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:

Equations:
o
o
FG 
Gm1m2
r2
g on Earth is 9.8 m/s2
Gm
r2
F1 r22

F2 r12
g
G  6.67 1011
N  m2
kg 2

Universal Gravitation Constant:

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
m1 and m2 who are separated by a distance r
(between their centres) is governed by the rule:
FG 
Gm1m2
r2
N  m2
kg 2
Where G is the universal gravitational constant:
G  6.67 1011
Note: There are two equal but opposite forces.
m1 pulls on m2 with a force equal in
magnitude to
m2 pulling on m1 .
The inverse square relationship between
FG and r means that the force of
attraction diminishes rapidly as the two objects move apart. But no matter
how large r gets, the FG never gets to zero.
Example
The mass
mass
m1 of one of the small spheres of a Cavendish balance is 0.01000 kg, the
m2 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
FG 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:
Gm1m2
r2
2

11 N  m 
6.67

10

  0.0100kg  0.500kg 
kg 2 


2
 0.0500m 
FG 
 1.33 1010 N
b)
The acceleration
a1 
FG 1.33 1010 N
m

 1.33 108 2
m1
0.0100kg
s
The acceleration
a2 
a1 of the smaller sphere has magnitude
a2 of the larger sphere has magnitude
FG 1.33 1010 N
m

 2.66 1010 2
m2
0.500kg
s
Example
What is a 70 kg persons weight
the local g?
1.234 107 m away from the Earth’s centre. What is
Solution:
mE  5.98 1024 kg , m2  70kg , r  1.234 107 m, G  6.67 1011
N  m2
kg 2
Method 1:
Gm1m2
r2
2

11 N  m 
6.67

10
5.98 1024 kg   70kg 

2 
kg


2
1.234 107 m 
FG 
 183N
You can calculate g two ways:
2

11 N  m 
6.67

10
5.98 1024 kg 

2 
kg 
Gm 
N
m
First g  2 
 2.62 or 2
7
r
kg s
1.234 10 m 
Second The g is
g
F 183N
N
m

 2.61 or 2
m 70kg
kg s
Method 2:
F1  mg  70kg  9.8
m
 686 N
s2
r1  rE  6.38 106 m
Using
F1 r22

F2 r12
7
686 N 1.234 10 m 

2
F2
 6.38 106 m 
2
F2  183 N
Therefore a 70kg persons weight
1.234 107 m from Earth is 183 N