Transcript Monday, Oct. 6, 2003

PHYS 1443 – Section 003
Lecture #11
Monday,Oct. 6, 2003
Dr. Jaehoon Yu
•Newton’s Law of Gravitation
•Kepler’s Laws
•Work Done by Constant Force
•Work Done by Varying Force
Deadline for Homework #5 is noon, Wednesday, Oct. 8!!
Deadline for Homework #6 is noon, Wednesday, Oct. 15!!
I will be out of town this Wednesday, Oct. 8  Dr. Sosebee will give the lecture.
Monday, Oct. 6, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
1
Free Fall Acceleration & Gravitational Force
Weight of an object with mass m is
mg. Using the force exerting on a
particle of mass m on the surface of
the Earth, one can get
What would the gravitational
acceleration be if the object is at
an altitude h above the surface of
the Earth?
mg
g
M Em
RE2
ME
G
RE2
G
M Em  G M Em
Fg  mg '  G
2
2


R

h
r
E
ME
g'  G
RE  h 2
What do these tell us about the gravitational acceleration?
•The gravitational acceleration is independent of the mass of the object
•The gravitational acceleration decreases as the altitude increases
•If the distance from the surface of the Earth gets infinitely large, the weight of the
object approaches 0.
Monday, Oct. 6, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
2
Example for Gravitation
Using the fact that g=9.80m/s2 at the Earth’s surface, find the average density of the Earth.
Since the gravitational acceleration is
g
So the mass of the Earth is
Therefore the density of the
Earth is
G
ME
11 M E

6
.
67

10
2
2
RE
RE
2
R g
ME  E
G
2


ME
VE
RE g
3g
G


4GRE
4
3
RE

3  9.80
3
3


5
.
50

10
kg
/
m
4  6.67 10 11  6.37 106
Monday, Oct. 6, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
3
Kepler’s Laws & Ellipse
a
b
F1
c
F2
Ellipses have two different axis, major (long) and
minor (short) axis, and two focal points, F1 & F2
a is the length of a semi-major axis
b is the length of a semi-minor axis
Kepler lived in Germany and discovered the law’s governing planets’
movement some 70 years before Newton, by analyzing data.
1. All planets move in elliptical orbits with the Sun at one focal point.
2. The radius vector drawn from the Sun to a planet sweeps out equal
area in equal time intervals. (Angular momentum conservation)
3. The square of the orbital period of any planet is proportional to the
cube of the semi-major axis of the elliptical orbit.
Newton’s laws explain the cause of the above laws. Kepler’s third law is
the direct consequence of law of gravitation being inverse square law.
Monday, Oct. 6, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
4
The Law of Gravity and Motions of Planets
•Newton assumed that the law of gravitation applies the same
whether it is on the Moon or the apple on the surface of the Earth.
•The interacting bodies are assumed to be point like particles.
Apple g
RE
aM
Moon
Newton predicted that the ratio of the Moon’s
acceleration aM to the apple’s acceleration g would be
2
aM

1 / rM 
 RE   6.37 106 
4


  


2
.
75

10
2  
g
1 / RE   rM   3.84 108 
2
v
2
Therefore the centripetal acceleration of the Moon, aM, is
aM  2.75 104  9.80  2.70 103 m / s 2
Newton also calculated the Moon’s orbital acceleration aM from the knowledge of its distance
from the Earth and its orbital period, T=27.32 days=2.36x106s
9.80
4   3.84 108
v 2 2rM / T 2 4  rM
3
2



2
.
72

10
m
/
s



aM r
2
6 2
2


60
r
2
.
36

10
T
M
M


This means that the Moon’s distance is about 60 times that of the Earth’s radius, its acceleration
is reduced
by 6,the
square of the ratio. PHYS
This 1443-003,
proves that
the inverse square law is valid.
Monday, Oct.
2003
Fall 2003
5
Dr. Jaehoon Yu
Kepler’s Third Law
It is crucial to show that Keper’s third law can be predicted from the
inverse square law for circular orbits.
v
r
Since the gravitational force exerted by the Sun is radially
directed toward the Sun to keep the planet circle, we can
apply Newton’s second law
GM s M P M p v 2

r2
r
Ms
2r
Since the orbital speed, v, of the planet with period T is v  T
2
GM s M P
M p 2r / T 
The above can be written

2
r
Solving for T one
can obtain
T
2
2  4 r 3  K r 3 and


s
 GM s 
 4 2
K s   GM
s

r

  2.97 10 19 s 2 / m3

This is Kepler’s third law. It’s also valid for ellipse for r being the length of the
semi-major axis. The constant Ks is independent of mass of the planet.
Monday, Oct. 6, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
6
Example of Kepler’s Third Law
Calculate the mass of the Sun using the fact that the period of the Earth’s orbit around the
Sun is 3.16x107s, and its distance from the Sun is 1.496x1011m.
Using Kepler’s third law.
The mass of the Sun, Ms, is
2


4

3
3
2 

T  GM s r  K s r
 4 2  3
r
M s  
 GT 


4 2
11




1
.
496

10
11
7 
6
.
67

10

3
.
16

10



 1.99 10 kg
30
Monday, Oct. 6, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
7

3
Kepler’s Second Law and Angular Momentum Conservation
Consider a planet of mass Mp moving around the Sun in an elliptical orbit.
D
S
C
r
Since the gravitational force acting on the planet is
A always toward radial direction, it is a central force
dr
B Therefore the torque acting on the planet by this
force is always 0.
  r  F  r  Frˆ  0
Since torque is the time rate change of angular
momentum L, the angular momentum is constant.


dL
0
dt
L  const
Because the gravitational force exerted on a
 r  p  r  M p v  M p r  v  const
planet by the Sun results in no torque, the
angular momentum L of the planet is constant.
Since the area swept by the
L
1
1
dA  L

dt
 const
dA  r  d r  r  vdt
motion of the planet is
2M p
2M p
2
2
dt
L
This is Keper’s second law which states that the radius vector from the Sun
to a planet sweeps our equal areas in equal time intervals.
Monday, Oct. 6, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
8
Work Done by a Constant Force
Work in physics is done only when a sum of forces
exerted on an object made a motion to the object.
F
M
y
q
FN
M
Free Body
Diagram
F
q
d
x
FG  M g
Which force did the work? Force F
 
How much work did it do? W   F  d  Fd cosq
What does this mean?
Monday, Oct. 6, 2003
Unit?
N m
 J (for Joule)
Physical work is done only by the component of of
the force along the movement of the object.
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
Work is energy transfer!!
9
Example of Work w/ Constant Force
A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F=50.0N at
an angle of 30.0o with East. Calculate the work done by the force on the vacuum
cleaner as the vacuum cleaner is displaced by 3.00m to East.
F
M
0o
W
M
 F  d   F  d cosq
W  50.0  3.00  cos 30  130 J
d
Does work depend on mass of the object being worked on?
Why don’t I see the mass
term in the work at all then?
Monday, Oct. 6, 2003
Yes
It is reflected in the force. If the object has smaller
mass, its would take less force to move it the same
distance as the heavier object. So it would take less
work. Which makes perfect sense, doesn’t it?
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
10
Scalar Product of Two Vectors
• Product of magnitude of the two vectors and the cosine of the
angle between them
A B cos q
A B 
• Operation is commutative A  B  A B cosq  B A cos q  B  A


• Operation follows distribution
A B  C  A B  A C
law of multiplication
 
 
 
 
 
 
• Scalar products of Unit Vectors i  i  j j  k  k  1 i  j  j k  k  i  0
• How does scalar product look in terms of components?



A  Ax i  Ay j  Az k



B  Bx i  By j  Bz k



C  Cx i  C y j  Cz k
 
 
 






 
  

A  B   Ax i  Ay j  Az k    Bx i  By j  Bz k    Ax Bx i  i  Ay B y j j  Az Bz k  k   cross terms

 



A  B  Ax Bx  Ay B y  Az Bz
Monday, Oct. 6, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
11
Example of Work by Scalar Product
A particle moving in the xy plane undergoes a displacement d=(2.0i+3.0j)m as a
constant force F=(5.0i+2.0j) N acts on the particle.
a) Calculate the magnitude of the displacement
and that of the force.
Y
d
F
X
d  d x2  d y2 
2.02  3.02  3.6m
F  Fx2  Fy2  5.02  2.02  5.4 N
b) Calculate the work done by the force F.
W  F d 


 
  
  2.0  5.0 i  i  3.0  2.0 j j  10  6  16( J )
 2.0 i  3.0 j    5.0 i  2.0 j 

 

Can you do this using the magnitudes and the angle between d and F?
W  F  d  F d cosq
Monday, Oct. 6, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
12