Chapter 12 Gravitation

C .

apple

mg

earth

m1 m2

F12 F21

r

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

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)

Polaris

Celestial sphere

Rotation Axis of the Celestial sphere

earth

N

S

Star

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 SystemIn 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)

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

t

t

A

A

constantA

t

Sun

P1

P2

R1

R2

Kepler’s third law

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:

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:

(12-8)

2

3

TC

R

2 21 2

3 31 2

T T

R R

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 is a constant) and applied Newton’s second law and calculus (which he had also discovered) he got orbits that are conic sections

3. A force gave Kepler’s third law:

(12-9)

2

3

TC

R

2

kF

r

2

kF

r

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)

Kepler’s third law for circular orbits

Sun

P

R

m

v

FM

(12-11)

2 2 22 2

2 2

2 32

2 2

3

2 4 Period

4Substitute v from eqs.3 into the last equation:

4 constant C Note: We know that C

does not depend on m.

(eqs.3)

The answe

mv k k R Rv T T

R mR vR v

R mT

k

T m

kR

r lies in the constant k

2

2 ,

Elimina

(eqs.1)

te F fro

(eqs.2

m eqs.1 and eqs

)

.2

k mvF F

RR

m1 m2

F12 F21

r

Newton’s Law of Gravitational Attraction

(12-12)

1 2

12 21

Two masses m and m placed at a distance r exert an

attractive force on each other. This force is known as the

" " and has the following characteristics:

1. F

gravitational for

and F

e

ac

c

1 2

1 212 21 2

1 22

t along the line that connects m and m

2. G is the gravitational constant

Previously we wrote Thus

Gm mF F

rk

F k Gm mr

1 212 21 2

Gm mF F

r

Revisit Kepler’s third law

Sun

P

R

m

v

FM

(12-13)

2 2

2 2

3

(eqs.1) (eqs.2)

(eqs.3)

4 (eqs.4)

k GmMF F

R R

T mk GmM C

kR

2 2 2

3

2

3

We substitute k from eqs.3 into eqs.4

4 4

Thus C does not contain m. The ratio does not depend

on the planet mass m. Instead it depends on the sun mass M and is

the

T mC

GmM GMR

T

R

same for all the planets of the solar system.

m1 m2

F12 F21

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

1 212 21 2

Gm mF F

r

(12-14)

(12-15)

F

r

F

/2 /2

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

betw2

2

een M and m is:

The gravitational force on the two

smaller masses create a torque

. This torque

twists the fiber by an angle

given by: = . c

GmMF

r

GmMF

r

2c rG

mM

r

C.

apple

mg

earth

RM

(12-16)

2

2

224

In his experiment Cavendish measured the mass

of the earth! Gravitational force on apple = mg

Gravitational force on apple =

Solve for M

= 5.96 10 kg

mMG

RmMG

mgR

gRM

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

m

C

RR

Sun’s rays

sAlexandria

Syene

well

h

Ground in Alexandria

Sun’s rays

Erathosthene’s stick

(12-17)

The distance s between Alexandria and Syene . 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

s R

angle was determined using Eratosthene's walking stick

and the length of the shadow it cast. tan . He got 7 h

Variation of g with height h

22

2

2

2

( ) ( )

(0) ( )

( )1

(0)

mMG

mg h RR hmMGmg R hR

g h h

g R

CR

A

B

m

mM

(12-18)

2

2

Gravitational force at point A:

( )( )

Gravitational force at point B:

(0)

Divide eqs.1 by eqs.2

(eqs.1)

(eqs.

2)

mMGmg h

R h

mMGmg

R

Note: As h increases, g(h) decreases

A

B

m

mR

C

(12-19)

2

2(1 ) 1

( )1

2 for

1(0)

2 ( ) (0) 1

1x x

g h h h

g R R

hg h g

R

x

Gravitational Potential close to the surface of the earth

In chapter 7 we saw that the potential U of the gravitational force close to the surface of the earth is:

U = mgh

mg

h

m

floor

U = 0

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)

rM m

. . mF

O x-axispath

M

xdx

Gravitational Potential U (12-21)

2

2 22

( ) ( ) ( ) ( ) Take ( ) 0

( )dx 1

1( )

x

r

r r

r

mMGU r U F x dx F x U

x

mMGdx dxU r mMG

x x

mMGU r mM

x

x

Gr

( )mMG

U rr

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

mM r=

v = 0

RM

m

ve

BeforeAfter

(12-22)

2

2

, 0 , 2

= 2 = 11 km/s The escape velocity2

does not depend on m! All objects large or small must be launched at

the same speed (1

ei f i f

ee

mv mMG mMGE E E E

R r

mv mMGv MGR

R

1 km/s) to escape from the gravitation of the earth

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

(12-23)

2

2

22

2

Eliminate F

between eqs.1 and eqs.2.

(eqs.1) (eqs.

2)

mMG mvF F

rr

mv mMG MGv

r rr

2

, 2 2

2

2

mMG mv m MGU K

r rmMG mMG

E U Kr

Er

rmMG

Gravitational force between extended spherical objects

.M mr

R

C

.M'

m

r

C

2

As long as the small sphere

is outside the larger sphere

the force between them is:

GmMF

r

2

If the small sphere is inside

the larger sphere the force

between them is given by:

M is the mass inside the

dotted line. The surrounding

shell sxerts zero force!

GmMF

r

(12-24)

Mm

Mm

r

M

m

r

2

GmMF

r

0F

0F

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)