Transcript Document

Chapter 12
apple
In chapter 2 we saw that close to the
surface of the earth the gravitational
force Fg is:
a. Constant in magnitude Fg = mg and
b. Is directed towards the center of the
earth
mg
C
.
earth
m1
m2
F12
F21
r
Gravitation
In this chapter we will give the general
form of the gravitational force between
two masses m1 and m2 . This force
explains in detail the motion of the
planets around the sun and of all celestial
bodies. This is a law truly at a cosmic
level
(12-1)
Final Exam:
Friday 5/2/03
8:00-11:00 am Rm. 215 NSC
3 long problems from chapters 9,10,11,12, and 13
6 mini problems from chapters 1-13
Bring with you:
2 pens
1 page with equations (both sides)
Your ID
Celestial objects are divided into two categories:
1. Stars They have fixed positions with respect to each other
That is the reason why we can group hem into
constellations that maintain the same shape
2. Planets They follow complicated paths among the stars
An example is given in the
picture.
Note: All stars rotate around
the star polaris every 24
hours
(12-2)
Rotation Axis
of the
Celestial
sphere
Polaris
Star
N
earth
S
Celestial
sphere
Geocentric System
1. The earth is at the center of the
universe
2. The stars are fixed on a sphere
(known as the “celestial sphere”)
which rotates about its axis every
24 hours. The axis connects the
center of the earth with the star
polaris. All stars move on
circular paths on the celestial
sphere and complete a rotation
every 24 hours
(12-3)
Ptolemaic System
Ptolemy in the 2nd century AD used
the following geocentric model to
describe the complicated motions of
the planets. The planets and the sun
move on small circular paths called
the epicycles. The centers of the
epicycles move around the earth on
larger circles called deferents
Ptolemy’s system gives a reasonable description of the
motion of the planets and it was accepted for 1400 years
(12-4)
The Heliocentric System
In 1543 Copernicus
introduced the heliocentric
system. According to this
scheme the sun is at the center
of the solar system. The
planets and the earth rotate
about the sun on circular
orbits. The immobility of the
stars was ascribed to their
great distance. The
heliocentric system was not
accepted for almost a century
(12-5)
Nicolaus Copernicus
(1473-1543)
He discussed his ideas
in the book:
De Revolutionibus
Orbitum Coelestium
published after his
death
Thycho Brache in Denmark constructed sophisticated
astronomical instruments and studied in detail the motion of
the planets with a accuracy of ½ minute (1 minute = 1/60
degree)
Brache died in 1601 before he had a chance to analyze his
data. This task was carried out by his assistant Johannes
Kepler for the next 20 years. His conclusions are
summarized in the form of three laws that bear his name
(Kepler’s laws)
(12-6)
t
A
A
t
Kepler’s first law
Planets move on elliptical
paths (orbits) with the sun at
one focus
(12-7)
Kepler’s second law
During equal time intervals
the vector r that points from
the sun to a planet sweeps
equal areas
A
 constant
t
Kepler’s third law
(12-8)
If T is the time that it takes
for a planet to complete one
revolution around the Sun
and R is half the major axis
of the ellipse then:
P1
R1
Sun
T2
C
3
R
R2
C is constant for all planets
of the solar system
If the planet moves on a circular path then R is simply the
orbit radius. For the two planets in the figure Kepler’s third
law can be written as:
2
2
P2
T1
T2
 3
3
R1
R2
Isaac Newton: Newton had formulated his three laws of
mechanics. It was natural to check and see if they apply beyond the
earth and be able to explain the motion of the planets around the sun.
Newton after studying Kepler’s laws came to the following
conclusions:
1. Kepler’s second law implies that the attractive force F exerted by
the sun on the planets is a central force
2. If he assumed that the magnitude of the attractive force F has the
form:
k
F
r2
(k is a constant) and applied Newton’s second law and calculus
(which he had also discovered) he got orbits that are conic sections
k
3. A force F  2
r
gave Kepler’s third law:
T2
C
3
R
(12-9)
Conic Section is one of the four possible curves (circle,
ellipse, parabola, hyperbola) we get when we cut the surface
of a cone with a plane, as shown in the figure below
(12-10)
v
Kepler’s third law for circular orbits
P
m
F
M
Sun
R
k
mv 2
F  2 (eqs.1) , F 
(eqs.2)
R
R
Eliminate F from eqs.1 and eqs.2
2 2
k
2

R
4

R
 v2 
(eqs.3) Period T 
T2 
mR
v
v2
2 3
4

Rm
2
Substitute v from eqs.3 into the last equation: T 

k
T 2 4 2 m

 constant C Note: We know that C
3
k
R
does not depend on m. The answer lies in the constant k
mv 2
k
 2
R
R
(12-11)
Newton’s Law of Gravitational Attraction
m1
m2
F12
F21
Gm1 m2
F12  F21 
r2
r
Two masses m1 and m 2 placed at a distance r exert an
attractive force on each other. This force is known as the
"gravitational force" and has the following characteristics:
1. F12 and F21 act along the line that connects m1 and m 2
Gm1 m2
2. F12  F21 
r2
k
Previously we wrote F  2
r
G is the gravitational constant
Thus k  Gm1 m2
(12-12)
v
Revisit Kepler’s third law
P
m
F
M
Sun
R
k
F  2 (eqs.1)
R
k  GmM (eqs.3)
We substitute k from eqs.3 into eqs.4
GmM
F
(eqs.2)
2
R
T 2 4 2 m

 C (eqs.4)
3
k
R

T 2 4 2 m 4 2
C 3 

GmM GM
R
T2
Thus C does not contain m. The ratio 3 does not depend
R
on the planet mass m. Instead it depends on the sun mass M and is
the same for all the planets of the solar system.
(12-13)
m1
m2
F12
F21
Gm1 m2
F12  F21 
r2
r
The gravitational constant G was
measured in 1798 by Henry Cavendish.
He used a balance with a quartz fiber. In
order to twist a quartz fiber by an angle 
one has to exert a torque  = c (this is
very similar to the spring force F = kx).
The constant c can be determined easily
(12-14)
Cavendish experiment: Two small
masses m are placed on either arm
of a quartz fiber balance. Two
larger masses M are placed at a
distance r from the smaller masses.
The gravitational force F
(12-15)
r
F
/2
GmM
between M and m is: F 
r2
The gravitational force on the two
smaller masses create a torque
GmM
 F 
. This torque
2
r
twists the fiber by an angle 
given by:  =

c
.
/2
Fr
c r 2
G
mM
apple
m
mg
M
C
.
earth
R
In his experiment Cavendish measured the mass
of the earth! Gravitational force on apple = mg
mMG
Gravitational force on apple =

2
R
mMG
mg 
Solve for M
2
R
gR 2
(12-16)
M 
= 5.96  1024 kg
G
How does one measure the radius of the earth? This was already
done by Cavendish's time by as librarian in Alexandria called
Eratosthenes (around 200 BC). Eratosthenes knew that ot a
particular day every year sunlight reached the bottom of a very
deep well in Syene (modern Aswan). He also knew the distance
between Alexandria and Syene. From this information he was
able to determine R
Sun’s rays
Sun’s rays


s
(12-17)
Alexandria
Syene
h 
Erathosthene’s stick
well
R
R


Ground in
Alexandria
C
The distance s between Alexandria and Syene s  R . Here 
is the angle between the sun's rays and the vertical in Alexandria
at the time when the sun's rays in Syene reach the bottom of the well.
The angle  was determined using Eratosthene's walking stick
and the length of the shadow it cast. tan  
h
. He got 7
A
m
M
m
R
C
B
Variation of g with height h
Gravitational force at point A:
mMG
mg (h) 
(eqs.1)
2
( R  h)
Gravitational force at point B:
mMG
(eqs.2)
2
R
Divide eqs.1 by eqs.2 
mg (0) 
mMG
mg (h) ( R  h) 2
R2


mMG
mg (0)
( R  h) 2
R2
g ( h)  h 
 1  
g (0)  R 
2
(12-18)
m
A
2
C
B
R
m
g ( h)  h 
h
 1  
 1
g (0)  R 
R
(1  x) 2  1  2 x for x  1
 2h 
 g (h)  g (0) 1  
R

Note: As h increases, g(h) decreases
(12-19)
m
Gravitational Potential close to the
surface of the earth
mg
h
U=0
floor
In chapter 7 we saw that the potential U of
the gravitational force close to the surface
of the earth is:
U = mgh
Note 1: Close to the surface of the earth the gravitational force
is constant and equal to mg
Note 2: The point at which U = 0 can be chosen arbitrarily
we will now remove the restriction that m is close to the
surface of the earth and determine U
(12-20)
M
M
m
r
F
.
.
O
Gravitational Potential U
(12-21)
m
x
x-axis
path
dx
r
mMG
F ( x)   2
x
U (r )  U ()    F ( x)dx

r
r
mMGdx
dx
U (r )  
 mMG  2
2
x
x


r
mMG
 1
U (r )  mMG     
r
 x 
dx
1
 x2   x
Take U ()  0

mMG
U (r )  
r
Escape velocity is the minimum speed with which we must
launch an object from the surface of the earth so that it leaves
the earth for ever
After
Before
m
m
r=

M
R
M
ve
v=0
mve2 mMG
mMG
Ei 

, Ef  
 0 , Ei  E f 
2
R
r
mve2
mMG
=
 ve  2 MGR = 11 km/s The escape velocity
2
R
does not depend on m! All objects large or small must be launched at
the same speed (11 km/s) to escape from the gravitation of the earth
(12-22)
Example (12-2) page 327 An object of mass m moves on a
circular orbit of radius r around a planet of mass M. Calculate
the energy E = K + U
mMG
F
(eqs.1)
2
r
mv 2
F
r
between eqs.1 and eqs.2. 
(eqs.2) Eliminate F
mv 2 mMG

r
r2
MG
 v 
r
2
mMG
mv 2 m MG
U 
, K

r
2
2 r
mMG mMG
E U  K  

r
2r
mMG
E
2r
(12-23)
Gravitational force between extended spherical objects
As long as the small sphere
is outside the larger sphere
M
r
.
C
m
the force between them is:
F
R
GmM
r2
If the small sphere is inside
the larger sphere the force
M'
.
C
(12-24)
r
m
between them is given by:
GmM 
F
r2
M is the mass inside the
dotted line. The surrounding
shell sxerts zero force!
M
m
r
GmM
F
r2
M
r
m
m
M
F 0
F 0
If m is outside the shell the
gravitational force F is as if all
the mass M of the shell is
concentrated at its center and all
the mass m of the sphere is
concentrated at it center
If m is anywhere inside the shell
then the gravitational force
between the shell and the sphere
is zero !
Special case: The sphere and the
shell are co-centric
(12-25)