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18/1/2004
Test Breakdown
A >= 93% (63)
A- >= 90% (60)
B+ >= 87% (57)
B >= 83% (54)
B- >= 80% (52)
C+ >= 77% (51)
C >= 70% (45)
D >= 60% (38)
F < 60% (37 or less)
Average: 63.2% + 10%
Std Dev: 21.5%
Scores will be adjusted up by 10%
Chapter 14
Gravitation
38/1/2004
Gravity
One of the fundamental forces of Nature
Not just the reason things fall….
Why the Earth is roundWhy the moon goes around the earthWhy the earth goes around the sunWhy there are ocean tides
48/1/2004
Gravitational Force
Fg is not a constant unless you have a small object near the surface of a big sphere
Fg is an attractive force between any two masses
221
r
mmGFg
G = 6.672 x 10-11 N m2/kg2
r
m1 m2
Fg Fg
58/1/2004
Multiple Objects
Obeys principle of superposition:
...23
22
21
r
mMG
r
mMG
r
mMGFg
r1 r3
M1 M3
M2
m
r2
68/1/2004
Extended Objects
1 2Fg Fg
Each bit of #1 attracts each bit of #2
Need to integrate over whole object to get Fg
FdFg
78/1/2004
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
88/1/2004
Gravitational Force Examples
Force between earth and sun
Force between two people (assumed spherical)
221
r
mmGF
N.
m.
kg.kg.
kg
mN.F
22
211
2430
2
211
10523
)10491(
)10985)(10991(106726
N.m
kgkg
kg
mN.F 7
22
211 1042
)1(
)60)(60(106726
98/1/2004
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?
108/1/2004
Example:
Find the mass of the object:
kgsm
N
g
Fm
Earth
Earthg 2.10/8.9
1002
,
On the Moon:
26
2211
2
)1074.1(
)2.10)(1035.7)(10672.6( 2
2
m
kgkg
r
mGMF
kgNm
Moon
Moong
NFg 5.16
118/1/2004
Example:
Distance from the earth with the same weight:
EarthEarth
kgNm
g
Earth
Rm
Rmr
N
kgkgr
F
mGMr
5.21038.6
11057.1
5.16
)2.10)(1097.5)(10672.6(
67
24112
2
2r
mGMF Earthg
128/1/2004
Objects Near Earth’s Surface
Re
As long as the distance above earth’s surface is small compared to RE, the force is approximately constant
g
E
Eg ma
R
MmGF 2 2
E
Eg
R
MGa
138/1/2004
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, gag
148/1/2004
Variation of g from Rotation
At the earth’s pole, there is no centripetal acceleration:
N
Fg
ac=0
0
g
cnet
FN
maFmg
R
MGmN
e
e
2
At the pole, g=ag
158/1/2004
Variation of g from Rotation
At the equator:
RmFN
maF
g
cnet
2
mgRR
MGmN
e
e
2
2
NFg
ac
At the equator, g < ag!
g = 9.801 m/s2 in Pittsburgh
g = 9.786 m/s2 in Jamaica!!
Effectively lower gravity
168/1/2004
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!
178/1/2004
“Inner” Gravity
Movie claims that earth’s core is a trillion trillion tons…
1012 1012 tons = 1027 kgMass of entire Earth: 61024 kg !!!
Real inner core: m = 1.7% MEarth
ginner = 0.017g
They walk as if under 1 g!!!
Only the mass of the inward sphere contributes to Fg
188/1/2004
Gravitational Potential Energy
x
mmGdx
x
mmGUUW
x21
221 )(-(x)
Gravity is a conservative force – what is the associated potential energy?
ΔU = -W and
So for point masses or spheres m1 and m2
Taking U = 0 at x =
2
1
)(x
xdxxFW
x
mmGUW 21(x)
198/1/2004
Gravitational Potential Energy
Note: U = at r = 0
For point masses or spheres m1 and m2
U
r F = - dU/dr Always attractive
r
mmGU 21 (r)
208/1/2004
Gravitational Potential Energy for Astronaut between Earth and Moon
rEM
x
x
mMGxU AE
Earth )(
xr
mMGxU
EM
AMMoon
)(
xr
M
x
MGmxUxUxU
EM
MEAMoonEarth )( )()(
Ftot can be zero at some x… What about Utot?
218/1/2004
Example:
rEM
x
Find where Fnet on the astronaut equals zero.
22 )( xr
mGM
x
mGM
FF
EM
AMAE
MoonEarth
2
x
mMGF AE
Earth
2)(
xr
mMGF
EM
AMMoon
)2( 222 xxrrMxM EMEMEM
02)( 22 EMEEMEME rMxrMxMM
EMME
E rMM
Mx
Which solution is real?
228/1/2004
Path Independence
The amount of work done against a gravitational potential does not depend on the path taken (conservative force)
x
y
A
B1
23
238/1/2004
Escape Velocity
What speed does an object need to escape the Earth’s gravity?
02
1 2 UR
mMGmvUKE
e
eesciii
mphsmvesc 000,25/1012.1 4
It needs just enough KE to get to r and stop
e
eesc R
GMv
2
Escape velocity from Earth is:
248/1/2004
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?
258/1/2004
Kepler’s 1st Law
All planets move in elliptical orbits, with the Sun at one focus
a
F F’
ea
r
Eccentricities, e, of planets are small (close to circular)
268/1/2004
Kepler’s 2nd Law
The rate at which a planet sweeps out an area A is constant. (Constant areal velocity)
F F’
A
A
278/1/2004
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
32
2 4r
GMT
288/1/2004
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.
298/1/2004
Example:
32
2 4r
GMT
Earth
Kepler’s 3rd law:
2611
382
2
32
)1036.2)(10672.6(
)1082.3(4
4
2
2
s
mM
TG
rM
kgNmEarth
Earth
s1036.2days3.27 6
kgM Earth24109.5
308/1/2004
Circular Orbits
Simplest case:
rFg
v
r
mvma
r
mMGF c
etot
2
2
esce v
r
GMv
318/1/2004
Geosynchronous Orbit
r
A satellite can stay over one location on earth.
2/322r
GMv
rT
e
3/2
2
T
GMr e
Period = 1 day
milesmr 000,261022.4 7
r
GMv e
328/1/2004
Why is there free fall on the orbiting space shuttle?
Both shuttle and occupants accelerating toward center of earth with same acceleration
R=Rorbit < 2Re so gravitational force is not negligible
a = GME/R2
a = GME/R2
338/1/2004
Energy in a Circular Orbit
r
MmGU
M
m
r
GMm
r
GMmmvK
22
1
2
12
2
r
GMm
r
GMm
r
GMmUKE
22
For an elliptical orbit, substitute a for r
a
GMmE
2
348/1/2004
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?
358/1/2004
Tides
The effects from the sun and moon can work together to
form a spring tide…
or against each other to form a neap tide