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Chapter 13 Gravitation

In this chapter we will explore the following topics:

-Newtons law of gravitation that describes the attractive force between

two point masses and its application to extended objects

-The acceleration of gravity on the surface of the earth, above it,

as well as below it.

-Gravitational potential energy

-Keplers three laws on planetary motion

-Satellites (orbits, energy , escape velocity)

(13-1)

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Claudius Ptolomeu

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In contrast, planets follow complicated paths

in the sky. An example of such motion is

given in the figure. Tycho Brahe made

very careful measurementsof planetary

motions but he died before he had

the chance to analyze his data.

(13-12)

This task was carried out by his assistant Johannes Kepler who summarized

the results into three empirical laws known by his name.

Later, Newton used his second law of motion with his gravitational law

and the newly developed methods of calculus and derived Kepler's laws

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Nicolau Coprnico

Giordano Bruno

Galileu Galilei

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The square of the period of any planet is proportional to

the cube of the semimajor For the sake

of simplicity we will consider

Keple

the c

r's Third

irculr orb

axis

it sh

la

ow

of

n i

its

n

t

or

he

w

bit.

figure.

( ) ( )2 2 22 3

A planet of mass moves on a circular orbit of radius around a star of mass .

We apply Newton's second law to the motion:

(eqs.1) The period can be expressed

in

g

m r M

GMm GM F ma m r m r T

r r

= = = = =

22

2

2

2

3

2 2

3

2 4terms of the angular speed . = (eqs.2)

If we substitute form eqs.1 into eqs.2 we get:

: The ratio does

4

not deNote pend on the mass of the1

p

T

r M

T

m

G

T

T

r

=

=

2

3Note 2:

lanet but only on the mass of the central star.

For elliptical orbits the ratio remains cons at nt

M

T

a (13-14)

2 2

3

4

T

r MG

=

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m1

m2

1 2

2m mF G

r=

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m1

m2

F12

F21

r

m1

m2r

F1

m2

m1

1 2

2

m mF G

r=

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(13-4)

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1. A Terra no uniformemente distribuda2. A Terra no esfrica

3. A Terra est girando

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y-axis

3. Terra est girando

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m1m2

r

F1

m2

m1

Newton proved that the net gravitational force on a particle by a shell depends

on the postion of the particle with respect to the shell

If the pa

Gravitation

rticle is i

insid

nside

e the ear

the s

th

hell,

1 21 2

the net force is zero

If the particle is outside the shell the force is given by:

Consider a mass inside the earth at a distance from the center of the earth

If we divide the earth in a

m r

m mF G

r=

2

series of concentric shells, only the shells with

radius less than exert a force on . The net force on is:

Here is the mass of the part of the earth inside a sphere os radius

ins

ins

i

GmMr m m F r

M r

M

=

34 4is linear with r

3 3ns ins

r GmV F r F

= = =

(13-7)

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(13-9)

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r

m

In chapter 8 we derived the potential energy of a mass

near the surface of the earth. We will remove this restriction

and assume that t

Gravitational P

he mass can

otential E

move away

ne

fr

rgy

om th

U m

m e surfaceof the earth, at a distance from the center of the earth as shown

in the figure. In this case the gravitational potential energy is:

The negative sign of expresses the fact thGmM

Ur

r

U= at

the corresponding gravitational force is

The gravitational potential energy is not only associated

with the mass but with as well i.e. w

att

ith

N

ot

bo

rac

th

tiv

o

e

b

:

s

e

jectm M

1 2 3

1 3 2 31 2

12 13 23

If we have three masses , , and positi oned as shown in the figure

the potential energy due to the gravitational forces among the objects is:

We take into

m m m

U

Gm m Gm mGm mU

r r r

= + +

account each paironce

(13-8)

GmM

Ur

=

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m

r =

B

v = 0

m

v

A

If a projectile of mass is fired upward

at point A as shown in the figure, the projectile

will stop momentarily and return t

Escape

o the e

Spe

.

ed

arth

m

There is however a minimum initial speed for which the projectile

will escape from the gravitational pull of the earth and will stop

at infinity (point B in the figure). This minimum speed is known as

es

2

2

We can determine the escape velocity using

energy conservation between point A and point B.

cape velo

02

02

The escape speed form the earth

2

city.

is

A B

A B

mv GMmE K U E K U

R

mv GMmE ER

GMvR

= + = = + =

= = =

11.2 km/s

The escape speed does depeNote: ndnot on m

(13-10)

2

GMv

R=

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Stars follow regular paths in the evening sky.

They rotate once every 24 hours about

an axis that passes through the

star polari

Planets and Satellites: Kepler'

s. Polaris is the only star

th

s Laws

at does not move in the sky.

The stars have fixed spatial relationships

among them. Humans have

classified them in groups

known as "constellations"

Polaris

Celestial

sphere

Rotation Axis

of the

Celestial

sphere

earth

N

S

Star

(13-11)

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22

2

Consider a satellite that follows a circular orbit of radius

around a planet of mass . We apply Newton's second

law and hav

Sate

e:

llites: Orbits and Ener

The kinetic ene

gy

rgy

r

M

GMm v GM

ma m vr r r= = =2

(eqs.1)2 2

The potential energy (eqs.2)

If we compare eqs.1 with eqs.2 we have: (eqs.3)2

The total energy2 2

The energies ,

mv GMmK

r

GMmU

r

UK

GMm GMm GMmE K U K

r r r

E K

= =

=

=

= + = = =

, and are plotted as function of rin the figure to the left.

For elliptical orbits2

Note:

U

GMmE

a=

(13-15)

2

GMmE

r=