note - chapter 13

8/11/2019 Note - Chapter 13

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Chapter 13

GRAVITATION

13.1 Newtons Law of Universal Gravitation

Everyparticle withmassm1in the universe

attractsevery other particle with massm2with a

force that is directly proportional to the product of

their masses and inversely proportional to the square

of the distance rbetween them.

2

21

r

mGmF =

G = 6.67 x 10 11! !2

2

kg

Nm Universal Gravitational

Constant

r= distance from the center of mass of one particle to

the center of mass of the other particle.


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13.2 Weight

The attractive force exerted by the Earth on an object

is called thegravitational force

r

Fg or

r

W . This forceis called the weightof the object, and its directionis

toward the center of the Earth. The weightof an

object of mass mon the Earth has a magnitudeequal

to:

r

Fg =Weight=GM

Em

RE2

The quantity2

E

E

R

MG appears so often that it is given

the name gr

or simply g. This quantity is also called

thegravitational fieldgenerated by the Earth at

locations near its surface. The gravitational field gr

is

a vectorquantity with a directiontoward the center of

the Earth, and with a magnitudedefined as

2

E

E

R

M

Ggg ==

r

using

G = 6.67x10-11

Nm2/kg

2


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ME= 5.98x1024

kg (mass of the Earth)

RE= 6.37x106m (radius of the Earth)

yields a value of g = 9.8 m/s2. One can thus say thatthe weightof an object of mass mon the surface of

the Earth is

r

W = mr

g

whereg

r

= the acceleration due to gravity (or the

gravitational field), always directed toward the center

of the Earth.

Mass and weight are thus related quantities. The

magnitudeof a bodys weight Wis directly

proportional to its mass m.


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13.3 Gravitational Potential Energy

1.Object of mass mat a heightynear the Earths

surface,y


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If y


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Escape Speed:

With what initial speedmust an object of mass mbe

thrown vertically upwardso it escapes from the

Earth?Neglect air friction effects.

Using conservation of mechanical energy yields

ffii UkEUkE +=+

( )!

"="mGM

mR

mGM

mv E

E

E 22

02121

E

E

escR

GM2=! =

sec2.11 km for the Earth.

13.4 The Motions of Satellites

Many artificial satellites have nearly circular orbits

around the Earth. The orbits of the planets around the

Sun are also nearly circular.

Consider the motion of a satellite of mass min acircularorbit of radius raround the Earth.

F"# =m$

2

r


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GMEm

r2

=

m"2

r

v =GM

E

r

(circular orbit)

use also

T

r!"

2=

where Tis the period of revolution of the satellite

around the Earth, then

GME

r=

2" r

T

#

$%

&

'(2

T2=

4"2

GME

#

$%

&

'(r3

(circular orbit)

These equations indicate that larger orbits

correspond to slower speeds and longer periods.


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13.5 Keplers (1571-1630) Laws and Planetary

Motion:

Geometry of an ellipse.

=a

R Aphelion distance

( )eaRa

+= 1

=pR Perihelion distance

eaRp != 1

Kepplers Laws of Planetary Motion:

1.All planets move in elliptical orbitshaving the Sun

at one focus.


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2.A line joining any planet to the Sun sweeps out

equal areas in equal time intervals. ! Results for

conservation of angular momentum of orbiting

planets

!"=L = constant

L =mr2"

dA =1

2r2d"

r2d"=2 dA

dt

dmrL !2

= so(dt

dA

m

L 2=

dA

dt

=

L

2m

Note that since the only force acting on the planet is

the gravitational force exerted by the sun, and the line

of action of this gravitational force passes through the


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center of the sun, then the torque of the gravitational

force acting on the planet about the sun is 0=! .

Hence

" =

dL

dt=0

ThusL = constant

==

m

L

dt

dA

2constant

!equal areas in equal time intervals.

3.The square of the period of revolution of any

planet about the Sun is proportional to the cube ofthe planets mean distance from the Sun. Newton

showed that for an elliptic orbit, the planets orbit

radius r should be replaced by the semi-major axis

a of the elliptical orbit.

T2=

4"2

GMS

#

$%

&

'(a

3

(elliptical orbit)


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Energy Considerations in Planetary and Satellite

Motion.

The mechanical energyEof a planet orbiting arounda star or a satellite orbiting around a planet is always

(a) Constant and

(b) Negative (bound).

In general, if the mechanical energy E is such that

E(r) < 0"elliptical orbit

E(r) = 0" parablic orbit

E(r) > 0"hyperbolic orbit

#

$%

&%

For an elliptical orbit with semi-major axis a , the

mechanical energyEbecomes

E="G M m

2a (elliptical orbit)

Lets use conservation of energy and show that the

speed of an object in an elliptical orbit satisfies

!"

#$%

&'=ar

GM 12

( (elliptical orbit)


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Proof: Use conservation of mechanical energy

gUkEE +=

r

GMmm

a

GMm!=

! 2

2

1

2"

a

GM

r

GM

22

2

!=

"

"(r) = GM 2

r#1a

$

%& '

()

Here r is the distance of the orbiting body from

the central body whose mass isM.

13.7 Apparent Weight and the Earths Rotation

Gravitation Near the Earths Surface

Assume Earth is spherical, homogeneous, and at

rest (not spinning) Then, the gravitational force (true

weight wo = mgo) acting on an object of mass mis

wo=

GMEm

RE

2


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2

E

Eo

R

mGMmg =

22 8.9

sm

R

GMg

E

Eo ==

But the values ofgovary with latitudebecause:

1.Earths crust is not uniform!density variations

provide information about oil prospects.

2.Earth is not a sphere. Earths equatorial radius

is 21 km longer than the radius at the poles. At

the north and south poles, you are closer to

Earths center sogoshould be higher.

3.

Earths rotation effectleads to a difference in g o of about 0.034

2s

m . Lets show this:

Consider someone at the equator holding a spring

scale with a body of mass mhanging from it. Each

scale applies a tension forcer

F to the body hanging

from it, and the reading on the scale is the magnitudeF of this force. If the observer is unaware of the

Earths rotation, then he thinks that the scale reading

equals the weight of the body because he thinks the

body is in equilibrium. But the body is rotating with


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the Earth and is notprecisely in equilibrium. Thus

the magnitude of the forcer

F is the objects apparent

weight W. Consider the Earth spinning with

"=2#T. Then

F"# =ma

mgo"F=m#

2R

E

r

F = apparent weight = mg

Eo Rmmgmg 2

!="


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Eo Rgg 2

!=" E

RT

2

2!"

#$%

&=

'

( )( )[ ]

( )62

2

10370.6606024

4!=

"

2034.0

s

mggo

=!

That is, the actual acceleration due to gravity at the

equator when the Earth is spinning is smaller by

0.034 2s

m than when the Earth is stationary. Spinning

Earth reduces g slightly.


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13.8 Black Holes

A black holeis an object that exerts a gravitational

force on other objects but does not allow any light ofits own to escape from it.

A body of massMwill act as a black hole if its radius

Ris less than or equal to a certain critical radius

called the Schwarzschild radiusRswhich equals

Rs=

2GM

c2

The surface of the sphere with radius Rssurrounding

a black hole is called the event horizon. Since light

cant escape from within that sphere, we cant see

events occurring inside. All that an observer outsidethe event horizon can know about a black hole are

its mass(from its gravitational effects on other

bodies

itselectric charge(from the electric forces it

exerts on other charged bodies)

its angular momentum(because a black hole

tends to drag space and everything in thatspace around with it).


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note - chapter 13

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