PHYSICS CHAPTER 7

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

PHYSICS CHAPTER 7

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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 Newton’s law of gravitation, the Newton’s law of gravitation,

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

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 athe change in agg with altitude with altitude and depth from the surface of the earth.and depth from the surface of the earth.

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

221

rmmGFg

gFam

2gMa Gr

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7.1 Newton’s law of gravitation7.1.1 Newton’s law of gravitation States that a magnitude of an …………………..between two point masses a magnitude of an …………………..between two point masses

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

2

1r

Fg 21mmFg and

221

rmmFg

221

rmmGFg

: 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

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

where

221

1221 rmmGFFF g

1m 2m

r12F

Figure 7.1Figure 7.1

21F

2 particleon 1 particleby force nalGravitatio:12F21 : Gravitational force by particle 2 on particle 1F

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

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.

gF

r0

gF

21r

0

21mGmgradient

Figure 7.2aFigure 7.2a Figure 7.2bFigure 7.2b

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A spaceship of mass 9000 kg travels from the Earth to the Moon along a line that passes through the Earth’s centre and the Moon’s 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=6.001024 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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Solution :Solution :

Given

kg; 107.35 kg; 106.00 22M

24E mm

m 103.84 kg; 0900 8EMs rm

Em Mmsm

x

EMrxr EM

EsF

MsF

MsEs F2F

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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 sphere’s acceleration.

(Given G = 6.671011 N m2 kg2)

Example 7.2 :

Figure 7.3Figure 7.3A B

C

cm 8kg 3.2 kg 2.5

g 50

cm 6

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Solution :Solution :

a.

The magnitude of the forces on mC,

kg 1050kg; .52 kg; 3.2 3CBA

mmmm 1010m; 106 2

AC2

BC rr

0.6sin θ0.8cos θ

A B

C

m 108 2-

m 10 6 2

θ

θ

m 10 10 2

AF

BF

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Solution :Solution :

2BC

CBB r

mGmF

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

AF θF cosA θF sinA

BF

BF0

kg 1050kg; .52 kg; 3.2 3CBA

mmmm 1010m; 106 2

AC2

BC rr

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Solution :Solution :

The magnitude of the nett force is

and its direction is

xF

22

yx FFF

yF

x

y

FF

θ 1tan

θ

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Solution :Solution :b. By using the Newton’s second law of motion, thus

and the direction of the acceleration in the …………………..of the nett force…………………..of the nett force on the mC i.e………………………………………

amF C

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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. :ANS. : 8.28.2101033 N; 45 N; 45

2. Three 5.0 kg spheres are located in the xy plane as shown in Figure 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 N N

Figure 7.4Figure 7.4

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Exercise 7.1 :3.

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 nett gravitational force on a central sphere with mass of m5 = 250 kg.

ANS. :ANS. : 1.681.68101022 N; 45 N; 45

Figure 7.5Figure 7.5

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is defined as a ……………………………surrounding a body that has the a ……………………………surrounding a body that has the property of massproperty of mass wherewhere the …………………….is experienced if a test the …………………….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

Figure 7.6.

7.1.2 Gravitational Field

M

Figure 7.6Figure 7.6

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The gravitational field in small region near the Earth’s surfacesmall region near the Earth’s surface are …………….. ……………..and can be drawn ……………………..to each other……………………..to each other as shown in Figure 7.7.

The field lines indicate two things: The arrowsarrows – the ……………….………………. of the field The spacingspacing – the ……………….………………. of the field

Figure 7.7Figure 7.7

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 path taken.

Note:Note:

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Gravitational field strength, is defined as the ……………………per ………………of a body (test the ……………………per ………………of a body (test

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

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

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

ga

mF

a gg

where strength field nalGravitatio:gaforce nalGravitatio:gF

mass)(test body a of mass:m

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

mF

a gg and

2g rGMmF

2g rGMa

masspoint and massst between te distance : r

2g

1r

GMmm

a

wheremasspoint theof mass :M

7.1.3 Gravitational Acceleration

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Figure 7.8 shows the direction of the gravitational field strength on a point S at distance r from the centre of the planet.

2rGMag

r

M

Figure 7.8Figure 7.8

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The gravitational field in the small region near the Earth’s 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 7.9Figure 7.9

2RGMgag

Earth theof radius :Rwhere2s m 9.81onaccelerati nalgravitatio : g

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Determine the Earth’s 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 :a.

Example 7.3 :

RM

gaRr g m; 1040.6 6

2RGMg

The gravitational field strength isrg

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Solution :Solution :b.

2g rGMa R M

hRr 36 103501040.6

m 1075.6 6r

ga

hr

The gravitational field strength is given by

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The gravitational field strength on the Earth’s surface is 9.81 N kg1.Calculate

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

the Earth’s 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 :a. The gravitational field strength on the Earth’s surface is

The distance of point C from the Earth’s centre is

Example 7.4 :

1kg N 81.9 g

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

b. Given The weight of the rock is

2Cr

GMag 25.2 R

GMag

gmaW kg 5.2m

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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.

Example 7.5 :

AB

km 5

Figure 7.10Figure 7.10

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Solution :Solution :

At point C,

BA3 4 m; 105 MMr

0nett

gar

ABC

xr x

2ga

1ga

21 gg aa

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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.

The gravitational field strength outside the Earthoutside the Earth is

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

R

rM

Figure 7.11Figure 7.11

2g rGMa 2g

1r

a

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On the Earth ( On the Earth ( r r = = RR)) Figure 7.12 shows a test mass on the Earth’s surface.

The gravitational field strength on the Earth’s surfaceon the Earth’s surface is

RrM

Figure 7.12Figure 7.12

22g s m 81.9 g

RGMa

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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.

The gravitational field strength inside the Earth is given by

Figure 7.13Figure 7.13

2g'

rGMa

where

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

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

Therefore the gravitational field strength inside the Earthinside the Earth is

VV

MM

''

3

3

334

334'

Rr

Rr

MM

MRrM 3

3

'

2

3

3

g r

MRrG

a

rR

GMa 3g ra g

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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.

Figure 7.14Figure 7.14

R

ga

r0 R

gR

GMa 2g

ra g

2g1r

a

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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,

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.

Learning Outcome:7.2 Gravitational potential (1 hour)

rGMV

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

OR mathematically, V is written as:

It is a ……………………..……………………...

7.2.1 Gravitational potential, V

mWV

wheremass test theof mass :m

point aat potential nalgravitatio :V

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

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The S.I unit for gravitational potential is …………………………or ……………….………………. Another formula for the gravitational potential at a point is given by

21

11rrm

GMmV

mWV and

21

11rr

GMmW

where 1r and rr 2

rm

GMmV 11

rGMV

where point ebetween th distance : rM mass,point theand

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The gravitational potential difference between point A and B (gravitational potential difference between point A and B (VVABAB)) in the Earth’s gravitational field is defined as the ………………..in bringing the ………………..in bringing a test mass from point B to point A per unit the test mass.a test mass from point B to point A per unit the test mass.

OR mathematically, VAB is written as:

BABA

AB -VVm

WV

where

A.point toBpoint frommass test thebringingin done work :BAW

Apoint at potential nalgravitatio : AV

Bpoint at potential nalgravitatio : BV

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Figure 7.15 shows two points A and B at a distance rA and rB from the centre of the Earth respectively in the Earth’s gravitational field.

M

A

BrA

rB

Figure 7.15Figure 7.15

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

BAAB VVV

BAAB r

GMr

GMV

ABAB

11rr

GMV

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The gravitational potential difference between point B and A in the Earth’s gravitational field is given by

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

mWVVV AB

ABBA

Note: The Gravitational potential at infinity

is zero. 0V

R

RGM

r0

V

rV 1

Figure 7.16Figure 7.16

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When in orbit, a satellite attracts the Earth with a force of 19 kN and the satellite’s gravitational potential due to the Earth is 5.45107 J kg1.a. Calculate the satellite’s distance from the Earth’s surface.b. Determine the satellite’s 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)Solution :Solution :

Example 7.6 :

R

gF

rh

173 kg J 10455 N; 1019 .VFg

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Solution :Solution :

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

Therefore the satellite’s distance from the Earth’s surface is:

rGMV

Rhr

173 kg J 10455 N; 1019 .VFg

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Solution :Solution :

b. From the Newton’s law of gravitation, hence:

2rGMmFg

173 kg J 10455 N; 1019 .VFg

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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 :

On the surface of the planet, the gravitational potential is

Example 7.7 :

Rrm 1 ;kg 0010

RRh 2M

m

2r

1r

1rGMV

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Solution :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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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 Moon’s surface. Calculate its change in gravitational potential energy.c. If all the gravitational potential energy lost is converted to kinetic 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 moon’s surface is

Example 7.8 :

m1074.1 ;kg 1035.7 622 RM

R

GMV

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Solution :Solution :b. Given

Hence the change in the gravitational potential energy is

m 101.74 kg; 100 62 Rrm

m1074.1 ;kg 1035.7 622 RM

Rm1000.1 3hM

m2r

1r hRr1

2112

11rr

GMmr

GMmr

GMmΔU

if UUΔU

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Solution :Solution :c. Given Gravitational potential energy lost = kinetic energy The speed of the probe when hit the moon’s surface is given by

KΔU 2

21 mvΔU

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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,

period, period,

Learning Outcome:

7.3 Satellite motion in a circular orbit (½ hour)

rGMv

GMrT

3

2

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7.3 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 …………………………….speed, v as shown in Figure 8.22.

Figure 8.22Figure 8.22

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The centripetal force, Fc is contributed by the gravitational force of

attraction, Fg exerted on the satellite by the Earth.

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

ccg maFF

rmv

rGMm 2

2

rGMv

where

Earth theof mass :M

from satellite theof distance :rEarth theof centre the

constant nalgravitatio universal : G

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For a satellite close to the Earth’s surfacesatellite close to the Earth’s surface,

Therefore

The relationship between tangential velocity and angular velocity is

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

Rr and 2gRGM

gRv

Trrv 2

rGM

Tr2

GMrT

3

2

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Figure 7.17 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 ……………………………as the Earth. It rotates with the …………………….of rotation as that of the Earth (24

hours). It moves directly above the ……………………... The centre of a synchronous satellite orbit is at the centre of the Earth.

It is used as a ……………………………...

7.3.2 Synchronous (Geostationary) Satellite

Figure 7.17Figure 7.17

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Exercise 7.2 :Given G = 6.671011 N m2 kg2

1. 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 s m s11

2. 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 rad s rad s11

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Exercise 7.2 :3. A geostationary satellite of mass 2400 kg is placed 35.92 Mm from

the Earth’s 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,e. the mass of the Earth.

(Given radius of the Earth , R = 6.38 106 m)ANS. :ANS. : 7.277.27101055 rad s rad s11; 3.08; 3.08101033 m s m s11; 0.224 m s; 0.224 m s22; ;

537 N ; 6.00537 N ; 6.0010102424 kg kg

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THE END…Next Chapter…

CHAPTER 8 :Rotational of A Rigid Body