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UNIT-IV
Kinetics of Particles:
Newton’s Second Law
Contents
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
12 - 2
Kinetics of Particles
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.
2-3
Introduction
F  ma
• Newton’s Second Law of Motion
• 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
12 - 4
Linear Momentum of a Particle
• Replacing the acceleration by the derivative of the velocity
yields


dv
F  m
dt

d
 dL
 m v  
dt
dt

L  linear momentum of the particle
• 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.
12 - 5
Systems
of
Units
• 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,
kg  m
 m
1 N  1 kg 1 2   1 2
 s 
s
• U.S. Customary Units: base units are the units of force
(lb), length (m), and time (second). The unit of mass is
derived,
1lb
1lb
lb  s 2
1lbm 
1slug 
1
2
2
ft
32.2 ft s
1ft s
12 - 6
Equations of Motion

• Newton’s second law

 F  ma
• Can use scalar component equations, e.g., for
rectangular components,






 Fx i  Fy j  Fz k   ma x i  a y j  a z k 
 Fx  max  Fy  ma y  Fz  maz
 Fx  mx  Fy  my  Fz  mz
12 - 7
Dynamic
Equilibrium
• Alternate expression of Newton’s second law,


F

m
a
0


 ma  inertial vector
• 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.
12 - 8
Free Body Diagrams and Kinetic
The free body diagram isDiagrams
the same as you have done in statics; we
will add the kinetic diagram in our dynamic analysis.
1. Isolate the body of interest (free body)
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)
5. Draw appropriate dimensions (usually angles for particles)
y
x
225 N
25o
N
Ff
mg
12 - 9
Free Body Diagrams and Kinetic
Diagrams
Put the inertial terms for the
body of interest on the kinetic diagram.
1. Isolate the body of interest (free body)
2. Draw in the mass times acceleration of the particle; if unknown,
do this in the positive direction according to your chosen axes
y
x
225 N
may

25o
N
max
Ff
mg
F

ma
12 - 10
Free Body Diagrams and Kinetic
Draw the FBD andDiagrams
KD for block A (note that the
massless, frictionless pulleys are attached to block A
and should be included in the system).
2 - 11
Free Body Diagrams and Kinetic
1. Isolate body
Diagrams
2. Axes
3.
4.
5.
6.
T
T
T
Applied forces
Replace supports with forces
Dimensions (already drawn)
Kinetic diagram
y
NB
T
x
Ff-B
T
mg
N1
Ff-1
=
may = 0
max
2 - 12
Free Body Diagrams and Kinetic
Diagrams
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 q is increasing
2 - 13
Free Body Diagrams and Kinetic
1. Isolate body
Diagrams
2. Axes
3.
4.
5.
6.
Applied forces
Replace supports with forces
Dimensions
Kinetic diagram
eq
er
maq
mar
q
Ff
=
q
mg
N
2 - 14
Sample Problem 12.1
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.
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 mk  0.25.
12 - 15
Sample Problem
12.1
SOLUTION:
• Resolve the equation of motion for the block
into two rectangular component equations.
 Fx  ma :

y
O

P cos 30  0.25N  6.21lb  s 2 ft 10 ft s 2
 62.1lb
x
W
200 lb
m 
g 32.2 ft s 2
lb  s 2
 6.21
ft
F  mk N
 0.25N
 Fy  0 :
N  P sin 30  200 lb  0
• Unknowns consist of the applied force P and
the normal reaction N from the plane. The two
equations may be solved for these unknowns.
N  P sin 30  200 lb
P cos 30  0.25P sin 30  200 lb  62.1lb
P  151lb
12 - 16

Sample Problem 12.3
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.
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.
12 - 17
Sample
Problem
12.3
SOLUTION:
O
x
y
• Write the kinematic relationships for the dependent
motions and accelerations of the blocks.
y B  12 x A
a B  12 a A
• Write equations of motion for blocks and pulley.
 Fx  m Aa A :
T1  100 kg a A
 Fy  mB aB :
mB g  T2  mB a B
300 kg 9.81m s 2  T2  300 kg a B
T2  2940N - 300 kg a B
 Fy  mC aC  0 :
T2  2T1  0
12 - 18
Sample
Problem
12.3
• Combine kinematic relationships with equations of
motion to solve for accelerations and cord tension.
O
x
y
y B  12 x A
a B  12 a A
T1  100 kg a A
T2  2940N - 300 kg a B

 2940N - 300 kg  12 a A

T2  2T1  0
2940 N  150 kg a A  2100 kg a A  0
a A  8.40 m s 2
a B  12 a A  4.20 m s 2
T1  100 kg a A  840 N
T2  2T1  1680 N
12 - 19
Sample Problem
12.4
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.
The 12-lb block B starts from rest and
slides on the 30-lb wedge A, which is
supported by a horizontal surface.
• Write the equations of motion for the
wedge and block.
• Solve for the accelerations.
Neglecting friction, determine (a) the
acceleration of the wedge, and (b) the
acceleration of the block relative to the
wedge.
12 - 20
SampleSOLUTION:
Problem 12.4
• The block is constrained to slide down the
wedge. Therefore, their motions are dependent.



aB  a A  aB A
• Write equations of motion for wedge and block.
 Fx  m Aa A :
N1 sin 30  m A a A
y
0.5 N1  W A g a A
x
 Fx  mB a x  mB a A cos 30  aB A :
 WB sin 30  WB g a A cos 30  a B
a B A  a A cos 30  g sin 30
 Fy  mB a y  mB  a A sin 30 :
N1  WB cos 30  WB g a A sin 30
12 - 21
A

Sample• Solve
Problem
12.4
for the accelerations.
0.5 N1  W A g a A
N1  WB cos 30  WB g a A sin 30
2W A g a A  WB cos 30  WB g a A sin 30
aA 
gWB cos 30
2W A  WB sin 30

32.2 ft s 2 12 lb cos 30
aA 
230 lb  12 lb sin 30
a A  5.07 ft s 2
a B A  a A cos 30  g sin 30




a B A  5.07 ft s 2 cos 30  32.2 ft s 2 sin 30
a B A  20.5 ft s 2
12 - 22
Group Problem Solving
SOLUTION:
• Write the kinematic relationships for the
dependent motions and accelerations of
the blocks.
• Draw the FBD and KD for each block
• 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.
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.
2 - 23
Group Problem Solving
SOLUTION:
• Write the kinematic relationships for the
dependent motions and accelerations of
the blocks.
xA
yB
This is the same problem worked last
chapter- write the constraint equation
x A  3 yB  constants  L
Differentiate this twice to get the
acceleration relationship.
v A  3vB  0
a A  3aB  0
a A  3aB
(1)
2 - 24
Group
Problem
Solving
• Draw the FBD and KD for each block
2T
T
B
mAg
+y
A
T
=
maBy
mBg
Fx  m A a A :
Fy  mB aB
9.81 m/s 2
aB 

 0.83136 m/s2
mA
30 kg
1 9
1 9
25 kg
mB
g
=
+x
NA
• Write the equation of motion for each block
WB  3T  mB aB (2)
• Solve the three equations, 3 unknowns
(3) (2) mB g  3(3mA aB )  mB aB
maAx
T  m A aB
From Eq (1)
T  3m A aB
(3)
T  3  30 kg  0.83136 m/s2
T  74.8 N
a A  2.49  2.49 m/s2
2 - 25
Concept Quiz
(1)
(2)
(3)
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)
2 - 26
Kinetics: Normal and Tangential
Coordinates
Aircraft and roller coasters can both experience large
normal forces during turns.
2 - 27
Equations of Motion
• Newton’s second law


 F  ma
• For tangential and normal components,
 F n  man
F  ma

F
t
t
t
m
dv
dt
F
n
m
v2

12 - 28
Sample Problem
12.5
SOLUTION:
• Resolve the equation of motion for the
bob into tangential and normal
components.
• Solve the component equations for the
normal and tangential accelerations.
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 acceleration of the bob in that position.
• Solve for the velocity in terms of the
normal acceleration.
12 - 29
Sample
Problem
12.5
SOLUTION:
• Resolve the equation of motion for the bob into
tangential and normal components.
• Solve the component equations for the normal and
tangential accelerations.
mg sin 30  mat
 Ft  mat :
at  g sin 30
 Fn  man :
at  4.9 m s 2
2.5mg  mg cos 30  man
an  g 2.5  cos 30
an  16.03 m s 2
• Solve for velocity in terms of normal acceleration.
an 
v2

v  an 
2 m 16.03 m s 2 
v  5.66 m s
12 - 30
Sample Problem
12.6
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.
Determine the rated speed of a
highway curve of radius  = 400 ft
banked through an angle q = 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.
• Resolve the equation of motion for
the car into vertical and normal
components.
• Solve for the vehicle speed.
12 - 31
Sample Problem
12.6
• Resolve the equation of motion for
the car into vertical and normal
components.
R cosq  W  0
 Fy  0 :
W
R
cosq
 Fn  man : R sin q 
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.
W
an
g
W
W v2
sin q 
cosq
g 
• Solve for the vehicle speed.
v 2  g tan q


 32.2 ft s 2 400 ft  tan 18
v  64.7 ft s  44.1 mi h
12 - 32
Group Problem Solving
v
SOLUTION:
• Draw the FBD and KD for the collar.
• Write the equations of motion for the
collar.
• Determine kinematics of the collar.
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.
• Combine the equations of motion with
kinematic relationships and solve.
2 - 33
Group Problem Solving
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
mat
et
=
en
T
man
N
Write the equations of motion
Fn  man
N m
v2

Ft  mat
dv
T m
dt
2 - 34
Group Problem
Solving
e
t
Kinematics : find vt, an, at
vt  0.2t 2  0.2(52 ) =5 m/s
v
2
mat
en
=
2
5
an  
 62.5 (m/s2 )
 0.4
T
q N
man
dv
at 
 0.4t  0.4(5)  2 m/s 2
dt
Substitute into equations of motion
Fn  man
Ft  mat
N  3.0(62.5)
T  3.0(2)
N  187.5 N
T  6.0 N
2 - 35
Group Problem
Solving
e
t
How would the problem
change if motion was in the
vertical plane?
mat
en
=
q N
T
You would add an mg term
and would also need to
calculate q
man
mg
When is the tangential force greater than the normal force?
Only at the very beginning, when starting to accelerate.
In most applications, an >> at
2 - 36
Concept Question
B
C
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
C – stepping on the brake
D
B – stepping on the accelerator
D – stepping on the accelerator
Draw the approximate direction of the car’s acceleration
at each point.
2 - 37
Kinetics: Radial and Transverse
Coordinates
Hydraulic actuators and
extending robotic arms are
often analyzed using radial
and transverse coordinates.
2 - 38
Eqs of Motion in Radial &
Transverse
Components
• Consider particle at r and q, in polar coordinates,


2

F

ma

m
r

r
q


 r
r
 Fq  maq  mrq  2rq 
12 - 39
Sample Problem
12.7
SOLUTION:
• Write the radial and transverse
equations of motion for the block.
• Integrate the radial equation to find an
expression for the radial velocity.
A block B of mass m can slide freely on
a frictionless arm OA which rotates in a
horizontal plane at a constant rate q0 .
• Substitute known information into the
transverse equation to find an
expression for the force on the block.
Knowing that B is released at a distance
r0 from O, express as a function of r
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.
12 - 40
Sample Problem
12.7
• Integrate the radial equation to find an
SOLUTION:
• Write the radial and transverse
equations of motion for the block.
 Fr  m ar :


0  m r  rq
 F q  m aq : F  mrq  2rq 
2
expression for the radial velocity.
dv
dv dr
dv
 vr r
r  vr  r  r
dt
dr dt
dr
dv
dv dr
dv
 vr r
r  vr  r  r
dt
dr dt
dr
vr dvr  rq 2 dr  rq02 dr
vr
r
2
 vr dvr  q0  r dr
0
r0

vr2  q 02 r 2  r02

• Substitute known information into the
transverse equation to find an expression
for the force on the block.
F
 2mq 02
r
2

2 12
 r0
12 - 41
Group Problem
Solving
SOLUTION:
• Draw the FBD and KD for the collar.
• Write the equations of motion for the
collar.
• Determine kinematics of the collar.
• Combine the equations of motion with
kinematic relationships and solve.
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 q  0.75t where q
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.
2 - 42
Group Problem Solving
SOLUTION: • Given: q  0.75t
r (0)  0
r  5 m/s
• Find: time when T = N
Draw the FBD and KD of the collar
maq
eq
er
T
=
mar
N
Write the equations of motion
Fr  mar
T  m(r  rq 2 )
Fq  mB aq
N  m(rq  2rq )
2 - 43
Group Problem Solving
r  5 m/s
Kinematics : find expressions for r and q

q  (0.75t ) rad/s
r
0
dr 

t
0.5 dt
0
r  (0.5t ) m
q  0.75 rad/s2
r 0
Substitute values into ar , aq
ar  r  rq 2  0  [(0.5t ) m][(0.75t ) rad/s]2  (0.28125t 3 ) m/s 2
aq  rq  2rq  [(0.5t ) m][0.75 rad/s2 ]  2(0.5 m/s)[(0.75t ) rad/s]
 (1.125t ) m/s2
Substitute into equation of motion
Set T = N
Fr  mar :  T  (3 kg)(0.28125t ) m/s
3
Fq  mB aq : N  (3 kg)(1.125t ) m/s
2
2
(0.84375t 3 )  (3.375t )
t 2  4.000
t  2.00 s
2 - 44
Concept Quiz
e2
Top View
e1
v
w
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 directions2 - 45
Angular Momentum of a Particle
Satellite orbits are analyzed using conservation
of angular momentum.
2 - 46
Eqs of Motion in Radial &
Transverse
Components
• Consider particle at r and q, in polar coordinates,


2

F

ma

m
r

r
q


 r
r
 Fq  maq  mrq  2rq 
• This result may also be derived from conservation
of angular momentum,
H O  mr 2q



d
mr 2q
dt
 m r 2q  2rrq
r  Fq 
 Fq  mrq  2rq 

12 - 47
Angular Momentum
of
a
Particle



• H  r  mV  moment of momentum or the angular
O
momentum of the particle about O.



• H O is perpendicular to plane containing r and mV



H O  rmV sin 
i
j
k

 rm vq
HO  x
y
z
mv x mv y mvz
 mr 2q
• Derivative of angular momentum with respect to time,

 
 
 


H O  r  mV  r  mV  V  mV  r  ma

 rF

  MO
• 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.
12 - 48
Conservation of Angular
• When only force acting on particle is directed
Momentum
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,  M O  H O  0 and
 

r  mV  H O  constant
• Position vector and motion
 of particle are in a
plane perpendicular to H O .
• Magnitude of angular momentum,
H O  rmV sin   constant
 r0 mV0 sin 0
or
H O  mr 2q  constant
HO
angular momentum
 r 2q  h 
m
unit mass
12 - 49
Conservation of Angular
Momentum
• Radius vector OP sweeps infinitesimal area
dA  12 r 2 dq
• Define
dA 1 2 dq 1 2 
 2r
 2 r q  areal velocity
dt
dt
• Recall, for a body moving under a central force,
h  r 2q  constant
• When a particle moves under a central force, its
areal velocity is constant.
12 - 50
Newton’s Law of Gravitation
• 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,
Mm
F G 2
r
G  constant of gravitatio n
 66.73  10
12
m3
kg  s
2
 34.4  10
9
ft 4
lb  s 4
• For particle of mass m on the earth’s surface,
MG
m
ft
W  m 2  mg g  9.81 2  32.2 2
R
s
s
12 - 51
Sample Problem 12.8
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.
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.
12 - 52
Sample Problem 12.8
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.
rm v sin   H O  constant
rA m v A  rB m v B
vB  v A
rA
rB
 18820mi h 
3960  240mi
3960  2340mi
v B  12550 mi h
12 - 53
Trajectory of a Particle Under a
• For particle movingCentral
under central force
directed towards force center,
Force


m r  rq 2   Fr   F
mrq  2rq    Fq  0
• Second expression is equivalent to r 2q  h  constant , from which,
q 
h
r2
h2 d 2  1 
and r   2
2  r 
r dq
• After substituting into the radial equation of motion and simplifying,
d 2u
F

u

dq 2
mh 2u 2
where u 
1
r
• If F is a known function of r or u, then particle trajectory may be
found by integrating for u = f(q), with constants of integration
determined from initial conditions.
12 - 54
Application to Space Mechanics
• Consider earth satellites subjected to only gravitational pull
of the earth,
d 2u
F

u

dq 2
mh 2u 2
where u 
1
r
F
GMm
r2
 GMmu 2
d 2u
GM

u

 constant
2
2
dq
h
• Solution is equation of conic section,
1 GM
u   2 1   cosq 
r
h
Ch2

 eccentricity
GM
• 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.
12 - 55
Application to Space Mechanics
• Trajectory of earth satellite is defined by
1 GM
 2 1   cosq 
r
h
Ch2

 eccentricity
GM
• hyperbola,  > 1 or C > GM/h2. The radius vector
becomes infinite for
1  1 
1  GM 
1   cosq1  0 q1   cos      cos 
 C h2 
 


• parabola,  = 1 or C = GM/h2. The radius vector
becomes infinite for
1  cosq 2  0 q 2  180
• ellipse,  < 1 or C < GM/h2. The radius vector is finite
for q and is constant, i.e., a circle, for  < 0.
12 - 56
Application to
Space Mechanics
• Integration constant C is determined by conditions
at beginning of free flight, q =0, r = r0 ,

1 GM  Ch 2
 2 1
cos 0 

r0
h  GM

1 GM 1
GM
C  2  
r0 h
r0 r0 v0 2
• Satellite escapes earth orbit for
  1 or C  GM h 2  GM r0 v0 2
vesc  v0 
2GM
r0
• Trajectory is elliptic for v0 < vesc and becomes
circular for  = 0 or C = 0,
GM
vcirc 
r0
12 - 57
Application• Recall
to Space
Mechanics
that for a particle moving under a central
force, the areal velocity is constant, i.e.,
dA 1 2  1
 2 r q  2 h  constant
dt
• Periodic time or time required for a satellite to
complete an orbit is equal to area within the orbit
divided by areal velocity,
 ab 2 ab


h2
h
where a  12 r0  r1 
b  r0 r1
12 - 58
Sample Problem
12.9
SOLUTION:
• Trajectory of the satellite is described by
1 GM
 2  C cosq
r
h
Evaluate C using the initial conditions
at q = 0.
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.
Determine:
a) the maximum altitude reached by
the satellite, and
b) the periodic time of the satellite.
• Determine the maximum altitude by
finding r at q = 180o.
• With the altitudes at the perigee and
apogee known, the periodic time can
be evaluated.
12 - 59
Sample Problem
12.9
SOLUTION:
• Trajectory of the satellite is described by
1 GM
 2  C cosq
r
h
Evaluate C using the initial conditions
at q = 0.
1 GM
C  2
r0 h
r0  6370  500 km
 6.87  106 m
km 1000 m/km
v 0  36900

h
3600 s/h
 10.25  103 m s



h  r0v0  6.87  106 m 10.25  103 m s
 70.4  109 m 2 s


GM  gR 2  9.81m s 2 6.37  106 m

1
6.87  10 m
6

398  1012 m3 s 2
70.4 m s
2
2
 65.3  109 m-1
2
 398  1012 m3 s 2
12 - 60
Sample •Problem
12.9
Determine the maximum altitude by finding r
at q =
1
180o.
1 GM
398  1012 m3 s 2
9 1
 2 C 

65
.
3

10
2 2
r1 h
m
70.4 m s


r1  66.7  106 m  66700 km
max altitude  66700 - 6370km  60300 km
• With the altitudes at the perigee and apogee known,
the periodic time can be evaluated.
a  12 r0  r1   12 6.87  66.7   106 m  36.8  106 m
b  r0 r1  6.87  66.7  106 m  21.4  106 m


2 ab 2 36.8  106 m 21.4  106 m


h
70.4  109 m 2 s

  70.3  103 s  19 h 31min
12 - 61
Kepler’s Laws of Planetary Motion
• 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.
12 - 62