LECTURE 30
GRAVITY
Instructor: Kazumi Tolich
Lecture 30
2
! 
Reading chapter 12-1 to 12-2
!  Newton’s
law of universal gravitation
!  Gravitational attraction of spherical bodies
!  Measuring the mass of Earth
Newton’s law of universal gravitation
3
! 
The force of gravity between any two point objects of mass m1
and m2 is attractive and of magnitude given by
m1m2
F12 = G 2
r
! 
G = 6.674 × 10-11 N·∙m2/kg2 is the universal gravitational
constant.
Newton’s law of universal gravitation: 2
4
! 
! 
The force of gravity is directed along the line connecting the
objects.
Each object experiences a force of the same magnitude, acting
in opposite direction.
F=G
! 
m1m2
r2
The force of gravity between two objects forms an actionreaction pair.
Inverse square dependence on distance
5
! 
The force of gravity decreases with an inverse
square dependence on distance, r.
m1m2
1
F=G 2 ∝ 2
r
r
Net gravitational force
6
! 
If a given mass is acted on by gravitational
interactions with a number of other masses, the net
force acting on it is the vector sum of each of the
forces.
!  Decompose
each of the forces in components.
!  Add the force components in each direction.
Attractive people?
7
! 
Newton’s law of gravity tells you that everything
pulls on everything else in the universe.
!  Do
you exert a gravitational pull on Andromeda
galaxy?
!  Does Andromeda galaxy exert a gravitational pull on
you?
!  How do these forces compare?
Example: 1
8
! 
In one hand you hold an apple of mass
ma = 0.11 kg, in the other hand an orange of mass
mo = 0.24 kg. The apple and orange are
separated by r = 0.85 m. What is the magnitude of
the force of gravity that
a) 
b) 
the orange exerts on the apple, and
the apple exerts on the orange?
Gravitational attraction of spherical bodies
9
! 
! 
Think about yourself standing on Earth. What
is the distance between you and Earth?
mm
F = G 12 2
r
For a finite-sized spherically symmetric object
exerting a force on a particle outside the
sphere, the object can be treated as a
particle with its entire mass located at the
center.
! 
If you are standing on the surface of
Earth, use the radius of Earth as r. The
force on you (weight) is directed towards
the center of Earth.
Gravitational and inertial mass
10
Gravitational masses of objects are responsible for
their gravitational attractions.
m1g m2g
F12 = G
r2
!  Inertial mass is the property of an object that
measures the object’s resistance to acceleration.
F = mi a
!  As far as we can tell, these two types of masses are
the same.
! 
Galileo’s experiment
11
! 
Galileo’s famous experiment
showed that gravitational mass and
inertial mass are equivalent.
!  Two
objects with different inertial
masses are dropped from the Leaning
Tower of Pisa.
!  They are accelerated at the same
rate due to the gravitational forces on
them.
mbg M E
F=G
= mbi a
2
RE
Clicker question: 1
Example: 2
13
! 
A spaceship is on a straight-line path between Earth
and the moon. At what distance from Earth is the net
gravitational force on the spaceship by Earth and
the moon zero?
Example: 3
14
! 
When the earth, moon, and sun form a right
triangle, with the moon located at the right angle as
shown, the moon is in its third-quarter phase. (The
earth is viewed here from above its North Pole.)
Find the magnitude and direction of the net force
acting on the sun. Give the direction relative to the
line connecting the sun and the moon.
Clicker question: 2
15
G and g
16
According to Newton’s law of gravity, the force on an
object near the surface of Earth is
mo M E
F=G
RE2
!  The gravitational force on an object (weight) near the
surface of Earth is
! 
F = mo g
mo M E
ME
F=G
= mo g → g = G 2
2
RE
RE
Variation of g on Earth
17
! 
g is largest in red areas and smallest in the dark
blue areas.
g on the moon
18
! 
Why can astronauts on the moon jump around as if
they are floating?
!  http://www.youtube.com/watch?v=HY-Vxj4YwSY
!  The
acceleration of gravity near the surface of the moon is
about 1/6th of that of Earth.
22
mM
7.35
×
10
kg
−11
2
2
2
gM = G 2 = 6.67 × 10 N ⋅ m /kg
=
1.62
m/s
2
6
rM
1.74 × 10 m
(
)
24
mE
5.97
×
10
kg
−11
2
2
2
gE = G 2 = 6.67 × 10 N ⋅ m /kg
=
9.81
m/s
2
6
rE
6.37 × 10 m
(
)
Example: 4
19
! 
The acceleration due to gravity on the moon’s
surface is known to be about one-sixth the
acceleration due to gravity on the earth. Given that
the radius of the moon is roughly one-quarter that
of the earth, find the mass of the moon in terms of
the mass of the earth.
g at various altitude
20
! 
g at various distance h above
Earth’s surface, or a distance r
away from Earth’s center:
r = RE + h.
ME
ME
g=G 2 =G
2
r
( RE + h )
Example: 5
21
! 
What is the acceleration due to Earth’s gravity at a
distance from the center of Earth equal to the
orbital radius of the moon?