Monday,Oct. 6, 2003
Dr. Jae hoon 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

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 mg g

G
M
E m
R
2
E

G
M
R
2
E
E
What would the gravitational acceleration be if the object is at an altitude h above the surface of the Earth?
F g
 mg ' 
G
M
E m r
2

G
M

R
E
E
 m h

2 g '

G 
R
E
M

E 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

Using the fact that g=9.80m/s 2 at the Earth’s surface, find the average density of the Earth.
Since the gravitational acceleration is g 
G
M
E
R
E
2

6 .
67

10

11
M
E
R
E
2
So the mass of the Earth is
M
E

R
E
2 g
G
Therefore the density of the
Earth is 

4


M
E
R
E
2 g
V
E

4

G

R
E
3

3 g
4

GR
E
3

9 .
80

6 .
67

10

11 
6 .
37

10
6

5 .
50

10
3 kg / m
3
Monday, Oct. 6, 2003 PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
3

F
1 b c a
F
2
Ellipses have two different axis, major (long) and minor (short) axis, and two focal points, F
1
& F
2 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

•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.
Apple
•The interacting bodies are assumed to be point like particles.
Moon
Newton predicted that the ratio of the Moon’s acceleration a
M to the apple’s acceleration g would be g
R
E a
M v a
M g



1
1
/
/ r
M
R
E


2
2



R
E r
M


2



6 .
37

10
6
3 .
84

10
8


2

2 .
75

10

4
Therefore the centripetal acceleration of the Moon, a
M, a
M

2 .
75

10

4 
9 .
80

2 .
70

10

3 m / s
2 is
Newton also calculated the Moon’s orbital acceleration a
M from the knowledge of its distance from the Earth and its orbital period, T=27.32 days=2.36x10
6 s a
M
 v
2 r
M


2
 r
M r
M
/ T

2

4
  r
M
T
2

4
  

2 .
36
3 .
84

10
6

10
8

2

2 .
72

10

3 m / s
2

9 .
80
 
2
This means that the Moon’s distance is about 60 times that of the Earth’s radius, its acceleration
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
5

It is crucial to show that Keper’s third law can be predicted from the inverse square law for circular orbits.
v
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 r
M s
GM s
M r
2
P 
M p v
2 r
Since the orbital speed, v, of the planet with period T is
The above can be written
GM r s
2
M
P 
M p

2
 r r
/ T

2
Solving for T one can obtain
T
2 


4

2
GM s

 r
3

K s r
3 and v

2
 r
T
K s



4

2
GM s



2 .
97

10

19 s
2
/ m
3
This is Kepler’s third law. It’s also valid for ellipse for r being the length of the semi-major axis. The constant K s is independent of mass of the planet.
Monday, Oct. 6, 2003 PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
6

Calculate the mass of the Sun using the fact that the period of the Earth’s orbit around the
Sun is 3.16x10
7 s, and its distance from the Sun is 1.496x10
11 m.
Using Kepler’s third law.
T
2 


4

2
GM s

 r
3 
K s r
3
The mass of the Sun, M s
, is
M s



4

2
GT

 r
3



4
6 .
67

10

11

2

3 .
16

10
7




1 .
496

10
11

3

1 .
99

10
30 kg
Monday, Oct. 6, 2003 PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
7

Kepler’s Second Law and Angular Momentum Conservation
Consider a planet of mass M p moving around the Sun in an elliptical orbit.
D
S r dr
A
B
Since the gravitational force acting on the planet is always toward radial direction, it is a central force
Therefore the torque acting on the planet by this force is always 0.
C

Since torque is the time rate change of angular momentum L , the angular momentum is constant.
 r

F
 
 d L dt r

F r
ˆ

0

L
0
 const
Because the gravitational force exerted on a planet by the Sun results in no torque, the angular momentum L of the planet is constant.
L
 r
 p
 r

M p v

M p r
 v
 const
Since the area swept by the motion of the planet is dA

1
2 r
 d r

1
2 r
 v dt

L
2 M p dt dA dt

L
2 M p
 const
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

y
F
M q
M
Free Body
Diagram
F
N q
F x d
F
G

M g
Which force did the work?
How much work did it do?
W
Force
 
F
 
 d

Fd cos q
Unit?
N

 m
J (for Joule)
What does this mean?
Physical work is done only by the component of of the force along the movement of the object.
Monday, Oct. 6, 2003 PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
Work is energy transfer!!
9

A man cleaning a floor pulls a vacuum cleaner with a force of magnitude F=50.0N at an angle of 30.0
o 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
0 o
W

  d cos q
M M
 
 d


50 .
0

3 .
00
 cos 30
 
130 J d
W
Does work depend on mass of the object being worked on?
Yes
Why don’t I see the mass term in the work at all then?
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?
Monday, Oct. 6, 2003 PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
10

• Product of magnitude of the two vectors and the cosine of the angle between them
A

B
 A B cos q
• Operation is commutative
A

B

A B cos q 
B A cos q 
B

A
• Operation follows distribution law of multiplication
• Scalar products of Unit Vectors
A

 

A

B

A

C i

 i


 j

 j

 k
 k

 1 i


 j

 j

 k

 k
 i

 0
• How does scalar product look in terms of components?
A

A x i


A y
 j

A z
 k B

B x i


B y
 j

B z
 k C

C x i


C y
 j

C z
 k
A

B




A x i


A y j


A z k








B x i


B y
 j

B z
 k




A x
B x i

 i


A y
B y
 j

 j

A z
B z
 k

 k
 cross terms
Monday, Oct. 6, 2003
A

B

A x
B x

A y
B y

A z
B z
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
11

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.
Y d
F a) Calculate the magnitude of the displacement and that of the force.
d
 d x
2  d y
2 
   
2 
3 .
6 m
X
F

F x
2 
F y
2 
   
2 
5 .
4 N b) Calculate the work done by the force F.
W

F
 d




2 .
0 i


3 .
0
 j







5 .
0 i


2 .
0 j





2 .
0

5 .
0 i


 i

3 .
0

2 .
0
 j

 j

10

6

16 ( J )
Can you do this using the magnitudes and the angle between d and F?
Monday, Oct. 6, 2003 PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
W

F
 d

F d cos q
12