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

Isaac Newton

Newton & Physics

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

Universal Gravitation - Apple

Universal Gravitation - Moon

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 Universal Gravitation

The Moon’s Orbital Speed radius of orbit: r = 3.8 x 10 8 m Circumference: 2  r = ???? m orbital period: T = 27.3 days = ???? sec orbital speed: v = (2  r)/T = ??? m/sec

The Moon’s Orbital Speed radius of orbit: r = 3.8 x 10 8 m Circumference: 2  r = 2.4 x 10 9 m orbital period: T = 27.3 days = 2.4 x 10 6 sec orbital speed: v = (2  r)/T = 10 3 m/sec = 1 km/s

The Moon’s Centripetal Acceleration The centripetal acceleration of the moon: orbital speed: v = 10 3 m/s orbital radius: r = 3.8 x 10 8 m centripetal acceleration: A c = v 2 / r = ???? m/s 2

The Moon’s Centripetal Acceleration The centripetal acceleration of the moon: orbital speed: v = 10 3 m/s orbital radius: r = 3.8 x 10 8 m centripetal acceleration: A c = v 2 / r A c = (10 3 m/s) 2 / (3.8 x 10 8 m) = m/s 2

At the surface of Earth (r = radius of Earth) g = 9.8 m/s 2 At the orbit of the Moon (r = 60x radius of Earth) a = m/s 2 What’s relation between them? The Moon’s Centripetal Acceleration

At the surface of Earth ( r = radius of Earth ) g = 9.8 m/s 2 At the orbit of the Moon ( r = 60x radius of Earth ) a = m/s m/s 2 / m/s 2 = 3600 / 1 = 60 2 / 1 The Moon’s Centripetal Acceleration

r 2r 3r 4r 5r 6r 60r g g g g g g g Bottom Line The Moon’s Centripetal Acceleration

Bottom Line 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’s Centripetal Acceleration

Bottom Line The Moon’s Centripetal Acceleration

Bottom Line The acceleration due to gravity is inverse proportional to the square of the distance g ~ 1/r 2 The gravity is inverse proportional to the square of the distance Fg = mg Fg ~g Fg ~ 1/r 2 Gravity’s Inverse Square Law

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

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

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 meters), 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

Location Distance from Earth's center (m) Value of g (m/s 2 ) Earth's surface 6.38 x 10 6 m km above 7.38 x 10 6 m km above 8.38 x 10 6 m km above 9.38 x 10 6 m km above 1.04 x 10 7 m km above 1.14 x 10 7 m km above 1.24 x 10 7 m km above 1.34 x 10 7 m km above 1.44 x 10 7 m km above 1.54 x 10 7 m km above 1.64 x 10 7 m km above 5.64 x 10 7 m0.13 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 F g = G m 1 m 2 / d 2 F g : gravitational force between objects G:universal gravitational constant m 1 :mass of one object m 2 :mass of the other object d: distance between their centers of mass

Bottom Line Law of Universal Gravitation p.83 m1m1 m2m2 d FgFg FgFg

Bottom Line Law of Universal Gravitation F g = 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 Law of Universal Gravitation Example Mass 1Mass 2DistanceRelative Force m1m1 m2m2 dF 2m 1 m2m2 d m1m1 3m 2 d 2m 1 3m 2 d m1m1 m2m2 2d m1m1 m2m2 3d 2m 1 2m 2 2d

Law of Universal Gravitation Example Mass 1Mass 2DistanceRelative Force m1m1 m2m2 dF 2m 1 m2m2 d2F m1m1 3m 2 d3F 2m 1 3m 2 d6F m1m1 m2m2 2dF/4 m1m1 m2m2 3dF/9 2m 1 2m 2 2dF

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

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

Cavendish’s Experiment

G = F g d 2 / m 1 m 2 = 6.67 x N·m 2 /kg 2 F g = G m 1 m 2 / d 2 Universal Gravitational Constant

G = 6.67 x N·m 2 /kg 2 F g = G M m / r 2 The force (F g ) 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 10 6 m Calculate the Mass of Earth

G = 6.67 x N·m 2 /kg 2 F g = G M m / r N = 6.67 x N·m 2 /kg 2 x 1 kg x M / (6.4 x 10 6 m) 2 where M is the mass of Earth M = 6 x kg Calculate the Mass of Earth

Universal Gravitational Force

G = 6.67 x N·m 2 /kg 2 Gravity is is the weakest of the presently known four fundamental forces Universal Gravitational Force

ForceStrong Electro- magnetic WeakGravity Strength11/ x Range m ∞ m ∞

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 F g = G m 1 m 2 / d 2 F g = (6.67 x N·m 2 /kg 2 )(55 kg)(45 kg)/(1 m) 2 = 1.65 x N

Universal Gravitation Example Calculate the force of gravity between Earth (mass = 6.0 x kg) and the moon (mass = 7.4 x kg). The Earth-moon distance is 3.8 x 10 8 m

Universal Gravitation Example Calculate the force of gravity between Earth (mass = 6.0 x kg) and the moon (mass = 7.4 x kg). The Earth-moon distance is 3.8 x 10 8 m F g = G m 1 m 2 / d 2 F g = (6.67 x N·m 2 /kg 2 )(6.0 x kg) (7.4 x kg)/(3.8 x 10 8 m) 2 = 2.1 x N

Acceleration Due to Gravity Law of Universal Gravitation: F g = G m M / r 2 Weight F g = m g Acceleration due to gravity g = G M / r 2 F g : 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 kg, radius = 6.37 × 10 6 m )

Universal Gravitation Example Calculate the acceleration due to gravity of Earth (mass = 6.0 x kg, radius = 6.37 × 10 6 m ) g = G M / r 2 g = (6.67 x N·m 2 /kg 2 )(6.0 x kg)/(6.37 x 10 6 m) 2 = 9.86 m/s 2

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 10 4 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 10 4 kg, what is the acceleration due to gravity on asteroid B612? g = G M / r 2 g = (6.67 x N·m 2 /kg 2 )(1.00 x 10 4 kg)/(20.0 m) 2 = 1.67 x 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? F g = G m M / r 2 F g ~ M / r 2 F g = 1000 N x 95 / (9.4) 2 = 1075 N

Relative Weight on Each Planet

Isaac Newton’s Influence 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