chapter 7 matriculation stpm

7/28/2019 Chapter 7 Matriculation STPM

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PHYSICS CHAPTER 7

1

CHAPTER 7:Gravitation

(2 Hours)

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PHYSICS CHAPTER 7

In this chapter, we learns about

7.1 Gravitational force and field strength

7.2 Gravitational potential

7.3 Satellite motion in a circular orbit

2


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PHYSICS CHAPTER 7

7.1 Gravitational Force and Field

Strength7.1.1 Newtons law of gravitation

7.1.2 Gravitational Field

7.1.3 Gravitational force and field strength

3


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PHYSICS CHAPTER 7

4

At the end of this chapter, students should be able to:

State and use the Newtons law of gravitation,

Learning Outcome:

7.1 Newtons law of gravitation (1 hour)

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PHYSICS CHAPTER 7

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7.1.1 Newtons law of gravitation

States that a magnitude of an attractive force between two

point masses is directly proportional to the product of their

masses and inversely proportional to the square of the

distance between them.

OR mathematically,

2

1

rFg 21mmFg and

2

21

r

mmFg

2and1particleofmasses:, 21 mm

2and1particlebetweendistance:r2211

kgmNx106.67ConstantnalgravitatioUniversal:

G

where


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PHYSICS CHAPTER 7

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The statement can also be shown by using the Figure 7.1.

where

2

211221

r

mmGFFF g

1m 2m

r12F

Figure 7.1

21F

2particleon1particlebyforcenalGravitatio:12F

1particleon2particlebyforcenalGravitatio:21F

Simulation 7.1
http://localhost/var/www/apps/conversion/tmp/scratch_8/AF_1301.htmlhttp://localhost/var/www/apps/conversion/tmp/scratch_8/AF_1301.html


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PHYSICS CHAPTER 7

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Figures 7.2a and 7.2b show the gravitational force,Fg varieswith the distance, r.

Notes: Every spherical object with constant density can be

reduced to a point mass at the centre of the sphere.

The gravitational forces always attractive in nature andthe forces always act along the line joining the two pointmasses.

gF

r0

gF

2

1

r0

21mGmgradient

Figure 7.2a Figure 7.2b


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PHYSICS CHAPTER 7

8

A spaceship of mass 9000 kg travels from the Earth to the Moon

along a line that passes through the Earths centre and the Moons

centre. The average distance separating Earth and the Moon is

384,000 km. Determine the distance of the spaceship from the

Earth at which the gravitational force due to the Earth twice the

magnitude of the gravitational force due to the Moon.(Given the mass of the Earth, mE=6.0010

24 kg, the mass of the

Moon, mM=7.351022 kg and the universal gravitational constant,

G=6.671011 N m2 kg2)

Example 7.1 :


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PHYSICS CHAPTER 7

9

Solution :

Given

kg;107.35kg;106.00 22M24

E mmm103.84kg;0900 8EMs rm

Em Mmsm

x

EMr

xr EM

EsF

MsF

MsEs F2F

2EM

sM

2

sE 2xr

mGm

x

mGm

ME

2

EM

2

2m

m

xr

x

2224

28

2

107.352

106.00

103.84

x

x

m103.32 8x


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PHYSICS CHAPTER 7

10

Two spheres of masses 3.2 kg and 2.5 kg respectively are fixed at

points A and B as shown in Figure 7.3. If a 50 g sphere is placed

at point C, determine

a. the resultant force acting on it.

b. the magnitude of the spheres acceleration.

(Given G = 6.671011 N m2 kg2)

Example 7.2 :

Figure 7.3

A B

C

cm8kg3.2 kg2.5

g50

cm6


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PHYSICS CHAPTER 7

11

Solution :

a.

The magnitude of the forces on mC,

22

311

2

AC

CAA

1010

10503.2106.67

r

mGmF

N101.07 9AF

kg1050kg;.52kg;3.2 3CBA mmm

m1010m;106 2AC2

BC rr

0.6sin 0.8cos

A B

C

m108 2-

m106 2

m1010 2

AF

BF


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PHYSICS CHAPTER 7

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Solution :

22

311

2

BC

CBB

106

10502.5106.67

r

mGmF

N10.322 9BF

Force x-component (N) y-component (N)

AF

F cosA F sinA

0.8101.07 910

108.56

0.6101.07 910

106.42

BF

BF09102.32

kg1050kg;.52kg;3.2 3CBA mmm

m1010m;106 2AC2

BC rr


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PHYSICS CHAPTER 7

13

Solution :

The magnitude of the nett force is

and its direction is

N108.56 10

x

F

22

yx FFF

N10.96210.322106.42 9910 yF

29210 10.962108.56

N10.083 9F

109

11

10.56810.962tantan

x

y

FF

.973 (254 from +x axis anticlockwise)


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PHYSICS CHAPTER 7

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Solution :

b. By using the Newtons second law of motion, thus

and the direction of the acceleration in the same direction of the

nett force on the mC i.e. 254 from +x axis anticlockwise.

amF C a39 1050103.08

28 sm10.166 a

3

9

1050

103.08

a


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PHYSICS CHAPTER 7

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is defined as a region of space surrounding a body that has

the property of mass where the attractive force isexperienced if a test mass placed in the region.

Field lines are used to show gravitational field around an object

with mass.

Forspherical objects (such as the Earth) the field is radial asshown in Figure 7.4.

7.1.2 Gravitational Field

M

Figure 7.4


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PHYSICS CHAPTER 7

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The gravitational field in small region near the Earths surfaceare uniform and can be drawn parallel to each otheras shown

in Figure 7.5.

The field lines indicate two things: The arrows the direction of the field

The spacing the strength of the field

Figure 7.5

The gravitational field is a conservative field in which the work donein moving a body from one point to another is independent of

the path taken.

Note:

New


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PHYSICS CHAPTER 7

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Exercise 7.1 :

Given G = 6.671011 N m2 kg2

1. Four identical masses of 800 kg each are placed at the corners

of a square whose side length is 10.0 cm. Determine the nett

gravitational force on one of the masses, due to the other three.

ANS. : 8.2103 N; 45

2. Three 5.0 kg spheres are located in thexy plane as shown inFigure 7.6.Calculate the magnitude

of the nett gravitational force

on the sphere at the origin due to

the other two spheres.

ANS. : 2.1108 N

Figure 7.6


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PHYSICS CHAPTER 7

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Exercise 7.1 :

3.

In Figure 8.7, four spheres form the corners of a square

whose side is 2.0 cm long. Calculate the magnitude and

direction of the nett gravitational force on a central sphere with

mass ofm5 = 250 kg.

ANS. : 1.68102 N; 45

Figure 7.7


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PHYSICS CHAPTER 7

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Define gravitational field strength as gravitational force perunit mass,

Derive and use the equation for gravitational field strength.

Sketch a graph ofag against rand explain the change in agwith altitude and depth from the surface of the earth.

Learning Outcome:

7.1.3 Gravitational force and field strength

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PHYSICS CHAPTER 7

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7.1.3 Gravitational field strength,

is defined as the gravitational force per unit massof a body (test mass) placed at a point.

OR

It is a vector quantity.

The S.I. unit of the gravitational field strength is N

kg1

orm s2

.

ga

where

strengthfieldnalGravitatio:ga

forcenalGravitatio:gF

mass)(testbodyaofmass:m


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PHYSICS CHAPTER 7

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It is also known as gravitational acceleration (the free-fallacceleration).

Its direction is in the same direction of the gravitational force. Another formula for the gravitational field strength at a point is

given by

m

Fa

g

g and 2g r

GMmF

where

masspointtheofmass:M


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PHYSICS CHAPTER 7

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Figure 7.8 shows the direction of the gravitational field strength

on a point S at distance rfrom the centre of the planet.

2r

GMag

r

M

Figure 7.8


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PHYSICS CHAPTER 7

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The gravitational field in the small region near the Earths

surface( r R) are uniform where its strength is 9.81 m s2 and

its direction can be shown by using the Figure 7.9.

Figure 7.9

2R

GMgag

Earththeofradius:Rwhere2sm9.81onacceleratinalgravitatio: g


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PHYSICS CHAPTER 7

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Determine the Earths gravitational field strength

a. on the surface.

b. at an altitude of 350 km.

(Given G = 6.671011 N m2 kg2, mass of the Earth,

M= 6.00 1024 kg and radius of the Earth,R = 6.40 106 m)

Solution :

a.

Example 7.3 :

R M

gaRr g m;1040.66

262411

21040.6

1000.61067.6

RGMg

The gravitational field strength is

1kgN77.9 g OR 2sm77.9

rg

(Towards the centre of the Earth)


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PHYSICS CHAPTER 7

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Solution :

b.

2g rGMa

26

2411

1075.6

106.001067.6

2g sm78.8

a(Towards the centre of the Earth)

R M

hRr 36 103501040.6 m1075.6 6r

ga

hr

The gravitational field strength is given by


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PHYSICS CHAPTER 7

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The gravitational field strength on the Earths surface is 9.81 N kg1.

Calculate

a. the gravitational field strength at a point C at distance 1.5R from

the Earths surface whereR is the radius of the Earth.

b. the weight of a rock of mass 2.5 kg at point C.Solution :

a. The gravitational field strength on the Earths surface is

The distance of point C from the Earths centre is

Example 7.4 :

1kgN81.9 g

1

2kgN81.9

R

GMg

RRRr 5.25.1


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PHYSICS CHAPTER 7

27

Solution :

a. Thus the gravitational field strength at point C is given by

b. Given

The weight of the rock is

N93.3W

2

Cr

GMag 25.2 RGMag

225.6

1

R

GM

gmaW

kg5.2m

57.15.2

1kgN57.181.925.6

1 ga




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PHYSICS CHAPTER 7

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Figure 8.10 shows an object A at a distance of 5 km from the object

B. The mass A is four times of the mass B. Determine the location

of a point on the line joining both objects from B at which the nett

gravitational field strength is zero.

Example 7.5 :

A

B

km5

Figure 7.10


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PHYSICS CHAPTER 7

29

Solution :

At point C,

BA3 4m;105 MMr

0nett ga

2B

23

B

105

4

x

M

x

M

m10.6713

x

r

A

BC

xr x

2ga

1ga

21 gg aa

2

B

2

A

x

GM

xr

GM


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PHYSICS CHAPTER 7

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Outside the Earth ( r> R)

Figure 8.11 shows a test mass which is outside the Earth and at

a distance rfrom the centre.

The gravitational field strength outside the Earth is

7.1.4 Variation of gravitational field strength on the

distance from the centre of the Earth

R

rM

Figure 8.11

2gr

GMa


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PHYSICS CHAPTER 7

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On the Earth ( r= R)

Figure 7.12 shows a test mass on the Earths surface.

The gravitational field strength on the Earths surface is

R

rM

Figure 7.12

2

2gsm81.9 g

R

GMa


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PHYSICS CHAPTER 7

32

R

r

M

'M

Inside the Earth ( r< R)

Figure 7.13 shows a test mass which is inside the Earth and at

distance r from the centre.

The gravitational field strength inside the Earth is given by

Figure 7.13

2g

'

r

GMa

where

portionsphericalofmassthe:'Mradius,ofEarththeof r


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PHYSICS CHAPTER 7

33

By assuming the Earth is a solid sphere and constant

density, hence

Therefore the gravitational field strength inside the Earth is

V

V

M

M

'' 33

3

34

334'Rr

Rr

MM

M

R

rM

3

3

'

2

3

3

g r

MR

rG

a

rR

GMa

3g


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PHYSICS CHAPTER 7

34

The variation of gravitational field strength, ag as a function of

distance from the centre of the Earth, ris shown in Figure 7.14.

Figure 7.14

R

ga

r0 R

gR

GMa

2g

ra g2g

1r

a


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PHYSICS CHAPTER 7

35


Define gravitational potential in a gravitational field.

Derive and use the formulae,

Sketch the variation of gravitational potential, Vwithdistance, rfrom the centre of the earth.

Learning Outcome:

7.2 Gravitational potential ( hour)

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PHYSICS CHAPTER


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PHYSICS CHAPTER 7

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7.2 Gravitational potential

7.2.1 Work done by the external force

Consider an external force, F

is required to bring a test

mass, m from r1

to r2

,

as shown in Figure 7.18.

At the distance r2 from the

centre of the Earth,

The work done by theexternal force through

the small displacement

dris

m

M

1r2r

F

gF

dr

Figure 7.18

gFF

0cosFdrdW

drFdW g

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

37

Therefore the work done by the external force to bring test

mass, m from r1 to r2 is

2

1

r

rgdrFdW

2

12

r

rdr

r

GMmW

2rGMmFg and

2

1

1r

rr

GMmW

2

12

1rr

drr

GMmW

where

distancefinal:2rdistanceinitial:1r

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

38

at a point is defined as the work done by an external force in

bringing a test mass from infinity to a point per unit thetest mass.

OR mathematically, Vis written as:

It is a scalar quantity.

Its dimension is given by

7.2.2 Gravitational potential, V

where

masstesttheofmass:mpointaatpotentialnalgravitatio:V

masstestabringingindonework:Wpointaoinfinity tfrom

mW

V M

TML 22 V

22TL V

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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The S.I unit for gravitational potential is m2 s2 orJ kg1. Another formula for the gravitational potential at a point is given

by

21

11

rrm

GMmV

m

WV and

21

11

rrGMmW

where 1r

and rr 2

rm

GMmV

11

where

pointebetween thdistance:rMmass,pointtheand

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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The gravitational potential difference between point A and B

(VAB) in the Earths gravitational field is defined as the work

done in bringing a test mass from point B to point A perunit the test mass.

OR mathematically, VAB is written as:

where

A.pointtoBpointfrom

masstestthebringingindonework:BAW

Apointatpotentialnalgravitatio:AV

Bpointatpotentialnalgravitatio:BV

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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Figure 7.19 shows two points A and B at a distance rA and rBfrom the centre of the Earth respectively in the Earths

gravitational field.

M

A

BrA

rB

Figure 7.19

The gravitational potential

difference between the points A

and B is given by

BAAB VVV

BA

ABr

GM

r

GMV

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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The gravitational potential difference between point B and A in

the Earths gravitational field is given by

The variation of gravitational potential, Vwhen the test mass, mmove away from the Earths surface is illustrated by the graph

in Figure 7.20.

R

R

GM

r0

V

rV

1

Note:

The Gravitational potential at infinityis zero. 0V

Figure 7.20

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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When in orbit, a satellite attracts the Earth with a force of 19 kN

and the satellites gravitational potential due to the Earth is5.45107 J kg1.

a. Calculate the satellites distance from the Earths surface.

b. Determine the satellites mass.

(Given G = 6.67

1011

N m2

kg2

, mass of the Earth,M= 5.981024 kg and radius of the Earth , R = 6.38106 m)

Solution :

Example 7.7 :

R

gF

rh

173 kgJ10455N;1019 .VFg

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

44

Solution :

a. By using the formulae of gravitational potential, thus

Therefore the satellites distance from the Earths surface is

rGMV

m1032.7 6r

66

1038.61032.7

hm104.9 5h

r

.. 24117 10985106761045.5

Rhr

173 kgJ10455N;1019 .VFg

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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Solution :

b. From the Newtons law of gravitation, hence

2rGMmFg

kg2552m

262411

3

10327

1098510676

1019

.

m..

173 kgJ10455N;1019 .VFg

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

46


Explain satellite motion with:

velocity,

period,

Learning Outcome:

7.3 Satellite motion in a circular orbit ( hour)

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PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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7.3 Satellite motion in a circular orbit

7.3.1 Tangential (linear/orbital) velocity,v

Consider a satellite of mass, m travelling around the Earth ofmass,M, radius, R, in a circular orbit of radius, rwith constanttangential (orbital) speed, v as shown in Figure 7.22.

Figure 7.22

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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The centripetal force, Fc is contributed by the gravitational force

of attraction,Fgexerted on the satellite by the Earth.

Hence the tangential velocity, vis given by

ccg maFF

r

mv

r

GMm 2

2

where

Earththeofmass:M

fromsatellitetheofdistance:rEarththeofcentrethe

constantnalgravitatiouniversal:G

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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For a satellite close to the Earths surface,

Therefore

The relationship between tangential velocity and angular

velocity is

Hence , the period, Tof the satellite orbits around the Earth is

given by

Rr and 2gRGM

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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Figure 8.23 shows a synchronous (geostationary) satellite which

stays above the same point on the equator of the Earth.

The satellite have the following characteristics:

It revolves in the same direction as the Earth.

It rotates with the same period of rotation as that of the Earth

(24 hours). It moves directly above the equator.

The centre of a synchronous satellite orbit is at the centre ofthe Earth.

It is used as a communication satellite.

7.3.2 Synchronous (Geostationary) Satellite

Figure 8.23

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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The weight of a satellite in a circular orbit round the Earth is half of

its weight on the surface of the Earth. If the mass of the satellite is800 kg, determine

a. the altitude of the satellite,

b. the speed of the satellite in the orbit,

(GivenG

= 6.671011 N m2 kg2, mass of the Earth,

M= 6.001024 kg, and radius of the Earth , R = 6.40106 m)

Example 7.12 :

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

52

Solution :

a. The satellite orbits the Earth in the circular path, thus

b. The speed of the satellite is given by

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

53

The radius of the Moons orbit around the Earth is 3.8 108 m and

the period of the orbit is 27.3 days. The masses of the Earth andMoon are 6.0 1024 kg and 7.4 1022 kg respectively. Calculate

the total energy of the Moon in the orbit.

Solution :

The period of the satellite is

The tangential speed of the satellite is

Example 7.13 :

sm50.9kg;120m;1050.8 26 gmr

T

rv

2

13 sm1024.4 v

36005.3Ts12600T

12600

1050.82 6

v

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

54

Solution :

A satellite orbits the planet in the circular path, thus

cg FF

2

2 r

mv

r

GMm

sm50.9kg;120m;1050.8 26 gmr

r

GMv

2

and2

gRGM

r

gRv

22

62

23

1050.850.91024.4

R

m1001.4 6R

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

55

Exercise 7.2 :

Given G = 6.671011 N m2 kg2

1. A rocket is launched vertically from the surface of the Earthat speed 25 km s-1. Determine its speed when it escapes from

the gravitational field of the Earth.

(Givengon the Earth = 9.81 m s2, radius of the Earth ,

R = 6.38

10

6

m)ANS. : 2.24104 m s12. A satellite revolves round the Earth in a circular orbit whose

radius is five times that of the radius of the Earth. The

gravitational field strength at the surface of the Earth is

9.81 N kg1

. Determinea. the tangential speed of the satellite in the orbit,

b. the angular frequency of the satellite.

(Given radius of the Earth , R = 6.38 106 m)

ANS. : 3538 m s1 ; 1.11104 rad s1

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

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Exercise 7.2 :

3. A geostationary satellite of mass 2400 kg is placed

35.92 Mm from the Earths surface orbits the Earth along acircular path.

Determine

a. the angular velocity of the satellite,

b. the tangential speed of the satellite,

c. the acceleration of the satellite,

d. the force of attraction between the Earth and the satellite,

e. the mass of the Earth.

(Given radius of the Earth , R = 6.38 106 m)

ANS. : 7.27105 rad s1; 3.08103 m s1; 0.224 m s2;537 N ; 6.001024 kg

PHYSICS CHAPTER 7


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PHYSICS CHAPTER 7

THE END

Next ChapterCHAPTER 8 :

Simple Harmonic Motion

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chapter 7 matriculation stpm

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