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

Universal Gravitation

Eleanor Roosevelt High School

Chin-Sung Lin

Lesson 12

Isaac Newton

1643 - 1727

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

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 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

At the surface of Earth (r = radius of Earth)

g = 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)

g = 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

r 2r 3r 4r 5r 6r 60r

g g g g g g g1 4 9 16 25 36 3600

Bottom LineThe 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

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 !!!

The Moon’s Centripetal Acceleration

Bottom Line

The acceleration due to gravity is inverse proportional to the square of the distance

g ~ 1/r2

The gravity is inverse proportional to the square of the distance

Fg = mg Fg ~g Fg ~ 1/r2

Gravity’s Inverse Square Law

Bottom Line

Gravity is reduced as the inverse square of its distance from its source increased

Fg ~ 1/r2

Gravity’s Inverse Square Law

r 2r 3r 4r 5r 6r 60r

Fg Fg Fg Fg Fg Fg Fg

1 4 9 16 25 36 3600

Bottom LineGravity’s Inverse Square Law

Fg ~ 1/r2

Bottom LineGravity’s Inverse Square Law

Bottom LineGravity’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 LineGravity’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

LocationDistance fromEarth's center

(m)

Value of g(m/s2)

Earth's surface

6.38 x 106 m 9.8

1000 km above

7.38 x 106 m 7.33

2000 km above

8.38 x 106 m 5.68

3000 km above

9.38 x 106 m 4.53

4000 km above

1.04 x 107 m 3.70

5000 km above

1.14 x 107 m 3.08

6000 km above

1.24 x 107 m 2.60

7000 km above

1.34 x 107 m 2.23

8000 km above

1.44 x 107 m 1.93

9000 km above

1.54 x 107 m 1.69

10000 km above

1.64 x 107 m 1.49

50000 km above

5.64 x 107 m 0.13

Bottom LineGravity’s Inverse Square Law

Bottom LineLaw 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 LineLaw 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 LineLaw of Universal Gravitation

p.83

m1m2

d

Fg Fg

Bottom LineLaw 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

Bottom LineLaw of Universal Gravitation Example

Mass 1 Mass 2 Distance Relative Force

m1 m2 d F

2m1 m2 d

m1 3m2 d

2m1 3m2 d

m1 m2 2d

m1 m2 3d

2m1 2m2 2d

Law of Universal Gravitation Example

Mass 1 Mass 2 Distance Relative Force

m1 m2 d F

2m1 m2 d 2F

m1 3m2 d 3F

2m1 3m2 d 6F

m1 m2 2d F/4

m1 m2 3d F/9

2m1 2m2 2d 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

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 = Fg d2 / m1 m2 = 6.67 x 10-11 N·m2/kg2

Fg = G m1 m2 / d2

Universal Gravitational Constant

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 106 m

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/kg2 x 1 kg x M / (6.4 x 106 m)2 where M is the mass of Earth M = 6 x 1024 kg

Calculate the Mass of Earth

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 Gravitational Force

Universal Gravitational Force

Force Strong Electro-magnetic Weak Gravity

Strength 1 1/137 10-6 6x10-39

Range 10-15 m ∞ 10-18 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

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)(6.0 x 1024 kg)/(6.37 x 106 m)2

= 9.86 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?

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 / r2 Fg ~ M / r2

Fg = 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

Defined the World

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