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Chapter 8
Gravity
Newton’s law of gravitation
• Any two (or more) massive bodies attract each other
• Gravitational force (Newton's law of gravitation)

m1m2
F  G 2 rˆ
r
• Gravitational constant G = 6.67*10 –11 N*m2/kg2 =
6.67*10 –11 m3/(kg*s2) – universal constant
Chapter 8
Problem 15
Two identical lead spheres with their centers 14 cm apart attract each other
with a 0.25-µN force. Find their mass.
Gravitation and the superposition
principle
• For a group of interacting particles, the net
gravitational force on one of the particles is
n 

F1,net   F1i
i2
• For a particle interacting with a continuous
arrangement of masses (a massive finite object) the
sum is replaced with an integral

F1,body 

 dF
body
Shell theorem
• For a particle interacting with a uniform spherical
shell of matter

F1,shell 

 dF
shell
• Result of integration: a uniform spherical shell of
matter attracts a particle that is outside the shell as if
all the shell's mass were concentrated at its center
Shell theorem
• For a particle inside a uniform spherical shell of
matter

F1,shell 

 dF
shell
• Result of integration: a uniform spherical shell of
matter exerts no net gravitational force on a particle
located inside it
Gravity force near the surface of Earth
• Earth can be though of as a nest of shells, one
within another and each attracting a particle outside
the Earth’s surface
• Thus Earth behaves like a particle located at the
center of Earth with a mass equal to that of Earth

mEarthm1 ˆ
 GmEarth  ˆ
F1, Earth  G 2
j   2
m1 j  g m1 ˆj
REarth
 REarth 
g = 9.8 m/s2
• This formula is derived for stationary Earth of ideal
spherical shape and uniform density
Gravity force near the surface of Earth
In reality g is not a constant because:
Earth is rotating,
Earth is approximately an ellipsoid
with a non-uniform density
Gravitational field
• A gravitational field exists at every point in space
• When a particle is placed at a point where there is
gravitational field, the particle experiences a force
• The field exerts a force on the particle
• The gravitational field is defined as:

g

Fg
m
• The gravitational field is the gravitational force
experienced by a test particle placed at that point
divided by the mass of the test particle
Gravitational field
• The presence of the test particle is not necessary
for the field to exist
• The source particle creates the field
• The gravitational field vectors point in the direction
of the acceleration a particle would experience if
placed in that field
GmEarth
g
2
REarth
• The magnitude is that of the freefall acceleration at
that location
Gravitational potential energy
• Gravitation is a conservative force (work done by it
is path-independent)
• For conservative forces (Ch. 7):

rf
rf
 
 Gm1mEarth 
dr 
U    F  d r     
2

r

ri 
ri
1 1 
 Gm1mEarth  
r r 
f 
 i
Gravitational potential energy
1 1 
U  U f  U i  Gm1mEarth  
r r 
f 
 i
• To remove a particle from initial position to infinity
 1 1  Gm1mEarth
U   U i  Gm1mEarth   
ri
 ri  
• Assuming U∞
=0
Gm1mEarth
U i (ri )  
ri
Gm1m2
U (r )  
r
Gravitational potential energy
Gm1m2
U (r )  
r
Orbits
• Accounting for the shape of Earth, projectile motion
(Ch. 3) has to be modified:
2
v
ac 
 g  v  gR
R
Orbits
• The total mechanical energy E = K + U determines
the type of orbit an object follows:
• E < 0: The object is in a bound, elliptical orbit
Orbits
• The total mechanical energy E = K + U determines
the type of orbit an object follows:
• Special cases include circular orbits and the
straight-line paths of falling objects
Orbits
• The total mechanical energy E = K + U determines
the type of orbit an object follows:
• E > 0: The orbit is unbound and hyperbolic
Orbits
• The total mechanical energy E = K + U determines
the type of orbit an object follows:
• E = 0: The borderline case gives a parabolic orbit
Orbits
• Elliptical orbits of planets are described by a
semimajor axis a and an eccentricity e
• For most planets, the eccentricities are very small
(Earth's e is 0.00167)
Orbits
• The “parabolic” trajectories of projectiles near
Earth’s surface are actually sections of elliptical
orbits that intersect Earth
Orbits
• The trajectories are parabolic only in the
approximation that we can neglect Earth’s curvature
and the variation in gravity with distance from Earth’s
center
Orbits
• For a circular orbit and the Newton’s Second law
2

GMm
v 
 (m) 
F  ma
2
r
 r 
• Kinetic energy of a satellite
2
GMm
U
mv


K
2r
2
2
• Total mechanical energy of a satellite
GMm
GMm
GMm
E  K U 


 K
2r
r
2r
Orbits
• For an elliptic orbit it can be shown
GMm
E
2a
• Orbits with different e but the same a have the same
total mechanical energy
Chapter 8
Problem 40
A white dwarf is a collapsed star with roughly the Sun’s mass compressed into
the size of Earth. What would be (a) the orbital speed and (b) the orbital period
for a spaceship in orbit just above the surface of a white dwarf?
Escape speed
• Escape speed: speed required for a particle to
escape from the planet into infinity (and stop there)
Ki  U i  K f  U f
2
m1v Gm1m planet

 00
2
R planet
vescape 
2Gm planet
R planet
Escape speed
• If for some astronomical object
vescape 
2Gmobject
Robject
 3 10 m / s  c
8
• Nothing (even light) can escape from the surface of
this object – a black hole
Chapter 8
Problem 54
A projectile is launched vertically upward from a planet of mass M and radius
R; its initial speed is twice the escape speed. Derive an expression for its
speed as a function of the distance r from the planet’s center.
Kepler’s laws
Tycho Brahe/
Tyge Ottesen
Brahe de Knudstrup
(1546-1601)
Johannes Kepler
(1571-1630)
Three Kepler’s laws
• 1. The law of orbits: All planets move in elliptical
orbits, with the Sun at one focus
• 2. The law of areas: A line that connects the planet
to the Sun sweeps out equal areas in the plane of the
planet’s orbit in equal time intervals
• 3. The law of periods: The square of the period of
any planet is proportional to the cube of the
semimajor axis of its orbit
Third Kepler’s law
• For a circular orbit and the Newton’s Second law
F  ma
GMm mv

2
r
r
2
GM
v 
r
2
• From the definition of a period
2r
4 r
2
T
T 
2
v
v
2 2
4 3
T 
r
GM
2
2
• For elliptic orbits
4 3
T 
a
GM
2
2
Chapter 8
Problem 23
The Mars Reconnaissance Orbiter circles the red planet with a 112-min period.
What’s the spacecraft’s altitude?
Questions?
Answers to the even-numbered problems
Chapter 8
Problem 16
4.62 × 10−8 N
Answers to the even-numbered problems
Chapter 8
Problem 20
1.87 y
Answers to the even-numbered problems
Chapter 8
Problem 26
1.67 × 106 m