Download Chap. 13: Gravitation

Chap. 13: Gravitation
Newton’s Law of Universal Gravity
– Attractive Force (Vector)
– Inverse Square Law
– Universality
Kepler’s Laws
Taken from Fig. 2.16, “the Universe in a Nutshell” by Stephen Hawking
1
Goals for Chapter 13
• To calculate the gravitational forces that bodies exert
on each other
• To relate weight to the gravitational force
• To use the generalized expression for gravitational
potential energy
• To study the characteristics of circular orbits
• To investigate the laws governing planetary motion
• To look at the characteristics of black holes
Introduction
• What can we say about the
motion of the particles that
make up Saturn’s rings?
• Why doesn’t the moon fall
to earth, or the earth into
the sun?
• By studying gravitation and
celestial mechanics, we will
be able to answer these and
other questions.
Newton’s Law of Gravitation
• Law of universal gravitation:
Every
particle
of
matter
attracts every other particle
with a force that is directly
proportional to the product of
their masses and inversely
proportional to the square of
the distance between them.
• The gravitational force can be
expressed mathematically as:

m1m2
Fg  G 2 r̂
r
where G is the gravitational
constant.
Newton’s Law of Universal Gravitation
Newoton's Law of Universal Gravitation
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Newton’s Law of Universal Gravitation

m1m2
F  G 2 r̂
r
Attractive Force
m1 = MM
r
Magnitude of the force
between any 2 masses
6.67x1011 Nm2/kg2
m2 = ME
Newoton's Law of Universal Gravitation
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Parameters
•
•
•
•
•
•
•
•
G = 6.67 x 1011 Nm2/kg2
ME = 5.98 x 1024 kg
RE = 6.38 x 103 km
MS = 1.99 x 1030 kg
RS = 6.96 x 105 km
rE-S = 150 x 106 km
Mmoon = 7.35 x 1022 kg
Rmoon = 1.74 x 103 km
Newoton's Law of Universal Gravitation
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Spherically Symmetric Bodies
• The gravitational force
between bodies with
spherically symmetric mass
distributions is the same as
if all their mass were
concentrated at their
centers.
• See Figure 13.2 at the
right.
Problem 1: Find FME
(Force on Moon due to Earth)

m1m2
F  G 2 r̂
r
•
•
•
•
•
•
3. Give the vector an
appropriate label.
G = 6.67 x 1011 Nm2/kg2
ME = 5.98 x 1024 kg
RE = 6.38 x 103 km
MS = 1.99 x 1030 kg
Mmoon = 7.35 x 1022 kg
Rmoon = 1.74 x 103 km
Earth
1. Represent the Moon as a
particle and Place the tail
of the force vector on the
Moon’s center.
FME
r
r = 3.80 x 108 m
Answer: FME = 2.03 x 1020 N
Moon
2. Draw the vector force
as an arrow pointing in the
Earth’s center.
Newoton's Law of Universal Gravitation
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Gravitational Forces: Vector
FMnet = FME + FMS
MM ME
MM
ME
FME = G
rME2
Univeral Gravitational Constant
vector
MM MS
FMS = G
MS
rMS2
Newoton's Law of Universal Gravitation
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Problem 2: Find the Net Force on Sun.
 Assume that the masses
and the coordinates are
known.
y
Mars
(x3, y3)
m3
NOTE
The gravitational force is:
• …
m1
(x2, 0)
x
m2
Sun Earth
Newoton's Law of Universal Gravitation
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More Example
The starship Enterprise (mass m) is investigating a new star of mass M1, and is
holding position a distance Y from the star. There is another star of mass M2 a
distance Y from the first along a line perpendicular to that connecting the
Enterprise with the first star. Suppose the Enterprise experiences total engine
failure. Express the x and y components of the net gravitational force on the
Enterprise. Use the coordinate system provided. Don’t forget that force is a
vector.
M2
y
Y
x
M1
Starship Enterprise
(not to scale)
X
Newoton's Law of Universal Gravitation
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Satellite Orbital Period : T = (2pr)/v
Try to relate between F and v.
v
FG
Newoton's Law of Universal Gravitation
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Satellite
Orbital
Newton’s Law of Universal
GravitationPeriod : T = (2pr)/v
Try to relate
between F and v.
m =M
G
FFGG == G
m11 m
m22
m
1
M
rr 22
Gravity = Center-seeking Force
Attractive Force
r
6.67x1011 Nm2/kg2
mS , ME , r = 2rE , G
v
Magnitude of the force
between any 2 masses
m2 = ME
Newoton's Law of Universal Gravitation
F
Dynamics G
of
FG
Uniform Circular Motion
Center-seeking Acceleration and Force
Frad = m arad
FG = mS v 2 / (2 rE)
Newoton's Law of Universal Gravitation
Circular Motion
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Problem 3: Find T
v
FME
Earth
r
Moon
r = 3.80 x 108 m
FME = 2.03 x 1020 N
Newoton's Law of Universal Gravitation
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Problem 4: Geosynchronous Satellite
 A geosynchronous satellite, which is one that
stays above the same point on the Earth, is
used for such purpose as cable TV
transmission, for weather forecasting, and as
communication relays.
a) Determine the height above the Earth’s
surface.
b) Determine the speed of the satellite.
Newoton's Law of Universal Gravitation
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Gravitational Acceleration:
Constant?
FG = m g (Newton’s 2nd Law)
m
∴ g = FG / m
FG
r
where
FG = G m M / r 2
= m (G M / r 2)
M
R
Newoton's Law of Universal Gravitation
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Gravity: Near the Earth’s Surface
m
g(h) = G M / r 2
= G M / (R+h)
h
2
FG
r
M
R
Newoton's Law of Universal Gravitation
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Gravity: Near the Earth’s Surface
h = altitude above the surface
MEarth
g(h) = G ----------------(REarth + h) 2
e.g., h = 0 m  g = 9.80 m/s2
Newoton's Law of Universal Gravitation
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Gravity: Near the Earth’s Surface
9.9
9.80
9.77
9.8
m/s2
9.7
9.6
9.5
7.33
g
9.4
9.3
9.2
9.1
9
New York
h=0m
Mt. Everest
h=1000km
h = 8848 m
Newoton's Law of Universal Gravitation
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Gravity: Near the Moon’s Surface
h = altitude above the surface
MMoon
g(h) = G ----------------(RMoon + h) 2
e.g., h = 0 m  g = 1.62 m/s2
Newoton's Law of Universal Gravitation
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Gravimeters
1. Accuracy: Dg/g ~ 1 part in 109
2. Applications:
Mineral deposits
r > rave  g > gave
Salt domes (i.e., petroleum)
r < rave  g < gave
Newoton's Law of Universal Gravitation
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Average Density of the Earth
g = 9.80 m/s2
RE = 6.37 x 106 m
ME = 5.96 x 1024 kg
 rE = 5.50 x 103 kg/m3
= 5.50 g/cm3 ~ 2 x rRock
 What can we conclude about the
structure of the Earth?
Newoton's Law of Universal Gravitation
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Interior of the Earth
• The earth is approximately
spherically symmetric, but it
is not uniform throughout its
volume, as shown in Figure
13.9 at the right.
• Follow Example 13.4, which
shows how to calculate the
weight of a robotic lander on
Mars.
International Space Station (1)
www.station.nasa.gov
Weight on the Earth
4.22 x 106 N
Weight at h = 350 km?
5 kg
M
=
4.31
x
10
S
Newoton's Law of Universal Gravitation
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International Space Station (2)
www.station.nasa.gov
Weight on the Earth
4.22 x 106 N (MS = 4.31 x 105 kg)
Weight at h = 350 km
F = MS g(h)
= MS G ME / (RE + h)2
Newoton's Law of Universal Gravitation
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Weight
• The weight of a body decreases with its distance from the
earth’s center, as shown in Figure 13.8 below.
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Apparent Weight (FN)
F.B.D.
y
a = (1/2)g (up)
Fnet = FN – mg = m a
Newoton's Law of Universal Gravitation
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Weightlessness
F.B.D. ?
y
a = g (down)
Newoton's Law of Universal Gravitation
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Circular Satellite Orbits
• For a circular orbit, the speed of a satellite is just right to keep its distance
from the center of the earth constant. (See Figure 13.15 below.)
• A satellite is constantly falling around the earth. Astronauts inside the
satellite in orbit are in a state of apparent weightlessness because they are
falling with the satellite. (See Figure 13.16 below.)
• Follow Example 13.6.
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Gravitational Potential Energy
• Derivation of
gravitational potential
energy using Fig. 13.10
at the right.
• The gravitational
potential energy of a
system consisting of a
particle of mass m and
the earth is
U = –GmEm/r.
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Gravitational Potential Energy
• The gravitational potential
energy of the earthastronaut system increases
(becomes less negative) as
the astronaut moves away
from the earth, as shown in
Fig. 13.11 at the right.
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From the Earth to the Moon to 
 To escape from
the earth, an
object must have
the escape speed.
 See Example 13.5
using Figure 13.12
Ki  U i  K f  U f
 K i  U G ,i  K f  U G , f


r  RE
r 
1 2
  M E  m 
  mvescape    
00
2

RE 
2
 
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Chap. 13: Gravitation
Newton’s Law of Universal Gravity
– Attractive Force (Vector)
– Inverse Square Law
– Universality
Kepler’s Laws
Taken from Fig. 2.16, “the Universe in a Nutshell” by Stephen Hawking
43
Kepler’s Laws of Planetary Motion
s = semi-major axis
F1Q + F1R = 2 xNewoton's
s Law of Universal Gravitation
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Elliptical and Circular Orbits
A special case of
elliptical orbit is
circular orbit:
e (eccentricity) = 0
Newoton's Law of Universal Gravitation
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Kepler’s 2nd Law
(2) A12 = A34
v12 > v34
A
d12
d v Dt
d34
Newoton's Law of Universal Gravitation
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Example 6
Use Kepler’s second law to show that the ratio
of the speeds of a planet at its nearest and
farthest points from the Sun is equal to the
inverse ratio of the nearest and farthest
distances: vN/vF = dF/dN
Newoton's Law of Universal Gravitation
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Kepler’s 3rd Law
e = 0 (circular orbit)  s = r
T=2pr/v
G MP MS / r 2 = MP v 2 / r
Frad
= m arad
 r 3 = (G MS /4 p 2) T 2
 T 2 = (4 p 2 / G MS ) r 3 = KSun r 3
 r (circular orbit)
s (elliptical orbit)
Newoton's Law of Universal Gravitation
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Satellite about Earth:
Circular Orbit
T=2pr/v
G MS ME / r 2 = MS v 2 / r
Frad
= m arad
 r 3 = (G ME /4 p 2) T 2
 T 2 = (4 p 2 / G ME ) r 3 = KE r
3
Newoton's Law of Universal Gravitation
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•
•
•
•
•
•
•
•
G = 6.67 x 1011 Nm2/kg2
ME = 5.98 x 1024 kg
RE = 6.38 x 103 km
MS = 1.99 x 1030 kg
2
RS = 6.96 x 105 km
rE-S = 150 x 106 km
s xp10a22ckge s h i p
MmoonA
= 7.35
Rmoon = 1.74 x 103 km
Example 1
T = (4 p 2 / G ME ) r 3 = KE r
3
(mass m = 300 kg) is in an
elliptical orbit around the Earth. At perigee (apogee) of its
orbit, it is 620 km (3620 km) above
P
the Earth’s surface. Calculate the
period of the orbit.
Note: RE = 6380 km.
Newoton's Law of Universal Gravitation
(1) s = …
(2) T 2 = KE s 3  T = …
Newoton's Law of Universal Gravitation
A
59
•
•
•
•
•
•
•
•
G = 6.67 x 1011 Nm2/kg2
ME = 5.98 x 1024 kg
RE = 6.38 x 103 km
MS = 1.99 x 1030 kg
RS = 6.96 x 105 km
rE-S = 150 x 106 km
Mmoon = 7.35 x 1022 kg
Rmoon = 1.74 x 103 km
Example 2
T 2 = (4 p 2 / G ME ) r 3 = KE r
3
The National Aggie Space Administration
(NASA) wants to send a satellite (300 kg)
into a circular orbit of radius R from the
center of the Earth at the rotational period
of T = 240 hours. Find R and v.
Newoton's Law of Universal Gravitation
(1) T = …
(2) T 2 = KE R 3  R = …
(3) v = 2 p R / T = …
Newoton's Law of Universal Gravitation
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•
•
•
•
•
•
•
•
G = 6.67 x 1011 Nm2/kg2
ME = 5.98 x 1024 kg
RE = 6.38 x 103 km
MS = 1.99 x 1030 kg
RS = 6.96 x 105 km
rE-S = 150 x 106 km
Mmoon = 7.35 x 1022 kg
Rmoon = 1.74 x 103 km
Example 3
T 2 = (4 p 2 / G ME ) r 3 = KE r
3
A geosynchronous satellite (m = 300 kg) is one
that stays above the same point on the Earth,
which is possible only if it is above a point of the
equator. Determine the height H above the
Earth’s surface such a satellite must orbit.
Newoton's Law of Universal Gravitation
(1) T = 1 day = … s
(2) T 2 = KE R 3  R = … m
(3) H = R – RE = … m (… km)
Newoton's Law of Universal Gravitation
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•
•
•
•
•
•
•
•
G = 6.67 x 1011 Nm2/kg2
ME = 5.98 x 1024 kg
RE = 6.38 x 103 km
MS = 1.99 x 1030 kg
RS = 6.96 x 105 km
rE-S = 150 x 106 km
Mmoon = 7.35 x 1022 kg
Rmoon = 1.74 x 103 km
Example 4
T 2 = (4 p 2 / G MS ) r 3 = KS r
3
Halley’s comet orbits the Sun roughly once
every 76 years. It comes very close to the
surface of the Sun on its closest approach
about how far out from the Sun is it at its
farthest ?
Newoton's Law of Universal Gravitation
Newoton's Law of Universal Gravitation
66
x
s
s
Newoton's Law of Universal Gravitation
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•
•
•
•
•
•
•
•
G = 6.67 x 1011 Nm2/kg2
ME = 5.98 x 1024 kg
RE = 6.38 x 103 km
MS = 1.99 x 1030 kg
RS = 6.96 x 105 km
rE-S = 150 x 106 km
Mmoon = 7.35 x 1022 kg
Rmoon = 1.74 x 103 km
Example 4
T 2 = (4 p 2 / G MS ) r 3 = KS r
3
Halley’s comet orbits the Sun roughly once
every 76 years. It comes very close to the
surface of the Sun on its closest approach
about how far out from the Sun is it at its
farthest ?
Newoton's Law of Universal Gravitation
(1) T = 76 yrs = … s
(2) T 2 = KS s 3  s = … m
(3) s >> RS  x = … m
Newoton's Law of Universal Gravitation
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Example 5
When it orbited the Moon, the Apollo 11
spacecraft’s mass was 9.979 x 103 kg, its
period was 7.140 x 103 s, and its mean
distance from the Moon’s center was
1.849 x 106 m. Find v and MMoon.
(1) v = 2 p R / T = … m/s
(2) arad = v2 / R = … m/s2
(3) G m MMoon / R2 = m aradMMoon= … kg
Newoton's Law of Universal Gravitation
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Newoton's Law of Universal Gravitation
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Appendix
Gravitation
Determining the Value of G
• In 1798 Henry Cavendish made the first measurement of the
value of G. Figure 13.4 below illustrates his method.
Some gravitational calculations
• Example 13.1 shows how to calculate the gravitational force
between two masses.
• Example 13.2 shows the acceleration due to gravitational force.
• Example 13.3 illustrates the superposition of forces, meaning that
gravitational forces combine vectorially. (See Figure 13.5 below.)
Weight
• The weight of a body is the total gravitational force exerted
on it by all other bodies in the universe.
• At the surface of the earth, we can neglect all other
gravitational forces, so a body’s weight is w = GmEm/RE2.
• The acceleration due to gravity at the earth’s surface is
g = GmE/RE2.
The motion of satellites
• The trajectory of a projectile fired from A toward B depends on
its initial speed. If it is fired fast enough, it goes into a closed
elliptical orbit (trajectories 3, 4, and 5 in Figure 13.14 below).
Kepler’s laws and planetary motion
• Each planet moves in an
elliptical orbit with the
sun at one focus.
• A line from the sun to a
given planet sweeps out
equal areas in equal
times (see Figure 13.19
at the right).
• The periods of the
planets are proportional
to the 3/2 powers of the
major axis lengths of
their orbits.
Some orbital examples
• Follow Conceptual Example 13.7 on orbital speeds.
• Follow Example 13.8 involving Kepler’s third law.
• Example 13.9 examines the orbit of Comet Halley. See Figure
13.20 below.
Spherical mass distributions
• Follow the proof that the gravitational interaction between two spherically
symmetric mass distributions is the same as if each one were concentrated at its
center. Use Figure 13.22 below.
A point mass inside a spherical shell
• If a point mass is inside a spherically symmetric shell, the
potential energy of the system is constant. This means that
the shell exerts no force on a point mass inside of it.
• Follow Example 13.10 using Figure 13.24 below.
Apparent weight and the earth’s rotation
• The true weight of an
object is equal to the
earth’s gravitational
attraction on it.
• The apparent weight of
an object, as measured
by the spring scale in
Figure 13.25 at the
right, is less than the
true weight due to the
earth’s rotation.
Black holes
• If a spherical nonrotating body has radius less than the Schwarzschild radius,
nothing can escape from it. Such a body is a black hole. (See Figure 13.26
below.)
• The Schwarzschild radius is RS = 2GM/c2.
• The event horizon is the surface of the sphere of radius RS surrounding a black
hole.
• Follow Example 13.11.
Detecting black holes
• We can detect black holes by looking for x rays emitted
from their accretion disks. (See Figure 13.27 below.)