Gravitational Field Strength
Gravity → Force of attraction between any 2 masses in the universe
Weight → Term used to describe the force of gravity that a celestial
body, (i.e. Earth) exerts on a mass.
 If we say that a person weighs 800N on Earth, we mean
that the Force of Gravity of Earth on the person is 800N
Weight = Force of Gravity on a Mass
Mass → Used to describe the amount of matter in an object
Measured in kilograms (kg)
The Earth is surrounded by a Gravitational Force Field
 Every mass feels a force pulling it towards the Earth
Gravitational Field Strength → The amount of force acting on
each kg of mass – measured in N/kg
NOTE *** GFS is not the same everywhere. It depends how close
the object is to the centre of the Earth
At its surface, the earth attracts a 50 kg girl with a certain force. At its
surface, the moon attracts the same 50 kg girl with about 1/6 the force.
*** Gravitational Field Strength = Gravitational Acceleration
g = 9.81 N/kg
or
g = 9.81 m/s2
Fg = m • g
Fg = Force (N)
m = Mass (kg)
g = GFS (N/kg)
Variations in Gravitational Field Strength
As the Earth rotates, it bulges at the equator.
 An object at the North Pole is approx. 21km closer to the centre of
the Earth than an object at the Equator.
 g varies 1) as distance to earth’s centre changes
2) on mountains & sea level
Distance from Earth’s Centre  = GFS 
Inverse Square Law → the force depends inversely on the
square of the distance from Earth’s centre
F1
r2
-or-
F1
F2
=
r22
r12
-or-
F2
F1
=
r12
r22
Example 1
At twice the distance from the centre, the force of Gravity would be:
(1/2)2 = 1/4 as much
The Law of Universal Gravitation
Newton – Everything attracts, & is attracted to, every other particle in
the universe. The strength of the attraction depends upon
the masses of the objects & the distance between them.
The equation Newton gave for this force:
Fg = Gravitational Attraction between any 2
Fg = Gm1m2
objects (N)
2
m1 = Mass of 1st object (kg)
r
m2 = Mass of 2nd object (kg)
r = Distance between centres of the objects (m)
G = Universal gravitational Constant = 6.67 x 10-11 N•m2
kg2
Observations:
1. We have Two Opposite but Equal Forces
i.e. Earth pulls on Moon –and- Moon pulls on Earth
2. d cannot be great enough to give a force = 0
* All objects exert a force on all objects within the universe
3. G is very small, force of Gravity is only significant for great masses.
i.e. It takes the Earth with a mass of 5.98 x 1024 kg to exert a force
of 9.8M of force on a 1kg mass.
Example 2
Mount Logan in the Yukon is 5959 m above sea level & is the highest
peak in Canada. Earth’s mass is 5.97 x 1024 kg & Earth’s equatorial
radius is 6.38 x 106 m. What would be the difference in the
magnitudes of a 65.0 kg person’s weight at the top of the mountain &
that person’s weight at sea level? Assume the Earth’s equatorial
radius is equal to the distance from Earth’s centre to sea level.
Given:
mP = 65.0 kg
h = 5959 m
mEarth = 5.97 x 1024 kg
rEarth = 6.38 x 106 m
Required:
ΔFg = Fsea - Fmount = ?
Fg (sea) = ?
Fg (mountain) = ?
Analysis:
Fg = Gm1m2 do 2 x, once for sea level and once for mountain
r2
Substitute:
Fg (sea) = Gm1m2 = (6.67 x 10-11 N•m2/kg2)(65.0 kg)(5.97 x 1024 kg)
r2
(6.38 x 106 m)2
= 636 N
Fg (mountain) = Gm1m2 = (6.67 x 10-11 N•m2/kg2)(65.0 kg)(5.97 x 1024 kg)
r2
(6.38 x 106 m + 5959 m)2
= 635 N
ΔFg = Fsea - Fmount = 636 N – 635 N
Solve:
= 1.00 N
Example 3
The force of gravity between 2 objects is 800 N. What will the force
be if: a) the mass of one object is doubled?
Given:
Required:
Fg = 800 N
Fg new = ?
if m1 = 2 x m1
Analysis:
Fg = Gm1m2
r2
Solve:
Substitute:
Fg new = G2m1m2 = 2(Gm1m2 ) = 2(Fg)
r2
r2
= 2(800 N)
= 1600 N
b) the mass of both objects is tripled?
Given:
Required:
Fg = 800 N
Fg new = ?
Analysis:
Fg = Gm1m2
r2
Solve:
if m1 = 3 x m1 & m2 = 3 x m2
Substitute:
Fg new = G3m13m2 = 9(Gm1m2 ) = 9(Fg)
r2
r2
= 9(800 N)
= 7200 N
c) the distance between the two objects is doubled?
Given:
Required:
Fg = 800 N
Fg new = ?
if r = 2 x r
Analysis:
Fg = Gm1m2
r2
Solve:
Substitute:
Fg new = Gm1m2 = (Gm1m2 ) = ¼(Fg)
(2r)2
4r2
= ¼(800 N)
= 200 N
d) the mass of one object is tripled & the other object is moved closer
so that it is only one quarter as far?
Given:
Required:
Fg = 800 N
Fg new = ?
if m1 = 3 x m1 & r = ¼ x r
Analysis:
Fg = Gm1m2
r2
Substitute:
Fg new = G3m1m2 = (G3m1m2 ) = 3 x 16(Fg)
1
(¼r)2
/16r2
= 48(800 N)
= 38,400 N
Solve:
Example 4
What is the force of gravity between the Earth and the Moon if the
Moon's mass is 0.013 times that of Earth?
Given:
Required:
24
mEarth = 5.98 x 10 kg
Fg = ?
24
22
mmoon = 0.013(5.98 x 10 kg) = 7.77 x 10 kg
rmoon’s orbit = 3.82 x 108 m
Analysis:
Substitute:
Fg = Gm1m2 = (6.67 x 10-11N•m2/kg2)(5.98 x 1024kg)( 7.77 x 1022kg)
(3.82 x 108 m)2
Solve:
= 2.1 x 1020 N
To find g on any planet we modify the universal
gravitation equation with Newton’s 2nd Law:
g = Gmplanet
r2
Example 5
Calculate the surface gravitational field strength of a white dwarf with
a radius of 7.0 x 106 m & a mass of 1.2 x 1030 kg.
Given:
Required:
Analysis:
30
mplanet = 1.2 x 10 kg
g=?
g = Gmplanet
6
rplanet = 7.0 x 10 m
r2
Substitute:
g = Gmplanet = (6.67 x 10-11 N•m2/kg2)( 1.2 x 1030 kg)
r2
(7.0 x 106 m)2
Solve:
= 1.6 x 106 N