Download Chapter 13

13.6 Gravitational
Potential Energy
Gravitational Potential
Energy


In Chapter 8: U = mgy (Particle-Earth).
Valid only when particle is near the
Earth’s surface.
Now with the introduction of the
Inverse Square Law (Fg  1/r2) we
must look for a general potential energy
U function without restrictions.
Gravitational Potential Energy, cont
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We will accept as a fact that:
The gravitational force is conservative (Sec 8.3)
The gravitational force is a central force
 It is directed along a radial line toward the center
 Its magnitude depends only on r
 A central force can be represented by
F  r  rˆ
Grav. Potential Energy–Work
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A particle moves from A to B
while acted on by a central
force F
The path is broken into a
series of radial segments and
arcs
Because the work done along
the arcs is zero (Fv), the
work done is independent of
the path and depends only on
rf and ri
Grav. Potential Energy–Work, 2

The work done by F along any radial segment
is
dW  F  dr  F ( r )dr


The work done by a force that is perpendicular
to the displacement is 0
r
Total work is
W   F ( r ) dr
f
ri

Therefore, the work is independent of the
path
Grav. Potential Energy–Work, final

As a particle moves from A to
B, its gravitational potential
energy changes by (Recall
Eqn 8.15)
rf
U  U f  U i    F ( r ) dr
ri

(13.11)
This is the general form, we
need to look at gravitational
force specifically
Grav. Potential Energy for the
Earth

Since: F(r) = - G(MEm)/r2  Eqn. (13.11) becomes
U f  U i  GM E m 
rf
ri

rf
 1 1
1
 1
dr  GM E m    GM E m  
2
r r 
r
 r  ri
i 
 f
(13.12)
Taking Ui = 0 at ri → ∞ we obtain the important
result:
GM E m
U (r)  
r
(13.13)
Gravitational Potential Energy
for the Earth, final
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Eqn. 13.13 applies to Earth-Particle
system, with r ≥ RE
(above the Earth’s surface).
Eqn. 13.13 is not valid for particle
inside the Earth r < RE
U is always negative (the force is
attractive and we took Ui = 0 at
ri → ∞
An external agent (you lifting a
book) must do + Work to increase
the height (separation between the
Earth’s surface and the book). This
work increases U  U becomes less
negative as r increases.
Gravitational Potential Energy,
General

For any two particles, the gravitational
potential energy function becomes
Gm1m2 (13.14)
U 
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The gravitational potential energy between any two
particles varies as 1/r
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r
Remember the force varies as 1/r 2
The potential energy is negative because the force is
attractive and we chose the potential energy to be
zero at infinite separation
Systems with Three or More
Particles
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The total gravitational
potential energy of the
system is the sum over all
pairs of particles
Gravitational potential
energy obeys the
superposition principle

Each pair of particles
contributes a term of U
Systems with Three or More
Particles, final
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Assuming three particles:
U total  U12  U13  U 23
 m1m2 m1m3 m2m3 
 G 



r
r
r
13
23 
 12

(13.15)
The absolute value of Utotal represents the
work needed to separate the particles by
an infinite distance
Binding Energy
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The absolute value of the potential
energy can be thought of as the binding
energy

If an external agent applies a force
larger than the binding energy, the
excess energy will be in the form of
kinetic energy of the particles when
they are at infinite separation
Example 13.8
The Change in Potential Energy
(Example 13.6 Text Book)
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A particle of mass m is displaced through a small vertical
displacement y near the Earth’s surface. Show that the
general expression for the change in gravitational potential
energy (Equation 13.12) reduces to the familiar relationship:
U = mgy
From Equation 13.12
 1 1
 ri  r f


U  U f  U i  GM E m

 GM E m
r

 rr
ri 
 f
 i f
 r f  ri 

U  GM E m
 rr 
 i f 




Example 13.8
The Change in Potential Energy, cont
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Since ri and rf are to close to the Earth’s surface,
then:
rf – ri ≈ y and rfri ≈ RE2
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Remember r is measured from center of the Earth
 rf  ri 
 y 
 GM E 




U  GM E m
 GM E m 2   m 2 y  mgy 
 rr 
 RE 
 RE 
 i f 
U = mgy
13.7 Energy Considerations in
Planetary and Satellite Motion
An object of mass m is moving
with speed v near a massive
object of mass M (M >> m).
 The Total Mechanical Energy
E, when the objects are separated
a distance r, will be:
E = K +U = ½ m v2 – GMm/r
(13.16)
 Depending on the value of v,
E could be +, - , 0

m
v
r
M
Energy and Satellite Motion, cont
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Newton’s 2nd Law for mass m:
GMm/r2 = m aR = mv2/r
Multiplying by r/2 both sides:
GMm/2r = mv2/2 
½mv2 = ½(GMm/r) 
K = ½(GMm/r) (13.17)
Kinetic Energy (K) is positive
and equal to half the absolute
value of the potential Energy
(U)!!!!
m
v
r
M
Energy in a Circular Orbit
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Substituting (13.17) into (13.16):
E = ½ (GMm/r) – GMm/r 
GMm
E
2r
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(13.18)
The total mechanical energy is
negative in the case of a circular orbit
The kinetic energy is positive and is
equal to half the absolute value of the
potential energy
The absolute value of E is equal to the
binding energy of the system
Energy in an Elliptical Orbit
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For an elliptical orbit, the radius is replaced by
the semimajor axis a
GMm
E
2a
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(13.19)
The total mechanical energy is negative
The total energy is constant if the system
is isolated
Summary of Two Particle
Bound System
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Conservation of Total Energy will be
written as:
1 2 GMm 1 2 GMm
E  mvi 
 mv f 
2
ri
2
rf
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(13.20)
Both the total energy and the total angular
momentum of a gravitationally bound,
two-object system are constants of the
motion
Escape Speed from Earth
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An object of mass m is projected
vertically upward from Earth’s
surface, with initial speed vi.
Let’s find the minimum value of vi
needed by the object to move
infinitely far away from Earth.
The energy of the system at
any point is:
E = ½ mv2 – GMEm/r
Escape Speed from Earth, 2
At Earth’s surface:
v = vi & r = ri = RE
 At Maximum height:
v = vf = 0 & r = rf = rmax
 Using conservation of the Energy (Eqn.13.20):
½ mvi2 – GMEm/ri = ½ mvf2 – GMEm/rf 
½ mvi2 – GMEm/RE = 0 – GMEm/rmax 
2
 Solving for vi :
vi2 = 2GME(1/RE – 1/rmax)
(13.21)
 Knowing vi we can calculate the maximum
altitude h: h = rmax – RE
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Escape Speed from Earth, 3
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Let’s calculate Escape Speed (vesc )
Minimum speed for the object must have
at the Earth’s surface in order to approach
an infinite separation distance from Earth.
Traveling at this minimum speed the
object continues to move farther away, such
that we take:
rmax → ∞ or 1/rmax → 0 
Equation (13.21) becomes:
vi2 = vesc2 = 2GME(1/RE – 0) 
vesc
2GM E

RE
(13.22)
Escape Speed From Earth, final
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vesc is independent of the mass of
the object (Question 8 Homework)
vesc is the same for a spacecraft or a
molecule.
The result is independent of the
direction of the velocity and ignores air
resistance
Escape Speed, General
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The Earth’s result can be
extended to any planet of
mass M
vesc

2GM

R
(13.22a)
The table at right gives
some escape speeds from
various objects
Escape Speed, Implications
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Complete escape from an object is not
really possible
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The gravitational field is infinite and so some
gravitational force will always be felt no matter
how far away you can get
This explains why some planets have
atmospheres and others do not
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Lighter molecules have higher average
speeds and are more likely to reach escape
speeds
Example 13.9
Escape Speed of a Rocket
(Example 13.8 Text Book)
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Calculate vesc from the Earth for a 5,000 kg
spacecraft.
vesc = (2GME/RE)½ = 1.12 x 104 m/s
vesc = 11.2 km/s ≈ 40,000 km/h
Find the kinetic energy K it must have at
Earth’s surface in order to move infinitely far
away from the Earth
K = ½ mvesc2
K = ½(5,000 kg)(1.12 x 104 m/s)2 
K = 3.14 x 1011 J ≈ 2,300 gal of Gasoline
Material for the Midterm
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Examples to Read!!!
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Example 13.7
(page 407)
Homework to be solved in Class!!!
Question: 16
 Problem: 27
 Problem: 44
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Satellites & “Weightlessness”
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What keeps a satellite in
orbit?
Centripetal Acceleration
Due to: Gravitational Force!
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Fg = mac 
Fg = (mv2)/r = G(mME)/r2
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Its high speed or
centripetal acceleration
keeps it in orbit!!
Satellites & “Weightlessness”, 2
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If the satellite stops it will fall directly
back to Earth
But at the very high speed it has, it will
quick fly out into space
In absence of gravitational force
(Fg), the satellite would move in a
straight line!
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Newton’s 1st Law
Due to Fg of the Earth the satellite is
pulling into orbit
In fact, a satellite is falling
(accelerating toward Earth), but its
high tangential speed keeps it from
hitting Earth
Satellites & “Weightlessness”, final
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Objects in a satellite, including people,
experience “EFFECTIVE weightlessness”.
What causes this?
NOTE: THIS DOES NOT MEAN THAT
THE GRAVITATIONAL FORCE ON THEM
IS ZERO!
To understand this, first, look at a simpler
problem: A person in an elevator
“Weightlessness” Case 1
Person in elevator. Mass m attached to
scale. Elevator at rest (a = 0) 
∑F = ma = 0
 W (weight) = upward force on mass m
 By Newton’s 3rd Law:
 W is equal & opposite to the reading
on the scale.
∑F = 0 = W – mg 
Scale reading is W = mg
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“Weightlessness” Case 2
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Elevator moves up with acceleration a
∑F = ma or W – mg = ma
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W (weight) = upward force on mass m
W is equal & opposite to the reading on
the scale.
Scale reading (apparent weight)
W = mg + ma > mg
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For a = ½g W = 3/2mg
Person is “heavier”
Person experiences: 1.5 g’s!
“Weightlessness” Case 3
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Elevator moves down with acceleration a
∑F = ma or W – mg = – ma
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½mg
W (weight) = upward force on mass m
W is equal & opposite to the reading on
the scale.
Scale reading (apparent weight)
W = mg – ma < mg

For a = ½g W = 1/2mg
Person is “lighter”
Person experiences: 0.5 g’s!

“Weightlessness” Case 4
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Elevator cable breaks! Free falls down
with acceleration a = g
∑F = ma or W – mg = – ma
W (weight) = upward force on mass m
W is equal & opposite to the reading
on the scale.
Scale reading (apparent weight)
W = mg – ma = mg – mg
W = 0!
Elevator & person are apparently
“weightless”.
Clearly, though: THE FORCE OF
GRAVITY STILL ACTS!
“Weightlessness” in a satellite
Apparent “Weightlessness” experienced by
people in a satellite orbit?
 Similar to the elevator case 4
 With gravity, satellite (mass m) free “falls”
continually due to
F = G(mME)/r2 = mv2/r
 Consider scale in satellite:
∑F = ma = – mv2/r (towards Earth center!)
 W – mg = – mv2/r 
W = mg – mv2/r << mg
 Objects in the satellite experience an apparent
weightlessness because they, and the
satellite, are accelerating as in free fall

Reporters are Wrong
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TV reporters are just plain WRONG when they say
things like:
“The space shuttle has escaped the Earth’s
gravity”
“The space shuttle has reached a point outside
the Earth’s gravitational pull”
Why? Because the gravitational force F = G(mME)/r2
exists (& is NOT zero) even for HUGE distances r
away from Earth
F  0 only for r  
Black Holes
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A black hole is the remains of a star that
has collapsed under its own gravitational
force
The escape speed for a black hole is very
large due to the concentration of a large
mass into a sphere of very small radius

If the escape speed exceeds the speed of light,
radiation cannot escape and it appears black
Black Holes, cont
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The critical radius at which
the escape speed equals c
is called the
Schwarzschild radius, RS
The imaginary surface of a
sphere with this radius is
called the event horizon

This is the limit of how close
you can approach the black
hole and still escape
Black Holes and Accretion Disks
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Although light from a black hole
cannot escape, light from events
taking place near the black hole
should be visible
If a binary star system has a black
hole and a normal star, the material
from the normal star can be pulled
into the black hole
This material forms an accretion
disk around the black hole
Friction among the particles in the
disk transforms mechanical energy
into internal energy
Black Holes and Accretion
Disks, cont
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The orbital height of the material above
the event horizon decreases and the
temperature rises
The high-temperature material emits
radiation, extending well into the x-ray
region
These x-rays are characteristics of black
holes
Black Holes at Centers of
Galaxies
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There is evidence that
supermassive black holes
exist at the centers of
galaxies
Theory predicts jets of
materials should be
evident along the
rotational axis of the black
hole
An HST image of the galaxy
M87. The jet of material in the
right frame is thought to be
evidence of a supermassive
black hole at the galaxy’s
center.
