Lesson 12 Universal Gravitation Eleanor Roosevelt High School

Lesson 12 Universal Gravitation Eleanor Roosevelt High School Chin-Sung Lin

Isaac Newton 1643 - 1727

Newton & Physics

Universal Gravitation Newton’s Law of Universal Gravitation states that gravity is an attractive force acting between all pairs of massive objects. Gravity depends on: q Masses of the two objects q Distance between the objects

Universal Gravitation - Apple

Universal Gravitation - Moon

Universal Gravitation - Moon

Universal Gravitation Newton’s question: Can gravity be the force keeping the Moon in its v orbit? v v v Newton’s approximation: Moon is on a circular orbit Even if its orbit were perfectly circular, the Moon would still be accelerated

The Moon’s Orbital Speed radius of orbit: r = 3. 8 x 108 m Circumference: 2 pr = ? ? m orbital period: T = 27. 3 days = ? ? sec orbital speed: v = (2 pr)/T = ? ? ? m/sec

The Moon’s Orbital Speed radius of orbit: r = 3. 8 x 108 m Circumference: 2 pr = 2. 4 x 109 m orbital period: T = 27. 3 days = 2. 4 x 106 sec orbital speed: v = (2 pr)/T = 103 m/sec = 1 km/s

The Moon’s Centripetal Acceleration The centripetal acceleration of the moon: orbital speed: orbital radius: v = 103 m/s r = 3. 8 x 108 m centripetal acceleration: Ac = v 2 / r = ? ? m/s 2

The Moon’s Centripetal Acceleration The centripetal acceleration of the moon: orbital speed: orbital radius: v = 103 m/s r = 3. 8 x 108 m centripetal acceleration: Ac = v 2 / r Ac = (103 m/s)2 / (3. 8 x 108 m) = 0. 00272 m/s 2

The Moon’s Centripetal Acceleration At the surface of Earth (r = radius of Earth) a = 9. 8 m/s 2 At the orbit of the Moon (r = 60 x radius of Earth) a =0. 00272 m/s 2 What’s relation between them?

The Moon’s Centripetal Acceleration At the surface of Earth (r = radius of Earth) a = 9. 8 m/s 2 At the orbit of the Moon (r = 60 x radius of Earth) a =0. 00272 m/s 2 9. 8 m/s 2 / 0. 00272 m/s 2 = 3600 / 1 = 602 / 1

Line. Acceleration The Moon’s. Bottom Centripetal r 2 r 3 r 4 r 5 r 6 r 60 r g 1 g 4 g 9 g 16 g 25 g 3600

Line. Acceleration The Moon’s. Bottom Centripetal 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”)

Line. Acceleration The Moon’s. Bottom Centripetal 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 Law Gravity’s Inverse Square The acceleration due to gravity is inverse proportional to the square of the distance Ac ~ 2 1/r The gravity is inverse proportional to the square of the distance Fg = Fc = m Ac Fg ~ 1/r 2

Bottom Line Law Gravity’s Inverse Square Gravity is reduced as the inverse square of its distance from its source increased Fg ~ r Fg 1 2 1/r 2 r 3 r 4 r 5 r 6 r 60 r Fg 4 Fg 9 Fg 16 Fg 25 Fg 3600

Bottom Line Law Gravity’s Inverse Square Fg ~ 1/r 2

Bottom Line Law Gravity’s Inverse Square

Bottom Line Law Gravity’s Inverse Square 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 Law Gravity’s Inverse Square 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 Law Gravity’s Inverse Square Location Distance from Earth's center (m) Value of g (m/s 2) Earth's surface 6. 38 x 106 m 9. 8 6000 km above 1. 24 x 107 m 2. 60 1000 km above 7. 38 x 106 m 7. 33 7000 km above 1. 34 x 107 m 2. 23 2000 km above 8. 38 x 106 m 5. 68 8000 km above 1. 44 x 107 m 1. 93 3000 km above 9. 38 x 106 m 4. 53 9000 km above 1. 54 x 107 m 1. 69 4000 km above 1. 04 x 107 m 3. 70 10000 km above 1. 64 x 107 m 1. 49 5000 km above 1. 14 x 107 m 3. 08 50000 km above 5. 64 x 107 m 0. 13

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 m 1 m 2 / d 2 F g: G: m 1: m 2: d: gravitational force between objects universal gravitational constant mass of one object mass of the other object distance between their centers of mass

Bottom Line Law of Universal Gravitation d m 1 Fg Fg m 2 p. 83

Bottom Line Law of Universal Gravitation Fg = G m 1 m 2 / d 2 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

Bottom Line Example Law of Universal Gravitation Mass 1 Mass 2 Distance Relative Force m 1 m 2 d F 2 m 1 m 2 d m 1 3 m 2 d 2 m 1 3 m 2 d m 1 m 2 2 d m 1 m 2 3 d 2 m 1 2 m 2 2 d

Law of Universal Gravitation Example Mass 1 Mass 2 Distance Relative Force m 1 m 2 d F 2 m 1 m 2 d 2 F m 1 3 m 2 d 3 F 2 m 1 3 m 2 d 6 F m 1 m 2 2 d F/4 m 1 m 2 3 d F/9 2 m 1 2 m 2 2 d F

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 q Use Torsion balance (Metal thread, 6 -foot wooden rod and 2” diameter lead sphere) q Two 12”, 350 lb lead spheres q The reason why Cavendish measuring the G is to “Weight the Earth” q The measurement is accurate to 1% and his data was lasting for a century

Cavendish’s Experiment

Universal Gravitational Constant G = Fg d 2 / m 1 m 2 = 6. 67 x 10 -11 N·m 2/kg 2 Fg = G m 1 m 2 / d 2

Calculate the Mass of Earth G = 6. 67 x 10 -11 N·m 2/kg 2 Fg = G M m / r 2 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 106 m

Calculate the Mass of Earth G = 6. 67 x 10 -11 N·m 2/kg 2 Fg = G M m / r 2 9. 8 N = 6. 67 x 10 -11 N·m 2/kg 2 x 1 kg x M / (6. 4 x 106 m)2 where M is the mass of Earth M = 6 x 1024 kg

Universal Gravitational Force G = 6. 67 x 10 -11 N·m 2/kg 2 Fg = G m 1 m 2 / d 2 Gravitational force is a VERY WEAK FORCE

Universal Gravitational Force G = 6. 67 x 10 -11 N·m 2/kg 2 Gravity is is the weakest of the presently known four fundamental forces

Universal Gravitational Force Strong Electromagnetic Strength 1 1/137 10 -6 6 x 10 -39 Range 10 -15 m ∞ 10 -18 m ∞ Weak Gravity

Universal Gravitation Example Calculate the force of gravity between two students with mass 55 kg and 45 kg, 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 45 kg, and they are 1 meter away from each other Fg = G m 1 m 2 / d 2 Fg = (6. 67 x 10 -11 N·m 2/kg 2)(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 m 1 m 2 / d 2 Fg = (6. 67 x 10 -11 N·m 2/kg 2)(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 / r 2 Weight Fg = m g Acceleration due to gravity g = G M / r 2 Fg : G: M: m: r: g: gravitational force / weight univ. gravitational constant mass of Earth mass of the object radius of Earth 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 / r 2 g = (6. 67 x 10 -11 N·m 2/kg 2)(5. 98 x 1024 kg)/(6. 37 x 106 m)2 = 9. 83 m/s 2

Universal Gravitation Example In The Little Prince, the Prince visits a small asteroid called B 612. If asteroid B 612 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 B 612?

Universal Gravitation Example In The Little Prince, the Prince visits a small asteroid called B 612. If asteroid B 612 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 B 612? g = G M / r 2 g = (6. 67 x 10 -11 N·m 2/kg 2)(1. 00 x 104 kg)/(20. 0 m)2 = 1. 67 x 10 -9 m/s 2

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?

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? Fg = G m M / r 2 Fg ~ M / r 2 Fg = 1000 N x 95 / (9. 4)2 = 1075 N

Relative Weight on Each Planet

Isaac Newton’s Influence Defined the World People could uncover the workings of the physical universe Moons, planets, stars, and galaxies have such a beautifully simple rule to govern them Phenomena of the world might also be described by equally simple and universal laws

Summary • • • Isaac Newton Universal gravitation – Apple and Moon? Moon’s centripetal acceleration Gravity’s inverse square law Law of universal gravitation Universal gravitational constant – Henry Cavendish Calculate the mass of Earth Weak gravitational force Acceleration due to gravity Newton’s influence