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Test Breakdown
A
AB+
B
B-
>= 93% (63)
>= 90% (60)
>= 87% (57)
>= 83% (54)
>= 80% (52)
C+
C
D
F
>= 77% (51)
>= 70% (45)
>= 60% (38)
< 60% (37 or less)
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Average: 63.2% + 10%
Std Dev: 21.5%
Scores will be
adjusted up by 10%
1
Chapter 14
Gravitation
Gravity
One of the fundamental forces of Nature
Not just the reason things fall….
Why the Earth is round
Why the moon goes around the earth
Why the earth goes around the sun
Why there are ocean tides
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3
Gravitational Force
m1
Fg
Fg
m2
r
m1m2
Fg  G 2
r
Fg is an attractive force between any two masses
Fg is not a constant unless you have a small object
near the surface of a big sphere
G = 6.672 x 10-11 N m2/kg2
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4
Multiple Objects
Obeys principle of superposition:
M1
M3
r1
r3
m
r2
M 1m
Fg  G 2
r
M 2m
G 2
r
M 3m
 G 2  ...
r
M2
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Extended Objects
1
Fg
Fg
2
Each bit of #1 attracts each bit of #2
Need to integrate over whole object to get Fg


Fg   dF
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Extended Objects: Special Case
Can treat uniform spherical shells (and thus
spheres) like point masses located at geometric
center
No gravitational force inside uniform spherical
shell (it integrates to zero)
Fg = 0
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7
Gravitational Force Examples
m1m2
F G 2
r
Force between earth and sun
2
30
24


m
(
1
.
99

10
kg
)(
5
.
98

10
kg)
11
F   6.672 10 N 2 
11
2
kg
(
1
.
49

10
m
)


 23.5 10 22 N
Force between two people (assumed spherical)
2

 (60 kg)(60 kg)
m
11
7
F   6.672 10 N 2 

2
.
4

10
N
2
kg 
(1 m)

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Example: (Problem 14.16)
a)What will an object weigh on the Moon’s
surface if it weighs 100 N on Earth’s surface?
b)How many Earth radii must this same object be
from the center of Earth if it is to weigh the
same as it does on the Moon?
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Example:
Find the mass of the object:
m
Fg , Earth
g Earth
100 N

 10.2 kg
2
9.8 m / s
On the Moon:
GM Moonm
Fg 
2
rMoon

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(6.672 10
11 Nm 2
kg 2
)(7.35 10 22 kg)(10.2 kg)
(1.74 106 m) 2
Fg  16.5 N
10
Example:
Distance from the earth with the same weight:
GM Earthm
Fg 
r2
GM Earthm
r
Fg
r
(6.672  10
11 Nm 2
kg 2
)(5.97  10 24 kg)(10.2 kg)
16.5 N
 1 REarth 
  2.5 REarth
r  1.57  10 m 
6
 6.38  10 m 
7
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Objects Near Earth’s Surface
Re
mME
Fg  G
 mag
2
RE
ME
ag  G 2
RE
As long as the distance above earth’s surface is
small compared to RE, the force is approximately
constant
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Variation of Gravitational Force on
Earth’s Surface
1)Earth is not uniformly dense
Variations in crust from region to region
2)Earth is not a sphere
Bulge at the equator
3)Apparent change from earth’s rotation
In this case, a g  g
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Variation of g from Rotation
At the earth’s pole, there is no
centripetal acceleration:
N
w
ac=0
Fg
Fnet  mac
N  Fg  0
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 Me 
N  m G 2   mg
 Re 
At the pole, g=ag
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Variation of g from Rotation
At the equator:
ac
Fg
N
w
Fnet  mac
N  Fg  mw 2 R
At the equator, g < ag!
 Me

2
N  m G 2  w R   mg
 Re

Effectively lower gravity
g = 9.801 m/s2 in Pittsburgh
g = 9.786 m/s2 in Jamaica!!
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Gravitation Inside a Sphere
Recall: No gravitational force exerted on an object
inside a spherical shell
As you travel further into a
sphere, the layers above can
be thought of as many
spherical shells, which exert
no gravitational force!
Only the mass of the sphere below you matters
for calculating Fg!
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“Inner” Gravity
Only the mass of the
inward sphere
contributes to Fg
Movie claims that earth’s core is a trillion trillion tons…
1012  1012 tons = 1027 kg
Mass of entire Earth: 61024 kg !!!
Real inner core: m = 1.7% MEarth
 ginner = 0.017g
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They walk as if under 1 g!!!
17
Gravitational Potential Energy
Gravity is a conservative force – what is the
associated potential energy?
ΔU = -W
x2
W   F ( x)dx
and
x1
So for point masses or spheres m1 and m2
m1m2
m1m2
W  U (x) - U ()     G 2 dx  G

x
x
Taking U = 0 at x = 
x
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m1m2
W  U (x)  G
x
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Gravitational Potential Energy
For point masses or spheres m1 and m2
m1m2
U (r)   G
r
Note: U =  at r = 0
U
r
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F = - dU/dr
Always attractive
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Gravitational Potential Energy for
Astronaut between Earth and Moon
M E mA
U Earth ( x)   G
x
rEM
x
M M mA
U Moon ( x)   G
rEM  x
 ME
MM 

U ( x)  U Earth ( x)  U Moon ( x)   GmA 

rEM  x 
 x
Ftot can be zero at some x… What about Utot?
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Example:
Find where Fnet on the
astronaut equals zero.
M E mA
FEarth  G
x2
M M mA
FMoon  G
(rEM  x) 2
FEarth  FMoon
GM E m A GM M m A

2
x
(rEM  x) 2
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rEM
x
M M x 2  M E (rEM  2rEM x  x 2 )
2
(M E  M M ) x  2M E rEM x  M E rEM  0
2
2
ME
x
rEM
ME  MM
Which solution is real?
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Path Independence
The amount of work done against a gravitational
potential does not depend on the path taken
(conservative force)
y
B
1
3
2
A
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x
22
Escape Velocity
What speed does an object need to escape the
Earth’s gravity?
It needs just enough KE to get to r  and stop
mM e
1 2
Ei  K i  U i  mvesc  G
 U  0
2
Re
vesc
2GM e

Re
Escape velocity from Earth is:
vesc  1.12 10 m / s  25,000 mph
4
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Example: (Problem 14.29)
The mean diameters of Mars and Earth are
6.9x103 km and 1.3x104 km, respectively. The
mass of Mars is 0.11 times Earth’s mass.
a)What is the ratio of the mean density of Mars
to that of Earth?
b)What is the value of the gravitational
acceleration on Mars?
c) What is the escape speed on Mars?
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Kepler’s 1st Law
All planets move in elliptical orbits, with the Sun
at one focus
r
F
q
F’
ea
a
Eccentricities, e, of planets are small (close to
circular)
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Kepler’s 2nd Law
The rate at which a planet sweeps out an area A
is constant. (Constant areal velocity)
A
F
F’
A
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Kepler’s 3rd Law
The square of the period of any planet is
proportional to the cube of the semi-major axis of
its orbit
2
4

T2 
r3
GM
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Example: (Problem 14.42)
Determine the mass of Earth from the period T
(27.3 days) and the radius r (3.82x105 km) of the
Moon’s orbit about Earth. Assume that the Moon
orbits the center of Earth rather than the center of
mass of the Earth-Moon system.
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Example:
27.3 days  2.36 106 s
Kepler’s 3rd law:
2
4

T2 
r3
GM Earth
4 2 r 3
M Earth 
GT 2
4 2 (3.82  108 m) 3
M Earth 
11 Nm 2
(6.672  10 kg 2 )( 2.36 106 s ) 2
M Earth  5.9 10 24 kg
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Circular Orbits
Fg
v
Simplest case:
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r

mM e
mv2
Ftot  G 2  mac 
r
r
GM e
v
 vesc
r
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Geosynchronous Orbit
A satellite can stay over one
location on earth.
Period = 1 day
r
GM e
v
r
T
2 r
2

r 3/ 2
v
GM e
 GM e 
r 
T
 2



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2/3
r  4.22 107 m  26,000 miles
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Why is there free fall on the orbiting
space shuttle?
R=Rorbit < 2Re so gravitational force is not negligible
a = GME/R2
a = GME/R2
Both shuttle and occupants accelerating toward
center of earth with same acceleration
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Energy in a Circular Orbit
m
Mm
U  G
r
2
M
1 2 1  GM  GMm
 
K  mv  m
2
2  r 
2r
GMm GMm
GMm
E  K U 


2r
r
2r
For an elliptical orbit, substitute a for r
GMm
E
2a
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Ocean Tides
Caused by difference in
gravitational force across
the extent of the earth.
r
Water closest to Moon
pulled “upward”
Water farthest from Moon pulled less and thus
bulges outward
Smaller effect from the sun even though it has
stronger gravitational pull. Why?
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Tides
The effects from the sun and
moon can work together to
form a spring tide…
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or against each other to form
a neap tide
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