Nicholas J. Giordano

www.cengage.com/physics/giordano

Chapter 5

Circular Motion and Gravitation

Chapter 5Circular Motion and

Gravitation

Introduction• Circular motion

• Acceleration is not constant• Cannot be reduced to a one-dimensional problem

• Examples• Car traveling around a turn• Parts of the motion of a roller coaster• Centrifuge• The Earth orbiting the Sun

• Gravitation• Explore gravitational force in more detail• Look at Kepler’s Laws of Motion• Further details about g

Introduction

Uniform Circular Motion

• Path of Motion: Circle• The direction of the

velocity is continually changing.• The vector is always

tangent to the circle• Uniform circular motion

(UCM) is circular motion at constant speed.

Section 5.1

Circular Motion - Period

v

v

v

v

r

r: radius of circle

Period T (s): time for a complete rotation

If n rotations happen in time t: T=t/n.

Frequency f (s-1, Hertz-Hz): number of rotations per unit of time - f=n/t.

T=1/f, f=1/T, Tf=1

Tangential (Linear) Velocity

rfv 2

C = 2πr = vT

Trv 2

or

TCv

Exercises Set 11. A car’s tire rotates at 1200 RPM.

a. What is the frequency? f = 1200/60s = 20 Hz

b. What is the period? T= 1/f = 1/20 Hz = 0.05s

c. What is the speed (R= 0.15 m)

v = 2πRf = 2 x 3.14 x 0.15m x 20 Hz = 18.85 m/s

Ex 2Question 1: An object is constrained to move in a circular orbit with radius 1 m. a) How far does the object travel after completing one orbit? b) If the orbital period (time to complete one orbit) is 0.5 s, what is its velocity?

a) C = 2πR = 2 x 3.14 x 1 m = 6.28 m

b) v = C/T = 6.28 m/0.5s = 12. 56 m/s

Ex3• Problem: The earth is 1.50 x1011 m (93 million miles) from the

sun. What is its speed in m/s (neglecting the motion of the sun through the galaxy)?

sec

Angular Measure

The position of an object can be described using polar coordinates—r and θ—rather than x and y. The figure at left gives the conversion between the two descriptions.

Angular MeasureIt is most convenient to measure the angle θ in radians:

Angular Speed and Velocity

In analogy to the linear case, we define the angular speed:

where θ is the angle swept by the radius in the time t.

rθ

[θ] = rad/s

Angular Speed and Velocity

The angular velocity is (+) positive when the rotation is ccw and (-)negative if the rotation is cw.

Angular Speed and VelocityRelationship between tangential and angular speeds:

This means that parts of a rotating object farther from the axis of rotation move faster.

rtr

tr

tsv

Angular Speed, Period, and Frequency

Trv 2

From:

we get the relation of the frequency to the angular speed:

and

Uniform Circular Motion and Centripetal Acceleration

A careful look at the change in the velocity vector of an object moving in a circle at constant speed shows that the acceleration is toward the center of the circle.

Uniform Circular Motion and Centripetal Acceleration

The change in the direction of the velocity vector is provided by the centripetal acceleration:

Exercises Set 4• 2) A car is traveling along a circular path of 50

m radius at 22 m/s.

222

/68.9)50()/22( sm

msm

rvacp

a. What is the car’s acceleration?

b. How much time to complete a circuit?

ssmmxx

vrT

Trv 28.14

/225014.3222

Example Set 5• 3) Satellite, radius 1.3 x 107 m, g =

2.5 m/s2.

smravrva cpcp /88.570010*3.1*5.2 72

a) What is the speed?

b) Period?

hssmmxxx

vrT

Trv 98.386.327,14

/88.5700103.114.3222 7

Ex. 6

smravrva cpcp /8.28.*8.92

Hzm

smrvfrfv

56.08.0*14.3*2

/8.22

2

n = 0.56 x 60 = 33.44 rev/min

Uniform Circular Motion and Centripetal Force

The centripetal force is the mass multiplied by the centripetal acceleration.

This force is the net force on the object. As the force is always perpendicular to the velocity, it does no work.

Centripetal Force Features:• A force causing a centripetal

acceleration acting toward the center of the circle.

• It causes a change in the direction of the velocity vector

• If the force vanishes, the object would move in a straight-line path tangent to the circle (“Centrifugal” effects which come from inertia)

! This is not a new force, it is a new application of a force …

Centripetal force The centripetal force can be

produced in any number of ways (for example):• Due to the tension in a string

(pendulum)

• Due to friction between tires and the road

• Due to gravity

Conical Pendulum

• A small object of mass m is suspended from a string of length L. The object revolves in a horizontal circle of radius r with constant speed v, as shown. – Find the speed of the object.– Find the tension, T, in the string.

Conical Pendulum

• Need a free body diagram:– T is the force

exerted by the string

x

y

Conical Pendulum

• The body does not move in the vertical direction …therefore no acceleration here!

– Rearranging gives

0cos

0

mgTF

maF

y

yy

mgT cosx

y

Conical Pendulum

• In the x direction, there is centripetal acceleration!

– Rearranging gives

rvmTF

mamaF

x

cxx

2

sin

rvmT2

sin

x

y

Conical Pendulum

• Consider the x and y equations:

• We have two equations in two unknowns (v, T)– T can be eliminated by dividing 2 by 1 to give

– and

rvmT2

sin

mgT cos 1

2

rgv

mgrvm

TT

2

2

tan

cossin

x

y

Conical Pendulum

• And solving for v

tanrgv

rgv2tan

Conical Pendulum

• To find tension, just go back to the y-component equation:

• And solve for TmgT cos

cosmgT

Problem Solving Strategy – Circular Motion• Recognize the principle

• If the object moves in a circle, then there is a centripetal force acting on it

• Sketch the problem• Show the path the object travels• Identify the circular part of the path• Include the radius of the circle• Show the center of the circle• Selecting a coordinate system that assigns the positive

direction toward the center of the circle is often convenient

Section 5.1

Problem Solving Strategy, cont.• Identify the principles

• Find all the forces acting on the object• A free body diagram is generally useful

• Find the components of the forces that are directed toward the center of the circle

• Find the components of the forces perpendicular to the center

• Apply Newton’s Second Law for both directions• The acceleration directed toward the center of the circle is a

centripetal acceleration

Section 5.1

• Solve for the quantities of interest• Check your answer

• Consider what the answer means• Does the answer make sense

Horizontal (Flat) Curve

• A 1500 kg car moving on a flat, horizontal road negotiates a curve with a radius of 35.0 m. If the static coefficient of friction between the tires and the dry pavement is 0.523, what is the maximum speed with which the car can negotiate the turn?

Centripetal Acceleration Example: Car

• A car rounding a curve travels in an approximate circle

• The radius of this circle is called the radius of curvature

• Forces in the y-direction• Gravity and the normal

force• Forces in the x-direction

• Friction is directed toward the center of the circle

Section 5.1

Horizontal (Flat) Curve• What is the maximum force of friction FFriction?

We can get that from the vertical force balance and the static friction coefficient:

• So

• And, mgNF ssFriction

0- mgNFy

mgN

Car Example, cont.• Since friction is the only force acting in the x-

direction, it supplies the centripetal force

• Solving for the maximum velocity at which the car can safely round the curve gives

Section 5.1

• Finish up,

– Does it look okay?• Increase μs, vmax increases √• Increase r, vmax increases √

– Put in numbers

Horizontal (Flat) Curve

rgv smax

smvmsmv

/4.13)0.35)(/80.9)(523.0(

max

2max

Horizontal (Flat) Curve• Reality check

• You might want a bit more of a margin of safety here …

hkmsmhkmsmsmv /2.48

/1/6.3/4.13/4.13max

hkmv /2.48max

Angular AccelerationThe average angular acceleration is the rate at which the angular speed changes:

In analogy to constant linear acceleration:

t

Angular Acceleration

If the angular speed is changing, the linear speed must be changing as well. The tangential acceleration is related to the angular acceleration:

Angular Acceleration

Circular Motion Example: Vertical Circle

• The speed of the rock varies with time

• At the bottom of the circle:• Tension and gravity are

in opposite directions

•

• The tension supports the rock (mg) and supplies the centripetal force Section 5.2

Vertical Circle Example, cont.

• At the top of the circle:• Tension and gravity are

in the same direction• Pointing toward the

center of the circle

•

Section 5.2

Vertical Circle Example, Final• There is a minimum value of v needed to keep the

string taut at the top• Let Ttop = 0

• • If the speed is smaller than this, the string will become slack

and circular motion is no longer possible

Circular Motion Example: Roller Coaster

• The roller coaster’s path is nearly circular at the minimum or maximum points on the track

• There is a maximum speed at which the coaster will not leave the top of the track:• • If the speed is greater

than this, N would have to be negative

• This is impossible, so the coaster would leave the track

Section 5.2

Newton’s Law of Gravitation, Equation

• Law states: There is a gravitational attraction between any two objects. If the objects are point masses m1 and m2, separated by a distance r the magnitude of the force is

Section 5.3

Law of Gravitation, cont.• Note that r is the distance between the objects• G is the Universal Gravitational Constant

• G = 6.67 x 10-11 N . m2/ kg2

• The gravitational force is always attractive• Every mass attracts every other mass

• The gravitational force is symmetric• The magnitude of the gravitational force exerted by

mass 1 on mass 2 is equal in magnitude to the force exerted by mass 2 on mass 1

• The two forces form an action-reaction pair

Section 5.3

Gravitation and the Moon’s Orbit

• The Moon follows an approximately circular orbit around the Earth

• There is a force required for this motion

• Gravity supplies the force

Section 5.3

Notes on the Moon’s Motion• We assumed the Moon orbits a “fixed” Earth

• It is a good approximation• It ignores the Earth’s motion around the Sun

• The Earth and Moon actually both orbit their center of mass• We can think of the Earth as orbiting the Moon• The circle of the Earth’s motion is very small compared

to the Moon’s orbit

Section 5.3

Gravitation and g

• Assuming a spherical Earth, we can consider all the mass of the Earth to be concentrated at its center

• The value of r in the Law of Gravitation is just the radius of the Earth

Section 5.3

Gravitation Force From The Earth – Assumptions • We assumed

• A spherical Earth • That the gravitational force could be calculated as if all

the mass of the Earth was located at its center• Assumptions are true as long as the density of the

object is spherically symmetric• The object has a constant density• The object’s density varies with depth as long as the

density depends only on the distance from the center

Section 5.3

Measuring G

• Henry Cavendish measured the force of gravity between two large lead spheres

• By an experimental set-up similar to the picture, he was able to determine the value of G

Section 5.3

More About Gravity – Newton’s Apple• G is a constant of nature• Gravity is a weak force

• Much smaller than typical normal or tension forces• Gravity is considered the weakest of the

fundamental forces of nature• Newton showed the motion of celestial bodies and

the motion of terrestrial motion are caused by the same force and governed by the same laws of motion

Section 5.3a

An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle in the time t, through what angle did it rotate in the time 1/2 t?

1) 1/2 2) 1/4 3) 3/4 4) 2 5) 4

ConcepTest 8.3a Angular Displacement I

An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle in the time t, through what angle did it rotate in the time 1/2 t?

1) 1/2 2) 1/4 3) 3/4 4) 2 5) 4

The angular displacement is = 1/2 t 2 (starting from rest), and there is a quadratic dependence on time. Therefore, in half the time, the object has rotated through one-quarter the angle.

ConcepTest 8.3a Angular Displacement I

An object at rest begins to rotate with a constant angular acceleration. If this object has angular velocity at time t, what was its angular velocity at the time 1/2 t?

1) 1/2 2) 1/4 3) 3/4 4) 2 5) 4

ConcepTest 8.3b Angular Displacement II

An object at rest begins to rotate with a constant angular acceleration. If this object has angular velocity at time t, what was its angular velocity at the time 1/2t?

1) 1/2 2) 1/4 3) 3/4 4) 2 5) 4

The angular velocity is = t (starting from rest), and there is a linear dependence on time. Therefore, in half the time, the object has accelerated up to only half the speed.

ConcepTest 8.3b Angular Displacement II

Example: Wheel And Rope A wheel with radius R = 0.4 m rotates freely about a fixed

axle. There is a rope wound around the wheel. Starting from rest at t = 0, the rope is pulled such that it has a constant acceleration a = 4 m/s2. How many revolutions has the wheel made after 10 seconds? (One revolution = 2 radians)

a

R

Wheel And Rope... Use a = R to find :

= a / R = 4 m/s2 / 0.4 m = 10 rad/s2

Now use the equations we derived above just as you would use the kinematic equations from the beginning of the semester.

rev80radrot

21x rad500

20 t

21t = 0 + 0(10) + (10)(10)2 = 500 rad2

10

a

R

Newton’s Law of GravitationNewton’s law of universal gravitation describes the force between any two point masses m1, m2 separated by the distance r:

G is called the universal gravitational constant:

m1m2r

Fg -Fg

Newton’s Law of Gravitation

Gravity provides the centripetal force that keeps planets, moons, and satellites in their orbits.We can relate the universal gravitational force to the local acceleration of gravity:

Newton’s Law of Gravitation

The gravitational potential energy is given by the general expression:

Kepler’s Laws and Earth Satellites(No!)

Kepler’s laws were the result of his many years of observations. They were later found to be consequences of Newton’s laws.Kepler’s first law:Planets move in elliptical orbits, with the Sun at one of the focal points.

Kepler’s Laws and Earth SatellitesKepler’s second law:A line from the Sun to a planet sweeps out equal areas in equal lengths of time.

Kepler’s Laws and Earth Satellites

Kepler’s third law:The square of the orbital period of a planet is directly proportional to the cube of the average distance of the planet from the Sun; that is, .

This can be derived from Newton’s law of gravitation, using a circular orbit.

Kepler’s Laws and Earth SatellitesIf a projectile is given enough speed to just reach the top of the Earth’s gravitational well, its potential energy at the top will be zero. At the minimum, its kinetic energy will be zero there as well.

Kepler’s Laws and Earth SatellitesThis minimum initial speed is called the escape speed.

Kepler’s Laws and Earth Satellites

Any satellite in orbit around the Earth has a speed given by

Kepler’s Laws and Earth Satellites

Kepler’s Laws and Earth SatellitesAstronauts in Earth orbit report the sensation of weightlessness. The gravitational force on them is not zero; what’s happening?

Kepler’s Laws and Earth SatellitesWhat’s missing is not the weight, but the normal force. We call this apparent weightlessness.“Artificial” gravity could be produced in orbit by rotating the satellite; the centripetal force would mimic the effects of gravity.

Summary of Chapter 5Angles may be measured in radians; the angle is the arc length divided by the radius.Angular kinematic equations for constant acceleration:

Summary of Chapter 5Tangential speed is proportional to angular speed.Frequency is inversely proportional to period.Angular speed:

Centripetal acceleration:

Summary of Chapter 5

Centripetal force:

Angular acceleration is the rate at which the angular speed changes. It is related to the tangential acceleration.

Newton’s law of gravitation:

Summary of Chapter 5

Gravitational potential energy:Kepler’s laws:1. Planetary orbits are ellipses with Sun at one

focus2. Equal areas are swept out in equal times.3. The square of the period is proportional to

the cube of the radius.

Summary of Chapter 5

Escape speed from Earth:

Energy of a satellite orbiting Earth:

ConcepTest 7.1 Tetherball1) Toward the top of the pole2) Toward the ground3) Along the horizontal component of

the tension force4) Along the vertical component of the

tension force5) Tangential to the circle

In the game of tetherball,

the struck ball whirls

around a pole. In what

direction does the net

force on the ball point?

W

T

The vertical component of the

tension balances the weight. The

horizontal component of tension

provides the centripetal force that

points toward the center of the

circle.

1) Toward the top of the pole2) Toward the ground3) Along the horizontal component of

the tension force4) Along the vertical component of the

tension force5) Tangential to the circle

In the game of tetherball,

the struck ball whirls

around a pole. In what

direction does the net

force on the ball point?

W T

W

T

ConcepTest 7.1 Tetherball

You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you?

1) You are thrown to the right2) You feel no particular change3) You are thrown to the left4) You are thrown to the ceiling5) You are thrown to the floor

ConcepTest 7.2a Around the Curve I

You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you?

1) You are thrown to the right2) You feel no particular change3) You are thrown to the left4) You are thrown to the ceiling5) You are thrown to the floor

ConcepTest 7.2a Around the Curve I

The passenger has the tendency to continue moving in a straight line. From your perspective in the car, it feels like you are being thrown to the right, hitting the passenger door.

1) centrifugal force is pushing you into the door

2) the door is exerting a leftward force on you

3) both of the above4) neither of the above

During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening?

ConcepTest 7.2b Around the Curve II

1) centrifugal force is pushing you into the door

2) the door is exerting a leftward force on you

3) both of the above4) neither of the above

During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening?

The passenger has the tendency to continue moving in a straight line. There is a centripetal force, provided by the door, that forces the passenger into a circular path.

ConcepTest 7.2b Around the Curve II

1) car’s engine is not strong enough to keep the car from being pushed out

2) friction between tires and road is not strong enough to keep car in a circle

3) car is too heavy to make the turn4) a deer caused you to skid5) none of the above

You drive your dad’s car too fast around a curve and the car starts to skid. What is the correct description of this situation?

ConcepTest 7.2c Around the Curve III

The friction force between tires and road provides the centripetal force that keeps the car moving in a circle. If this force is too small, the car continues in a straight line!

1) car’s engine is not strong enough to keep the car from being pushed out

2) friction between tires and road is not strong enough to keep car in a circle

3) car is too heavy to make the turn4) a deer caused you to skid5) none of the above

You drive your dad’s car too fast around a curve and the car starts to skid. What is the correct description of this situation?

ConcepTest 7.2c Around the Curve III

Follow-up: What could be done to the road or car to prevent skidding?

ConcepTest 7.3 Missing Link

A ping-pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the ping pong ball leaves the track, which path will it follow?

ConcepTest 7.3 Missing Link

Once the ball leaves the tube, there is no longer

a force to keep it going in a circle. Therefore, it

simply continues in a straight line, as Newton’s

First Law requires!

A ping-pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the ping pong ball leaves the track, which path will it follow?

Follow-up: What physical force provides the centripetal force?

ConcepTest 7.4 Ball and String1) T2 = 1/4 T1

2) T2 = 1/2 T1

3) T2 = T1

4) T2 = 2 T1

5) T2 = 4 T1

Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1?

The centripetal force in this case is given by the

tension, so T = mv2/r. For the same period, we find

that v2 = 2v1 (and this term is squared). However, for

the denominator, we see that r2 = 2r1 which gives us

the relation T2 = 2T1.

ConcepTest 7.4 Ball and String

Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1?

1) T2 = 1/4 T1

2) T2 = 1/2 T1

3) T2 = T1

4) T2 = 2 T1

5) T2 = 4 T1

ConcepTest 7.5 Barrel of FunA rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?

1 2 3 4 5

The normal force of the wall on the rider provides the centripetal force needed to keep her going around in a circle. The downward force of gravity is balanced by the upward frictional force on her, so she does not slip vertically.

ConcepTest 7.5 Barrel of FunA rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?

1 2 3 4 5

Follow-up: What happens if the rotation of the ride slows down?

ConcepTest 7.6a Going in Circles I

1) N remains equal to mg

2) N is smaller than mg

3) N is larger than mg

4) none of the above

You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion?

ConcepTest 7.6a Going in Circles I

1) N remains equal to mg

2) N is smaller than mg

3) N is larger than mg

4) none of the above

You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion?

You are in circular motion, so there

has to be a centripetal force pointing

inward. At the top, the only two

forces are mg (down) and N (up), so

N must be smaller than mg.

Follow-up: Where is N larger than mg?

R

v

1) Fc = N + mg

2) Fc = mg – N

3) Fc = T + N – mg

4) Fc = N

5) Fc = mg

A skier goes over a small round hill with radius R. Since she is in circular motion, there has to be a centripetal force. At the top of the hill, what is Fc of the skier equal to?

ConcepTest 7.6b Going in Circles II

R

vFc points toward the center of the circle, i.e., downward in this case. The weight vector points down and the normal force (exerted by the hill) points up. The magnitude of the net force, therefore, is: Fc = mg – N

mg N

A skier goes over a small round hill with radius R. Since she is in circular motion, there has to be a centripetal force. At the top of the hill, what is Fc of the skier equal to?

ConcepTest 7.6b Going in Circles II

Follow-up: What happens when the skier goes into a small dip?

1) Fc = N + mg

2) Fc = mg – N

3) Fc = T + N – mg

4) Fc = N

5) Fc = mg

R

vtop

1) Fc = T – mg

2) Fc = T + N – mg

3) Fc = T + mg

4) Fc = T

5) Fc = mg

You swing a ball at the end of string in a vertical circle. Since the ball is in circular motion there has to be a centripetal force. At the top of the ball’s path, what is Fc equal to?

ConcepTest 7.6c Going in Circles III

R

v Tmg

You swing a ball at the end of string in a vertical circle. Since the ball is in circular motion there has to be a centripetal force. At the top of the ball’s path, what is Fc equal to?

Fc points toward the center of the circle, i.e. downward in this case. The weight vector points down and the tension (exerted by the string) also points down. The magnitude of the net force, therefore, is: Fc = T + mg

ConcepTest 7.6c Going in Circles III

Follow-up: What is Fc at the bottom of the ball’s path?

1) Fc = T – mg

2) Fc = T + N – mg

3) Fc = T + mg

4) Fc = T

5) Fc = mg

ConcepTest 7.7a Earth and Moon I1) the Earth pulls harder on the Moon2) the Moon pulls harder on the Earth3) they pull on each other equally4) there is no force between the Earth and

the Moon5) it depends upon where the Moon is in its

orbit at that time

Which is stronger,

Earth’s pull on the

Moon, or the

Moon’s pull on

Earth?

By Newton’s 3rd Law, the forces are

equal and opposite.

ConcepTest 7.7a Earth and Moon I1) the Earth pulls harder on the Moon2) the Moon pulls harder on the Earth3) they pull on each other equally4) there is no force between the Earth and

the Moon5) it depends upon where the Moon is in its

orbit at that time

Which is stronger,

Earth’s pull on the

Moon, or the

Moon’s pull on

Earth?

ConcepTest 7.7b Earth and Moon II

1) one quarter2) one half3) the same4) two times5) four times

If the distance to the Moon were

doubled, then the force of

attraction between Earth and

the Moon would be:

The gravitational force depends inversely on the distance squared. So if you increase the distance by a factor of 2, the force will decrease by a factor of 4.

ConcepTest 7.7b Earth and Moon II

1) one quarter2) one half3) the same4) two times5) four times

If the distance to the Moon were

doubled, then the force of

attraction between Earth and

the Moon would be:

2RMmGF

Follow-up: What distance would increase the force by a factor of 2?

You weigh yourself on a scale inside an airplane that is flying with constant speed at an altitude of 20,000 feet. How does your measured weight in the airplane compare with your weight as measured on the surface of the Earth?

1) greater than2) less than3) same

ConcepTest 7.8 Fly Me Away

You weigh yourself on a scale inside an airplane that is flying with constant speed at an altitude of 20,000 feet. How does your measured weight in the airplane compare with your weight as measured on the surface of the Earth?

1) greater than2) less than3) same

At a high altitude, you are farther away from the center of Earth. Therefore, the gravitational force in the airplane will be less than the force that you would experience on the surface of the Earth.

ConcepTest 7.8 Fly Me Away

ConcepTest 7.9 Two Satellites1) 1/8

2) 1/4

3) 1/2

4) it’s the same

5) 2

Two satellites A and B of the same mass are going around Earth in concentric orbits. The distance of satellite B from Earth’s center is twice that of satellite A. What is the ratio of the centripetal force acting on B compared to that acting on A?

Using the Law of Gravitation:

we find that the ratio is 1/4.

ConcepTest 7.9 Two Satellites

2RMmGF

1) 1/8

2) 1/4

3) 1/2

4) it’s the same

5) 2

Two satellites A and B of the same mass are going around Earth in concentric orbits. The distance of satellite B from Earth’s center is twice that of satellite A. What is the ratio of the centripetal force acting on B compared to that acting on A?

Note the 1/r2 factor

ConcepTest 7.10 Averting Disaster1) It’s in Earth’s gravitational field2) the net force on it is zero3) it is beyond the main pull of Earth’s

gravity4) it’s being pulled by the Sun as well as by

Earth5) none of the above

The Moon does not

crash into Earth

because:

The Moon does not crash into Earth because of its high

speed. If it stopped moving, it would, of course, fall

directly into Earth. With its high speed, the Moon would

fly off into space if it weren’t for gravity providing the

centripetal force.

ConcepTest 7.10 Averting Disaster

The Moon does not

crash into Earth

because:

Follow-up: What happens to a satellite orbiting Earth as it slows?

1) It’s in Earth’s gravitational field2) the net force on it is zero3) it is beyond the main pull of Earth’s

gravity4) it’s being pulled by the Sun as well as by

Earth5) none of the above

ConcepTest 7.11 In the Space Shuttle

Astronauts in the

space shuttle

float because:

1) they are so far from Earth that Earth’s gravity doesn’t act any more

2) gravity’s force pulling them inward is cancelled by the centripetal force pushing them outward

3) while gravity is trying to pull them inward, they are trying to continue on a straight-line path

4) their weight is reduced in space so the force of gravity is much weaker

Astronauts in the space shuttle float because

they are in “free fall” around Earth, just like a

satellite or the Moon. Again, it is gravity that

provides the centripetal force that keeps them

in circular motion.

ConcepTest 7.11 In the Space Shuttle

Astronauts in the

space shuttle

float because:

Follow-up: How weak is the value of g at an altitude of 300 km?

1) they are so far from Earth that Earth’s gravity doesn’t act any more

2) gravity’s force pulling them inward is cancelled by the centripetal force pushing them outward

3) while gravity is trying to pull them inward, they are trying to continue on a straight-line path

4) their weight is reduced in space so the force of gravity is much weaker

If you weigh yourself at the equator of Earth, would you get a bigger, smaller or similar value than if you weigh yourself at one of the poles?

1) bigger value2) smaller value3) same value

ConcepTest 7.12 Guess My Weight

If you weigh yourself at the equator of Earth, would you get a bigger, smaller or similar value than if you weigh yourself at one of the poles?

1) bigger value2) smaller value3) same value

The weight that a scale reads is the normal force exerted by the floor (or the scale). At the equator, you are in circular motion, so there must be a net inward force toward Earth’s center. This means that the normal force must be slightly less than mg. So the scale would register something less than your actual weight.

ConcepTest 7.12 Guess My Weight

ConcepTest 7.13 Force Vectors

A planet of mass m is a distance d from Earth. Another planet of mass 2m is a distance 2d from Earth. Which force vector best represents the direction of the total gravitation force on Earth?

1 23

4

5

2d

d

2m

m

Earth

1 23

4

5

2d

d

2m

mThe force of gravity on the Earth due to m is greater than the force due to 2m, which means that the force component pointing down in the figure is greater than the component pointing to the right.

F2m = GME(2m) / (2d)2 = 1/2 GMm / d2

Fm = GME m / d2 = GMm / d2

A planet of mass m is a distance d from Earth. Another planet of mass 2m is a distance 2d from Earth. Which force vector best represents the direction of the total gravitation force on Earth?

ConcepTest 7.13 Force Vectors