Focus SPM Physics samplebook

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PHYSICS

KSSM

üInfographics üComprehensive Notes üConcept Maps üActivities & Experiments üSPM Tips

REINFORCEMENT & ASSESSMENT

üSPM Practices üCheckpoint

REVISION

üSPM Model Paper üComplete Answers

4∙5

KSSM

ASSESSMENT EXTRA

üQR Codes

• Matematik • Matematik Tambahan • Sains • Biologi • Fizik • Kimia • Prinsip Perakaunan • Perniagaan

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SPM FORM

REINFORCEMENT

EXTRA FEATURES

üSPM Highlights üExamples

PHYSICS

PHYSICS

FOCUS SPM KSSM Form 4 • 5 – a complete and precise series of reference books with special features to enhance students’ learning as a whole.

This series covers the latest Kurikulum Standard Sekolah Menengah (KSSM) and integrates Sijil Pelajaran Malaysia (SPM) requirements.

A great resource for every student indeed!

REVISION REVISI

SPM

4∙5

FOCUS

FORM

CC038542

SPM

PELANGI BESTSELLER

Yew Kok Leh Chang See Leong Abd Halim Bin Jama’in

NEW SPM ASSESSMENT FORMAT

2021

Format: 190mm X 260mm TP Focus SPM Phy BI 2022_pgi_2IMP

PHYSICS

SPM

Yew Kok Leh Chang See Leong Abd Halim Bin Jama’in

© Penerbitan Pelangi Sdn. Bhd. 2022

All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, photocopying, mechanical, recording or otherwise, without the prior permission of Penerbitan Pelangi Sdn. Bhd. ISBN: 978-967-2779-53-7 eISBN: 978-967-2779-54-4 (eBook) First Published 2022

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FORM

4∙5

KSSM

Exclusive Features of This Book Ch

ap

te r

Form 5

7

OCUS TER F CHA7.1P Quantum Theory of

Physics SPM

Chapter 1 Measu

rement

SPM Tips

Light Effect 7.2 Photoelectric ectric 7.3 Einstein’s Photoel Theory

Prefixes 10 9

Form 1. The dista nce of Pluto from the Earth 6 000 000 is about 000 hydrogen atom 000 m and the radius of a is about 0.000 These quan 000 000 05 tities are eithe m. r too large small and a or too simp is by using stand ler way of expressing them ard form of representa scientific notat tion or ion.

mega (M)

10 3

kilo (k)

10 –1 10 –2 10 –3

4

Standard Bigger

giga (G)

10 6

Form

Chapter Focus lists the learning objectives for an overview of the chapter.

Quantum Physics

deci (d) centi (c) mili (m)

10 –6

micro (μ)

10 –9

Electron

nano (n)

SPM Tips points out important tips for students to take note of.

Smaller

Proton

1.1

EXAMPLE

Hydrogen atom

Form

Convert (a) 0.0042 kg (b) 5 800 g to g to kg (c) 10 cm to m Solution

5

10 9

and quantum?

• What is light quantum matter? physics? associated with • How is light important in modern effects? m theory of light and photoelectric • Why is quantu by quantum energy • What is meant

giga (G)

10 6

mega (M)

10 3

(a) 10 3

EXAMPLE

kilo (k) (b)

SPM Highlights provides exposure to the frequentlytested questions that appear in the actual exam.

(c) 10 –2 10 –1 10 –2 10 –3

454

10 –6

micro (µ)

nics

lights

SPM High

in connected thermistor to control a shows an NTC Figure 5.54 circuit is used circuit. The a transistor ioner.

240 V air condit

S2

termin on Figure (a) Label the type of transistor shown the (b) Name .

5.55. transistor circuit shows a simple Figure 5.56

R Air condition

9V

Figure 5.54

Tips is high (hot Examiner’s thermistor es rature of the istor becom When the temperesistance of therm e base voltag the condition), r R. Then the the red to resisto current to flow through low compa tor current causing the becomes high tor to turn it on. The collec air condition. base of transis relay to turn on the h the flows throug Answer: C

!

Checkpoint Figure 5.55

S1

shows the

5 .2

transistor. symbol of a

__________

P : ______

Form

__________

R : ______

__________

Q : ______

5

Physics SPM

Chapter 4 Heat

transistor T. circuit and the type of (a) Name , label the base figure above (b) On the . that the emitter circuit V and V c so l dry cell at s (c) Draw symbolights up.

Objective Ques bulb M can tions resistor R? function of the is (d) What 1. A studen fan t to turn on a block fromtransfers a metal circuit tor a cold beake transis 5.57 shows r designed a P into beake atically. Figure S3 Azmin r Q at room . oom autom temperature control system in the classr . the electronic the circuit for Beaker

SPM Practic e

X

a.c / a.u Relay

P

Fan

P

Form

condition cause the air situations can Which of these e. base voltag istor and low high base to turn on? rature therm A Low tempe rature thermistor and B Low tempe high base istor and voltage. therm rature C High tempe base and low istor voltage. rature therm D High tempe voltage.

Checkpoint provides questions to test students’ understanding of the subtopic and reinforce learning.

VC

T

Figure 5.56

R

Example provides solution for example of questions in the subtopics.

VS

Resistor Resistor

4

M

240 V a.u ~

Relay Thermistor

nano (n)

(a) 0.0042 kg (b) 5 800 g = 0.0042 × 10 3 g = 4.2 g = 5 800 × 10 –3 (c) 10 cm kg = 5.8 kg = 10 × 10 –2 m = 0.1 m R. al P, Q and

Chapter 5 Electro

1.2

For each of the using scientific following, express the magnitude notation. (a) The lengt h (b) The mass of a virus = 0.000 000 08 m of a ship = 75 000 000 kg Solution (a) The lengt h of a virus = 0.000 000 08 m = 8 × 10 –8 m (b) The mass of a ship = 75 000 000 kg = 7.5 × 10 7 kg

(b) 10 –3

deci (d) centi (c) mili (m)

10 –9

Physics SPM

2. In a stand ard form or scientific notat numerical magn ion, a itude can be written as: A × 10 n, wher e 1 < A  10 and n is an integer Hence, the distance of Pluto from can be writt the Earth en as 6 × 10 12 m a hydrogen atom as 5 × –11and the radius of 10 m.

4

Metal block

Which of the following statements is true about metal block?

10 k

I

II

Figure 5.57

• • •

immersed in melting ice is 4.5 cm, put in boiling water is 22.0 cm, placed in warm water is 11.5 cm.

What is the tempe the warm water? rature of A 30 oC B 40 oC C 50 oC D 55 oC 3. Which of the following is the concept used in the temperature measuring of a human body using a thermometer?

422

Figure 2

Which of the following is the most suitab le characteristic the material used as base of the pan? of A Low meltin g B High densit point C Low specifi y D Low thermac heat capacity l conductivity 5. In an experi ment, Ramly found that 100 liquid require g of a s 12 thermal energy 600 J of to raise the temperature from 30°C to 60°C. The heat 100 g of liquid capacity of A 42 J °C –1 is .

B 420 J °C –1 .

C 4 200 J –1 °C .

D 42 000 –1 J °C .

6. Figure 3 shows a glass containing 200 ml of hot water at 80 o C. A studen t filled in 200 ml of cold water at 0 oC and stirred a spoon. Assum it with ing that there is no heat loss to the surrounding, what is the final temperature of the water in the glass?

SPM Practice provides ample qustions to test students’ mastery of the chapter.

7. A milk seller boils his milk in a pot. The temperature the milk is of raised from 10°C to 130° C with of heat energy 12 000 kJ . If the mass of milk is 25 kg, what is the specific heat capacity of the milk?

A 4 000 J –1 –1 kg B 8 000 J –1 °C kg °C –1 C 12 000 J kg –1 °C –1 D 16 000 J kg –1 °C –1 8. Figure 4 shows radiator. Water a car is used in the radiator as a cooling agent.

The tempe rature of metal block will rise to 28 oC. The tempe rature of metal block increa ses but is lower

meter in an unmarked glass. The length of the mercury colum n when

Figure 3

A 20 oC B 30 oC C 40 oC D 60 oC

Base of pan

a

than o P. in your 28 C. III The the component of P? Expla metal block (a) Name on releases is the functi heat in beake (b) What r Q. A warm I and, IIthe only answer. becomes B the classroom to 6 kΩ.II and III only (c) When C I, of P decreases II and XZ. across nce III resistance potential differe nce across 2.differe An experi (i) State the tial ment poten the is conducted to calibrate (ii) Calculate tor. a mercury thermo the themis

Figure 5.55

200 ml of hot water at 80°C

Water at 28°C

Figure 1

Transistor

Z

Beaker Q

Water at 0°C

9V

R1 Y

4

A Specific heat capac ity B Specific latent heat C Thermal equilibrium D Heat conve ction 4. Alisa’s mother cooking pan chose a as Figure 2, and shown in asked Alice about the charac teristics of the material used as the base of the pan.

Figure 4

Water is used as cooling agent becau se A water evapo rates easily.

B water is not C water has corrosive. a D water can high density. absorb a lot of heat.

134

Physics SPM

Chapter 1 Forces

and

Motion II

2.1 entn 1.1 ime erim Expper Eks

Form

Activity/ Experiment helps students to master handson scientific knowledge and skills.

of a spring the extension the force and spring. p between sion of the the relationshi affects the exten Aim: To find on a spring of the spring force applied the extension Inference: The force, the larger the applied : The larger Hypothesis ) weight added , F (slotted Force Variable: x le: , ed variab the spring (a) Manipulat variable: Extension of of spring h and type (b) Responding with clamp. variable: Lengt retort stand r, metre rule, (c) Constant d mass hange Materials: d mass, slotte Metre rule Apparatus and cm long), five 50 g slotte (20 Steel spring Figure as shown in Procedure: atus is set up (a) The appar of the at the end 1.42. , l o is hanger is fixed Spring (b) The mass initial length of the spring spring. The a metre ruler. hanger measured using = 50 g is slotted on the m red. mass, spring is measu (c) A slotted length, l of the is calculated as, lo and the new of the spring Extensionx The extension Original length of d mass of spring x = l – l o. by adding slotte Mass holder l is repeated 250 g. (d) Step 3 Slotted weight lated. g, 200 g and 100 g, 150 d mass is calcu the t of the slotte act on F (e) The weigh represents the force, F –1 ] Weight The weight kg = mg g g = 10 N . spring. [Usin ated as below tabul are s Figure 1.42 (f) The result

5

Result: = 25 cm g, of spring, l o Initial length Length of sprin d l / cm Weight of slotte added, mass, F / N Slotted mass m / kg 27.5 0.5 0.05 30.0 1.0 0.10 33.0 1.5 0.15 35.0 2.0 0.20 38.0 2.5 0.25 shown. as d plotte F is against force extension, x of graph A

SPM MODE L PAPER One hour 1. Which of the following quantities has the same unit as impulse?

A Work B Momentum C Weight D Kinetic energy 2. A car move s from a rest state with an acceleration 1.5 m s –2 for of 8 velocity after s. What is the 8 s?

A 6.0 m s –1 C 10.0 m –1 s B 8.0 m s –1 D 12.0 m –1 s 3. Which of the following is a vector quant ity?

A Speed B Weight C Energy D Density

Extension / cm x = (l – l o) 2.5 5.0 8.0 10.0

4. An object initially move s with a constant accele then with consta ration and nt velocity.

Which of the following graph shows this s motion?

A

13.0

Velocity

0

B

266

0

474

0

Time

Velocity

0

Time

PAPER 1 and fiften

Answer all C

minutes

questions.

Velocity

Velocity

C 0

D

0

Velocity

B Time

0

D

A 0

Figure 1

0

0

Time

5. A stude nt jumps down from a height. He bends his legs upon landin g. Which of the following statements is correct?

A The stude nt wanted to reduce the impulse on his feet.

B The stude nt wanted to reduce the impulsive force that acted on his feet.

C The stude nt wanted to increase his velocity upon landing.

D The stude nt wanted to reduce his time of impac t between his feet and the floor.

6. A car is travelling along a straight horizo ntal road. The speed-time graph is shown in Figure 1.

In the labelle d part of the journey, which indicates the one resultant force acting on the car is zero?

Time

7. Figure 2 shows flight. The bird a bird in horizontal directis flying in a right. In which ion to the direction does air resistance act on the bird?

SPM Model Paper prepares students for the SPM with the actual exam format.

Figure 2

8. The escap e velocity, v of an object on the surface of the Earth does not depend on A the mass of the object .

B the mass of C the object Earth. ’s distance from the centre of the Earth.

D the Unive rsal Gravitationa l Constant.

9. The value of free fall acceleration , g, surface of plane on the t A and planet B is the same. Which of the follow ing quantities the same for is planet A and planet B?

ii

00 Prelims Physics.indd 2

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CONTENTS FORM 4 ter

ap

Ch

ter

ap

1

Measurement

1

1.1

Physical Quantities

2

1.2

Scientific Investigation

6

SPM Practice 1

15

2

Force and Motion I

19

2.1

Linear Motion

20

2.2

Linear Motion Graphs

28

2.3

Free Fall Motion

33

2.4

Inertia

36

2.5

Momentum

41

2.6

Force

49

2.7

Impulse and Impulsive Force

54

2.8

Weight

58

SPM Practice 2

60

Ch

ter

4

ap

Ch

er

Ch

3

Gravitation

3.1

Newton’s Law of Universal Gravitation 68

3.2

Kepler’s Laws

80

3.3

Man-made Satellites

83

Praktis SPM 3

67

88

93

4.1 Thermal Equilibrium 4.2 Specific Heat Capacity 4.3 Specific Latent Heat 4.4 Gas Laws SPM Practice 4

94 98 110 119 134

5

140

ter

ap

Ch

Waves

5.1 Fundamentals of Waves 5.2 Damping and Resonance 5.3 Reflection of Waves 5.4 Refraction of Waves 5.5 Diffraction of Waves 5.6 Interference of Waves 5.7 Electromagnetic Waves SPM Practice 5

141 148 151 155 160 165 174 179

6

Light and Optics

187

6.1 Refraction of Light 6.2 Total Internal Reflection 6.3 Image Formation by Lenses 6.4 Thin Lens Formula 6.5 Optical Instruments 6.6 Image Formation by Spherical Mirrors SPM Practice 6

188 200 207 212 217

ter

ap

Ch t ap

Heat

222 230

iii

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FORM 5 ter

ap

Ch

ter

ap

1

Force and Motion II

235

1.1

Resultant Force

236

1.2

Resolution of Forces

252

1.3

Forces in Equilibrium

256

1.4

Elasticity

264

SPM Practice 1

274

Ch

4

Electromagnetism

366

Force on a Current-carrying Conductor in a Magnetic Field

367

4.2

Electromagnetic Induction

375

4.3

Transformer

382

4.1

SPM Practice 4

388

5

Electronics

397

5.1

Electron

398

5.2

Semiconductor Diode

404

5.3

Transistor

413

SPM Practice 5

424

6

Nuclear Physics

431

6.1

Radioactive Decay

432

6.2

Nuclear Energy

436

ter

ap

Ch er

t ap

Ch

2

Pressure

2.1

Pressure in Liquids

281

2.2

Atmospheric Pressure

289

2.3

Gas Pressure

293

2.4

Pascal’s Principle

295

2.5

Archimedes’ Principle

297

2.6

Bernoulli’s Principle

304

SPM Practice 2

ter

ap

Ch

310

ter

ap

Ch

280

SPM Practice 6

444

7

Quantum Physics

454

ter

ap

3

Electricity

317

3.1

Current and Potential Difference

318

7.1

Quantum Theory of Light

455

3.2

Resistance

329

7.2

Photoelectric Effect

460

Electromotive Force (e.m.f) and Internal Resistance

344

7.3

Einstein’s Photoelectric Theory

463

Electrical Energy and Power

353

3.3

3.4

SPM Practice 3

360

Ch

SPM Practice 7

470

SPM MODEL PAPER

474

ANSWERS

487

iv

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Important Formulae Period of oscillation of pendulum, T = 2π Velocity,

Gm1m2 r2

Gravitational force, F = Displacement,

1 s = 2 (u + v)t 1 s = ut + 2 at2

v = u + at v2 = u2 + 2as

Gravitational acceleration, g =

 l

g

GM r2

Centripetal acceleration,

Centripetal force, Fc =

mv2 r

Kepler’s III law:

4π2 T2 =  GM r3 → T 2 ∝ r3

v ac = r

2

T21 r31 2 = 3 T2 r2

 GM

Orbital velocity, v =

 r

Escape velocity, v =

 r

Kinetic energy, KE =

 2GM

1 2 2 mv

Gravitational potential energy, U = – GMm

r

Q

Specific heat capacity, c = mDq

1

Q

Specific latent heat, l = m Charles’ law: V a T

Boyle’s law: P a V Speed of waves, v = fl

Refraction of wave:

v1

v2 = l1 l2

Critical angle, c :

Snell’s law:

1

n1 sin q1 = n2 sin q2

n = Interference of waves: l=

n = sin c

sin i sin r

Lens formula,

ax D

1 1 1 f = u + v

v

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Nuclear energy,

Quantum energy, E = mc2

E = hf

Properties of wave-particle duality, Momentum, p = mv (Particle property) h p = (Wave property) l

Einstern’s photoelectric theory,

Hooke’s law,

Elastic potential energy, F = kx

Ep = Ep =

Liquid pressure, P = hρg

1

hf = W +

2

1 2

1 2

mv2

Fx kx2

Pascal’s principle: Force multiplier,

F1 F2 A1 = A2 Archimedes’s principle: Buoyant force, FB = ρVg

Bernoulli’s principle: Lift force, F = (P2 – P1)A

Electric field strength,

Electric current,

E=

F q

Potential difference,

W V= Q

Ohm’s law, R=

V I

I=

Q t

R=

ρl A

Resistance,

Total resistance in a circuit, RT: series: RT = R1 + R2 + R3 +... parallel:

1 1 1 1 = + + + ...

RT R1 R2 R3

Electromotive force, ε = V + Ir

Electrical power,

Efficiency of transformer,

Kinetic energy of electron,

Output power × 100% η= Input power VI = s s × 100% VpIp

Ek = eV =

P = VI = I2R =

V2

R

1

mv 2 2 m vm is the maximum velocity of an electron.

vi

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r

a

Ch

Form 4

e pt

3

Physics SPM Chapter 3 Gravitation

Gravitation OCUS

CHAPTER F

Form

3.1 Newton’s Law of Universal Gravitation 3.2 Kepler’s Laws 3.3 Man-made Satellites

4

What keeps the satellite’s motion in its orbit?

How do planets orbit around the Sun? How is satellite and planetary motion related to Kepler’s Law?

67

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Physics SPM Chapter 3 Gravitation

Law of Universal 3.1 Newton’s Gravitation

6. Then, according to the Newton’s law of Universal Gravitation:

Gravitational Force Between Two Bodies in The Universe 1. Force of gravity exists among all the objects in the universe.

2. The gravitational force between objects near us is so small that it is almost undetected.

Form

4

3. This gravitational force is only large enough to be detected if one of the objects or both are large enough such as Earth, Moon, Sun and planets in the Solar System. m1

F

F

Every object in the universe attracts every other object with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centre. m1 × m2 F=G r2

Solving Problems involving Newton’s Law of Universal Gravitation (I) Two bodies at rest on Earth

m2 m1

r

5. According to Newton, the force of gravity, F between the two objects is proportional to the product of their masses.

F ∝ m1 × m2 ………

This gravitational force, F is also inversely proportional to the square of the distance between the centre of the objects.

F ∝ 12 ……… r Combining  and , we will get m1 × m2

F∝ r2

Then,

F2

m2

r

Figure 3.1

4. Figure 3.1 shows two objects of mass m1 and m2. The distance between the two objects is r. The gravitational force acting between two objects is F.

F1

Figure 3.2

1. For two bodies at rest m1 and m2 at a distance r apart on Earth’s surface, the magnitude of gravitational on each other, F1 and F2, are equal and in opposite directions.

EXAMPLE

3.1

Determine the force of gravitational attraction between two students of masses 70 kg and 80 kg respectively standing at 2 m apart. Given that G = 6.67 × 10–11 N m2 kg–2 Solution

m1 × m2 r2 70 × 80

= (6.67 × 10–11) × 22

= 9.34 × 10–8 N Using F = G

Therefore, the force of gravitational attraction between the two students is 9.34 × 10–8 N.

where G = the universal gravitational constant

= 6.67 × 10–11 N m2 kg–2

This is the formula for Newton’s Law of Universal Gravitation.

2. This gravitational attraction is very small compared to the students’ weight, which is 700 N

and 800 N respectively. (Taken the value of gravitational acceleration = 10 m s–2).

68

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Physics SPM Chapter 3 Gravitation

3. This indicates that the two students will not touch each other due to the gravitational force that exists between them.

The Effect of Mass and Distance Between Two Bodies on Gravitational Force 4. From the equation of Newton’s law of universal m m gravitation, F = G 1 2 2 , two factors that r affect the force of gravitational attraction

between two bodies are mass, m and distance between their centre r.

5. If mass increases, the force of gravity increases.

If distance between their centre increases, the force of gravity decreases.

6. The effect of mass and the distance between the centre of two bodies on force of gravity can be illustrated as shown in the Table 3.1.

Form

Table 3.1

Effect of mass on F F ∝ m1 × m 2 Distance between the centre of two bodies is fixed Gravitational force

m

F

m

Effect of distance between the centre of two bodies on F 1 F∝ r2 Two bodies with same mass m

m

r

Gravitational force

m

Gravitational force 1 –– F 4 2r

m

2m

m

Gravitational force 4F 1 –– r 2

m

3m

2m

2m

r Gravitational force 4F r Gravitational force

m

3F

(II) Body on Earth’s surface Body m

mg

M

Gravitational force F

1. On the surface of the Earth, the force of gravity between the body and the Earth is the object’s weight.

2. The two forces are related by the acceleration due to gravity, F = mg.

3. From Newton’s law of universal gravitation, the force of gravitational attraction between the body and the Earth is given by F=G

Figure 3.3

2m

2r

r

Earth’s radius R

m

r

2F

2m

Gravitational force F

4

Mm R2

69

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Physics SPM Chapter 3 Gravitation

3.2

EXAMPLE

EXAMPLE

What is the magnitude of the gravitational force between the Earth and 1 kg mass object on its surface?

[Mass of Earth, M = 5.98 × 1024 kg; G = 6.67 × 10–11 N m2 kg–2; Radius of Earth, R = 6.38 × 106 m]

Form

Solution Applying

(5.98 × 1024) × 1 = (6.67 × 10–11) × (6.38 × 106)2 = 9.8 N

4

A satellite of mass 900 kg orbits round the Earth at a height of 300 km above the surface of the Earth.

The mass of Earth is 5.98 × 1024 kg and the radius of the Earth is 6.38 × 106 m. What is the gravitational force between the satellite and the earth?

Solution Applying F=G

Mm F=G 2 R

The gravitational force of attraction between the Earth and the object is the same as the weight of 1 kg object. (Taken the value of gravitational acceleration = 9.8 m s–2)

3.3

Mm (R + h)2

= (6.67 × 10–11) × = 8 045 N

(5.98 × 1024) × 900 (6.38 × 106 + 300 × 103)2

Therefore, the gravitational force between the satellite and the Earth is 8 045 N.

(IV) Earth and the Sun (III) Earth and Satellite

M Sun

Satellite m

r

m Earth

h R

Figure 3.5

Earth

EXAMPLE M Center of Earth

Figure 3.4

1. A satellite of mass m orbits the Earth at a height of h km from the Earth of mass M and radius R. 2. The gravitational force, F, between the satellite and the Earth can be determined by using the formula: F=G

Mm (R + h)2

3.4

What is the gravitational force between the Earth and the Sun, given that the mass of Earth = 5.98 × 1024 kg and the mass of the Sun = 2 × 1030 kg. The average distance between the Earth and the Sun is 1.5 × 1011 m.

Solution Applying F= G

Mm R2

= (6.67 × 10–11) × = 3.54 × 1022 N

(2 × 1030) × 5.98 × 1024 (1.5 × 1011)2

Therefore, the gravitational force between the Earth and the Sun is 3.54 × 1022 N.

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Physics SPM Chapter 3 Gravitation

Relationship Between Acceleration due to Gravity, g at Earth’s Surface and Universal Gravitational Constant, G Body

m

mg

R

Variation of Acceleration due to Gravity 5. At a given place, the value of acceleration due to gravity is constant but it varies from one place to another place on the Earth’s surface.

6. This is due to this fact that Earth is not a perfect sphere. It is flattened at the Poles and bulges out at the Equator as shown in Figure 3.7.

Pole

O M

Rp

2. If the size of the body is very small as compared to that of the Earth, then the distance between the centre of the body and the centre of the Earth will be approximately equal to R (radius of Earth). Therefore, the gravitational force of attraction on the body due to the Earth is given by: Mm

F=G …… R2 3. The force F acting on the body of mass m due to Earth’s gravity produces an acceleration, g. This force is the weight of the body and is given as

F = mg ……

S

Figure 3.7

7. In Figure 3.7, the polar radius, Rp is not equal to the equatorial radius, Re. M 8. From the expression g = 2 , as G and M are R 1 constant, then g ∝ 2 . As Re > Rp, thus, the R value of g is smallest at the Equator and largest at the poles. It means, g increases as we move from the Equator to the Pole of the Earth.

Variation of g with altitude, h 9. The value of acceleration due to gravity also varies with altitude and depth of the Earth.

Object

m

g´

h

Therefore, weight of body = gravitational force, Mm

mg = G 2 R

r

g

R O

GM g= …… R2 4. Equation  gives the value of acceleration due to gravity on the surface of Earth. This is clear from the expression that the value of ‘g’ is independent of mass, shape and size of the body but depends on the mass and radius of the Earth.

4

Re

Figure 3.6

1. Consider the Earth is a perfect sphere of mass M and radius R. The whole mass of the Earth is supposed to be concentrated at the centre O. Consider a body of mass m lying on the surface of Earth as shown in Figure 3.6.

Form

Earth

N

Earth, M

Figure 3.8

10. For a body on Earth’s surface, h = 0, the gravity is given by the following equation:

g=

GM …… R2 71

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Physics SPM Chapter 3 Gravitation

11. On going away from the Earth’s surface, the value of g decreases.

If ρ is the Earth’s density, then, the mass of the Earth is,

12. For a body of mass m raised to a height h above the surface of the Earth where r . R, the acceleration g due to gravity is given as:

Mass = Volume × density 4 M = pR3ρ 3

GM …… r2

g' =

Therefore,

where r = (R + h), distance between the centre of the Earth and the centre of the body.

Form

4

13. Dividing  by , we get GM (r)2 g' = GM g R2

Therefore, g' =

GM R2 4 G pR3ρ 3 = R2 4

g = GpRρ …… 3

The acceleration due to gravity at Q where radius, r of the inner sphere of the Earth is GM'

g' = (r)2

2

 Rr  g → g' ∝ r1

2

where M' = the mass of inner sphere.

This shows that g’ is independent of the mass of the body.

and

Variation of g with Depth

Then,

M' = g' =

4 3 pr ρ 3

4 Gprρ …… 3

Dividing  by ,

Q

r <R O

Earth’s surface

Earth M

g =

R

P

Figure 3.9

14. Consider the Earth to be a homogeneous sphere with uniform density. The radius of the Earth is R and the mass of Earth is M as shown in Figure 3.9.

15. Let P be a point at the surface of the Earth and Q be the point at a distance r from the centre of the Earth, where r < R. The acceleration due to gravity at P on the surface is GM

g= R2

4 Gprρ g' 3 = g 4 GpRρ 3 g' r g = R r g' = R g → g' ∝ r

 

r g' = R g → g' ∝ r

 

This shows that the value of acceleration due to gravity increases with increasing r.

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Physics SPM Chapter 3 Gravitation

16. The variation of g with distance from Earth’s core is shown in Figure 3.10.

r

R

Gravitational acceleration, g´

h

g

r R

r=0 at Earth’s centre

Form

1 g´ α –– above the Earth’s surface r2

g´ α r below the Earth’s surface

r R

4

r = R = radius of Earth, i.e. at the surface

r, distance from centre of Earth

Figure 3.10

The Importance of Knowing the Value of the Gravitational Acceleration of Planets in the Solar System 1. The value of gravitational acceleration, g for any planet in the Solar System can be solved so long as we know the planet’s size and mass, using the equation: GM

g= 2 r

where G = universal gravitational constant = 6.67 × 10–11 N m2 kg–2, M = mass of the planet r = radius of the planet

2. The acceleration due to gravity of Earth is about 9.81 m s–2 can be determined by substituting the Earth’s mass (ME) and Earth’s radius (RE) into the equation, GM

g = 2 r GME

gE = (RE)2

ME = 5.98 × 1024 kg RE = 6.375 × 106 m 6.67 × 10–11 × 5.98 × 1024

g = (6.375 × 106)2 = 9.81 m s–2

3. Table 3.2 shows the comparison of acceleration due to gravity for planets in the Solar System.

Table 3.2 Comparison of acceleration due to gravity for planets in the Solar System Acceleration due to gravity g / m s–2

In ratio g : gE

3.29 × 1023

3.72

0.38

6.05 × 10

24

4.87 × 10

8.87

0.90

6.37 × 106

5.97 × 1024

9.81

1.00

Mars

6

3.39 × 10

6.39 × 10

3.71

0.38

Jupiter

7

6.99 × 10

1.90 × 10

25.94

2.64

Saturn

5.82 × 10

26

5.68 × 10

11.18

1.14

Uranus

2.54 × 107

8.68 × 1025

8.97

0.91

Neptune

2.46 × 107

1.02 × 1026

11.24

1.15

Planets in Solar System

Radius /m

Mass / kg

Mercury

2.43 × 106

Venus Earth

6

7

23 27

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Kepler's Laws

I N FO G R A P H Chapter 3

Form

IC

4

Elipse

F1

F2 Focus

First Law All planets move about the Sun in elliptical orbits, having the Sun as one of the focus.

Kepler's Law

Second Law A straight line joining any planet to the Sun sweeps out equal areas in equal lengths of time as the planet orbits the Sun.

The orbit of the planet is ellipse in shape Distance x

Johannes Kepler (1571-1630)

Area A1

Third Law The squares of the periods of the planets are directly propotional to the cubes of their orbital radius, T2 ∝ r3.

Sun

Area A2 Distance y

Planet

Time taken by the planet to move the distance x = Time taken by the planet to move the distance y Therefore, Area A1 = Area A2

Geostationary Satellite

A geostationary satellite is always at a fixed place when observed by an observer on the surface of the Earth.

Orbit of geostationary satellite

Earth

00a Infografik Physics.indd 4

Satellite

A geostationary satellite is located above the Equator. Its period of orbit is the same as the period of the Earth’s rotation. Because of that, the satellite is always at the same location when observed by an observer on the surface of the Earth.

27/12/2021 12:16 PM

Force and Motion II

I N FO G R A P H Chapter 1

IC

Form

5

θ β

Tension, T2

Up further...

Engineer

F1 = Weight F2 = Normal reaction F3 = Frictional force θ

F3

F1

Tension, T1

F2

Weight, W

Forces in equilibrium

T1 cos Ө + T2 cos ß = W

Resolution of forces y

F2

θ

θ

F3

x

F1

Total

Resolve along the x-axis

Resolve along the y-axis

0

-F1

F2 sin Ө

F2 cos Ө

-F3 cos Ө

F3 sin Ө

0

0

Because of equilibrium of forces

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