29 Average Density of the Earth g = 9.80 m/s2 RE = 6.37 x 106 m ME = 5.96 x 1024 kg rE = 5.50 x 103 kg/m3 = 5.50 g/cm3 ~ 2 x rRock What can we conclude about the structure of the Earth? Newoton's Law of Universal Gravitation 30 Interior of the Earth • The earth is approximately spherically symmetric, but it is not uniform throughout its volume, as shown in Figure 13.9 at the right. • Follow Example 13.4, which shows how to calculate the weight of a robotic lander on Mars. International Space Station (1) www.station.nasa.gov Weight on the Earth 4.22 x 106 N Weight at h = 350 km? 5 kg M = 4.31 x 10 S Newoton's Law of Universal Gravitation 32 International Space Station (2) www.station.nasa.gov Weight on the Earth 4.22 x 106 N (MS = 4.31 x 105 kg) Weight at h = 350 km F = MS g(h) = MS G ME / (RE + h)2 Newoton's Law of Universal Gravitation 33 Weight • The weight of a body decreases with its distance from the earth’s center, as shown in Figure 13.8 below. 34 Apparent Weight (FN) F.B.D. y a = (1/2)g (up) Fnet = FN – mg = m a Newoton's Law of Universal Gravitation 35 Weightlessness F.B.D. ? y a = g (down) Newoton's Law of Universal Gravitation 36 Circular Satellite Orbits • For a circular orbit, the speed of a satellite is just right to keep its distance from the center of the earth constant. (See Figure 13.15 below.) • A satellite is constantly falling around the earth. Astronauts inside the satellite in orbit are in a state of apparent weightlessness because they are falling with the satellite. (See Figure 13.16 below.) • Follow Example 13.6. 37 Gravitational Potential Energy • Derivation of gravitational potential energy using Fig. 13.10 at the right. • The gravitational potential energy of a system consisting of a particle of mass m and the earth is U = –GmEm/r. 38 Gravitational Potential Energy • The gravitational potential energy of the earthastronaut system increases (becomes less negative) as the astronaut moves away from the earth, as shown in Fig. 13.11 at the right. 39 From the Earth to the Moon to To escape from the earth, an object must have the escape speed. See Example 13.5 using Figure 13.12 Ki U i K f U f K i U G ,i K f U G , f r RE r 1 2 M E m mvescape 00 2 RE 2 40 Chap. 13: Gravitation Newton’s Law of Universal Gravity – Attractive Force (Vector) – Inverse Square Law – Universality Kepler’s Laws Taken from Fig. 2.16, “the Universe in a Nutshell” by Stephen Hawking 43 Kepler’s Laws of Planetary Motion s = semi-major axis F1Q + F1R = 2 xNewoton's s Law of Universal Gravitation 44 Elliptical and Circular Orbits A special case of elliptical orbit is circular orbit: e (eccentricity) = 0 Newoton's Law of Universal Gravitation 48 Kepler’s 2nd Law (2) A12 = A34 v12 > v34 A d12 d v Dt d34 Newoton's Law of Universal Gravitation 49 Example 6 Use Kepler’s second law to show that the ratio of the speeds of a planet at its nearest and farthest points from the Sun is equal to the inverse ratio of the nearest and farthest distances: vN/vF = dF/dN Newoton's Law of Universal Gravitation 52 Kepler’s 3rd Law e = 0 (circular orbit) s = r T=2pr/v G MP MS / r 2 = MP v 2 / r Frad = m arad r 3 = (G MS /4 p 2) T 2 T 2 = (4 p 2 / G MS ) r 3 = KSun r 3 r (circular orbit) s (elliptical orbit) Newoton's Law of Universal Gravitation 53 Satellite about Earth: Circular Orbit T=2pr/v G MS ME / r 2 = MS v 2 / r Frad = m arad r 3 = (G ME /4 p 2) T 2 T 2 = (4 p 2 / G ME ) r 3 = KE r 3 Newoton's Law of Universal Gravitation 58 • • • • • • • • G = 6.67 x 1011 Nm2/kg2 ME = 5.98 x 1024 kg RE = 6.38 x 103 km MS = 1.99 x 1030 kg 2 RS = 6.96 x 105 km rE-S = 150 x 106 km s xp10a22ckge s h i p MmoonA = 7.35 Rmoon = 1.74 x 103 km Example 1 T = (4 p 2 / G ME ) r 3 = KE r 3 (mass m = 300 kg) is in an elliptical orbit around the Earth. At perigee (apogee) of its orbit, it is 620 km (3620 km) above P the Earth’s surface. Calculate the period of the orbit. Note: RE = 6380 km. Newoton's Law of Universal Gravitation (1) s = … (2) T 2 = KE s 3 T = … Newoton's Law of Universal Gravitation A 59 • • • • • • • • G = 6.67 x 1011 Nm2/kg2 ME = 5.98 x 1024 kg RE = 6.38 x 103 km MS = 1.99 x 1030 kg RS = 6.96 x 105 km rE-S = 150 x 106 km Mmoon = 7.35 x 1022 kg Rmoon = 1.74 x 103 km Example 2 T 2 = (4 p 2 / G ME ) r 3 = KE r 3 The National Aggie Space Administration (NASA) wants to send a satellite (300 kg) into a circular orbit of radius R from the center of the Earth at the rotational period of T = 240 hours. Find R and v. Newoton's Law of Universal Gravitation (1) T = … (2) T 2 = KE R 3 R = … (3) v = 2 p R / T = … Newoton's Law of Universal Gravitation 62 • • • • • • • • G = 6.67 x 1011 Nm2/kg2 ME = 5.98 x 1024 kg RE = 6.38 x 103 km MS = 1.99 x 1030 kg RS = 6.96 x 105 km rE-S = 150 x 106 km Mmoon = 7.35 x 1022 kg Rmoon = 1.74 x 103 km Example 3 T 2 = (4 p 2 / G ME ) r 3 = KE r 3 A geosynchronous satellite (m = 300 kg) is one that stays above the same point on the Earth, which is possible only if it is above a point of the equator. Determine the height H above the Earth’s surface such a satellite must orbit. Newoton's Law of Universal Gravitation (1) T = 1 day = … s (2) T 2 = KE R 3 R = … m (3) H = R – RE = … m (… km) Newoton's Law of Universal Gravitation 64 • • • • • • • • G = 6.67 x 1011 Nm2/kg2 ME = 5.98 x 1024 kg RE = 6.38 x 103 km MS = 1.99 x 1030 kg RS = 6.96 x 105 km rE-S = 150 x 106 km Mmoon = 7.35 x 1022 kg Rmoon = 1.74 x 103 km Example 4 T 2 = (4 p 2 / G MS ) r 3 = KS r 3 Halley’s comet orbits the Sun roughly once every 76 years. It comes very close to the surface of the Sun on its closest approach about how far out from the Sun is it at its farthest ? Newoton's Law of Universal Gravitation Newoton's Law of Universal Gravitation 66 x s s Newoton's Law of Universal Gravitation 67 • • • • • • • • G = 6.67 x 1011 Nm2/kg2 ME = 5.98 x 1024 kg RE = 6.38 x 103 km MS = 1.99 x 1030 kg RS = 6.96 x 105 km rE-S = 150 x 106 km Mmoon = 7.35 x 1022 kg Rmoon = 1.74 x 103 km Example 4 T 2 = (4 p 2 / G MS ) r 3 = KS r 3 Halley’s comet orbits the Sun roughly once every 76 years. It comes very close to the surface of the Sun on its closest approach about how far out from the Sun is it at its farthest ? Newoton's Law of Universal Gravitation (1) T = 76 yrs = … s (2) T 2 = KS s 3 s = … m (3) s >> RS x = … m Newoton's Law of Universal Gravitation 68 Example 5 When it orbited the Moon, the Apollo 11 spacecraft’s mass was 9.979 x 103 kg, its period was 7.140 x 103 s, and its mean distance from the Moon’s center was 1.849 x 106 m. Find v and MMoon. (1) v = 2 p R / T = … m/s (2) arad = v2 / R = … m/s2 (3) G m MMoon / R2 = m aradMMoon= … kg Newoton's Law of Universal Gravitation 70 Newoton's Law of Universal Gravitation 73 Appendix Gravitation Determining the Value of G • In 1798 Henry Cavendish made the first measurement of the value of G. Figure 13.4 below illustrates his method. Some gravitational calculations • Example 13.1 shows how to calculate the gravitational force between two masses. • Example 13.2 shows the acceleration due to gravitational force. • Example 13.3 illustrates the superposition of forces, meaning that gravitational forces combine vectorially. (See Figure 13.5 below.) Weight • The weight of a body is the total gravitational force exerted on it by all other bodies in the universe. • At the surface of the earth, we can neglect all other gravitational forces, so a body’s weight is w = GmEm/RE2. • The acceleration due to gravity at the earth’s surface is g = GmE/RE2. The motion of satellites • The trajectory of a projectile fired from A toward B depends on its initial speed. If it is fired fast enough, it goes into a closed elliptical orbit (trajectories 3, 4, and 5 in Figure 13.14 below). Kepler’s laws and planetary motion • Each planet moves in an elliptical orbit with the sun at one focus. • A line from the sun to a given planet sweeps out equal areas in equal times (see Figure 13.19 at the right). • The periods of the planets are proportional to the 3/2 powers of the major axis lengths of their orbits. Some orbital examples • Follow Conceptual Example 13.7 on orbital speeds. • Follow Example 13.8 involving Kepler’s third law. • Example 13.9 examines the orbit of Comet Halley. See Figure 13.20 below. Spherical mass distributions • Follow the proof that the gravitational interaction between two spherically symmetric mass distributions is the same as if each one were concentrated at its center. Use Figure 13.22 below. A point mass inside a spherical shell • If a point mass is inside a spherically symmetric shell, the potential energy of the system is constant. This means that the shell exerts no force on a point mass inside of it. • Follow Example 13.10 using Figure 13.24 below. Apparent weight and the earth’s rotation • The true weight of an object is equal to the earth’s gravitational attraction on it. • The apparent weight of an object, as measured by the spring scale in Figure 13.25 at the right, is less than the true weight due to the earth’s rotation. Black holes • If a spherical nonrotating body has radius less than the Schwarzschild radius, nothing can escape from it. Such a body is a black hole. (See Figure 13.26 below.) • The Schwarzschild radius is RS = 2GM/c2. • The event horizon is the surface of the sphere of radius RS surrounding a black hole. • Follow Example 13.11. Detecting black holes • We can detect black holes by looking for x rays emitted from their accretion disks. (See Figure 13.27 below.)