Unit 4.3 Universal Gravitation

Physics Tool Box

Newton’s Law of Gravitation: states that the force of gravitational

attraction between any two objects is directly proportional to the product of

their masses and inversely proportional to the square of the distance

between their centres.

Law: 1 2

2G

Gm mF

r

Equations:

o 2

Gmg

r

o

2

1 2

2

2 1

F r

F r

Universal Gravitation Constant:

211

26.67 10

N mG

kg

Gravitational Field Strength: then strength of the gravitational field at a

given point inspace, represented by the variable g in F mg

In Newton’s Principia (1687), he describes how he used data about the solar system

(moon) to discover the factors that affect the force of gravity.

Law of Universal Gravitation

The force between any two objects 1m and 2m who are separated by a distance r

(between their centres) is governed by the rule:

1 2

2G

Gm mF

r

Where G is the universal gravitational constant:

211

26.67 10

N mG

kg

Note: There are two equal but opposite forces. 1m pulls on 2m with a force equal in

magnitude to 2m pulling on 1m .

The inverse square relationship between GF and r means that the force of

attraction diminishes rapidly as the two objects move apart. But no matter

how large r gets, the GF never gets to zero.

g on Earth is 9.8 m/s2

Example

The mass 1m of one of the small spheres of a Cavendish balance is 0.01000 kg, the

mass 2m of one of the large spheres is 0.500 kg, and the centre to centre distance

between each large sphere and the nearer small one is 0.0500 m.

a) Find the gravitational force GF on each sphere due to the nearest other sphere

b) What is the magnitude of the acceleration of each, relative to an inertial system?

Solution:

1 2

2

211

2

2

10

6.67 10 0.0100 0.500

0.0500

1.33 10

G

Gm mF

r

N mkg kg

kg

m

N

b)

The acceleration 1a of the smaller sphere has magnitude

10

8

1 2

1

1.33 101.33 10

0.0100

GF N ma

m kg s

The acceleration 2a of the larger sphere has magnitude

10

10

2 2

2

1.33 102.66 10

0.500

GF N ma

m kg s

Example

What is a 70 kg persons weight 71.234 10 m away from the Earth’s centre. What is

the local g?

Solution:

2

24 7 11

2 25.98 10 , 70 , 1.234 10 , 6.67 10E

N mm kg m kg r m G

kg

Method 1:

1 2

2

211 24

2

27

6.67 10 5.98 10 70

1.234 10

183

G

Gm mF

r

N mkg kg

kg

m

N

You can calculate g two ways:

First

211 24

2

2 27

6.67 10 5.98 10

2.621.234 10

N mkg

kgGm N mg or

r kg sm

Second The g is 2

1832.61

70

F N N mg or

m kg kg s

Method 2:

1 270 9.8 686

mF mg kg N

s

6

1 6.38 10Er r m

Using

2

1 2

2

2 1

27

26

2

2

1.234 10686

6.38 10

183

F r

F r

mN

F m

F N

Therefore a 70kg persons weight 71.234 10 m from Earth is 183 N