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PHY 113 C Fall 2013-- Lecture 13 110/10/2013
PHY 113 C General Physics I11 AM - 12:15 PM MWF Olin 101
Plan for Lecture 13:
Chapter 13 – Fundamental gravitational force law and planetary motion
1. Gravitational force law; relationship with g near Earth’s surface
2. Orbital motion of planets, satellites, etc.
3. Energy associated with gravitation and planetary motion
PHY 113 C Fall 2013-- Lecture 13 210/10/2013
PHY 113 C Fall 2013-- Lecture 13 310/10/2013
Question from Webassign Assignment #11
PHY 113 C Fall 2013-- Lecture 13 410/10/2013
X
t1
t3
t2
Frτ
PHY 113 C Fall 2013-- Lecture 13 510/10/2013
X
t1
t3
t2
Frτ
PHY 113 C Fall 2013-- Lecture 13 610/10/2013
Question from Webassign Assignment #11
X
vmvmL
m
21 22
:caseour In
vrL
PHY 113 C Fall 2013-- Lecture 13 710/10/2013
Question from Webassign Assignment #11
2
2
2
12
1
fff
iii
fffiii
IK
IK
ILIL
iclicker question: A. Ki=Kf ? B. Ki=Kf ?
PHY 113 C Fall 2013-- Lecture 13 810/10/2013
Universal law of gravitation Newton (with help from Galileo, Kepler, etc.) 1687
212
122112
ˆ
r
mGm rF
2
211
kg
mN 10674.6
G
PHY 113 C Fall 2013-- Lecture 13 910/10/2013
Newton’s law of gravitation: m2 attracts m1 according to:
x
y
m1
m2
r2
r1
r2r1
212
122112
ˆ
r
mGm rF
G=6.67 x 10-11 N m2/kg2
NNF
rmm
82
11
1221
1017.82
70701067.6
:m 2 kg; 70 :Example
PHY 113 C Fall 2013-- Lecture 13 10
F12
10/10/2013
Vector nature of Gravitational law:
m1 m2
m3
x
y
d
d
jiF1 3221 mm
d
mG
F13
PHY 113 C Fall 2013-- Lecture 13 1110/10/2013
Gravitational force of the Earth
RE m
2226
2411
2
2
m/s8.9m/s)1037.6(
1098.51067.6
E
E
E
E
R
GMg
R
mGMF
Note: Earth’s gravity acts as a point mass located at the Earth’s center.
PHY 113 C Fall 2013-- Lecture 13 1210/10/2013
Question:
Suppose you are flying in an airplane at an altitude of 35000ft~11km above the Earth’s surface. What is the acceleration due to Earth’s gravity?
2226
2411
2
2
m/s796.9m/s)10)011.037.6((
1098.51067.6
)(
)(
hR
GMa
mahR
mGMF
E
E
E
E
a/g = 0.997
PHY 113 C Fall 2013-- Lecture 13 1310/10/2013
Attraction of moon to the Earth:
N1099.1N)1084.3(
1036.71098.51067.6 2028
222411
2
EM
ME
R
MGMF
Acceleration of moon toward the Earth:
F = MM a a = 1.99x2020 N/7.36x1022kg =0.0027 m/s2
REM
PHY 113 C Fall 2013-- Lecture 13 1410/10/2013
PHY 113 C Fall 2013-- Lecture 13 1510/10/2013
Gravity on the surface of the moon
EM
M
MM
M
gg
g
mgmR
GmMF
165.0
m/s 62.1)1074.1(
1035.71067.6
N)1074.1(
1035.71067.6
226
2211
26
2211
2
Gravity on the surface of mars
EMars
Mars
MarsMars
gg
R
GMg
380.0
m/s 73.3)1039.3(
1042.61067.6 2
26
2311
2
PHY 113 C Fall 2013-- Lecture 13 1610/10/2013
iclicker question:In estimating the gravitational acceleration near the surfaces of the Earth, Moon, or Mars, we used the full mass of the planet or moon, ignoring the shape of its distribution. This is a reasonable approximation because:
A. The special form of the gravitational force law makes this mathematically correct.
B. Most of the mass of the planets/moon is actually concentrated near the center of the planet/moon.
C. It is a very crude approximation.
PHY 113 C Fall 2013-- Lecture 13 1710/10/2013
days 27.4 s 32367353.951098.51067.6
)1084.3(π2
π2
π2ω
:Earth todueMoon for law sNewton'
2411
38
3
2
2
E
EM
EMEM
EM
E
EM
MMM
GM
RT
RT
Rv
R
GM
R
va
M aF
Stable circular orbit of two gravitationally attracted objects (such as the moon and the Earth)
REM
F
a
v
PHY 113 C Fall 2013-- Lecture 13 1810/10/2013
iclicker question:
In the previous discussion, we saw how the moon orbits the Earth in a stable circular orbit because of the radial gravitational attraction of the moon and Newton’s second law: F=ma, where a is the centripetal acceleration of the moon in its circular orbit. Is this the same mechanism which stabilizes airplane travel? Assume that a typical cruising altitude of an airplane is 11 km above the Earth’s surface and that the Earth’s radius is 6370 km.
(a) Yes (b) No
PHY 113 C Fall 2013-- Lecture 13 1910/10/2013
!mi/h! 18000km/s 7.9v
hours 1.4 s 07051098.51067.6
)1038.6(π2
π2
π2ω
2411
36
3
2
2
E
Ea
EaEa
Ea
E
Ea
GMR
T
RT
Rv
RGM
Rv
a
Stable (??) circular orbit of two gravitationally attracted objects (such as the airplane and the Earth)
REa
F
a
v
PHY 113 C Fall 2013-- Lecture 13 2010/10/2013
2226
2411
2
2
m/s8.9m/s)1037.6(
1098.51067.6
E
E
E
E
R
GMg
R
mGMF
212
122112
ˆ
r
mGm rF Newton’s law of gravitation:
Earth’s gravity:
Stable circular orbits of gravitational attracted objects:
RE m
REMF
a
vMM
Summary:
PHY 113 C Fall 2013-- Lecture 13 2110/10/2013
More details
If we examine the circular orbit more carefully, we find that the correct analysis is that the stable circular orbit of two gravitationally attracted masses is about their center of mass.
m1
R2
R1
m2
2211
:mass ofCenter
RmRm
PHY 113 C Fall 2013-- Lecture 13 2210/10/2013
m1
R2
R1
m2
Radial forces on m1:
2
2211
1
1
11
1
21
111221
211
π2
π2
GmRRR
T
TR
v
Rv
mamRR
mGmF rr
T2 ?
PHY 113 C Fall 2013-- Lecture 13 2310/10/2013
iclicker question:
What is the relationship between the periods T1 and T2 of the two gravitationally attracted objects rotating about their center of mass? (Assume that m1 < m2.)
(A) T1=T2 (B) T1<T2 (C) T1>T2
m1
R2
R1
m2
PHY 113 C Fall 2013-- Lecture 13 24
21
321
21
21
22222
21
212111
1
21
1
2211
2
mmG
RRTT
RmRR
mGmRm
R
vm
RmRm
10/10/2013
m1
R2
R1
m2
1
2212
2
2211
21
321
21
22
2
: thatNote
Gm
RRR
Gm
RRR
mmG
RRTT
PHY 113 C Fall 2013-- Lecture 13 2510/10/2013
iclicker questions:How is it possible that all of these relations are equal?
A. Magic.B. Trick.C. Algebra.
2211
2
2211
2
2211
21
321
since 11 because
21
122
2
1
1
2
2
1
1
2
RmRm
Gm
RRR
Gm
RRR
mmG
RRT
mm
RR
mm
RR
1
2212
2
2211
21
321 222
Gm
RRR
Gm
RRR
mmG
RRT
PHY 113 C Fall 2013-- Lecture 13 26
21
321
21
21
22222
21
212111
1
21
1
2211
2
mmG
RRTT
RmRR
mGmRm
R
vm
RmRm
10/10/2013
m1
R2
R1
m2
For m2>>m1 R2 << R1:
m2
m1
R12
31
1 2Gm
RT
Larger mass
PHY 113 C Fall 2013-- Lecture 13 2710/10/2013
What is the physical basis for stable circular orbits?
1. Newton’s second law?
2. Conservation of angular momentum? L = (const)
Note: Gravitational forces exert no torque
0
0
ˆ
121212
212
122112
dt
d
r
mGm
L
Frτ
rF
PHY 113 C Fall 2013-- Lecture 13 2810/10/2013
(const)
0
L
Lτ
dtd
m1
R2
R1
m2
v1
v2
L1=m1v1R1
L2=m2v2R2
L = L1 + L2
Question:
How are the magnitudes of L1 and L2 related?
2
2
1
1
R
L
R
L
Note: More generally, stable orbits can be elliptical.
PHY 113 C Fall 2013-- Lecture 13 2910/10/2013
Satellites orbiting earth (approximately circular orbits):
RE ~ 6370 km
Examples:
sRhRhGM
RT EE
E
E 23
23
/15058/1π23
Satellite h (km) T (hours) v (mi/h)
Geosynchronous 35790 ~24 6900
NOAA polar orbitor 800 ~1.7 16700
Hubble 600 ~1.6 16900
Inter. space station* 390 ~1.5 17200
PHY 113 C Fall 2013-- Lecture 13 3010/10/2013
Planets in our solar system – orbiting the sun
Planet Mass (kg) Distance to Sun (m)
Period of orbit (years)
Mercury 3.30x1023 5.79x1010 0.24
Venus 4.87x1024 1.08x1011 0.61
Earth 5.97x1024 1.496x1011 1.00
Mars 6.42x1023 2.28x1011 1.88
Jupter 1.90x1027 7.78x1011 11.85
Saturn 5.68x1026 1.43x1012 29.43
PHY 113 C Fall 2013-- Lecture 13 3110/10/2013
m1
R2
R1
m2
v1
v2
Review: Circular orbital motion about center of mass
CM
21
321
21
2
111
1
2
1
11
1
21
1
2211
2
22
2221
21
1
21
1
2
212
mmG
RRTT
TRm
RT
Rm
R
vm
RmRm
R
vm
RR
mGm
R
vm
2
31
21
321
21
1212
22
then if that Note
Gm
R
mmG
RRTT
RRmm
PHY 113 C Fall 2013-- Lecture 13 3210/10/2013
m1
R1
m2
v1
Review: Circular orbital motion about center of mass
2
31
21
321
21
1212
22
then if that Note
Gm
R
mmG
RRTT
RRmm
PHY 113 C Fall 2013-- Lecture 13 3310/10/2013
Example: Satellite in circular Earth orbit
kgm
kgM
GM
hRT
S
E
E
E
3
24
3
10
1097.5 :Note
2
onous)(geosynchr
11053.8
1083.35 If4
6
daysT
mh
PHY 113 C Fall 2013-- Lecture 13 3410/10/2013
Work of gravity:
ri
rf
f
i
f
i
f
i
r
r
r
r
fi
r
r
fi
r
mGm dr
r
mGmW
r
mGmdW
212
21
221 ˆ
r
FrF
ifif
fi rUrUr
mGm
r
mGmW
2121
PHY 113 C Fall 2013-- Lecture 13 3510/10/2013
Gravitational potential energy
r
mGm dr
r
mGmrU
r
mGmdrU
r
gravity
r
r
gravity
ref
212
21
221
''
)(
ˆ )(
rFrF
hR
mGMhRrU
E
SEEgravity
)(
Example:
PHY 113 C Fall 2013-- Lecture 13 3610/10/2013
hU=mgh
iclicker exercise:We previously have said that the gravitational potential of an object of mass m at a height h isU=mgh. How is this related to
A. No relationB. They are approximately equalC. They are exactly equal
? hR
mGMU
E
SEgravity
PHY 113 C Fall 2013-- Lecture 13 3710/10/2013
h
hR
mGM
R
mGM
hR
mGMRUhRU
R
h
RRh
RRhRhR
E
E
E
E
E
EEgravityEgravity
EEE
EEEE
2
2
)()(
1/1
1
/1
111
gNear the surface of the Earth U=mgh is a good approximation to the gravitation potential.
PHY 113 C Fall 2013-- Lecture 13 3810/10/2013
iclicker exercise:How much energy kinetic energy must be provided to an object of mass m=1000kg, initially on the Earth’s surface to outer space?
A. This is rocket science and not a fair question.B. It is not possible to escape the Earth’s
gravitational field.C. We can estimate the energy by simple
conservation of energy concepts.
J
JR
mGMK
KR
mGMU
UKUKE
E
Ei
iE
Ei
ffii
10
6
2411
1025.6
1037.6
10001097.510674.6
0
PHY 113 C Fall 2013-- Lecture 13 3910/10/2013
Total energy of a satellite in a circular Earth orbit
hR
mGMhRrU
E
SEEgravity
)(
hR
mGMvmK
hR
mGM
hR
vm
vmKUKE
E
SES
E
SE
ES
Sgravity
22
1
:orbitcircular aFor
2
1
2
2
2
2
PHY 113 C Fall 2013-- Lecture 13 4010/10/2013
Total energy of a satellite in a circular Earth orbit
hR
mGME
hR
mGMU
hR
mGMK
UKE
E
SE
E
SEgravity
E
SE
gravity
2
2
iclicker question:Compared to the energy needed to escape the Earth’s gravitational field does it take
A. moreB. less
energy to launch a satellite to orbit the Earth?
PHY 113 C Fall 2013-- Lecture 13 4110/10/2013
JJ
JJK
kmh
R
mGM
hR
mGM
R
mGMK
R
mGMU
hR
mGMUKUKE
i
E
E
E
E
E
Ei
E
Ei
E
Effii
1010
6
2411
6
2411
1038.393.62
37.611025.6
1093.62
10001097.510674.6
1037.6
10001097.510674.6
Telescope) Hubble assuch orbit earth (low 560 example,For
2
2
PHY 113 C Fall 2013-- Lecture 13 4210/10/2013
Energy involved with changing orbits
h’
J
E
kmhkmh
hR
mGM
hR
mGMEEE
E
E
E
E
8
6
2411
1065.1
97.6
1
93.6
1
102
10001097.510674.6
600' and 560 example,For
2'2'
PHY 113 C Fall 2013-- Lecture 13 4310/10/2013
Stable Elliptical Orbits
r
MGM
rM
LvME
vv
M
pS
prp
r
p
2
22
22
1
ˆˆ θrv
vrL