Kinetics of Particles: Newton’s Second Law

VECTOR MECHANICS FOR ENGINEERS: DYNAMICSDYNAMICS

Tenth Tenth EditionEdition

Ferdinand P. BeerFerdinand P. Beer

E. Russell Johnston, Jr.E. Russell Johnston, Jr.

Phillip J. CornwellPhillip J. Cornwell

Lecture Notes:Lecture Notes:

Brian P. SelfBrian P. SelfCalifornia Polytechnic State UniversityCalifornia Polytechnic State University

CHAPTER

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.

12Kinetics of Particles:

Newton’s Second Law


Vector Mechanics for Engineers: DynamicsVector Mechanics for Engineers: Dynamics

Te

nth

Ed

ition

Contents

12 - 2

Introduction

Newton’s Second Law of Motion

Linear Momentum of a Particle

Systems of Units

Equations of Motion

Dynamic Equilibrium

Sample Problem 12.1

Sample Problem 12.3

Sample Problem 12.4

Sample Problem 12.5

Sample Problem 12.6

Angular Momentum of a Particle

Equations of Motion in Radial & Transverse Components

Conservation of Angular Momentum

Newton’s Law of Gravitation

Sample Problem 12.7

Sample Problem 12.8

Trajectory of a Particle Under a Central Force

Application to Space Mechanics

Sample Problem 12.9

Kepler’s Laws of Planetary Motion



Te

nth

Ed

ition

Kinetics of Particles

2 - 3

We must analyze all of the forces acting on the wheelchair in order to design a good ramp

High swing velocities can result in large forces on a swing chain or rope, causing it to break.



Te

nth

Ed

ition

Introduction

12 - 4

• Newton’s Second Law of Motion

m F a

• If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of resultant and in the direction of the resultant.

• Must be expressed with respect to a Newtonian (or inertial) frame of reference, i.e., one that is not accelerating or rotating.

• This form of the equation is for a constant mass system



Te

nth

Ed

ition

Linear Momentum of a Particle

12 - 5

• Replacing the acceleration by the derivative of the velocity yields

particle theof momentumlinear

L

dt

Ldvm

dt

d

dt

vdmF

• Linear Momentum Conservation Principle: If the resultant force on a particle is zero, the linear momentum of the particle remains constant in both magnitude and direction.



Te

nth

Ed

ition

Systems of Units

12 - 6

• Of the units for the four primary dimensions (force, mass, length, and time), three may be chosen arbitrarily. The fourth must be compatible with Newton’s 2nd Law.

• International System of Units (SI Units): base units are the units of length (m), mass (kg), and time (second). The unit of force is derived,

22 s

mkg1

s

m1kg1N1

• U.S. Customary Units: base units are the units of force (lb), length (m), and time (second). The unit of mass is derived,

ft

slb1

sft1

lb1slug1

sft32.2

lb1lbm1

2

22



Te

nth

Ed

ition

Equations of Motion

12 - 7

• Newton’s second law amF

• Can use scalar component equations, e.g., for rectangular components,

zmFymFxmF

maFmaFmaF

kajaiamkFjFiF

zyx

zzyyxx

zyxzyx



Te

nth

Ed

ition

Dynamic Equilibrium

12 - 8

• Alternate expression of Newton’s second law,

ectorinertial vam

amF

0

• With the inclusion of the inertial vector, the system of forces acting on the particle is equivalent to zero. The particle is in dynamic equilibrium.

• Methods developed for particles in static equilibrium may be applied, e.g., coplanar forces may be represented with a closed vector polygon.

• Inertia vectors are often called inertial forces as they measure the resistance that particles offer to changes in motion, i.e., changes in speed or direction.

• Inertial forces may be conceptually useful but are not like the contact and gravitational forces found in statics.



Te

nth

Ed

ition

Free Body Diagrams and Kinetic Diagrams

12 - 9

The free body diagram is the same as you have done in statics; we will add the kinetic diagram in our dynamic analysis.

2. Draw your axis system (e.g., Cartesian, polar, path)

3. Add in applied forces (e.g., weight, 225 lb pulling force)

4. Replace supports with forces (e.g., normal force)

1. Isolate the body of interest (free body)

5. Draw appropriate dimensions (usually angles for particles) x y

225 N

FfNmg

25o



Te

nth

Ed

ition


12 - 10

Put the inertial terms for the body of interest on the kinetic diagram.

2. Draw in the mass times acceleration of the particle; if unknown, do this in the positive direction according to your chosen axes

1. Isolate the body of interest (free body)

x y225 N

FfNmg

25o

may

max

m F a



Te

nth

Ed

ition


2 - 11

Draw the FBD and KD for block A (note that the massless, frictionless pulleys are attached to block A and should be included in the system).



Te

nth

Ed

ition


2 - 12

1. Isolate body2. Axes

3. Applied forces

4. Replace supports with forces

5. Dimensions (already drawn)

x

y

mg

Ff-1N1

TT

T

T

T

Ff-B

NB

may = 0

max

6. Kinetic diagram

=



Te

nth

Ed

ition


2 - 13

Draw the FBD and KD for the collar B. Assume there is friction acting between the rod and collar, motion is in the vertical plane, and is increasing



Te

nth

Ed

ition


2 - 14

1. Isolate body2. Axes

3. Applied forces

4. Replace supports with forces

5. Dimensions6. Kinetic diagram

mgFf

N

mar

mae er

=



Te

nth

Ed

ition

Sample Problem 12.1

12 - 15

A 200-lb block rests on a horizontal plane. Find the magnitude of the force P required to give the block an acceleration of 10 ft/s2 to the right. The coefficient of kinetic friction between the block and plane is k0.25.

SOLUTION:

• Resolve the equation of motion for the block into two rectangular component equations.

• Unknowns consist of the applied force P and the normal reaction N from the plane. The two equations may be solved for these unknowns.



Te

nth

Ed

ition

Sample Problem 12.1

12 - 16

N

NF

g

Wm

k

25.0

ft

slb21.6

sft2.32

lb200

2

2

x

y

O

SOLUTION:

• Resolve the equation of motion for the block into two rectangular component equations.

:maFx

lb1.62

sft10ftslb21.625.030cos 22

NP

:0 yF

0lb20030sin PN

• Unknowns consist of the applied force P and the normal reaction N from the plane. The two equations may be solved for these unknowns.

lb1.62lb20030sin25.030cos

lb20030sin

PP

PN

lb151P



Te

nth

Ed

ition

Sample Problem 12.3

12 - 17

The two blocks shown start from rest. The horizontal plane and the pulley are frictionless, and the pulley is assumed to be of negligible mass. Determine the acceleration of each block and the tension in the cord.

SOLUTION:

• Write the kinematic relationships for the dependent motions and accelerations of the blocks.

• Write the equations of motion for the blocks and pulley.

• Combine the kinematic relationships with the equations of motion to solve for the accelerations and cord tension.



Te

nth

Ed

ition

Sample Problem 12.3

12 - 18

• Write equations of motion for blocks and pulley.

:AAx amF AaT kg1001

:BBy amF

B

B

BBB

aT

aT

amTgm

kg300-N2940

kg300sm81.9kg300

2

22

2

:0 CCy amF

02 12 TT

SOLUTION:


ABAB aaxy21

21

x

y

O



Te

nth

Ed

ition

Sample Problem 12.3

12 - 19

N16802

N840kg100

sm20.4

sm40.8

12

1

221

2

TT

aT

aa

a

A

AB

A

• Combine kinematic relationships with equations of motion to solve for accelerations and cord tension.

ABAB aaxy21

21

AaT kg1001

A

B

a

aT

21

2

kg300-N2940

kg300-N2940

0kg1002kg150N2940

02 12

AA aa

TT

x

y

O



Te

nth

Ed

ition

Sample Problem 12.4

12 - 20

The 12-lb block B starts from rest and slides on the 30-lb wedge A, which is supported by a horizontal surface.

Neglecting friction, determine (a) the acceleration of the wedge, and (b) the acceleration of the block relative to the wedge.

SOLUTION:

• The block is constrained to slide down the wedge. Therefore, their motions are dependent. Express the acceleration of block as the acceleration of wedge plus the acceleration of the block relative to the wedge.

• Write the equations of motion for the wedge and block.

• Solve for the accelerations.



Te

nth

Ed

ition

Sample Problem 12.4

12 - 21

SOLUTION:

• The block is constrained to slide down the wedge. Therefore, their motions are dependent.

ABAB aaa

• Write equations of motion for wedge and block.

x

y

:AAx amF

AA

AA

agWN

amN

1

1

5.0

30sin

:30cos ABABxBx aamamF

30sin30cos

30cos30sin

gaa

aagWW

AAB

ABABB

:30sin AByBy amamF

30sin30cos1 ABB agWWN



Te

nth

Ed

ition

Sample Problem 12.4

12 - 22

AA agWN 15.0

• Solve for the accelerations.

30sinlb12lb302

30coslb12sft2.32

30sin2

30cos

30sin30cos2

30sin30cos

2

1

A

BA

BA

ABBAA

ABB

a

WW

gWa

agWWagW

agWWN

2sft07.5Aa

30sinsft2.3230cossft07.5

30sin30cos

22AB

AAB

a

gaa

2sft5.20ABa



Te

nth

Ed

ition

Group Problem Solving

2 - 23

The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the horizontal surface, determine (a) the acceleration of each block, (b) the tension in the cable.

SOLUTION:


• Write the equations of motion for the blocks and pulley.

• Combine the kinematic relationships with the equations of motion to solve for the accelerations and cord tension.

• Draw the FBD and KD for each block



Te

nth

Ed

ition


2 - 24

xA

yB

const nts3 aA Bx y L

3 0A Bv v

3 0A Ba a

3A Ba a

SOLUTION:


This is the same problem worked last chapter- write the constraint equation

Differentiate this twice to get the acceleration relationship.

(1)



Te

nth

Ed

ition


2 - 25

: x A AF m a

A BT m a

3 A BT m a

y B BF m a

3(3 )B A B B Bm g m a m a 2

29.81 m/s0.83136 m/s

30 kg1 91 9

25 kg

BA

B

ga

m

m

22.49 2.49 m/sA a

23 30 kg 0.83136 m/sT

74.8 NT

• Draw the FBD and KD for each block

mAg

T

NA

maAx

mBg

2T T

maBy

• Write the equation of motion for each block

==

From Eq (1)(2) (3)3B B BW T m a

(2) (3)

BA

• Solve the three equations, 3 unknowns

+y

+x



Te

nth

Ed

ition

Concept Quiz

2 - 26

The three systems are released from rest. Rank the accelerations, from highest to lowest.

a) (1) > (2) > (3)

b) (1) = (2) > (3)

c) (2) > (1) > (3)

d) (1) = (2) = (3)

e) (1) = (2) < (3)

(1) (2) (3)



Te

nth

Ed

ition

Kinetics: Normal and Tangential Coordinates

2 - 27

Aircraft and roller coasters can both experience large normal forces during turns.



Te

nth

Ed

ition

Equations of Motion

12 - 28

• For tangential and normal components,

t t

t

F ma

dvF m

dt

• Newton’s second law amF

2

n n

n

F ma

vF m



Te

nth

Ed

ition

Sample Problem 12.5

12 - 29

The bob of a 2-m pendulum describes an arc of a circle in a vertical plane. If the tension in the cord is 2.5 times the weight of the bob for the position shown, find the velocity and accel-eration of the bob in that position.

SOLUTION:

• Resolve the equation of motion for the bob into tangential and normal components.

• Solve the component equations for the normal and tangential accelerations.

• Solve for the velocity in terms of the normal acceleration.



Te

nth

Ed

ition

Sample Problem 12.5

12 - 30

SOLUTION:

• Resolve the equation of motion for the bob into tangential and normal components.

• Solve the component equations for the normal and tangential accelerations.

:tt maF

30sin

30sin

ga

mamg

t

t

2sm9.4ta

:nn maF

30cos5.2

30cos5.2

ga

mamgmg

n

n

2sm03.16na

• Solve for velocity in terms of normal acceleration.

22

sm03.16m2 nn avv

a

sm66.5v



Te

nth

Ed

ition

Sample Problem 12.6

12 - 31

Determine the rated speed of a highway curve of radius = 400 ft banked through an angle = 18o. The rated speed of a banked highway curve is the speed at which a car should travel if no lateral friction force is to be exerted at its wheels.

SOLUTION:

• The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path.The forces acting on the car are its weight and a normal reaction from the road surface.

• Resolve the equation of motion for the car into vertical and normal components.

• Solve for the vehicle speed.



Te

nth

Ed

ition

Sample Problem 12.6

12 - 32

SOLUTION:

• The car travels in a horizontal circular path with a normal component of acceleration directed toward the center of the path.The forces acting on the car are its weight and a normal reaction from the road surface.

• Resolve the equation of motion for the car into vertical and normal components.

:0 yF

cos

0cos

WR

WR

:nn maF

2sin

cos

sin

v

g

WW

ag

WR n

• Solve for the vehicle speed.

18tanft400sft2.32

tan2

2 gv

hmi1.44sft7.64 v



Te

nth

Ed

ition


2 - 33

The 3-kg collar B rests on the frictionless arm AA. The collar is held in place by the rope attached to drum D and rotates about O in a horizontal plane. The linear velocity of the collar B is increasing according to v= 0.2 t2 where v is in m/s and t is in sec. Find the tension in the rope and the force of the bar on the collar after 5 seconds if r = 0.4 m.

v SOLUTION:

• Write the equations of motion for the collar.

• Combine the equations of motion with kinematic relationships and solve.

• Draw the FBD and KD for the collar.

• Determine kinematics of the collar.



Te

nth

Ed

ition


2 - 34

SOLUTION: • Given: v= 0.2 t2, r = 0.4 m

• Find: T and N at t = 5 sec

Draw the FBD and KD of the collar

manT N

et

en

mat

Write the equations of motion

=

n nF ma t tF ma 2v

N m

dvT m

dt



Te

nth

Ed

ition


2 - 35

manT N

et

en

mat

=2 2

2562.5 (m/s )

0.4n

va

20.4 0.4(5) 2 m/st

dva t

dt

Kinematics : find vt, an, at

2 20.2 0.2(5 ) =5 m/stv t

Substitute into equations of motion

3.0(62.5)N 3.0(2)T

187.5 NN 6.0 NT

n nF ma t tF ma



Te

nth

Ed

ition


2 - 36

manT N

et

en

mat

=How would the problem change if motion was in the vertical plane?

mg

You would add an mg term and would also need to calculate

When is the tangential force greater than the normal force?

Only at the very beginning, when starting to accelerate. In most applications, an >> at



Te

nth

Ed

ition

Concept Question

2 - 37

D

BC

A

A car is driving from A to D on the curved path shown. The driver is doing the following at each point:

A – going at a constant speed B – stepping on the acceleratorC – stepping on the brake D – stepping on the accelerator

Draw the approximate direction of the car’s acceleration at each point.



Te

nth

Ed

ition

Kinetics: Radial and Transverse Coordinates

2 - 38

Hydraulic actuators and extending robotic arms are often analyzed using radial and transverse coordinates.



Te

nth

Ed

ition

Eqs of Motion in Radial & Transverse Components

12 - 39

rrmmaF

rrmmaF rr

2

2

• Consider particle at r and , in polar coordinates,



Te

nth

Ed

ition

Sample Problem 12.7

12 - 40

A block B of mass m can slide freely on a frictionless arm OA which rotates in a horizontal plane at a constant rate.0

a) the component vr of the velocity of B along OA, and

b) the magnitude of the horizontal force exerted on B by the arm OA.

Knowing that B is released at a distance r0 from O, express as a function of r

SOLUTION:

• Write the radial and transverse equations of motion for the block.

• Integrate the radial equation to find an expression for the radial velocity.

• Substitute known information into the transverse equation to find an expression for the force on the block.



Te

nth

Ed

ition

Sample Problem 12.7

12 - 41

SOLUTION:

• Write the radial and transverse equations of motion for the block.

:

:

amF

amF rr

rrmF

rrm

2

0 2

• Integrate the radial equation to find an expression for the radial velocity.

r

r

v

rr

rr

rr

rrr

drrdvv

drrdrrdvv

dr

dvv

dt

dr

dr

dv

dt

dvvr

r

0

20

0

20

2

dr

dvv

dt

dr

dr

dv

dt

dvvr r

rrr

r

20

220

2 rrvr

• Substitute known information into the transverse equation to find an expression for the force on the block.

2120

2202 rrmF



Te

nth

Ed

ition


2 - 42

• Write the equations of motion for the collar.

• Combine the equations of motion with kinematic relationships and solve.

• Draw the FBD and KD for the collar.

• Determine kinematics of the collar.

SOLUTION:

The 3-kg collar B slides on the frictionless arm AA. The arm is attached to drum D and rotates about O in a horizontal plane at the rate where and t are expressed in rad/s and seconds, respectively. As the arm-drum assembly rotates, a mechanism within the drum releases the cord so that the collar moves outward from O with a constant speed of 0.5 m/s. Knowing that at t = 0, r = 0, determine the time at which the tension in the cord is equal to the magnitude of the horizontal force exerted on B by arm AA.

0.75t



Te

nth

Ed

ition


2 - 43

mar

T N

eer

ma

=

r rF ma BF m a

Draw the FBD and KD of the collar

Write the equations of motion

SOLUTION: • Given:

• Find: time when T = N

2( )T m r r ( 2 )N m r r

0.75t

5 m/sr

(0) 0r



Te

nth

Ed

ition


2 - 44

3 2: (3 kg)( 0.28125 ) m/sr rF ma T t 2: (3 kg)(1.125 ) m/sBF m a N t

0 00.5

r tdr dt

(0.5 ) mr t

0r

2

(0.75 ) rad/s

0.75 rad/s

t

2 2 3 20 [(0.5 ) m][(0.75 ) rad/s] (0.28125 ) m/sra r r t t t

2

2

2 [(0.5 ) m][0.75 rad/s ] 2(0.5 m/s)[(0.75 ) rad/s]

(1.125 ) m/s

a r r t t

t

Kinematics : find expressions for r and

Substitute values into ar , a

Substitute into equation of motion3(0.84375 ) (3.375 )t t

2 4.000t 2.00 st

Set T = N

5 m/sr



Te

nth

Ed

ition

Concept Quiz

2 - 45

Top View

e1

e2

v

The girl starts walking towards the outside of the spinning platform, as shown in the figure. She is walking at a constant rate with respect to the platform, and the platform rotates at a constant rate. In which direction(s) will the forces act on her?

a) +e1 b) - e1 c) +e2 d) - e2

e) The forces are zero in the e1 and e2 directions



Te

nth

Ed

ition


2 - 46

Satellite orbits are analyzed using conservation of angular momentum.



Te

nth

Ed

ition

Eqs of Motion in Radial & Transverse Components

12 - 47

rrmmaF

rrmmaF rr

2

2

• Consider particle at r and , in polar coordinates,

rrmF

rrrm

mrdt

dFr

mrHO

2

22

2

2

• This result may also be derived from conservation of angular momentum,



Te

nth

Ed

ition


12 - 48

• moment of momentum or the angular momentum of the particle about O.

VmrHO

• Derivative of angular momentum with respect to time,

O

O

M

Fr

amrVmVVmrVmrH

• It follows from Newton’s second law that the sum of the moments about O of the forces acting on the particle is equal to the rate of change of the angular momentum of the particle about O.

zyx

O

mvmvmv

zyx

kji

H

• is perpendicular to plane containingOH

Vmr

and

2

sin

mr

vrm

rmVHO



Te

nth

Ed

ition


12 - 49

• When only force acting on particle is directed toward or away from a fixed point O, the particle is said to be moving under a central force.

• Since the line of action of the central force passes through O, and 0 OO HM

constant OHVmr

• Position vector and motion of particle are in a plane perpendicular to .OH

• Magnitude of angular momentum,

000 sin

constantsin

Vmr

VrmHO

massunit

momentumangular

constant

2

2

hrm

H

mrH

O

O

or



Te

nth

Ed

ition


12 - 50

• Radius vector OP sweeps infinitesimal area

drdA 221

• Define 2212

21 r

dt

dr

dt

dAareal velocity

• Recall, for a body moving under a central force,

constant2 rh

• When a particle moves under a central force, its areal velocity is constant.



Te

nth

Ed

ition

Newton’s Law of Gravitation

12 - 51

• Gravitational force exerted by the sun on a planet or by the earth on a satellite is an important example of gravitational force.

• Newton’s law of universal gravitation - two particles of mass M and m attract each other with equal and opposite force directed along the line connecting the particles,

4

49

2

312

2

slb

ft104.34

skg

m1073.66

ngravitatio ofconstant

Gr

MmGF

• For particle of mass m on the earth’s surface,

222 s

ft2.32

s

m81.9 gmg

R

MGmW



Te

nth

Ed

ition

Sample Problem 12.8

12 - 52

A satellite is launched in a direction parallel to the surface of the earth with a velocity of 18820 mi/h from an altitude of 240 mi. Determine the velocity of the satellite as it reaches it maximum altitude of 2340 mi. The radius of the earth is 3960 mi.

SOLUTION:

• Since the satellite is moving under a central force, its angular momentum is constant. Equate the angular momentum at A and B and solve for the velocity at B.



Te

nth

Ed

ition

Sample Problem 12.8

12 - 53

SOLUTION:

• Since the satellite is moving under a central force, its angular momentum is constant. Equate the angular momentum at A and B and solve for the velocity at B.

mi23403960

mi2403960hmi18820

constantsin

B

AAB

BBAA

O

r

rvv

vmrvmr

Hvrm

hmi12550Bv



Te

nth

Ed

ition

Trajectory of a Particle Under a Central Force

12 - 54

• For particle moving under central force directed towards force center,

022 FrrmFFrrm r

• Second expression is equivalent to from which,,constant 2 hr

rd

d

r

hr

r

h 1and

2

2

2

2

2

• After substituting into the radial equation of motion and simplifying,

ru

umh

Fu

d

ud 1where

222

2

• If F is a known function of r or u, then particle trajectory may be found by integrating for u = f(), with constants of integration determined from initial conditions.



Te

nth

Ed

ition


12 - 55

constant

1where

22

2

22222

2

h

GMu

d

ud

GMmur

GMmF

ru

umh

Fu

d

ud

• Consider earth satellites subjected to only gravitational pull of the earth,

• Solution is equation of conic section,

tyeccentricicos11 2

2

GM

hC

h

GM

ru

• Origin, located at earth’s center, is a focus of the conic section.

• Trajectory may be ellipse, parabola, or hyperbola depending on value of eccentricity.



Te

nth

Ed

ition


12 - 56

tyeccentricicos11 2

2

GM

hC

h

GM

r

• Trajectory of earth satellite is defined by

• hyperbola, > 1 or C > GM/h2. The radius vector becomes infinite for

211

11 cos1

cos0cos1hC

GM

• parabola, = 1 or C = GM/h2. The radius vector becomes infinite for

1800cos1 22

• ellipse, < 1 or C < GM/h2. The radius vector is finite for and is constant, i.e., a circle, for < 0.



Te

nth

Ed

ition


12 - 57

• Integration constant C is determined by conditions at beginning of free flight, =0, r = r0 ,

20002

0

2

20

11

0cos11

vr

GM

rh

GM

rC

GM

Ch

h

GM

r

00

200

2

2

or 1

r

GMvv

vrGMhGMC

esc

• Satellite escapes earth orbit for

• Trajectory is elliptic for v0 < vesc and becomes circular for = 0 or C = 0,

0r

GMvcirc



Te

nth

Ed

ition


12 - 58

• Recall that for a particle moving under a central force, the areal velocity is constant, i.e.,

constant212

21 hr

dt

dA

• Periodic time or time required for a satellite to complete an orbit is equal to area within the orbit divided by areal velocity,

h

ab

h

ab 2

2

where

10

1021

rrb

rra



Te

nth

Ed

ition

Sample Problem 12.9

12 - 59

Determine:

a) the maximum altitude reached by the satellite, and

b) the periodic time of the satellite.

A satellite is launched in a direction parallel to the surface of the earth with a velocity of 36,900 km/h at an altitude of 500 km.

SOLUTION:

• Trajectory of the satellite is described by

cos1

2C

h

GM

r

Evaluate C using the initial conditions at = 0.

• Determine the maximum altitude by finding r at = 180o.

• With the altitudes at the perigee and apogee known, the periodic time can be evaluated.



Te

nth

Ed

ition

Sample Problem 12.9

12 - 60

SOLUTION:

• Trajectory of the satellite is described by

cos1

2C

h

GM

r

Evaluate C using the initial conditions at = 0.

2312

2622

29

3600

3

0

6

0

sm10398

m1037.6sm81.9

sm104.70

sm1025.10m106.87

sm1025.10

s/h3600

m/km1000

h

km36900

m106.87

km5006370

gRGM

vrh

v

r

1-9

22

2312

6

20

m103.65

sm4.70

sm10398

m1087.6

1

1

h

GM

rC



Te

nth

Ed

ition

Sample Problem 12.9

12 - 61

• Determine the maximum altitude by finding r1 at = 180o.

km 66700m107.66

m

1103.65

sm4.70

sm103981

61

922

2312

21

r

Ch

GM

r

km 60300km6370-66700 altitudemax

• With the altitudes at the perigee and apogee known, the periodic time can be evaluated.

sm1070.4

m1021.4m1036.82

h

2

m1021.4m107.6687.6

m1036.8m107.6687.6

29

66

6610

6621

1021

ab

rrb

rra

min31h 19s103.70 3



Te

nth

Ed

ition

Kepler’s Laws of Planetary Motion

12 - 62

• Results obtained for trajectories of satellites around earth may also be applied to trajectories of planets around the sun.

• Properties of planetary orbits around the sun were determined astronomical observations by Johann Kepler (1571-1630) before Newton had developed his fundamental theory.

1) Each planet describes an ellipse, with the sun located at one of its foci.

2) The radius vector drawn from the sun to a planet sweeps equal areas in equal times.

3) The squares of the periodic times of the planets are proportional to the cubes of the semimajor axes of their orbits.

Date post:	22-Jan-2016
Category:	Documents
Upload:	noreen
View:	105 times
Download:	2 times

Kinetics of Particles: Newton’s Second Law

Documents