Transcript 08._Gravity

8. Gravity
1.
2.
3.
4.
5.
Toward a Law of Gravity
Universal Gravitation
Orbital Motion
Gravitational Energy
The Gravitational Field
This TV dish points at a satellite in a fixed position in the sky.
How does the satellite manage to stay at that position?
period = 24 h
Ptolemaic (Geo-Centric) System
epicycle
equant
deferent
swf
8.1. Toward a Law of Gravity
1543: Copernicus – Helio-centric theory.
1593: Tycho Brahe – Planetary obs.
1592-1610: Galileo – Jupiter’s moons,
sunspots, phases of Venus.
1609-19: Kepler’s Laws
1687: Newton – Universal gravitation.
Phases of Venus:
Size would be constant
in a geocentric system.
Kepler’s Laws
Explains retrograde motion
T 2  a3
dA
 const
dt
Mathematica
8.2. Universal Gravitation
Newton’s law of universal gravitation:
F12  G
m1 m2
rˆ12
2
r
m1 & m2 are 2 point masses.
r12 = position vector from 1 to 2.
F12 = force of 1 on 2.
G = Constant of universal gravitation
= 6.67  1011 N m2 / kg2 .
F12
m2
m1
r12
Law also applies to spherical masses.
Example 8.1. Acceleration of Gravity
Use the law of gravitation to find the acceleration of gravity
(a) at Earth’s surface.
(b) at the 380-km altitude of the International Space Station.
(c) on the surface of Mars.
F m g G
(a)
(b)
(c)
m mE
r2
g   6.67 1011
g   6.67 1011
g   6.67 1011

g G
mE
r2
5.97 10 kg 

 9.81 m / s
N m / kg 
 6.37 10 m 
5.97 10 kg 

N m / kg 
 6.37 10 m  380 10 m 
0.642 10 kg 

 3.7 m / s
N m / kg 
3.38 10 m 
24
2
2
6
2
2
24
2
2
6
3
2
 8.74 m / s 2
24
2
2
2
6
2
see App.E
TACTICS 8.1. Understanding “Inverse Square”
Given Moon’s orbital period T & distance R from Earth,
Newton calculated its orbital speed v and hence acceleration a = v2 / R.
He found a ~ g / 3600.
 Moon-Earth distance is about 60 times Earth’s radius.
Cavendish Experiment: Weighing the Earth
g G
ME
R 2E
ME can be calculated if g, G, & RE are known.
Cavendish:
G determined using two 5 cm &
two 30 cm diameter lead spheres.
Gravity is weakest & long ranged
always attractive
 dominates at large range.
EM is strong & long ranged,
can be attractive & repulsive
 cancelled out in neutral objects.
Weak & strong forces: very short-range.
8.3. Orbital Motion
Orbital motion: Motion of object due to gravity from another larger body.
E.g. Sun orbits the center of our galaxy with a period of ~200 million yrs.
Newton’s “thought experiment”
g=0
Condition for circular orbit
Speed for circular orbit
orbit
projectiles
Orbital period
Mm
v2
G 2 m
r
r
v
GM
r
2 r
 2
T
v
r3
GM
Kepler’s 3rd law
Example 8.2. The Space Station
The ISS is in a circular orbit at an altitude of 380 km.
What are its orbital speed & period?
Orbital speed:
Orbital period:
v
T  2
GM

r
 6.67 10
11
6.37 10 m  380 10 m
6
r3
 2  3.1416
GM
 5.5 103 s
N m 2 / kg 2   5.97 10 24 kg 
3
 7.7 km / s
 6.37 106 m  380 103 m 
3
 6.67 10
 90 min
Near-Earth orbit T ~ 90 min.
Moon orbit T ~ 27 d.
Geosynchronous orbit T = 24 h.
11
N m2 / kg 2   5.97 1024 kg 
Example 8.3. Geosynchronous Orbit
What altitude is required for geosynchronous orbits?
T  2
 T 
r 

 2 
r3
GM
2/3
G M 
1/3
 24  3600 s 


 2  3.1416 
2/3
 6.67 1011 N m 2 / kg 2   5.97 1024 kg  


1/3
 4.22 107 m
Altitude = r  RE
 4.22 107 m  6.37 106 m  35.80 106 m
 35,800 km
Earth circumference =  2   6.37 106 m  40, 000 km
Earth not perfect sphere  orbital correction required every few weeks.
Elliptical Orbits
Projectile trajectory is parabolic only
if curvature of Earth is neglected.
ellipse
Orbits of most known comets, are highly elliptical.
Perihelion: closest point to sun.
Aphelion: furthest point from sun.
Open Orbits
Open
(hyperbola)
Closed
(circle)
Borderline
(parabola)
Closed
(ellipse)
Mathematica
8.4. Gravitational Energy
How much energy is required to boost a satellite to geosynchronous orbit?
U12  
r2
r1
M m

U12     G 2  d r
r1
r 

r2
F  dr
1 1
G M m  
 r1 r2 
U12 depends only on radial positions.
U = 0
on this path
… so U12 is
the same as if
we start here.
Example 8.4. Steps to the Moon
Materials to construct an 11,000-kg lunar observatory are boosted from Earth to geosyn orbit.
There they are assembled & launched to the Moon, 385,000 km from Earth.
Compare the work done against Earth’s gravity on the 2 legs of the trip.
1 1
W  U12  G M E m   
 r1 r2 
1st leg:


1
1
W   6.67 1011 N m2 / kg 2  5.97 1024 kg  11, 000 kg  


6
7
 6.37 10 m 4.22 10 m 
 5.8 1011 J
2nd leg:


1
1
W   6.67 1011 N m2 / kg 2  5.97 1024 kg  11,000 kg  


7
8
 4.22 10 m 3.85 10 m 
 9.2 1010 J
Zero of Potential Energy
1 1
U12  G M m   
 r1 r2 
U   0

U r   
GM m
r
Gravitational potential energy
E > 0, open orbit
Open
E < 0, closed orbit
Bounded motion
Turning point
Closed
Example 8.5. Blast Off !
A rocket launched vertically at 3.1 km/s.
How high does it go?
Initial state:
E0 
1
GM m
m v2 
2
RE
Final state:
E
GM m
r
E  K U
Energy conservation:
r
GM
1 2 GM
v 
2
RE

1
v2
1


2 G M RE


3.1 m / s 

1

r  

 2  6.67 1011 N m2 / kg 2  5.97 1024 kg  6.37 106 m 


2
Altitude = r  RE
 6.90 106 m  6.37 106 m  530 km
1
 6.90 Mm
Escape Velocity
Body with E  0 can escape to 
Open
1
GM m
m v2 
0
2
RE
Closed
vesc 

vesc 
2  6.67 1011 N m 2 / kg 2  5.97 1024 kg 
6.37 10 m
6
Moon trips have v < vesc .
2G M
RE
Escape velocity
 11.2 km / s
 40,300 km / h
Energy in Circular Orbits
Circular orbits:
U 

v2  a r 
GM
r
K

GM m
r
0
GM m
E  K U  
2r
1
 K  U
2
1
GM m
m v2 
2
2r
0
K
E
K
Higher K or v  Lower E & orbit (r) .
U
Conceptual Example 8.1. Space Maneuvers
Astronauts heading for the International Space Station find themselves
in the right circular orbit, but well behind the station.
How should they maneuver to catch up?
1. Fire rocket backward to
decrease energy & drop to
lower, & faster orbit.
2. Fire to circularize orbit.
3. After catching up with the
station, fire to boost to up to
its level.
4. Fire to circularize orbit.
Mathematica
U 
GMm
h
K E 
1
GMm
U 
2
2h
energy
Altitude

0
K
E = K+U = U / 2
E = K+U = U / 2
K>K
h
h<h
E = K+U < E
( K < K )
U
U<U
UG
0
GOT IT? 8.3.
Spacecrafts A & B are in circular orbits about Earth, with B at higher altitude.
Which of the statements are true?
 (a) B has greater energy.
 (b) B is moving faster.
 (c) B takes longer to complete an orbit.
 (d) B has greater potential energy.
 (e) a larger proportion of B’s energy is potential energy.
8.5. The Gravitational Field
Two descriptions of gravity:
1. body attracts another body (action-at-a-distance)
2. Body creates gravitational field.
Field acts on another body.
near earth
Near Earth:
Large scale:
g   g ˆj
g
GM
rˆ
2
r
g  9.8 m / s 2
 N / kg 
Action-at-a-distance  instantaneous messages
Field theory  finite propagation of information
Only field theory agrees with relativity.
in space
Great advantage of the field approach:
No need to know how the field is produced.
Application: Tide
Moon’s tidal (differential) force field at Earth’s surface
Moon’s tidal (differential) force field near Earth
F r   f r   f rE 
Two tidal bulges
Mathematica
Sun + Moon  tides with varying strength.
Tidal forces cause internal heating of Jupiter’s moons.
They also contribute to formation of planetary rings.