CHAPTER-7

PHYSICS CHAPTER 7CHAPTER 7: CHAPTER 7: GravitationGravitation(2 Hours)(2 Hours)

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PHYSICS CHAPTER 7PHYSICS CHAPTER 8Overview:

GravitationGravitation

Newtons law Newtons law Gravitational Gravitational GravitationalGravitational Satellite Satellite

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Newtons law Newtons law of gravitationof gravitation

Gravitational Gravitational Field strengthField strength

GravitationalGravitationalpotentialpotential

Satellite Satellite motionmotion

PHYSICS CHAPTER 7

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: State and useState and use the Newtons law of gravitation,the Newtons law of gravitation,

Define gravitational field strength as Define gravitational field strength as gravitational force per unit massgravitational force per unit mass

Learning Outcome:7.1 Gravitational Force and Field Strength(1 hour)

221

r

mmGFg =

F

3

Derive and use gravitational field strength,Derive and use gravitational field strength,

Sketch a graph of Sketch a graph of aagg against r against r and explain and explain the change in the change in aagg with altitude with altitude and depth from the surface of the earth.and depth from the surface of the earth.

gF

am

=

2gM

a Gr

=

PHYSICS CHAPTER 7

7.1 Newtons law of gravitation7.1.1 Newtons law of gravitation States that a magnitude of an ..between two point a magnitude of an ..between two point

masses is directly proportional to the product of their ... and masses is directly proportional to the product of their ... and inversely proportional to the square of the .between theminversely proportional to the square of the .between them.OR mathematically,

21r

Fg 21mmFg and

4

2r

g21mmFg

221

r

mmFg 221

r

mmGFg =

: Gravitational forcegF2 and 1 particle of masses:, 21 mm

2 and 1 particlebetween distance: r2211 kg m N x106.67 Constant nalgravitatio Universal: =G

where

PHYSICS CHAPTER 7 The statement can also be shown by using the Figure 7.1.

21mmGFFF ===

1m 2m

r12F

Figure 7.1Figure 7.1

21F

5

where

221

1221r

mmGFFF g ===

2 particleon 1 particleby force nalGravitatio:12F

21 : Gravitational force by particle 2 on particle 1F

PHYSICS CHAPTER 7 Figures 7.2a and 7.2b show the gravitational force, Fg varies with the

distance, r.

gF gF

mGm=gradient

6

Notes: Every spherical objectspherical object with constant densityconstant density can be reducedreduced to a

point masspoint mass at the centre of the sphere. The gravitational forcesgravitational forces always attractiveattractive in nature and the forces

always act along the line joining the two point masses.act along the line joining the two point masses.

r0 21r

0

21mGm=gradient

Figure 7.2aFigure 7.2a Figure 7.2bFigure 7.2b

PHYSICS CHAPTER 7

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, m =6.001024 kg, the mass of the

Example 7.1 :

7

(Given the mass of the Earth, mE=6.001024 kg, the mass of the Moon, mM=7.351022 kg and the universal gravitational constant, G=6.671011 N m2 kg2)

PHYSICS CHAPTER 7Solution :Solution :

Given

kg; 107.35 kg; 106.00 22M24E == mmm 103.84 kg; 0900 8EMs == rm

Em Mmsm

x

EMr

xr EM

EsF

MsF

MsEs F2=F

8

Given MsEs F2=F

( )2EMsM

2sE 2

xr

mGmx

mGm

=

( ) ME

2EM

2

2mm

xr

x=

PHYSICS CHAPTER 7Example 7.2 :

A B

C

cm 8kg 3.2 kg 2.5

g 50

cm 6

9

Two spheres of masses 3.2 kg and 2.5 kg respectively are fixed at points A and B as shown in Figure 8.3. If a 50 g sphere is placed at point C,determinea. the resultant force acting on it.b. the magnitude of the spheres acceleration. (Given G = 6.671011 N m2 kg2)

Figure 7.3Figure 7.3A B


a.

kg 1050kg; .52 kg; 3.2 3CBA === mmmm 1010m; 106 2AC

2BC

== rr

0.6sin =0.8cos =

C

m 10 6 2

m 10 10 2AF

BF

10

The magnitude of the forces on mC,

A CA 2

AC

Gm mFr

= =

A Bm 1082-


B CB 2

BC

Gm mFr

= =

kg 1050kg; .52 kg; 3.2 3CBA === mmmm 1010m; 106 2AC

2BC

== rr

11

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

AF

BF


The magnitude of the nett force is

xF =

( ) ( ) += 22 yx FFF

yF =

12

and its direction is 1tan y

x

F

F

= =

PHYSICS CHAPTER 7Solution :Solution :b. By using the Newtons second law of motion, thus

= amF C

13

and the direction of the acceleration in the same direction of the same direction of the nettnettforceforce on the mC i.e. 254254 from +x axis anticlockwise.from +x axis anticlockwise.

PHYSICS CHAPTER 7Exercise 7.1 :

Given G = 6.671011 N m2 kg21. 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. :ANS. : 8.28.2101033 N; 45N; 452. Three 5.0 kg spheres are located in the xy plane as shown in Figure

7.4.Calculate the magnitude

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7.4.Calculate the magnitude of the nett gravitational force on the sphere at the origin due to the other two spheres.

ANS. :ANS. : 2.12.1101088 NN


PHYSICS CHAPTER 7Exercise 7.1 :3.

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In Figure 7.5, four spheres form the corners of a square whose side is 2.0 cm long. Calculate the magnitude and direction of the nettgravitational force on a central sphere with mass of m5 = 250 kg.ANS. :ANS. : 1.681.68101022 N; 45N; 45


PHYSICS CHAPTER 7

is defined as a region of space surrounding a body that has the a region of space surrounding a body that has the property of massproperty of mass wherewhere the attractive force is experienced if a test the attractive force is experienced if a test mass placed in the region.mass placed in the region.

Field lines are used to show gravitational field around an object with mass.

For sphericalspherical objects (such as the Earth) the field is radialradial as shown in

7.1.2 Gravitational Field

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For sphericalspherical objects (such as the Earth) the field is radialradial as shown in Figure 7.6.

M


PHYSICS CHAPTER 7 The gravitational field in small region near the Earths surfacesmall region near the Earths surface are uniformuniform and

can be drawn parallel to each otherparallel to each other as shown in Figure 7.7.

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The field lines indicate two things: The arrowsarrows the directiondirection of the field The spacingspacing the strengthstrength of the field


The gravitational field is a conservative fieldconservative field in which the work done in work done in moving a body from one point to another is independent of the pathmoving a body from one point to another is independent of the pathtaken.

Note:Note:

PHYSICS CHAPTER 7

Gravitational field strength, is defined as the gravitational force per unit mass of a body (test the gravitational force per unit mass of a body (test

mass) placed at a point.mass) placed at a point.OR

ga

m

Fa

gg

=

force nalGravitatio:gF

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

It is a The S.I. unit of the gravitational field strength is ..or ......

where strength field nalGravitatio:gaforce nalGravitatio:gF

mass)(test body a of mass:m

PHYSICS CHAPTER 7

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

by

m

Fa

gg =

and2g

r

GMmF =

1 GMm

7.1.3 Gravitational Acceleration

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2gr

GMa =

masspoint and massst between te distance : r

= 2g

1r

GMmm

a

wheremasspoint theof mass :M

PHYSICS CHAPTER 7 Figure 7.8 shows the direction of the gravitational field strength on a point

S at distance r from the centre of the planet.

2r

GMag =

M

20

rFigure 7.8Figure 7.8

PHYSICS CHAPTER 7 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.

2RGMgag ==

21

Figure 7.9Figure 7.9Earth theof radius :Rwhere

2s m 9.81onaccelerati nalgravitatio : =g

PHYSICS CHAPTER 7

Determine the Earths gravitational field strengtha. 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 :Solution :

Example 7.3 :

22

Solution :Solution :a.

RM

gaRr g === m; 1040.6 6

2GMgR

= =

The gravitational field strength isrg

PHYSICS CHAPTER 7Solution :Solution :b.

GMa =

R M

hRr +=ga

hr

The gravitational field strength is given by

23

2gr

GMa =

R M

PHYSICS CHAPTER 7

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 where R is the radius of the Earth.b. the weight of a rock of mass 2.5 kg at point C.Solution :Solution :

Example 7.4 :

1kg N 81.9 =g

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Solution :Solution :a. The gravitational field strength on the Earths surface is

The distance of point C from the Earths centre is

1kg N 81.9 =g

12 kg N 81.9

==

RGMg

PHYSICS CHAPTER 7Solution :Solution :a. Thus the gravitational field strength at point C is given by

2Cr

GMag = ( )25.2 R

GMag =

25

b. Given The weight of the rock is gmaW =

kg 5.2=m

PHYSICS CHAPTER 7Example 7.5 :

AB

km 5


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Figure 7.10 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.


At point C,

BA3 4 m; 105 MMr ==

( ) 0nett

=ga

r

ABC

xr x

2ga

1ga

27

At point C, ( ) 0nett

=ga

21 gg aa =

( ) 2B

2A

x

GMxr

GM=

PHYSICS CHAPTER 7

Outside the Earth ( Outside the Earth ( r r > > RR)) Figure 7.11 shows a test mass which is outside the Earth and at a

distance r from the centre.

7.1.4 Variation of gravitational field strength on the distance from the centre of the Earth

rM

28

The gravitational field strength outside the Earthoutside the Earth is

R


2gr

GMa = 2g

1r

a

PHYSICS CHAPTER 7On the Earth ( On the Earth ( r r = = RR)) Figure 7.12 shows a test mass on the Earths surface.

Rr

M

29

The gravitational field strength on the Earths surfaceon the Earths surface is


22g s m 81.9

=== gR

GMa

PHYSICS CHAPTER 7

Rr

M'M

Inside the Earth ( Inside the Earth ( r r < < RR)) Figure 7.13 shows a test mass which is inside the Earth and at distance r

from the centre.where

portion spherical of mass the: 'M radius, ofEarth theof r

30

R

The gravitational field strength inside the Earth is given by


2g'

r

GMa =

PHYSICS CHAPTER 7 By assuming the EarthEarth is a solid spheresolid sphere and constant densityconstant density, hence

VV

MM

''

=

( )( ) 3

3

334

334

'

Rr

Rr

MM

==

pipi

MRrM 3

3

'=

31

Therefore the gravitational field strength inside the Earthinside the Earth is

2

3

3

gr

MRrG

a

=

rR

GMa 3g = ra g

PHYSICS CHAPTER 7 The variation of gravitational field strength, ag as a function of distance

from the centre of the Earth, r is shown in Figure 7.14.

R

a

32Figure 7.14Figure 7.14

ga

r0 R

gR

GMa == 2g

ra g

2g1r

a

PHYSICS CHAPTER 7

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DefineDefine gravitational potential in a gravitational field. gravitational potential in a gravitational field.

Derive and useDerive and use the formulae,the formulae,

Learning Outcome:7.2 Gravitational potential (1 hour)

GM

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SketchSketch the variation of gravitational potential, the variation of gravitational potential, VV with distance, with distance, rr from the centre of the earth.from the centre of the earth.

r

GMV =

PHYSICS CHAPTER 7

at a point is defined as the work done by an external force in the work done by an external force in bringing a test mass from infinity to a point per unit the test massbringing a test mass from infinity to a point per unit the test mass.

OR mathematically, V is written as:

7.2.1 Gravitational potential, V

where

34

It is a ..

m

WV =where

mass test theof mass :m point aat potential nalgravitatio :V

mass test a bringingin done work :W point a oinfinity t from

PHYSICS CHAPTER 7 The S.I unit for gravitational potential is mm22 ss22 or J kgJ kg11. Another formula for the gravitational potential at a point is given by

=

11GMmV

m

WV = and

=

21

11rr

GMmW

where =r and rr =

35

=

21 rrmV where =1r and rr =2

=

rm

GMmV 11

r

GMV =where point ebetween th distance : r

M mass,point theand

PHYSICS CHAPTER 7 The gravitational potential difference between point A and B (gravitational potential difference between point A and B (VVABAB)) in

the Earths gravitational field is defined as the work done in bringing a the work done in bringing a test mass from point B to point A per unit the test mass.test mass from point B to point A per unit the test mass.

OR mathematically, VAB is written as:

W

36

BABA

AB -VVm

WV ==

where

A.point toBpoint frommass test thebringingin done work :BAW

Apoint at potential nalgravitatio : AV

Bpoint at potential nalgravitatio : BV

PHYSICS CHAPTER 7 Figure 7.15 shows two points A and B at a distance rA and rB from the

centre of the Earth respectively in the Earths gravitational field.

A

BrA

r

The gravitational potential difference between the points A and B is given by

BAAB VVV =

37

MrB


BAAB VVV =

=

BAAB

r

GMr

GMV

=

ABAB

11rr

GMV

PHYSICS CHAPTER 7 The gravitational potential difference between point B and A in the

Earths gravitational field is given by

The variation of gravitational potential, V when the test mass, m move away from the Earths surface is illustrated by the graph in Figure 7.16.

m

WVVV ABABBA ==

38

Note: The Gravitational potential at infinity

is zero. 0=

V

R

RGM

r0

V

rV 1


PHYSICS CHAPTER 7

When in orbit, a satellite attracts the Earth with a force of 19 kN and the satellites gravitational potential due to the Earth is 5.45107 J kg1.a. Calculate the satellites distance from the Earths surface.b. Determine the satellites mass.(Given G = 6.671011 N m2 kg2, mass of the Earth, M = 5.981024 kg and radius of the Earth , R = 6.38106 m)

Example 7.6 :

39

M = 5.9810 kg and radius of the Earth , R = 6.3810 m)Solution :Solution :

R

gF

rh

173 kg J 10455 N; 1019 == .VFg


a. By using the formulae of gravitational potential, thus:

r

GMV =

173 kg J 10455 N; 1019 == .VFg

40

Therefore the satellites distance from the Earths surface is:

Rhr +=


b. From the Newtons law of gravitation, hence:

2r

GMmFg =

173 kg J 10455 N; 1019 == .VFg

41

PHYSICS CHAPTER 7

The gravitational potential at the surface of a planet of radius R is 12.8 MJ kg1. Determine the work done in overcoming the gravitational force when a space probe of mass 1000 kg is lifted to a height of 2R from the surface of the planet.Solution :Solution :

Example 7.7 :

Rrm == 1 ;kg 0010

Rmr

42

On the surface of the planet, the gravitational potential is

RRh 2=M

m

2r

1r

1r

GMV =

PHYSICS CHAPTER 7Solution :Solution :The final distance of the space probe from the centre of the Earth isThe work done required is given by

=

21

11rr

GMmW

RhRr 32 =+=Rrm == 1 ;kg 0010

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PHYSICS CHAPTER 7

The Moon has a mass of 7.351022 kg and a radius of 1740 km.a. Determine the gravitational potential at its surface.b. A probe of mass 100 kg is dropped from a height 1 km onto the

Moons surface. Calculate its change in gravitational potential energy.

c. If all the gravitational potential energy lost is converted to kinetic

Example 7.8 :

44

energy, calculate the speed at which the probe hits the surface.(Given G = 6.671011 N m2 kg2)Solution :Solution :a. The gravitational potential on the moons surface is

m1074.1 ;kg 1035.7 622 == RM

GMVR

= =

PHYSICS CHAPTER 7Solution :Solution :b. Given m 101.74 kg; 100 62 === Rrm

m1074.1 ;kg 1035.7 622 == RM

Rm1000.1 3=hM

m2r

1r36

1 1000.11074.1 +=+= hRrm10741.1 6=r

45

Hence the change in the gravitational potential energy is

1m10741.1 61 =r

=

=

2112

11rr

GMmr

GMmr

GMmU

if UUU =

PHYSICS CHAPTER 7Solution :Solution :c. Given

Gravitational potential energy lost = kinetic energyThe speed of the probe when hit the moons surface is given by

KU =2

21

mvU =

46

2

PHYSICS CHAPTER 7

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: ExplainExplain satellite motion with:satellite motion with:

velocity, velocity,

Learning Outcome:

7.3 Satellite motion in a circular orbit ( hour)

r

GMv =

47

period, period,

rv =

GMrT

3

2pi=

PHYSICS CHAPTER 77.3 Satellite motion in a circular orbit7.3.1 Tangential (linear/orbital) velocity, v Consider a satellite of mass, m travelling around the Earth of mass,

M, radius, R, in a circular orbit of radius, r with constant tangential (orbital) speed, v as shown in Figure 8.22.

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PHYSICS CHAPTER 7 The centripetal force, Fc is contributed by the gravitational force of

attraction, Fg exerted on the satellite by the Earth.

ccg maFF ==

r

mv

r

GMm 22 =

49

Hence the tangential velocity, tangential velocity, vv is given by

r

GMv =

where

Earth theof mass :M

from satellite theof distance :rEarth theof centre the

constant nalgravitatio universal : G

PHYSICS CHAPTER 7 For a satellite close to the Earths surfacesatellite close to the Earths surface,

Therefore

The relationship between tangential velocity and angular velocity is

Rr and 2gRGM =

gRv =

50

Hence , the period, period, TT of the satellite orbits around the Earth is given byT

rrv

pi

2==

r

GMT

r2=

pi

GMrT

3

2pi=

PHYSICS CHAPTER 7

Figure 7.17 shows a synchronous (geostationary) satellite which stays above the same point on the equator of the Earth.

7.3.2 Synchronous (Geostationary) Satellite


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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 of the

Earth. It is used as a communication satellite.


Given G = 6.671011 N m2 kg21. A rocket is launched vertically from the surface of the Earth at

speed 25 km s-1. Determine its speed when it escapes from the gravitational field of the Earth. (Given g on the Earth = 9.81 m s2, radius of the Earth , R = 6.38 106 m)

ANS. :ANS. : 2.242.24101044 m sm s11

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ANS. :ANS. : 2.242.24101044 m sm s112. 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. :ANS. : 3538 m s3538 m s11 ; 1.11; 1.11101044 radrad ss11


3. A geostationary satellite of mass 2400 kg is placed 35.92 Mm from the Earths surface orbits the Earth along a circular path. Determinea. 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,

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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. :ANS. : 7.277.27101055 radrad ss11; 3.08; 3.08101033 m sm s11; 0.224 m s; 0.224 m s22; ; 537 N ; 6.00537 N ; 6.0010102424 kgkg

PHYSICS CHAPTER 7PHYSICS CHAPTER 8

THE END

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54

Next ChapterCHAPTER 8 :

Rotational of A Rigid Body

CHAPTER-7

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