Create Presentation
Download Presentation

Download Presentation
## Universal Gravitation

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Universal Gravitation**Chin-Sung Lin**Isaac Newton**1643 - 1727**Universal Gravitation**Newton’s Law of Universal Gravitation states that gravity is an attractive force acting between allpairs of massive objects. Gravity depends on: • Masses of the two objects • Distance between the objects**Universal Gravitation**Newton’s question: Can gravity be the force keeping the Moon in its orbit? Newton’s approximation: Moon is on a circular orbit Even if its orbit were perfectly circular, the Moon would still be accelerated v v v v**The Moon’s Orbital Speed**radius of orbit: r = 3.8 x 108 m Circumference: 2pr = ???? m orbital period: T = 27.3 days = ???? sec orbital speed: v = (2pr)/T = ??? m/sec**The Moon’s Orbital Speed**radius of orbit: r = 3.8 x 108 m Circumference: 2pr = 2.4 x 109 m orbital period: T = 27.3 days = 2.4 x 106 sec orbital speed: v = (2pr)/T = 103 m/sec = 1 km/s**The Moon’s Centripetal Acceleration**The centripetal acceleration of the moon: orbital speed: v = 103 m/s orbital radius: r = 3.8 x 108 m centripetal acceleration: Ac = v2 / r = ???? m/s2**The Moon’s Centripetal Acceleration**The centripetal acceleration of the moon: orbital speed: v = 103 m/s orbital radius: r = 3.8 x 108 m centripetal acceleration: Ac = v2 / r Ac = (103 m/s)2 / (3.8 x 108 m) = 0.00272 m/s2**The Moon’s Centripetal Acceleration**At the surface of Earth (r = radius of Earth) a = 9.8 m/s2 At the orbit of the Moon (r = 60x radius of Earth) a =0.00272 m/s2 What’s relation between them?**The Moon’s Centripetal Acceleration**At the surface of Earth (r = radius of Earth) a = 9.8 m/s2 At the orbit of the Moon (r = 60x radius of Earth) a =0.00272 m/s2 9.8 m/s2 / 0.00272 m/s2 = 3600 / 1 = 602 / 1**The Moon’s Centripetal Acceleration**Bottom Line 60r 5r 4r 2r 6r 3r r g g g g g g g 9 16 4 3600 25 1 36**Bottom Line**The Moon’s Centripetal Acceleration If the acceleration due to gravity is inverse proportional to the square of the distance, then it provides the right acceleration to keep the Moon on its orbit (“to keep it falling”)**Bottom Line**The Moon’s Centripetal Acceleration If the acceleration due to gravity is inverse proportional to the square of the distance, then it provides the right acceleration to keep the Moon on its orbit (“to keep it falling”) The moon is falling as the apple does !!! Triumph for Newton !!!**Bottom Line**Gravity’s Inverse Square Law The acceleration due to gravity is inverse proportional to the square of the distance Ac ~ 1/r2 The gravity is inverse proportional to the square of the distance Fg = Fc = m Ac Fg ~Ac Fg ~ 1/r2**Bottom Line**Gravity’s Inverse Square Law Gravity is reduced as the inverse square of its distance from its source increased Fg ~ 1/r2 60r 5r 4r 2r 6r 3r r Fg Fg Fg Fg Fg Fg Fg 9 16 4 3600 25 1 36**Bottom Line**Gravity’s Inverse Square Law Fg ~ 1/r2**Bottom Line**Gravity’s Inverse Square Law**Bottom Line**Gravity’s Inverse Square Law Gravity decreases with altitude, since greater altitude means greater distance from the Earth's centre If all other things being equal, on the top of Mount Everest (8,850 metres), weight decreases about 0.28%**Bottom Line**Gravity’s Inverse Square Law Astronauts in orbit are NOT weightless At an altitude of 400 km, a typical orbit of the Space Shuttle, gravity is still nearly 90% as strong as at the Earth's surface**Bottom Line**Gravity’s Inverse Square Law**Bottom Line**Law of Universal Gravitation Newton’s discovery Newton didn’t discover gravity. In stead, he discovered that the gravity is universal Everything pulls everything in a beautifully simple way that involves only mass and distance**Bottom Line**Law of Universal Gravitation • Universal gravitation formula • Fg = G m1 m2 / d2 • Fg: gravitational force between objects • G: universal gravitational constant • m1: mass of one object • m2: mass of the other object • d: distance between their centers of mass**Bottom Line**Law of Universal Gravitation d Fg Fg m2 m1 p.83**Bottom Line**Law of Universal Gravitation • Fg = G m1 m2 / d2 • Gravity is always there • Though the gravity decreases rapidly with the distance, it never drop to zero • The gravitational influence of every object, however small or far, is exerted through all space**Law of Universal Gravitation Example**Bottom Line**Universal Gravitational Constant**The Universal Gravitational Constant (G) was first measured by Henry Cavendish 150 years after Newton’s discovery of universal gravitation**Henry Cavendish**1731 - 1810**Universal Gravitational Constant**• Cavendish’s experiment • Use Torsion balance (Metal thread, 6-foot wooden rod and 2” diameter lead sphere) • Two 12”, 350 lb lead spheres • The reason why Cavendish measuring the G is to “Weight the Earth” • The measurement is accurate to 1% and his data was lasting for a century**Universal Gravitational Constant**G = Fg d2 / m1 m2 = 6.67 x 10-11 N·m2/kg2 Fg = G m1 m2 / d2**Calculate the Mass of Earth**G = 6.67 x 10-11 N·m2/kg2 Fg = G M m / r2 The force (Fg) that Earth exerts on a mass (m) of 1 kg at its surface is 9.8 newtons The distance between the 1-kg mass and the center of Earth is Earth’s radius (r), 6.4 x 106m**Calculate the Mass of Earth**G = 6.67 x 10-11 N·m2/kg2 Fg = G M m / r2 9.8 N = 6.67 x 10-11 N·m2/kg2x 1 kg x M / (6.4 x 106 m)2 where M is the mass of EarthM = 6 x 1024 kg**Universal Gravitational Force**G = 6.67 x 10-11 N·m2/kg2 Fg = G m1 m2 / d2 Gravitational force is a VERY WEAK FORCE**Universal Gravitational Force**G = 6.67 x 10-11 N·m2/kg2 Gravity is is the weakest of the presently known four fundamental forces**Universal Gravitation Example**• Calculate the force of gravity between two students with mass 55 kg and 45kg, and they are 1 meter away from each other**Universal Gravitation Example**• Calculate the force of gravity between two students with mass 55 kg and 45kg, and they are 1 meter away from each other • Fg = G m1 m2 / d2 • Fg = (6.67 x 10-11 N·m2/kg2)(55 kg)(45 kg)/(1 m)2 • = 1.65 x 10-7 N**Universal Gravitation Example**• Calculate the force of gravity between Earth (mass = 6.0 x 1024 kg) and the moon (mass = 7.4 x 1022 kg). The Earth-moon distance is 3.8 x 108 m**Universal Gravitation Example**• Calculate the force of gravity between Earth (mass = 6.0 x 1024 kg) and the moon (mass = 7.4 x 1022 kg). The Earth-moon distance is 3.8 x 108 m • Fg = G m1 m2 / d2 • Fg = (6.67 x 10-11 N·m2/kg2)(6.0 x 1024 kg) • (7.4 x 1022 kg)/(3.8 x 108 m)2 • = 2.1 x 1020 N**Acceleration Due to Gravity**• Law of Universal Gravitation: • Fg = G m M / r2 • Weight • Fg = m g • Acceleration due to gravity • g = G M / r2 • Fg: gravitational force / weight • G: univ. gravitational constant • M: mass of Earth • m: mass of the object • r: radius of Earth • g: acceleration due to gravity**Universal Gravitation Example**• Calculate the acceleration due to gravity of Earth (mass = 6.0 x 1024 kg, radius = 6.37 × 106 m )**Universal Gravitation Example**• Calculate the acceleration due to gravity of Earth (mass = 6.0 x 1024 kg, radius = 6.37 × 106 m ) • g = G M/ r2 • g = (6.67 x 10-11 N·m2/kg2)(5.98 x 1024 kg)/(6.37 x 106 m)2 • = 9.83 m/s2**Universal Gravitation Example**• In The Little Prince, the Prince visits a small asteroid called B612. If asteroid B612 has a radius of only 20.0 m and a mass of 1.00 x 104 kg, what is the acceleration due to gravity on asteroid B612?**Universal Gravitation Example**• In The Little Prince, the Prince visits a small asteroid called B612. If asteroid B612 has a radius of only 20.0 m and a mass of 1.00 x 104 kg, what is the acceleration due to gravity on asteroid B612? • g = G M/ r2 • g = (6.67 x 10-11 N·m2/kg2)(1.00 x 104 kg)/(20.0 m)2 • = 1.67 x 10-9 m/s2**Universal Gravitation Example**• The planet Saturn has a mass that is 95 times as massive as Earth and a radius that is 9.4 times Earth’s radius. If an object is 1000 N on the surface of Earth, what is the weight of the same object on the surface of Saturn?