Chapter 6 - Gravitation
• Newton’s Law of Gravitation
(1687)
• Kepler’s Laws
• Implications of Newton’s Law
of Gravitation
– Gravitation near the Earth’s
surface
– Superposition of forces
– Justification for Kepler’s second
and third law
• Gravitational Potential Energy
Newton’s law of (universal) gravitation
1
FG  2
r
Gm1m 2
FG 
r2
FG  m1m2
FG acts along the line
joining the two objects
Cavendish Experiment (1798)
2
N

m
11
G  6.67x10
kg 2
Vector Form of Newton’s Law of Gravitation
by
Gm1m 2
F12  
rˆ12
2
r21
on
F12
r̂12
m2
m1
r21
F12  force on mass m2 due to mass m1
r21  distance from mass m2 to mass m1
r̂12  unit vector oriented from mass m1 to mass m2
Superposition of Forces
n
F0  F10  F20  F30  ....   Fi0
i 1
m1
m2
m3
r10
r20
r30
F10
F20
m0
F30
Gm 0 m1
Gm0 m 2
Gm0 m3
F0  
rˆ10 
rˆ20 
rˆ30  ....
2
2
2
r10
r20
r30
n
 m1

m3
m2
mi
F0  Gm0  2 rˆ10  2 rˆ20  2 rˆ30  ....   Gm0  2 rˆi0
r20
r30
i 1 ri0
 r10

Problem 1
• Three masses are each at a vertex of an isosceles
right triangle as shown. Write an expression for
the force on mass three due to the other two.
m1
r
m2
r
m3
Gravity near the earth’s surface
Gm1m 2
FG 
2
r
Gm E m
mg 
2
r
Gm E
g 2
r
Kepler’s Laws
• The Law of Orbits
– All Planets move in elliptical
orbits, with the sun at one
focus.
• The Law of Areas
– A line that connects a planet
to the sun sweeps out equal
areas in equal times.
• The Law of Periods
– The square of the period of
any planet is proportional to
the cube of the semi major
axis of its orbit
T R
2
3
Kepler’s 2nd Law
• The Law of Areas
– A line that connects a
planet to the sun
sweeps out equal areas
in equal times.
1
dA  rvdt
2
v
L  I  mr 2  mrv  Constant
r
1L
dA 
dt
2m
Justification of Kepler’s third law
(for circular orbits around the sun)
v
v2
 FR  ma R  m r
aR
GMS m E
v2
 mE
2
r
r
2

 3
4

2
T 
r
 GMS 
Problem 2
• Verify Kepler’s third law for the earth revolving
around the sun.
• distance from sun to earth = 1.496 x 1011 m
• mass of sun = 1.99 x 1030 kg.
r
Ms
Problem 3
• What would be the height of a satellite with a
period of one day
• mass of earth = 5.98 x 1024 kg.
h
Re
r
me
Gravitational potential energy again
Gm E m
F
rˆ
2
r
2
W   F  dl
1
W  U
GM E m
U r  
r
Escape velocity
GM E m
1
2
mvesc 
0
2
rE
vesc
2GM E

rE