phy 113 a general physics i 9-9:50 am mwf olin 101 plan for lecture 23:
Post on 22-Feb-2016
27 Views
Preview:
DESCRIPTION
TRANSCRIPT
PHY 113 A Fall 2012 -- Lecture 23 110/26/2012
PHY 113 A General Physics I9-9:50 AM MWF Olin 101
Plan for Lecture 23:
Chapter 13 – Fundamental force of gravity1. Relationship with g near earth’s
surface2. Orbital motion due to gravity3. Kepler’s orbital equation4. Note: We will probably not
emphasize elliptical orbits in this chapter.
PHY 113 A Fall 2012 -- Lecture 23 210/26/2012
PHY 113 A Fall 2012 -- Lecture 23 310/26/2012
Universal law of gravitation Newton (with help from Galileo, Kepler, etc.) 1687
212
122112
ˆrmGm rF
2
211
kgmN 10674.6
G
PHY 113 A Fall 2012 -- Lecture 23 410/26/2012
Newton’s law of gravitation: m2 attracts m1 according to:
x
y
m1
m2
r2
r1
r2r1
212
122112
ˆrmGm rF
G=6.67 x 10-11 N m2/kg2
NNF
rmm
82
111221
1017.82
70701067.6
:m 2 kg; 70 :Example
PHY 113 A Fall 2012 -- Lecture 23 510/26/2012
Vector nature of Gravitational law:
m1 m2
m3
x
y
d
d
jiF1 3221 mmdmG
PHY 113 A Fall 2012 -- Lecture 23 610/26/2012
Gravitational force of the Earth
RE m
2226
2411
2
2
m/s8.9m/s)1037.6(
1098.51067.6
E
E
E
E
RGMg
RmGMF
PHY 113 A Fall 2012 -- Lecture 23 710/26/2012
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)(
)(
hRGMa
mahRmGMF
E
E
E
E
a/g = 0.997
PHY 113 A Fall 2012 -- Lecture 23 810/26/2012
Attraction of moon to the Earth:
N1099.1N)1084.3(
1036.71098.51067.6 2028
222411
2
EM
ME
RMGMF
Acceleration of moon toward the Earth:
F = MM a a = 1.99x2020 N/7.36x1022kg =0.0027 m/s2
PHY 113 A Fall 2012 -- Lecture 23 910/26/2012
PHY 113 A Fall 2012 -- Lecture 23 1010/26/2012
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
ggRGMg
380.0
m/s 73.3)1039.3(
1042.61067.6 226
2311
2
PHY 113 A Fall 2012 -- Lecture 23 1110/26/2012
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 A Fall 2012 -- Lecture 23 1210/26/2012
days 27.4 s 32367353.951098.51067.6
)1084.3(π2
π2
π2ω
2411
38
3
2
2
E
EM
EMEM
EM
E
EM
GMRT
RT
Rv
RGM
Rva
Stable circular orbit of two gravitationally attracted objects (such as the moon and the Earth)
REM
F
a
v
PHY 113 A Fall 2012 -- Lecture 23 1310/26/2012
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 A Fall 2012 -- Lecture 23 1410/26/2012
!mi/h! 18000km/s 7.9vhours 1.4 s 0705
1098.51067.6)1038.6(π2
π2
π2ω
2411
36
3
2
2
E
Ea
EaEa
Ea
E
Ea
GMRT
RT
Rv
RGM
Rva
Stable (??) circular orbit of two gravitationally attracted objects (such as the airplane and the Earth)
REa F
a
v
PHY 113 A Fall 2012 -- Lecture 23 1510/26/2012
2226
2411
2
2
m/s8.9m/s)1037.6(
1098.51067.6
E
E
E
E
RGMg
RmGMF
212
122112
ˆrmGm rF Newton’s law of gravitation:
Earth’s gravity:
Stable circular orbits of gravitational attracted objects:
RE m
REMF
a
vMM
PHY 113 A Fall 2012 -- Lecture 23 1610/26/2012
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 A Fall 2012 -- Lecture 23 1710/26/2012
m1
R2
R1
m2
Radial forces on m1:
2
2211
1
1
11
1
21
111221
211
π2
π2
GmRRRT
TRv
Rvmam
RRmGmF rr
T2 ?
PHY 113 A Fall 2012 -- Lecture 23 1810/26/2012
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 A Fall 2012 -- Lecture 23 19
21
321
21
21
22222
21
212111
1
21
1
2211
2
mmGRRTT
RmRRmGmRm
Rvm
RmRm
10/26/2012
m1
R2
R1
m2
1
2212
2
2211
21
321
21
22
2
: thatNote
GmRRR
GmRRR
mmGRRTT
PHY 113 A Fall 2012 -- Lecture 23 2010/26/2012
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
GmRRR
GmRRR
mmGRRT
mm
RR
mm
RR
1
2212
2
2211
21
321 222
GmRRR
GmRRR
mmGRRT
PHY 113 A Fall 2012 -- Lecture 23 2110/26/2012
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
dtd
rmGm
LFrτ
rF
PHY 113 A Fall 2012 -- Lecture 23 2210/26/2012
(const)
0
L
Lτdtd
m1
R2
R1
m2
v1
v2
L1=m1v1R1
L2=m2v2R2
Question:
How are the magnitudes of L1 and L2 related?
2
2
1
1
RL
RL
Note: More generally, stable orbits can be elliptical.
PHY 113 A Fall 2012 -- Lecture 23 2310/26/2012
Satellites orbiting earth (approximately circular orbits):
RE ~ 6370 km
Examples:
sRhRhGMRT EE
E
E 23
23
/15058/1π23
Satellite h (km) T (hours) v (mi/h)Geosynchronous 35790 ~24 6900NOAA polar orbitor 800 ~1.7 16700Hubble 600 ~1.6 16900Inter. space station* 390 ~1.5 17200
PHY 113 A Fall 2012 -- Lecture 23 2410/26/2012
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
top related