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Transcript
UNIVERSAL GRAVITATION AND CIRCULAR MOTION Two distinct types of motion have been central to our understanding of the universe. Terrestrial motion - motion of objects on the Earth. Celestial motion - motion of objects in the heavens. Early theories of the universe As far back as the 6th and 7thcenturies B.C., Greek philosophers proposed that the heavens were composed of eight concentric, transparent spheres. Each sphere rotated on a different axis and at a different rate, but all were centered about the Earth. This geocentric, or Earth-centered, view of the universe persisted with minor modifications, for over 2000 years. Nicolaus Copernicus thought that the planetary position could be more easily explained if the sun was the center of the universe. Thus the heliocentric system was born. His theory predicted better the position of the planets than did the geocentric system. Tycho Brahe (1546-1601) plotted the paths of planets for more than 20 years to an accuracy of 1/1000th of a degree. Johann Kepler (1571-1630) was a mathematician. He came up with three laws: 1. The planets move about the sun in elliptical paths with the sun at one focus of the ellipse. 2. The straight line joining the sun and a given planet sweeps out equal areas in equal intervals of time. 3. The square of the period of revolution of a planet about the sun is proportional to the cube of its mean distance from the sun. R3/T2 = k = Kepler’s constant. Newton’s Law of Universal Gravitation Newton determined that all bodies (masses) anywhere in the universe attract each other. The force of attraction is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. Formula, Fg = GmM R2 Fg = Force of gravity (N) G = Universal Gravitational Constant = 6.67x10-11 N.m2/kg2 M and m = masses in kilograms. R = distance between the masses from center to center in m. Fg M R m Example Find the force between two masses of 50 kg and 60 kg separated by a distance of 50 cm. Fg= GmM R2 = (6.67x10-11)(50)(60) 0.502 = 8.0x10-7 N Example The force of attraction between two masses of 500 kg is 6.0x10-7 N. Find the distance between them. Fg=GmM R2 = GmM R2 F R2 = 6.67x10-11(500)(500) 6.0 x 10-7 R = 5.3 m Example A person whose mass is 70.0 kg stands on the surface of the Earth. (a) What does the person weigh using the law of universal gravitation? (b) Use another formula to calculate her weight. (a) F = GmM 11 (70)(5.98 x 1024) R2 x 106)2 F = 6.67x10(6.37 F = 688 N (b) F = W = mg = (70)(9.81) = 687 N F = 687 N Example The force between two equal masses is 6.5x10-3 N. The distance between the masses is 1.5 m. Find each mass. F = GmM R2 F = Gx2 14,808 kg R2 m=x M=x x2 = FR2 G Answer: 1.5x104 kg Assignment Text. p.172 #8 - 13 p.694 #4 - 9 x= Example What is the gravitational force on a 35.0 kg object standing on the Earth’s surface? Example The gravitational force between two objects is 3.25x10-6 N. The mass of one object is twice the mass of the second object. If the distance between the masses is 2.00 m, find the mass of each object. Ratio method of calculating the gravitational force m1 m2 R Fg Example The gravitational force between two small masses A and B when placed at a short distance apart is 3.24x10-7 N. What is the gravitational force between these objects if the masses of both A and B are doubled and distance between them is tripled? Example The force between two masses is 3.5x106 N. Find the force between them if the masses are tripled and the distance between them is halved. Example The force of gravity between two equal masses separated by a distance of 2.0 m is 5.0x10-7N. Find the masses. m=173kg Cavendish Experiment In 1798 Cavendish measured the value of G using a torsion balance. . Circular Motion v ac Fc Characteristics of circular motion: 1. Object is pulled toward center of circle by a force called CENTRIPETAL (centerseeking) FORCE 2. Acceleration is also directed to center and is called CENTRIPETAL ACCELERATION. 3. The speed of the object is constant 4. Velocity of the object is constantly changing. 5. Velocity is calculated from the tangent of the circle at any given instant Fc = mv2 (F = ma) R Where: (tension) in N ac = v2 R Fc = centripetal force ac = centripetal acceleration in m/s2 m = mass in kg R = radius of circle in m v = speed in m/s Example A 2.0 kg rubber stopper travels at 20 m/s in a circular path of radius 10 m. Find the centripetal acceleration and force. ac = v2 Fc = mac R = (2)(40) = (20)2 10 ac = 40 m/s2 Fc = 80 N The time for one revolution is called the PERIOD (T) You know that speed = distance/time Distance for a circle is the circumference = 2R v = 2R T Where: v = speed in m/s R = radius in m T = period in s T = Time for one rotation or revolution Example If a car travels in a circular path with a radius of 30 m and it takes 60 s, what is the speed of the car? R = 30 m v = 2R = 2(3.14)(30) T = 60 s T 60 = 3.14 = 3.1 m/s v=? Example An object with a mass of 2.0 kg travels in a circular path of radius 20 m in 15 s. Find the centripetal force. m = 2.0 kg 2π(20) = 8.4 m/s R = 20 m 15 Fc = mv2 R v = 2πR = T = (2)(8.4)2 20 T = 15 s Fc = ? Fc = 7.0 N Period (T) = time for one revolution FREQUENCY (f) = number of revolutions per second f=1 T T=1 f f = frequency r/s or Hz (Hertz) Example 1 A 10 kg object travels in a circular path at 30 rps. If the radius is 10 m find the centripetal force. m = 10 kg Fc = mv2 v = 2πR = 1,903 m/s R = 10 m f = 30 rps T = 1/30 = 0.033 s Fc = ? R T = (10)(1903)2 10 = 3.6 106 N Example What is the centripetal acceleration of a yo-yo being swung in a horizontal circle with a period of 0.20 s? The string is 10 cm long. T = 0.20 s ac = v2 v=? R = 10 cm = 0.1 m R ac = ? = 99 m/s2 Example An 8.0 kg object travels in a circular path at 1,800 rpm. If the radius of the path is 10 m, find a) the centripetal acceleration m = 8.0 kg f = 1,800 rpm = 1,800/60 = 30 rps T = 1/30 = 0.033 s R = 10 m ac = ? ac = v2 v= 2(3.14)(10) R 0.033 = 3.6 105 m/s2 b) the centripetal force Fc = mac = (8.0)(3.6 x 105) = 2.9 106 N p. 145 #9-11 p.151 – key terms p. 166 #5 p. 173 #20 - 24 Banking Curves Banking curves often occur so that vehicles do not need to rely only on friction to keep them on the road tan = v2 Rg where: = banking angle v = speed in m/s R = radius of circle in m g = acceleration due to gravity = 9.81 m/s2 Example 1 A car travels around a curve of radius 60.0 m at a speed of 22.0 m/s. At what angle must the curve be banked so the car doesn’t have to rely on friction to stay on the road? R = 60.0 m tan = v2 v = 22.0 m/s Rg g = 9.81 m/s2 = (22.0 m/s)2 (60.0 m)(9.81 m/s2) = 0.8222901801 = 39.43012526 = 39.4 we have looked at circular motion as horizontal motion now lets look at vertical circular motion…we need to incorporate gravity v = Rg where: v = speed in m/s Equation #20 R = radius in m g = 9.81 m/s2 Example 1 An amusement park ride spins in a vertical circle. If the diameter of the ride is 5.80 m, what minimum speed must the ride travel at the top? R = 5.80 m 2 v = Rg = 2.90 m = (2.90 m)(9.81 m/s2) g = 9.81 m/s2 = 5.3337604 m/s = 5.33 m/s Vertical Circular Motion now we can look at the forces exerted when an object is being swung in a vertical circle on a string orF rod: g Fc FT Fc FT Fg Fc = centripetal force (net force) and it always acts towards the centre FT = tension and it also always acts towards the centre Fg = force of gravity and it always acts towards the centre of the Earth Fc remains constant (as long as speed remains constant) and Fg remains constant FT changes depending on where the mass is in it’s circular path tension on the string/rod is greatest at the bottom Fc(net) = FT – Fg (vectors in opposite direction) FT = Fc + Fg tension on the string/rod is smallest at the top Fc(net) = FT + Fg (vectors in same direction) FT = Fc – Fg Example 1 A 3.0 kg mass moves on a string in a vertical circle of radius 2.5 m with a constant speed of 7.2 m/s. a) Where is the tension in the string the greatest? Calculate it. Tension is the greatest at the bottom. m = 3.0 kg FT = Fc + Fg r = 2.5 m FT = mv2 + mg v = 7.2 m/s g = 9.81 m/s2 (3.0 kg)(9.81 m/s2) r = (3.0 kg)(7.2 m/s)2 + 2.5 m = 91.638 N = 92 N b) Where is the tension in the string the smallest? Calculate it. Tension is smallest at the top. m = 3.0 kg FT = Fc – Fg r = 2.5 m FT = mv2 - mg v = 7.2 m/s r g = 9.81 m/s2 = (3.0 kg)(7.2 m/s)2 - (3.0 kg)(9.81 m/s2) 2.5 = 32.778 N = 33 N B. Gravitation in 1666, Newton used mathematical arguments to look at the force of attraction between two objects the falling apple made him realize that the earth was attracting the apple, and that this attraction probably extended far above the earth’s surface into space he recognized that the force acting on the apple was proportional to its mass and that, according to his own 3rd law, the apple must also be attracting the earth if the earth was involved in the attraction, then the force of attraction was also proportional to the earth’s mass m1 m2 r it all comes down to this: 1. all objects in the universe attract each other 2. the force of attraction (Fg) between 2 masses (m1, m2)is directly proportional (if one goes up the other goes up) to the product of the masses: Fg m1m2 3. the force of attraction between 2 masses is indirectly proportional (if one goes up the other goes down) to the square of the distance (r) between them: Fg 1 r2 combining these 2 proportionalities gives us: Fg m1m2 r2 now, whenever you change from to =, there is a constant involved the formula is: Fg = Gm1m2 r2 Equation #14 where: G = universal gravitation constant = 6.67 x 10-11 Nm2/kg2 m1m2 = mass 1 and mass 2 in kg Fg = gravitational force of attraction in N r = distance between centres in m Example 1 What is the force of attraction between a 20 kg mass and a 50 kg mass separated by a distance of 50 cm? m1 = 20 kg Fg = Gm1m2 m2 = 50 kg r2 r = 50 cm = (6.67x10-11 Nm2/kg2)(20 kg)(50 kg) = 0.50 m (0.50 m)2 G = 6.67x10-11 Nm2/kg2 = 2.668 x 10-7 N = 2.7 x 10-7 N Example 2 A person has a mass of 60 kg. a) what is his weight on the surface of the earth? m = 60 kg Fg = mg g = 9.81 m/s2 = (60 kg)(9.81 m/s2) = 588.6 N = 5.9 x 102 N b) what is the gravitational force of attraction between the person and the earth? m1 = 60 kg Fg = Gm1m2 m2 = 5.98x1024 kg r2 r = 6.37 x 106 m = (6.67x10-11 Nm2/kg2)(60 kg)(5.98 x 1024 kg) G = 6.67x10-11 Nm2/kg2 (6.37 x 106 m)2 = 589.7927146 N = 5.9 x 102 N *****NOTE: This is the same number as part a)!!!!!!!!!!!!!!!!!! Example 3 Two masses, m1 = 5.0 x 106 kg and m2 = 3.0 x 108 kg, exert a force of 5.0 x 10-7 N. Find the distance between them. m1 = 5.0 x 106 kg Fg = Gm1m2 m2 = 3.0 x 108 kg r2 Fg = 5.0 x 10-7 N r2 = Gm1m2 G = 6.67x10-11 Nm2/kg2 Fg = (6.67x10-11 Nm2/kg2)(5.0x106 kg)(3.0x108 kg) 5.0 x 10-7 N = 2.001 x 1011 N r = 447325.3849 N = 4.5 x 105 m Example 4 Two equal masses are attracted by a gravitational force of 3.8 x 10-8 N. If they are separated by 2.3 m, what is each mass? m1 = m2 Fg = Gm1m2 = Gm2 G = 6.67x10-11 Nm2/kg2 r2 r2 r = 2.3 m m2 = Fgr2 Fg = 3.8 x 10-8 N G = (3.8 x 10-8 N)(2.3 m)2 (6.67x10-11 Nm2/kg2) = 3013.793103 kg2 m = 54.89802459 kg = 55 kg Example 5 Two masses exert a force of 3.8 x 10-6 N on each other and they are 5.2 m apart. If one mass is 2 times the size of the other, find the mass of each object. Fg = 3.8 x 10-6 N Fg = Gm1m2 r = 5.2 m r2 m1 = m = G(m)(2m) m2 = 2m r2 G = 6.67x10-11 Nm2/kg2 = 2m2G r2 m2 = Fgr2 2G = (3.8 x 10-6 N)(5.2 m)2 2(6.67x10-11 Nm2/kg2) = 770254.8726 kg2 m1 = 877.6416538 kg = 8.8 x 102 kg m2 = 2(877.6416538 kg) = 1755.283308 kg = 1.8 x 103 kg Your Assignment: 1. pg 1 #1-10 in workbook 2. pg 172 #8-12 now we must look at the relationship between force, mass and distance if you change one, what happens to the others? F m1m2 r2 m1 X2 X2 X2 X3 X6 X8 m2 X2 X2 X2 X4 X4 r X2 X2 X3 X4 F 10 N 20 N 40 N 10 N 30 N 27 N 20 N X2 2.5 N you can use a chart like this for any formula let’s look at circular motion: Fc = mv2 r m X2 X2 X4 X2 v X2 X2 X3 r X2 - Fc 2.0 N 4.0 N 16 N 16 N 36 N Satellites Newton’s “thought experiment” Cannon on top of mountain – fire it horizontally Cannon is a projectile that falls 4.9 m every second If the mountain is high enough, the cannon will fall at the same rate that the earth curves away….orbit!!!!!! Mountain would have to be at least 150 km high to be above most of the atmosphere for no air resistance Satellites are objects that are projected into space They are at a constant height above the planet and are in uniform circular motion The force of gravity keeps the satellite in orbit around the earth (or any other planet) Fg = Fc lets look at how to calculate the speed and the period of an orbiting satellite: mP R Fg Fc satellite ms earth Speed: Fg = Fc GmsMP = msv2 R2 R GMP = v2 R solve for v v = GMP R this formula is used for satellites only. where: v = speed in m/s G = 6.67 x 10-11 Nm2/kg2 MP = mass of planet in kg R = distance between centre of planet and satellite (radius of planet + altitude of the satellite above the surface of the planet) in m If orbital radius is given, it is the distance from the centre of the planet to the satellite. Example ***A satellite is 100 km above the surface of the earth. What is the speed of the satellite? G = 6.67 x 10-11 Nm2/kg2 v = GMP MP = 5.98 x 1024 kg R R = 6.37 x 106 m + 1.0 x 105 m = 6.47 x 106 m v=? = (6.67 x 10-11)(5.98 x 1024) (6.47 x 106) v = 7.85 x 103 m/s Period: time taken to complete one revolution. some satellites are geosynchronous or geostationary, which means they have a period of 24 hours and they always maintain a specific longitude and latitude used for global communications and data collection and transmission eg. weather, minerals, agriculture, astronomy, land and ocean surveillance, solar radiation, the environment etc. other satellites are not geosynchronous and you can use a formula to calculate the period (time for one revolution) v = 2πR T T = 2πR v Example *****Find the period of a satellite which is 200 km above the surface of the earth. G = 6.67 x 10-11 Nm2/kg2 MP = 5.98 x 1024 kg R = 6.37 x 106 m + 2.0 x 105 m v2 = GM R v = 7791 m/s T = 2πR v T = 5.3 x 103 s D. Planetary Mechanics two different motions have been central to our understanding of the universe: 1. Terrestrial motion – motion of objects on earth 2. Celestial motion – motion of objects in space Geocentric view – people used to think that everything in the solar system revolved around the earth; proposed by Greeks Heliocentric view – sun-centred motion of planets; proposed by Copernicus (although Galileo was probably the first to propose this, he was thrown in jail) Tycho Brahe (1546 – 1601) – Danish astronomer who plotted the paths of planets for more than 20 years to an accuracy of 1/1000 of a degree Johannes Kepler (1571 – 1630) – German astronomer who was one of Brahe’s assistants he wanted to use a sun-centred system (instead of geocentric) to explain Brahe’s precise data he came up with 3 laws: 1. the planets move about the sun is elliptical paths, with the sun at one focus of the ellipse planet sun focus 2. the straight line joining the sun and a given planet sweeps out equal areas in equal time intervals planet faster speed closer to sun slower speed farther from sun 3. the square of the period of revolution of a planet about the sun is proportional to the cube of its mean distance from the sun K = R3 not on formula sheet but T2 comes from Equation #12 where: R = mean orbital radius in m T = orbital period in s K = Kepler’s constant = 3.315 x 1018 m3/s2 Example 1 An asteroid has a period of 8.1 x 107 s. What is its mean radius of orbit around the sun? T = 8.1 x 107 s K = R3 K = 3.315 x 1018 m3/s2 T2 R3 = KT2 = (3.315 x 1018 m3/s2)(8.1 x 107 s)2 = 2.1749715 x 1034 R = 2.791372894 x 1011 m = 2.8 x 1011 m Example 2 A planets mean distance from the sun is 2.0 x 1011 m. What is the orbital period? R = 2.0 x 1011 m K = R3 K = 3.315 x 1018 m3/s2 T2 T 2 = R3 K = (2.0 x 1011 m)3 (3.315 x 1018 m3/s2) = 2.413273002 x 1015 s2 T = 49125075.08 s = 4.9 x 107 s Gravitational Field Gravitational field surrounds a mass of any size as it is this field that provides the force of attraction on another body Knowing the weight (force gravity) and mass of an object, the gravitational field strength can be calculated: g = Fg m where: g = gravitational field in N/kg or m/s2 Fg = gravitational force or weight in N m = mass in kg Example 1 On a planet an object weighs 20 N. If the mass of the object is 4.5 kg, what is the gravitational field strength? Fg = 20 N g = Fg m = 4.5 kg m = 20 N 1 N = 1 kg m/s2 4.5 kg = 4.4 N/kg or m/s2 Example 2 What is the mass of a 40 N object experiencing a gravitational field of 13 N/kg? Fg = 40 N g = Fg g = 13 N/kg m 13 = 40 m 13m=40 m = 3.1 kg Gravitational field strength is simply the acceleration due to gravity at the location of the mass e.g. on earth, in orbit, on another planet etc. Strength varies inversely with the square of the distance from the center of a planet g = GM r2 where: g = gravitational field strength in N/kg or m/s2 G = 6.67 x 10-11 Nm2/kg2 M = mass of planet in kg r = distance from center of planet in m Example 1 Find the gravitational field strength on the surface of Mercury (rM = 2.43 x 106 m and M = 3.2 x 1023). G = 6.67 x 10-11 Nm2/kg2 r = 2.43 x 106 m M = 3.2 x 1023 kg 10-11)(3.2 x 1023) g = GM r2 g = (6.67 x (2.43 x 106)2 = 3.6 N/kg or m/s2 Example 2 Find the gravitational field strength experienced by astronauts if they are on the space shuttle 500 km above the earth’s surface. G = 6.67 x 10-11 Nm2/kg2 GM M = 5.98 x 1024 kg r2 r = 6.37 x 106 + 500,000 10-11)(5.98 x 1024) g= = (6.67 x (6.87 x 106)2 r = 6.87 x 106 m g = 8.45 N/kg or m/s2 m (electron) = 9.11 x 10-31 kg m (proton) = 1.67 x 10-27 kg F. Einstein’s Theory of Relativity use pages 169 – 170 of your text to make short point-form notes on Einstein’s Theory of Relativity Einstein proposed that gravity is not a force but an effect of space itself He believed that a mass changes the space around it and causes it to be curved acceleration takes place because other bodies move in a curved space (centripetal acceleration) masses set up a gravitational field around themselves which distorts the space around it Some Cool Stuff to Think About the gravitational field strength of a black hole is so strong that it bends light back into the hole…nothing can leave it black holes are formed when a star collapses in on itself when the radius is small enough and the collapsing star is dense enough, light can no longer escape the earth would have to collapse to a radius of 9 mm (0.009 m), without losing any mass, to become a black hole so… *****that’s 50 trillion times higher than the g we experience right now g = 5.0 x 1013 x 9.81 m/s2 = 4.905 x 1014 m/s2