Transcript Slide 1

2/21 Do Now
• A particle of mass m rotates with a uniform speed on the
inside of a bowl’s parabolic frictionless surface in a
horizontal circle of radius R = 0.4 meters as shown below.
At the position of the particle the surface makes an angle θ
= 17o with the vertical. What is the angular velocity of the
particle?
v
R = 0.4 m
θ
m
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Chapter 12
Gravitation
PowerPoint® Lectures for
University Physics, Twelfth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by James Pazun
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Goals for Chapter 12
• To study Newton’s Law of Gravitation
• To consider gravitational force, weight, and
gravitational energy
• To compare and understand the orbits of satellites
and celestial objects
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
12.1 Newton’s Law of Gravitation
• Every particle of matter in the universe attracts every other
particle with a force that is directly proportional to the product
of the masses of the particles and inversely proportional to the
square of the distance between them.
G  m1  m2
Fg 
2
r
CAUTION
Don’t confuse g and G
• Lowercase g is the acceleration due to gravity, which relates
the weight w of a body to its mass m: w = mg. The value of g
is different at different locations on the earth's surface and
on the surfaces of different planets.
• By contrast, capital G relates the gravitational force between
any two bodies to their masses and the distance between them.
Henry Cavendish measured the value of G in 1798. We call G
a universal constant because it has the same value for any tow
bodies, no matter where in space they are located.
11
G  6.6710 N  m / kg
2
2
Example 12.1 Calculating gravitational force
The mass m1 of one of the small spheres of a Canvendish balance is
0.0100 kg, the mass m2 of one of the large spheres is 0.500 kg. And
the center-to-center distance between each large sphere and the
nearer small one is 0.0500 m. Find the gravitational force Fg on each
sphere due to the nearest other sphere.
Example 12.2 Acceleration due to
gravitational attraction
Suppose one large sphere and one small sphere are detached from
the apparatus in Example 12.1 and placed 0.0500 m from each other
at a point in space far removed from all other bodies. What is the
magnitude of the acceleration of each?
Example 12.3 superposition of gravitational forces
Many stars in the sky are actually systems of two or more stars held
together by their mutual gravitational attraction. The figure shows a
three-star system at an instant when the stars are at the vertices of a
45o right triangle. Find the magnitude and direction of the total
gravitational force exerted on the small star by the two large ones.
12.2 Weight
• We defined the weight of a body as the attractive gravitational
force exerted on it by the earth.
• The broaden definition of weight is: The weight of a body is
the total gravitational force exerted on the body by all other
bodies in the universe.
• When the body is near the surface of the earth, we can neglect
all other gravitational forces and consider the weight as just
the earth’s gravitational attraction.
• At the surface of the moon we consider a body’s weight to be
the gravitational attraction of the moon, so on.
• If we model the earth as a spherically symmetric body with radius
RE and mass mE, the weight w of a small body of mass m at the
earth’s surface is
G  mE  m
Fg 
2
RE
We also know that weight w of a body is the force that causes the
acceleration g of free fall, w = mg. Equating this with the above
equation, we find
G  mE
g
2
RE
The acceleration due to gravity g is independent of the mass m of
the body. This equation allows us to calculate the mass of the
earth:
Example 12.4 Gravity on Mars
An unmanned Lander is sent to the surface of the planet Mars, which
has radius RM = 3.40 x 106 m and mass mM = 6.42 x 1023 kg. The earth
weight of the Mars Lander is 3290 N. Calculate its weight Fg and the
acceleration gM due to the gravity of Mars:
a. 6.0 x 106 m above the surface of Mars
b. At the surface of Mars.
“Journey to the center of the earth”
• Suppose we drill a hole through the earth (radius RE, mass mE)
along a diameter and drop a mail pouch (mass m) down the
hole. Derive an expression for the gravitational force on the
pouch as a function of its distance r from the center. Assume
that the density of the earth is uniform.
Weight
• Gravity (and hence, weight) decreases as altitude rises outside Earth
and also decrease as approaching the center of Earth and becomes
zero at the center of Earth.
• The apparent weight of a body
on earth differs slightly from
the earth’s gravitational force
because the earth rotates and is
therefore not precisely an
inertial frame of reference.
• We have used the fact that the
earth is an approximately
spherically symmetric
distribution of mass. But this
does not mean that the earth is
uniform. In fact, the density of
the earth decreases with
increasing distance from its
center.
Test Your Understanding 12.2
•
1.
2.
3.
4.
Rank the following hypothetical planets in order from
highest to lowest surface gravity:
Mass = 2 times the mass of earth radius = 2 times the radius
of the earth;
Mass = 4 times the mass of the earth, radius = 4 times the
radius of the earth
Mass = 4 times the mass of the earth, radius = 2 times the
radius of the earth;
Mass = 2 times the mass of the earth, radius = 4 times the
radius of the earth.
3, 1, 2, 4
2/24 Do Now
• Ball B (moment of inertial about its center ⅔ MR2), rolls
down the distance x along the inclined plane without
slipping. In terms of acceleration due to earth’s gravity g,
what is the acceleration of ball B along the inclined plane?
A
B
x
h
30o
horizontal
Extra credit is due
Test is on Tuesday, 3/4
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Brief review
• MC packet: #21, 23, 24, 25, 32,36
12.3 Gravitational Potential Energy
• To determine the gravitational potential energy of an object at a
height that is way above the earth’s surface, we need to
calculate work done by the gravitational force on the object and
use equation:
GM m
r2
Wgrav   Fg  dr
Fg  
r
E
2
r1
r2
Wgrav
GM E m
GM E m GM E m
  (
)dr 

2
r
r2
r1
r1
Wgrav = - ∆U = - (U2 – U1 ) = U1 – U2
GM E m
U 
r
• Gravitational potential energy
depends on the distance r between
the body of mass m and the center
of the earth. When the body
moves away from the earth, r
increases, the gravitational force
does negative work, and U
increases (becomes less negative).
When the body “falls” toward
earth, r decreases, the
gravitational work is positive,
energy decreases (becomes more
negative).
• Note that U is relative. Only ∆U
is significant.
Example 12.5 “From the earth to the moon”
In Jules Verne’s 1865 story with this title, three men were sent to the
moon in a shell fired from a giant cannon sunk in the earth in Florida.
a. Find the muzzle speed needed to shoot the shell straight up to a
height above the earth equal to the earth’s radius.
b. Find the escape speed – that is, the muzzle speed that would
allow the shell to escape from the earth completely. Neglect air
resistance, the earth’s rotation, and the gravitational pull of the
moon. (RE = 6380 km, ME = 5.97 x 1024 kg)
More on Gravitational Potential Energy
U 
GM E m
r
• Extra credit on test – use the above equation to prove that near
Earth’s surface the potential energy is U  m gh
Test Your Understanding 12.3
Is it possible for a planet to have the same surface gravity as the
earth (that is the same value of g at the surface) and yet have a
greater escape speed?
Example
•
1.
2.
3.
4.
5.
The planet Saturn has about 100 times the mass of the earth
and is about 10 times farther from the sun than the earth is.
Compared to the acceleration of the earth caused by the
sun’s gravitational pull, how great is the acceleration of
Saturn due to the sun’s gravitation? [show work]
100 times greater
10 times greater
The same
1/10 as great
1/100 as great.
12.4 Satellite motion
Trajectories 1 through 7 show the
effect of increasing the initial speed.
Trajectories 1 through 5 close on
themselves and are called closed
orbits. All closed orbits are ellipses;
Trajectory 4 is a circle, a special
case of an ellipse.
Trajectories 6 and 7 are open orbits. For these paths the projectile
never returns to its starting point but travels ever farther away from
the earth.
Satellites: Circular Orbits
A circular orbit is the simplest and
also an important case because
artificial satellites as well as planets
around the sun have nearly circular
orbits.
Since the only force acing on a
satellite in circular orbit around the
earth is the earth’s gravitational
attraction, which is directed
toward the center of the earth and
hence toward the center of the
orbit, the satellite is in uniform
circular motion with constant
speed.
To find the constant speed v of a satellite in a circular orbit with
radius r measured from the center of the earth, we need to use
Newton’s 2nd law: Fnet = ma
GM E m m v2

2
r
r
GM E
v
r
This relationship shows that we can’t chose the orbit radius r and the
speed v independently; for the given radius r, the speed v for a
circular orbit is determined. The larger the orbit, the slower the
speed.
The equation also shows that the motion of a satellite does not
depend on its mass. Any satellite launched in the same orbit will
have the same speed.
An astronaut on board a space shuttle is a satellite of the earth in
the same orbit, has the same velocity and acceleration as the shuttle.
The only force acting on the astronaut is gravity, so the astronaut is
experiencing apparent weightlessness as in a freely falling elevator;
Both the international space station and the moon are satellites of the
earth. The moon orbits much farther from the center of the earth than
does the space station, so its has a slower orbital speed and a longer
orbital period.
International space station
R = 6800 km
V = 7.7 km/s
T = 93 min
Moon
Distance from center of earth = 384,000 km
V = 1.0 km/s
T = 27.3 days
The relationship between r and T
In circular motion:
2r
v
T
2r
T

v
2r
2r 3 / 2

GM E
GM E
r
The equation shows that larger orbits longer periods.
Comparing circular orbit speed with
escaping speed
Circular orbit speed:
escaping speed:
GM E
v
r
2GM E
v
r
The escape speed from a spherical body with radius R is 2 times
greater than the speed of a satellite in a circular orbit at that
radius.
If our spacecraft is in circular orbit around any planet, we have to
multiply our speed by a factor of 2 to escape to infinity,
regardless of the planet’s mass.
The total mechanical energy E = K + U
Circular orbit speed:
GM E
v
r
1 GM E
GM E m
GM E m
E m
 (
)
2
r
r
2r
The total mechanical energy in a circular orbit is negative and equal
to ½ the potential energy.
Increasing the orbit radius r means increasing the mechanical
energy (making E less negative).
If the satellite is in a relatively low orbit that encounters the outer
fringes of earth’s atmosphere, mechanical energy decreases due to
negative work done by the force of air resistance; as a result, the
orbit radius decreases until the satellite hits the ground or burns up
in the atmosphere.
Example 12.6 A satellite orbit
Suppose you want to place a 1000 kg weather satellite into a
circular orbit 300 km above the earth’s surface.
a. What speed, period and radial acceleration must it have?
b. How much work has to be done to place this satellite into orbit?
c. How much additional work would have to be done to make this
satellite escape the earth?
(RE=6380 km, mE=5.97 x 1024 kg)
Test Your Understanding 12.4
• Your personal spacecraft is in a low-altitude circular orbit around
the earth. Air resistance from the outer regions of the atmosphere
does negative work on the spacecraft, causing the orbital radius to
decrease slightly. Does the speed of the spacecraft
1. remain the same
2. increase
3. decrease
12.5 Kepler’s laws for planetary motion
• 1st Law: Each planet moves
in an elliptical orbit with the
sun at one focus.
• 2nd Law: A line connecting
the sun to a given planet
sweeps out equal areas in
equal times.
• Perihelion is the point
closest to the sun;
• Aphelion is the point
furthest from the sun.
• Semi-Major axis is the
length of a
• ea = e∙a: the distance
from each focus to the
center of the ellipse.
• Eccentricity (e): a
dimensionless number
between 0 and 1. If e=0,
ea=0, the ellipse is a
circle.
Kepler’s Third Law
The periods of the planets are proportional to the 3/2 powers of
the semi-major axis lengths in their orbits
2a
T
Gm s
3/ 2
Where ms is the mass of the sun. a is the semi major axis.
Note that the period does not depend on the eccentricity e.
An asteroid in an elongated elliptical orbit with semi-major axis a
will have the same orbital period as a planet in a circular orbit of
radius a. The key difference is that the asteroid moves at different
speeds at different points in its elliptical orbit, while the planet’s
speed is constant around its circular orbit.
Example 12.7 Orbital Speeds
At what point in an elliptical orbit does a planet have the greatest
speed?
The speed v is maximum
at perihelion.
Example 12.8 Kepler’s third Law
The asteroid Pallas has an orbital period of 4.62 years and an orbital
eccentricity of 0.233. Find the semi-major axis of its orbit.
(ms=1.99x1030 kg)
Example 12.9 Comet Halley
Comet Halley moves in an elongated elliptical orbit around the sun.
at perihelion, the comet is 8.75 x 107 km from the sun; at aphelion,
it is 5.26 x 109 km from the sun. Find the semi-major axis,
eccentricity, and period of the orbit.
Planetary Motions and the Center of Mass
Test Your Understanding 12.5
• The orbit of Comet X has a semi-major axis that is
four times larger than the semi-major axis of Comet
Y . What is the ratio of the orbital period of X to the
orbital period of Y?
T1 / T2 = (a1/a2)3/2 = (4)3/2 = 8
Example
•
1.
2.
3.
4.
5.
The planet Saturn has about 100 times the mass of the earth
and is about 10 times farther from the sun than the earth is.
Compared to the acceleration of the earth caused by the
sun’s gravitational pull, how great is the acceleration of
Saturn due to the sun’s gravitation? [show work]
100 times greater
10 times greater
Fg = GmEms/r2
The same
2
a
=
Gm
/r
E
s
E
1/10 as great
1/100 as great.
aSaturn = Gms/rSaturn2
aSaturn = Gms/(10rE)2
aSaturn = (1/100) aE
Example
•
How far from a very
small 150 kg ball
would a particle
have to be placed
so that the ball
pulled on the
particle just as hard
as the earth does?
Example
•
What is the escape
speed from an
asteroid of diameter
255 km with a
density of 2720
kg/m3 ?
88.7 m/s
Example
a.
b.
How far from a very
small 120 kg ball would
a particle have to be
placed so that the ball
pulled on the particle
just as hard as the
earth does?
Is it reasonable that you
could actually set up
this as an experiment?
Why?
2.86 x10-5 m
More on Gravitational Potential Energy
GM E m
U 
r
GM E m
GM E m
U  (
)  (
)
r2
r1
GM E m GM E m GM E m(r2  r1 )
U 


r1
r2
r1  r2
• On Earth’s surface, r1 ≈ r2 = RE; (radius of the earth) and
r2 – r1 = h (height above Earth’s surface)
GM E  m  h
U 
2
RE
GM E
g
2
RE
U  m gh