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SECTION 3
Preview Vocabulary
Greek and Latin Word Origins The
word epicycle is rooted in both Latin
and Greek. In both languages, it is used
in the context of math and astronomy
concepts. This term characterizes a circle
whose center is moving around the
circumference of a larger circle.
Objectives
Describe Kepler’s laws of
planetary motion.
Relate Newton’s mathematical
analysis of gravitational force to
the elliptical planetary orbits
proposed by Kepler.
Motion in Space
Kepler's Laws
Solve problems involving orbital
speed and period.
In the second century CE, Claudius Ptolemy developed an elaborate
theory of planetary motion. Ptolemy’s theory attempted to reconcile
observation with theory and to keep Earth at the center of the universe.
In this theory, planets travel in small circles called epicycles while
simultaneously traveling in larger circular orbits. Even Ptolemy’s complex
model did not fully agree with observation, although the model did
explain more than previous theories.
Teach
Misconception Alert!
Students may believe the Copernican
heliocentric model solved all of the
problems inherent in the geocentric
model of the solar system. Point out
that even Copernicus’s model was not as
simple or as accurate as it sounded. He
was not able to eliminate the need for
epicycles. In an attempt to rectify the
problem, Tycho Brahe introduced a
modified geocentric model in which
all the planets except Earth revolved
around the sun, which in turn revolved
around Earth. All of these models failed
because they were dependent on the
ancient belief that orbits must be
perfectly circular.
People have studied the motions of the planets since ancient times. Until
the middle of the 16th century, most people believed that Earth was at the
center of the universe. Originally, it was believed that the sun and other
planets orbited Earth in perfect circles. However, this model did not
account for all of the observations of planetary motion.
FIGURE 3.1
Planetary Model This elaborate
planetary model—called an orrery—
shows the motions of Mercury, Venus,
and Earth around the sun. The model
also shows the moon’s inclined orbit
around Earth.
In 1543, the Polish astronomer Nicolaus Copernicus (1473–1543)
published On the Revolutions of the Heavenly Spheres, in which he
proposed that Earth and other planets orbit the sun in perfect circles.
Figure 3.1 shows a sun-centered planetary model that is believed to have
been made for King George III of England. The idea of a sun-centered
universe was not completely new in the 16th century. A Greek named
Aristarchus theorized 1700 years before Copernicus did that Earth revolved
around the sun, but most other scientists did not accept his theory.
Kepler’s three laws describe the motion of the planets.
The astronomer Tycho Brahe (1546–1601) made many precise observations of the planets and stars. However, some of Brahe’s data did not agree
with the Copernican model. The astronomer Johannes Kepler (1571–1630)
worked for many years to reconcile Copernican theory with Brahe’s data.
Kepler’s analysis led to three laws of planetary motion, which were
developed a generation before Newton’s law of universal gravitation.
Kepler’s three laws can be summarized as shown on the next page.
Kepler’s Laws of Planetary Motion
First Law: Each planet travels in an elliptical orbit around the sun,
and the sun is at one of the focal points.
Second Law: An imaginary line drawn from the sun to any planet
sweeps out equal areas in equal time intervals.
Third Law: The square of a planet’s orbital period (T 2) is
proportional to the cube of the average distance (r 3)
between the planet and the sun, or T 2 ∝ r 3.
©Adler Planetarium & Astronomy Museum
Plan and Prepare
SECTION 3
238
Chapter 7
Differentiated
Instruction
Pre-AP
Students may think that Aristarchus’s
heliocentric model was rejected because it was
unorthodox. Explain that his theory, though
unconventional, was actually tested. If Earth
were moving about the sun, the apparent
position of the foreground stars should shift
with respect to the apparent position of the
background stars. The same shift happens when
you are looking at a nearby object with respect
to the horizon and then you move to the side.
Untitled-53 238
238 Chapter 7
This shifting, called parallax, was not
observed; however, the failure to observe
stellar parallax was due simply to the very
great distance of the stars compared to the
size of Earth’s orbit. The first measurement of
stellar parallax was not made until 1838, when
telescopes were large enough to detect small
stellar motions.
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Untitled-53 239
Kepler’s first law states that the planets’ orbits are
ellipses rather than circles. Kepler discovered this law
while working with Brahe’s data for the orbit of Mars.
While trying to explain the data, Kepler experimented
with 70 different circular orbits and generated numerous pages of calculations. He finally realized that if the
orbit is an ellipse rather than a circle and the sun is at
one focal point of the ellipse, the data fit perfectly.
Kepler’s second law states that an imaginary line
from the sun to any planet sweeps out equal areas in
equal times, as shown in Figure 3.2. In other words, if
the time a planet takes to travel the arc on the left (∆t1)
is equal to the time the planet takes to cover the arc on
the right (∆t2), then the area A1 is equal to the area A2.
It is easy to see from Figure 3.2 that planets travel faster
when they are closer to the sun.
FIGURE 3.2
Kepler’s Second Law This diagram illustrates a planet
moving in an elliptical orbit around the sun. If ∆t1 equals ∆t2 ,
then the two shaded areas are equal. Thus, the planet travels
faster when it is closer to the sun and slower when it is farther
away.
t1
A1
Sun
A2
t2
While Kepler’s first two laws describe the motion of each planet
individually, his third law relates the orbital periods and distances of one
planet to those of another planet. The orbital period (T) is the time a
planet takes to finish one full revolution, and the distance (r) is the mean
distance between the planet and the sun. Kepler’s third law relates the
orbital period and mean distance for two orbiting planets as follows:
QuickLab
Teacher’s Notes
2
Students should find that T 2 = ​ ___
​ 4π   ​   ​r 3.
Gm
2
This equation expresses that T is
2
proportional to r 3, and that ___
​ 4π    ​is the
Gm
proportionality constant. This is a more
detailed form of Kepler’s third law
(T 2 ∝ r 3). Be sure students understand
that m is the mass of the object being
orbited.
Homework Options This QuickLab can
easily be performed outside of the
physics lab room.
( )
r3
T12 _
_
= 13 , or T 2 ∝ r 3
2
T2
r2
This law also applies to satellites orbiting Earth, including our moon.
In that case, r is the distance between the orbiting satellite and Earth.
The proportionality constant depends on the mass of the central object.
Kepler’s laws are consistent with Newton’s law of gravitation.
Newton used Kepler’s laws to support his law of gravitation. For example,
Newton proved that if force is inversely proportional to distance squared,
as stated in the law of universal gravitation, the resulting orbit must be an
ellipse or a circle. He also demonstrated that his law of gravitation could
be used to derive Kepler’s third law. (Try a similar derivation yourself in
the QuickLab below.) The fact that Kepler’s laws closely matched observations gave additional support for Newton’s theory of gravitation.
KEPLER’S THIRD LAW
You can mathematically show
how Kepler’s third law can be
derived from Newton’s law of
universal gravitation (assuming
circular orbits). To begin, recall
that the centripetal force is
provided by the gravitational force.
Set the equations for gravitational
and centripetal force equal to one
another, and solve for v 2t . Because
speed equals distance divided by
time and because the distance for
Problem Solving
one period is the circumference
(2πr), vt = 2πr/T. Square this
value, substitute the squared value
into your previous equation, and
then isolate T 2. How does your
result relate to Kepler’s third law?
Circular Motion and Gravitation
239
Take it Further
Show students how the mass of a celestial
object can be calculated if its radius and
period are known.
5/9/2011 11:01:06 AM
GmT 2 = 4π 2r 3
Rearrange the variables as follows to develop a
formula for finding the mass of the satellite:
2
2 3
_
​ GmT  ​ 
 = _
​ 4π r 2 ​ 
 
2
GT GT 2 3
m=_
​ 4π r 2 ​ 
GT Circular Motion and Gravitation 239
Kepler’s third law describes orbital period.
Teach continued
Teaching Tip
Explain how Kepler’s third law provides
a fairly simple and powerful means of
measuring the mass of stars and planets.
In general, the mass used in the third law
should equal the sum of the orbiting
body’s mass and the mass of the body
that is orbited. While in most cases the
orbiting body’s mass is so small that it
can be ignored, other cases, such as the
two stars in a binary star system, require
that one of the masses be known or
realistically estimated. The masses of
many stars are determined from the
stars’ colors and levels of brightness.
Kepler’s third law is especially useful
for binary star systems in which one of
the companions is small and thus hard
to see but is very massive. Examples of
small but massive bodies include white
dwarfs and neutron stars. A more
dramatic example is a black hole. The
mass of the invisible companion can be
estimated by observing the orbital
motions of a visible companion, whose
mass can be estimated from its light.
Did YOU Know?
We generally speak of the moon
orbiting Earth, but they are actually
both in orbit around a point in the
Earth-moon system. Because Earth
is so much more massive than the
moon, the point they orbit around lies
inside Earth. Thus, the moon appears
to orbit Earth. The point around which
two bodies orbit does not always lie
inside one of the bodies, however. For
example, Pluto and its moon, Charon,
orbit a point that lies between them.
Also, many binary star systems have
two stars that orbit a common center
between the stars.
According to Kepler’s third law, T 2 ∝ r 3. The constant of proportionality
between these two variables turns out to be 4π 2/Gm, where m is the mass
of the object being orbited. (To learn why this is the case, try the QuickLab
on the previous page.) Thus, Kepler’s third law can also be stated as follows:
( )
_2    r 3
T 2 = 4π Gm
The square root of the above equation, which is shown below on the left,
describes the period of any object that is in a circular orbit. The speed of an
object that is in a circular orbit depends on the same factors that the period
does, as shown in the equation on the right. The assumption of a circular
orbit provides a close approximation for real orbits in our solar system
because all planets except Mercury have orbits that are nearly circular.
Period and Speed of an Object in Circular Orbit

r   v = 
    Gm
 _
m
√G √_
r

(mean radius)
 
orbital period = 2π √____
 
 
 
(constant)(mass of central object)
T = 2π
3
t
3
√ 
(
)

mass of central object
___
orbital speed = (constant)  
 
 
mean radius
Note that m in both equations is the mass of the central object that is
being orbited. The mass of the planet or satellite that is in orbit does not
affect its speed or period. The mean radius (r) is the distance between the
centers of the two bodies. For an artificial satellite orbiting Earth, r is
equal to Earth’s mean radius plus the satellite’s distance from Earth’s
surface (its “altitude”). Figure 3.3 gives planetary data that can be used to
calculate orbital speeds and periods.
Figure 3.3
Planetary Data
Planet
Mass
(kg)
Mean
Mean distance Planet
radius (m) from sun (m)
Mass
(kg)
Mean
Mean distance
radius (m) from sun (m)
Earth
5.97 × 1024
6.38 × 106
1.50 × 1011
Neptune 1.02 × 1026 2.48 × 107
4.50 × 1012
Earth’s
moon
7.35 × 1022
1.74 × 106
——
Saturn
5.68 × 1026 6.03 × 107
1.43 × 1012
Jupiter
1.90 × 1027
7.15 × 107
7.79 × 1011
Sun
1.99 × 1030 6.96 × 108
——
Mars
6.42 × 1023
3.40 × 106
2.28 × 1011
Uranus
8.68 × 1025 2.56 × 107
2.87 × 1012
Mercury 3.30 × 1023
2.44 × 106
5.79 × 1010
Venus
4.87 × 1024 6.05 × 106
1.08 × 1011
240
Chapter 7
Problem
Solving
Take It Further
Have students use values in the table to
confirm Kepler’s third law. (They should get the
2
same ratio of __
​ T    ​for each planet.) Students who
r 3
did the QuickLab on Kepler’s third law can also
see how the orbital speed equation is derived.
They simply need to take the square root of
their expression for ​v ​2t​ ​ .
PH_CNLESE586694_C07S3.indd 240
240 Chapter 7
4/1/2013 9:52:57 PM
Period and Speed of an Orbiting Object
Classroom Practice
Sample Problem D The color-enhanced image of Venus
shown here was compiled from data taken by Magellan, the first
planetary spacecraft to be launched from a space shuttle. During
the spacecraft’s fifth orbit around Venus, Magellan traveled at a
mean altitude of 361 km. If the orbit had been circular, what
would Magellan’s period and speed have been?
AnAlyze
Given:
r1 = 361 km = 3.61 × 105 m
Unknown:
T=?
Period and Speed of an
Orbiting Object
A large planet orbiting a distant star is
discovered. The planet’s orbit is nearly
circular and fairly close to the star. The
planet’s orbital distance is 7.50 × 1010 m,
and its period is 105.5 days. Calculate the
mass of the star.
Answer: 3.00 × 1030 kg
vt = ?
PlAn
Choose an equation or situation: Use the equations for the period and
speed of an object in a circular orbit.
T = 2π

r
√_
Gm
3
vt =
PROBLEM guide D
m
G_
√
r
Use this guide to assign problems.
SE = Student Edition Textbook
PW = Sample Problem Set I (online)
PB = Sample Problem Set II (online)
Solving for:
Use Figure 3.3 to find the values for the radius (r2) and mass (m) of Venus.
r2 = 6.05 × 106 m
m = 4.87 × 1024 kg
Find r by adding the distance between the spacecraft and Venus’s
surface (r1) to Venus’s radius (r2).
r = r1 + r2 = (3.61 × 105 m) + (6.05 × 106 m) = 6.41 × 106 m
Solve
vt =
©NASA/JPL/USGS
ntitled-53 241
SE Sample, 1;
Ch. Rvw. 27, 49b
PW 3–4
PB 5–6
T
SE Sample, 1;
Ch. Rvw. 27, 49b
PW 5–6
PB 7–8
r
SE 2; Ch. Rvw. 28
PW Sample, 1–2
PB 3–4
m
SE Ch. Rvw. 29
PW 7–8
PB Sample, 1–2
Substitute the values into the equations and solve:
T = 2π
CHeCK
yoUR WoRK
vt
√(

(6.41 × 106 m)3
____
= 5.66 × 103 s
2
N
•
m
6.673 × 10-11 _
(4.87 × 1024 kg)
kg2
√(
)
)(
)

× 1024 kg
N•m2 4.87
__
6.673 × 10-11 _
= 7.12 × 103 m/s
2
6.41 × 106 m
kg
Magellan takes (5.66 × 103 s)(1 min/60 s) ≈ 94 min to complete
one orbit.
1. Find the orbital speed and period that the Magellan satellite from Sample
Problem D would have at the same mean altitude above Earth, Jupiter,
and Earth’s moon.
*Challenging Problem
2. At what distance above Earth would a satellite have a period of 125 min?
Circular Motion and Gravitation
241
Reality Check
Ask students if they see a pattern in their
answers for orbital speed from Practice D item 1.
Should the orbital speed be greater or less
when Magellan orbits a more massive body?
Why? Orbital speeds of satellites are greater for
objects orbiting more massive bodies. This
makes sense. In the equation for orbital speed,
the speed increases as mass (and thus the
square root of the mass) increases.
5/9/2011 11:01:08 AM
Answers
Practice D
1. Earth: 7.69 × 103 m/s, 5.51 × 103 s;
Jupiter: 4.20 × 104 m/s, 1.08 × 104 s;
moon: 1.53 × 103 m/s, 8.63 × 103 s
2. 1.90 × 106 m
Circular Motion and Gravitation 241
Weight and Weightlessness
Teach continued
QuickLab
Teacher’s Notes
To perform this lab, select an elevator
that accelerates through most of its trip
between the top and ground floors.
Slow elevators tend to move at a
constant speed, so no change in the
scale reading will be observed after the
moment when the elevator begins to
move. The measurements will be easier
to make if the building has more than
five floors.
In the chapter about forces, you learned that weight is the magnitude of
the force due to gravity. When you step on a bathroom scale, it does not
actually measure your weight. The scale measures the downward force
exerted on it. When your weight is the only downward force acting on the
scale, the scale reading equals your weight. If a friend pushes down on
you while you are standing on the scale, the scale reading will go up.
However, your weight has not changed; the scale reading equals your
weight plus the extra applied force. Because of Newton’s third law, the
downward force you exert on the scale equals the upward force exerted
on you by the scale (the normal force). Thus, the scale reading is equal to
the normal force acting on you.
MATERIALS
• elevator
• bathroom scale
• watch or stopwatch
ELEVATOR
ACCELERATION
In this activity, you will stand
on a bathroom scale while
riding an elevator up to the top
floor and then back. Stand on
the scale in a first-floor elevator, and record your weight. As
the elevator moves up, record
the scale reading for every
two-second interval. Repeat
the process as the elevator
moves down.
For example, imagine you are standing in an elevator, as illustrated in
Figure 3.4. When the elevator is at rest, as in Figure 3.4(a), the magnitude of
the normal force is equal to your weight. A scale in the elevator would
record your weight. When the elevator begins accelerating downward, as
in Figure 3.4(b), the normal force will be smaller. The scale would now
record an amount that is less than your weight. If the elevator’s acceleration were equal to free-fall acceleration, as shown in Figure 3.4(c), you
would be falling at the same rate as the elevator and would not feel the
force of the floor at all. In this case, the scale would read zero. You still
have the same weight, but you and the elevator are both falling with
free-fall acceleration. In other words, no normal force is acting on you.
This situation is called apparent weightlessness.
Now, find the net force for
each time interval, and then use
Newton’s second law to
calculate the elevator’s acceleration for each interval. How
does the acceleration change?
How does the elevator’s
maximum acceleration compare with free-fall acceleration?
Astronauts in orbit experience apparent weightlessness.
Astronauts floating in a space shuttle are experiencing apparent
weightlessness. Because the shuttle is accelerating at the same rate as the
astronauts are, this example is similar to the elevator in Figure 3.4(c).
FIGURE 3.4
Normal Force in an Elevator When this elevator accelerates,
the normal force acting on the person changes. If the elevator were in
a =0
a =0
600 N
600 N
a =0
a
a
a
600 N
300 N
300 N
Fn
Fn
Fn
Fg
Fg
Fg
(a)
free fall, the normal force would drop to zero and the person
would experience a sensation of apparent weightlessness.
(a)
(a)
300 N
Fn
Fn
Fn
Fg
Fg
Fg
(b)
(b)
(b)
a=g
a=g
(b)
a=g
0N
0N
Fn = 0
Fn = 0
Fn = 0
Fg
Fg
Fg
0N
(c)
(c)
(c)
242
Chapter 7
Differentiated
Instruction
Pre-AP
A means to overcome apparent weightlessness
in space is to have a spacecraft or portion of
the spacecraft that is a cylinder rotating about
its central axis. Objects or people standing
against the walls experience a normal force
that is proportional to the square of the speed
of the cylinder’s rotation and is inversely
proportional to the cylinder’s radius. If the
cylinder is rotated at the proper speed, the
cylinder creates an “artificial gravity” that is
Untitled-53 242
242 Chapter 7
equal to the gravitational field strength at
Earth’s surface. Have students investigate the 5/9/2011
relationship between the centripetal force and
the normal force that creates the sensation
of gravity.
11:01:12 AM
ntitled-53 243
The force due to gravity keeps the astronauts and shuttle in orbit, but the
astronauts feel weightless because no normal force is acting on them.
The Language of
Physics
The human body relies on gravitational force. For example, this
force pulls blood downward so that the blood collects in the veins of your
legs when you are standing. Because the body of an astronaut in orbit
accelerates along with the space shuttle, gravitational force has no effect
on the body. This state can initially cause nausea and dizziness. Over
time, it can pose serious health risks, such as weakened muscles and
brittle bones. When astronauts return to Earth, their bodies need time to
readjust to the effects of the gravitational force.
Apparent weightlessness is sometimes
called microgravity.
To understand microgravity, think
about a person who jumps from a height
of 4.6 meters. The person will reach the
ground in 1 second. In a microgravity
environment, where the gravitational
force is reduced to one percent of that
on Earth, it will take the person ten
seconds to reach the ground.
So far, we have been describing apparent weightlessness. Actual
weightlessness occurs only in deep space, far from stars and planets.
Gravitational force is never entirely absent, but it can become negligible
at distances that are far enough away from any masses. In this case, a star
or astronaut would not be pulled into an orbit but would instead drift in a
straight line at constant speed.
Assess and Reteach SECTION 3 FORMATIVE ASSESSMENT
Reviewing Main Ideas
Assess Use the Formative Assessment
on this page to evaluate student
mastery of the section.
1. Compare Ptolemy’s model of the solar system with Copernicus’s. How
does Kepler’s first law of planetary motion refine Copernicus’s model?
2. Does a planet in orbit around the sun travel at a constant speed? How do
you know?
Reteach For students who need
additional instruction, download the
Section Study Guide.
3. Suppose you know the mean distance between both Mercury and the sun
and Venus and the sun. You also know the period of Venus’s orbit around
the sun. How can you find the period of Mercury’s orbit?
4. Explain how Kepler’s laws of planetary motion relate to Newton’s law of
universal gravitation.
Response to Intervention To reassess
students’ mastery, use the Section Quiz,
available to print or to take directly
online at HMDScience.com.
5. Find the orbital speed and period of Earth’s moon. The average distance
between the centers of Earth and of the moon is 3.84 × 108 m.
Critical Thinking
6. An amusement park ride raises people high into the air, suspends them
for a moment, and then drops them at the rate of free-fall acceleration. Is
a person in this ride experiencing apparent weightlessness, true weightlessness, or neither? Explain.
7. Suppose you went on the ride described in item 6, held a penny in front
of you, and released the penny at the moment the ride started to drop.
What would you observe?
Answers to Section Assessment
1. Both Ptolemy’s and Copernicus’s
models used circular orbits to describe
the motions of planets. However, in
Ptolemy’s model, the planets and the sun
moved around a motionless Earth,
whereas in Copernicus’s model, the
planets and Earth moved around the sun.
Kepler replaced Copernicus’s circular
orbits with ellipses.
2. no; a planet moves faster as it comes
closer to the sun because, in accordance
Circular Motion and Gravitation
243
with Kepler’s second law, it sweeps out
equal areas in equal time intervals.5/9/2011 11:01:13 AM
3. Use Kepler’s third law:
​​T m​ ​ ​2​ ​​r ​m​ ​3​
​ _2 ​ = _
​   ​  
​​T v​ ​ ​ ​ ​​r ​v​ ​3​
so
​ ​ ​3​
  
​​r M
TM = Tv ​   _
​  3 ​ ​   
​​r ​V​ ​ ​
4. Newton demonstrated that an inversesquare force is necessary to produce an
elliptical orbit, as predicted by Kepler’s
√ 
first law. It also explains Kepler’s third law
for circular orbits.
5. vt = 1.02 × 103 m/s;
T = 2.37 × 106 s = 27.4 days
6. apparent weightlessness; The normal force
is zero because the person is in free fall.
7. You would see the penny float in front
of you.
Circular Motion and Gravitation 243