1443-501 Spring 2002
Lecture #21
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
Kepler’s Laws
The Law of Gravity & The Motion of Planets
The Gravitational Field
Gravitational Potential Energy
Energy in Planetary and Satellite Motions
Today’s Homework Assignment would have been #10 but I will assign
next Monday.
Example 14.3
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
ME
11 M E
g  G 2  6.67 10
2
RE
RE
So the mass of the Earth is
Therefore the density of the
Earth is
Apr. 17, 2002
2
R g
ME  E
G
2
RE g
M
3g
 E  G 
4
3
VE
4GRE
RE

3  9.80
3
3


5
.
50

10
kg
/
m
4  6.67  10 11  6.37  10 6
1443-501 Spring 2002
Dr. J. Yu, Lecture #21
2
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.
•All planets move in elliptical orbits with the Sun at one focal point.
•The radius vector drawn from the Sun to a planet sweeps out equal
area in equal time intervals. (Angular momentum conservation)
•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.
Apr. 17, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #21
3
The Law of Gravity and the Motion 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
Moon
aM
Newton predicted that the ratio of the Moon’s
acceleration aM to the apple’s acceleration g would be
2
 RE 
 6.37 106 

aM
1 / rM 
  
  2.75 10  4

 
2
8 
g
1 / RE   rM   3.84 10 
2
v
RE
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
v 2 2rM / T 
4  rM
4   3.84 108
9.80
3
2





2
.
72

10
m
/
s

rM
rM
T
2.36 106
602
2
aM
This means that the Moon’s distance is about 60 times that of the Earth’s radius, its acceleration
is reduced
by the square of the ratio. This
proves
that
the inverse square law is valid.
1443-501
Spring
2002
Apr. 17, 2002
4
Dr. J. Yu, Lecture #21
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
2
GM s M P M p v

2
r
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
2
Solving for T one T   4 r 3  K r 3
and
s
 GM 
can obtain
s 

r
 4 2 
  2.97 1019 s 2 / m3
K s  
 GM s 
This is Keper’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.
Apr. 17, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #21
5
Example 14.4
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.
 4 2  3
3


T 
r  Ksr

 GM s 
The mass of the Sun, Ms, is
 4 2  3
r
M s  
 GT 


4 2
11




1
.
496

10
11
7 
6
.
67

10

3
.
16

10


 1.99 1030 kg

Apr. 17, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #21

3
6
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
dL


 0; L  const
momentum L, the angular momentum is constant.
dt
Because the gravitational force exerted on a
L  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
1
L
dA
L
dA  r  d r  r  vdt 
dt

 const
motion of the planet is
2
2M p
dt 2M p
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.
Apr. 17, 2002
1443-501 Spring 2002
Dr. J. Yu, Lecture #21
7
The Gravitational Field
The gravitational force is a field force.
The force exists every point in the space.
If one were to place a test object of mass m at a any point in the
space in the existence of another object of mass M, the test object
will fill the gravitational force, F g  m g , exerted by M
Therefore the gravitational field g is defined as
g
Fg
m
In other words, the gravitational field at a point in space is the gravitational force
experienced by a test particle placed at the point divided by the mass of the test particle.
So how does the Earth’s
gravitational field look like?
Far away from the
Earth’s surface
Apr. 17, 2002
g
E
Fg
GM E

rˆ
2
m
RE
Where r̂ is the unit vector pointing
outward from the center of the Earth
Close to the
Earth’s surface
1443-501 Spring 2002
Dr. J. Yu, Lecture #21
8
The Gravitational Potential Energy
What is the potential energy of an object at the
height y from the surface of the Earth?
Do you think this would work in general cases?
Why not?
U  mgy
No, it would not.
Because this formula is only valid for the case where the gravitational force
is constant, near the surface of the Earth and the generalized gravitational
force is inversely proportional to the square of the distance.
OK. Then how would we generalize the potential energy in the gravitational field?
Because gravitational force is a central force and a
central force is a conservative force, the work done by
the gravitational force is independent of the path.
m
Fg
rf
RE
Apr. 17, 2002
m
ri
Fg
The path can be looked at as consisting of
many tangential and radial motions.
1443-501 Spring 2002
Dr. J. Yu, Lecture #21
9
More on The Gravitational Potential Energy
Since the gravitational force is a radial force, it only performed work while the
path was radial direction only. Therefore, the work performed by the gravitational
force that depends on the position becomes
dW  F  d r  F r dr    
W   F r dr
For the whole path
rf
ri
Therefore the potential energy is the
negative change of work in the path
U  U f  U i    F r dr
Since the Earth’s gravitational force is
F r   
So the potential energy
function becomes
rf
ri
U f Ui  
rf
ri
GM E m
r2
 1 1
GM E m
dr  GM E m   
2
r
ri 
 r f
Since potential energy only matters for differences, by taking the
infinite distance as the initial point of the potential energy, we get
For any two
particles?
Apr. 17, 2002
Gm1m2
U 
r
The energy needed
to take the particles
infinitely apart.
1443-501 Spring 2002
Dr. J. Yu, Lecture #21
For many
particles?
U 
GM E m
r
U  U i, j
i, j
10
Example 14.6
A particle of mass m is displaced through a small vertical distance y near the Earth’s
surface. Show that in this situation the general expression for the change in gravitational
potential energy is reduced to the U=mgy.
Taking the general expression of
gravitational potential energy
The above
formula becomes
 1 1
U  GM E m  
r
ri 
 f

r f  ri 
y
U  GM E m
 GM E m
r f ri
r f ri
Since the situation is close to
the surface of the Earth
Therefore, U becomes
ri  RE and rf  RE
U  GM E m
GM E
Since on the surface of the
g
RE2
Earth the gravitational field is
Apr. 17, 2002
y
RE2
The potential
energy becomes
1443-501 Spring 2002
Dr. J. Yu, Lecture #21
U  mgy
11
Energy in Planetary and Satellite Motions
v
r
M
Consider an object of mass m moving at a speed
v near a massive object of mass M (M>>m).
What’s the
total energy?
1
GMm
2
E  K  U  mv 
2
r
Systems like Sun and Earth or Earth and Moon whose motions
are contained within a closed orbit is called Bound Systems.
For a system to be bound, the total energy must be negative.
Assuming a circular orbit, in order for the object to be kept in
GM E m
v2
the orbit the gravitational force must provide the radial
 ma  m
2
r
r
acceleration. Therefore from Newton’s second law of motion
The kinetic energy for this system is
Therefore the total
mechanical energy
of the system is
Apr. 17, 2002
1
GM E m
mv 2 
2
2r
GMm
E  K U  
2r
1443-501 Spring 2002
Dr. J. Yu, Lecture #21
Since the gravitational
force is conservative, the
total mechanical energy of
the system is conserved.
12
Example 14.7
The space shuttle releases a 470kg communication satellite while in an orbit that is
280km above the surface of the Earth. A rocket engine on the satellite boosts it into a
geosynchronous orbit, which is an orbit in which the satellite stays directly over a single
location on the Earth, How much energy did the engine have to provide?
What is the radius of the geosynchronous orbit?
From Kepler’s
3rd
law
T K r
2
3
E GS
T  1day  8.64 104 s
Where KE is
4 
KE 
 9.89 10 14 s 2 / m3
GM E
4 2
4 2
Therefore the



  4.23 107 m
T2
8
.
64

10
8
.
64

10
3
rGS  3
3

geosynchronous radius is
KE
9.89 10 14
9.89 10 14
Because the initial position
before the boost is 280km
The total energy needed to
boost the satellite at the
geosynchronous radius is the
difference of the total energy
before and after the boost
Apr. 17, 2002
ri  RE  2.80 105 m  6.65 106 m
 1 1

 
r
 GS ri 
6.67 1011  5.98 10 24  470 
1
1



 1.19 1010 J

7
6 
2
 4.23 10 6.65 10 
E  
GM E ms
2
1443-501 Spring 2002
Dr. J. Yu, Lecture #21
13
Escape Speed
vf=0 at h=rmax
m
h
vi
RE
Consider an object of mass m is projected vertically from the surface of
the Earth with an initial speed vi and eventually comes to stop vf=0 at
the distance rmax.
Because the total
energy is conserved
ME
E  K U 
Solving the above equation
for vi, one obtains
vi 
1
GM E m
GM E m
mvi2 

2
RE
rmax
 1
1

2GM E 

R
 E rmax
Therefore if the initial speed vi is known one can use
this formula to compute the final height h of the object. h  rmax  RE 
In order for the object to escape
Earth’s gravitational field completely, vesc 
the initial speed needs to be
2GM E

RE



vi2 RE2
2GM E  vi2 RE
2  6.67 10 11  5.98 10 24
6.37 106
 1.12 10 4 m / s  11.2km / s
This is called the escape speed. This formula is
valid for any planet or large mass objects.
Apr. 17, 2002
How does this depend
on the mass of the
1443-501 Spring 2002escaping object?
Dr. J. Yu, Lecture #21
Independent of
the mass of the
escaping14object