Focus SPM 2022 - physics

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3Chapter Gravitation Form 4

CHAPTER FOCUS Physics SPM  Chapter 3 Gravitation
3.1 Newton’s Law of Universal Gravitation
3.2 Kepler’s Laws 4
3.3 Man-made Satellites
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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

  Physics SPM  Chapter 3 Gravitation

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

Gravitational Force Between Two Bodies Every object in the universe attracts every
in The Universe
other object with a force that is proportional
1. Force of gravity exists among all the objects
in the universe. to the product of their masses and inversely

2. The gravitational force between objects near proportional to the square of the distance
us is so small that it is almost undetected.
between their centre.
3. This gravitational force is only large enough
to be detected if one of the objects or both are

4 large enough such as Earth, Moon, Sun and
planets in the Solar System.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. F = G m1 × m2
r2


Form Solving Problems involving Newton’s Law
of Universal Gravitation

(I) Two bodies at rest on Earth

m1 F F m2

r mF F m
11 2 2

Figure 3.1 r

4. Figure 3.1 shows two objects of mass m1 and Figure 3.2
m2. The distance between the two objects is
r. The gravitational force acting between two 1. For two bodies at rest m1 and m2 at a distance
objects is F. r apart on Earth’s surface, the magnitude of
gravitational on each other, F1 and F2, are equal
5. According to Newton, the force of gravity, F and in opposite directions.
between the two objects is proportional to the
product of their masses. EXAMPLE 3.1

F ∝ m1 × m2 ……… Determine the force of gravitational attraction
between two students of masses 70 kg and 80 kg
This gravitational force, F is also inversely respectively standing at 2 m apart. Given that
G = 6.67 × 10–11 N m2 kg–2
proportional to the square of the distance

between the centre of the objects. Solution

F ∝ 1 ……… m1 × m2
r2 r2
Using F = G
Combining  and , we will get 70 × 80
= (6.67 × 10–11) × 22
m1 × m2
F∝ r2 = 9.34 × 10–8 N

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

2. This gravitational attraction is very small
compared to the students’weight, which is 700 N
where G = the universal gravitational constant
= 6.67 × 10–11 N m2 kg–2 and 800 N respectively. (Taken the value of
gravitational acceleration = 10 m s–2).
This is the formula for Newton’s Law of
Universal Gravitation.

68

3. This indicates that the two students will not Physics SPM  Chapter 3 Gravitation  
touch each other due to the gravitational force
that exists between them. between two bodies are mass, m and distance
between their centre r.
The Effect of Mass and Distance Between
Two Bodies on Gravitational Force 5. If mass increases, the force of gravity increases.
If distance between their centre increases, the
4. From the equation of Newton’s law of universal force of gravity decreases.
m1 m2
r2 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.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.gravitation, F=G,twofactorsthat

affect the force of gravitational attraction

Table 3.1 Form

Effect of mass on F Effect of distance between the centre of two 4
F ∝ m1 × m2 bodies on F

Distance between the centre of two bodies is F ∝ 1
fixed r2

Two bodies with same mass

Gravitational Gravitational
force
m force m
m Fm F

r r

Gravitational Gravitational
force
m force 2m m
2F m –41–F
2r
r

Gravitational Gravitational

2m force 2m m force m
4F 4F

r –12– r

Gravitational Gravitational

m force 3m 2m force 2m
3F F

r 2r

(II) Body on Earth’s surface 1. On the surface of the Earth, the force of gravity
between the body and the Earth is the object’s
Bomdy weight.
mg
2. The two forces are related by the acceleration
Earth’s radius due to gravity, F = mg.
R
M 3. From Newton’s law of universal gravitation,
the force of gravitational attraction between
Figure 3.3 the body and the Earth is given by

F=G Mm
R2


69

  Physics SPM  Chapter 3 Gravitation

EXAMPLE 3.2 EXAMPLE 3.3

What is the magnitude of the gravitational force A satellite of mass 900 kg orbits round the Earth at
between the Earth and 1 kg mass object on its a height of 300 km above the surface of the Earth.
surface? The mass of Earth is 5.98 × 1024 kg and the radius of
[Mass of Earth, M = 5.98 × 1024 kg; the Earth is 6.38 × 106 m. What is the gravitational
G = 6.67 × 10–11 N m2 kg–2; force between the satellite and the earth?
Radius of Earth, R = 6.38 × 106 m]
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. Solution

Solution Applying

Form Applying Mm F = G Mm
R2 (R + h)2
F = G
(5.98 × 1024) × 900
4 = (6.67 × 10–11) × (5.98 × 1024) × 1 = (6.67 × 10–11) × (6.38 × 106 + 300 × 103)2
(6.38 × 106)2
= 8 045 N
= 9.8 N
Therefore, the gravitational force between the
The gravitational force of attraction between the satellite and the Earth is 8 045 N.
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)

(IV) Earth and the Sun

(III) Earth and Satellite M
Sun
Samtellite r m
Earth

h Figure 3.5
R
Earth EXAMPLE 3.4

M What is the gravitational force between the
Center of Earth
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.

Figure 3.4 Solution

1. A satellite of mass m orbits the Earth at a Applying Mm
height of h km from the Earth of mass M and R2
radius R. F = G

2. The gravitational force, F, between the satellite = (6.67 × 10–11) × (2 × 1030) × 5.98 × 1024
and the Earth can be determined by using the (1.5 × 1011)2
formula:
= 3.54 × 1022 N

Mm Therefore, the gravitational force between the Earth
(R + h)2 and the Sun is 3.54 × 1022 N.

F = G



70

Relationship Between Acceleration due Physics SPM  Chapter 3 Gravitation  
to Gravity, g at Earth’s Surface and
Universal Gravitational Constant, G Variation of Acceleration due to Gravity
5. At a given place, the value of acceleration due
Body m
to gravity is constant but it varies from one
mg R 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
N

Rp

4Re
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Earth M

Figure 3.6

1. Consider the Earth is a perfect sphere of mass S
M and radius R. The whole mass of the Earth Figure 3.7
is supposed to be concentrated at the centre O.
Consider a body of mass m lying on the surface 7. In Figure 3.7, the polar radius, Rp is not equal
of Earth as shown in Figure 3.6. to the equatorial radius, Re.

2. If the size of the body is very small as compared 8. From the expression g = M , as G and M are
R2
to that of the Earth, then the distance between 1
R2
the centre of the body and the centre of the constant, then g ∝ . As Re > Rp, thus, the

Earth will be approximately equal to R (radius value of g is smallest at the Equator and largest

of Earth). Therefore, the gravitational force of at the poles. It means, g increases as we move

attraction on the body due to the Earth is given from the Equator to the Pole of the Earth.

by: F=G Mm …… Variation of g with altitude, h
R2
9. The value of acceleration due to gravity also
3. The force F acting on the body of mass m due varies with altitude and depth of the Earth.

to Earth’s gravity produces an acceleration, Object m g´

g. This force is the weight of the body and is

given as

F = mg …… h
r
Therefore, g
weight of body = gravitational force, R

mg = G Mm O
R2 Earth, M

g= GM ……
R2

4. Equation  gives the value of acceleration Figure 3.8
due to gravity on the surface of Earth. This
is clear from the expression that the value of 10. For a body on Earth’s surface, h = 0, the
‘g’ is independent of mass, shape and size of gravity is given by the following equation:
the body but depends on the mass and radius
of the Earth. g= GM ……
R2

71

  Physics SPM  Chapter 3 Gravitation

11. On going away from the Earth’s surface, the If ρ is the Earth’s density, then, the mass of the
value of g decreases. Earth is,

12. For a body of mass m raised to a height h Mass = Volume × density
above the surface of the Earth where r . R,
the acceleration g due to gravity is given as: M = 4 pR3ρ
3

g' = GM …… Therefore,
r2
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. g = GM
where r = (R + h), distance between the centre R2
of the Earth and the centre of the body.
G 4 pR3ρ
Form 13. Dividing  by , we get 3
=
R2

GM g = 4 GpRρ ……
3
4 g' = (r)2
g GM
Therefore, The acceleration due to gravity at Q where
R2 radius, r of the inner sphere of the Earth is

 g' =R 2 1 g' = GM'
r r2 (r)2
g → g' ∝

where M' = the mass of inner sphere.

This shows that g’ is independent of the mass and M' = 4 pr3ρ
of the body. 3
Then,
Variation of g with Depth g' = 4 Gprρ ……
3
Earth’s surface

Q Dividing  by ,

r <R g' 4 Gprρ
O g = 3

Earth RP 4 GpRρ
M 3

g' r
g = R

 g' = r g → g' ∝ r
R

Figure 3.9  

14. Consider the Earth to be a homogeneous sphere g' = r g → g' ∝ r
with uniform density. The radius of the Earth R
is R and the mass of Earth is M as shown in
Figure 3.9. This shows that the value of acceleration due
to gravity increases with increasing r.
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

g= GM
R2

72

Physics SPM  Chapter 3 Gravitation  

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

Gravitational R r
acceleration, h
g´

Penerbitan Pelangi Sdn Bhd. All Rights Reserved.g

g´ α r g´ α –r1–2 above the Earth’s surface Form
below the rϽR rϾR
Earth’s
surface

4

r=0 r = R = radius of Earth, r, distance from centre
at Earth’s i.e. at the surface of Earth
centre
Figure 3.10

The Importance of Knowing the Value of 2. The acceleration due to gravity of Earth
the Gravitational Acceleration of Planets
in the Solar System is about 9.81 m s–2 can be determined by

substituting the Earth’s mass (ME) and Earth’s

1. The value of gravitational acceleration, g for radius (RE) into the equation,

any planet in the Solar System can be solved g = GM
r2
so long as we know the planet’s size and mass,

using the equation: gE = GME
(RE)2
g= GM
where r2 ME = 5.98 × 1024 kg

G = universal gravitational constant RE = 6.375 × 106 m

= 6.67 × 10–11 N m2 kg–2, g = 6.67 × 10–11 × 5.98 × 1024
(6.375 × 106)2
M = mass of the planet  
= 9.81 m s–2
r = radius of the planet

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

Planets in Solar Radius Mass Acceleration due In ratio
System /m / kg to gravity g : gE
g / m s–2

Mercury 2.43 × 106 3.29 × 1023 3.72 0.38

Venus 6.05 × 106 4.87 × 1024 8.87 0.90

Earth 6.37 × 106 5.97 × 1024 9.81 1.00

Mars 3.39 × 106 6.39 × 1023 3.71 0.38

Jupiter 6.99 × 107 1.90 × 1027 25.94 2.64

Saturn 5.82 × 107 5.68 × 1026 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

73

  Physics SPM  Chapter 3 Gravitation

4. The last column of the table compares the 12. Figure 3.11 shows the weight of a person on
acceleration due to gravity of the planet to that the Moon and the planets in the Solar System.
of Earth.

5. A small planet like Mercury only exerts a

gravitational acceleration about 19 as strong
50
as that of Earth, while the massive Jupiter
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generates an acceleration nearly 2.6 times that

of Earth.

Form 6. Planets Mars and Mercury are of different 170 N on Moon 380 N on Mercury
sizes but have almost the same gravitational
acceleration.

4 7. This information is useful because we can
know the weight of an object on the surface
of each planet.

8. As we have seen, the mass of an object
remains constant whether it is on the surface
of Earth, the surface of the Mercury or in the
depths of space.

9. The quantity that does change depending on 900 N on Venus 380 N on Mars
the strength of the gravitational field is the
object’s weight. These variables are related by
the equation:

W = mg

where

W = weight force in, N 2 640 N on Jupiter 1 140 N on Satum
m = mass in, kg  

g = acceleration due to gravity in, m s–²

10. Since the acceleration due to gravity is different
for different planets, it is easy to determine the
weight of an object on that planet.

11. For example, the value of gm on Mercury is

19 g on Earth, so an object weighing W on
50 will have on Mercury. Let’s say
19
Earth 50 W

someone weighs 1000 N on Earth, then his 910 N on Uranus 1 450 N on Neptune

weight on Mercury is 380 N. Figure 3.11

74

Physics SPM  Chapter 3 Gravitation  

Use gravity as a thrust Orbit the planets Exploration of the surface of
Space ships can use the Space satellites can orbit the planets
gravity of planets to push them planet and provide information
forward like a catapult. This on the atmosphere and the Rockets can land safely on the
can save energy and can make
surface of the planet. surface of the planets.
spaceships go further.

Space exploration The importance of knowing SurvivalReserved. Form
the gravitational acceleration
of the planets in Solar System.

Rights 4

Progress in Study and forecast of weather Development of medical
telecommunications Earth satellite can provide research
The Earth’s satellite in orbit can information on Earth’s
atmospheric conditions and Research in the field of medicine
receive and transmit information can be enhanced by obtaining
further help to study and forecast research results from zero or
across the Earth. weather. weak gravity.
All

Bhd.
Figure 3.12  The importance of knowing the gravitational acceleration of the planets in the Solar System.

Table 3.3  Impact of gravity on human growth.
Sdn
Breathing is related to air pressure and not gravity. Gravity affects blood flow to the lungs
Lung size but does not change the size of the lungs.
human growth

Pelangi
Density change Density depends on mass and volume.
The mass and volume do not change with gravity. Moving from the gravity of the Earth
ThePeeffectneofrbgravityitaonn to the gravity of the Moon, one’s weight changes, but the mass remains the same. Thus,

density does not change with gravity.

Fracture of the Gravity smaller than gravity of the Earth can cause bone density to decrease because
bones bones do not need to support the body against larger gravity. As a result, the bones
become more fragile and easily broken, especially on the hips and spine.

Circulatory In microgravity, the output of the heart, that is, the amount of blood pumped out of the
system and blood heart decreases. Blood circulation is slower and blood pressure is lower than on the

pressure Earth.

75

  Physics SPM  Chapter 3 Gravitation

Centripetal Force in Satellite and Planetary Motion Systems

The activity below describes and explains the centripetal force in circular motion. This activity can help
you understand the centripetal force in satellite and planetary motion systems.

ACTIVITY 3.1

Aim: Understanding centripetal force
Materials and apparatus: Two holes rubber stopper, nylon string (1.5 m long), glass tube about 15 cm long

(wrapped by rubber tubing), labels, 2 × 50 g slotted weights, meter rule, paper clip,
stopwatch.
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r

4

Rubber stopper

Glass tube
Nylon string

Paper clip
as marker

Mass M

Figure 3.13

Instructions:

1. The apparatus is set up as shown in Figure 3.13.

2. A rubber stopper is fastened securely at one end of the nylon string. The other end of the string
is passed through the glass tube and a 50 g mass is fastened to it.

3. The length of the string is adjusted so that it is about 50 cm between the top of the tube and the
stopper. A paper clip is attached to the string just below the bottom of the tube as a marker.

4. The 50 g mass is supported with one hand and the glass tube is hold in the other. The rubber
stopper is whirled by moving the tube in a circular motion.

5. Slowly the 50 g mass is released, and the speed of the rubber stopper is adjusted so that the paper
clip always stays just below the bottom of the tube.

6. When the motion of the rubber stopper is stable, the time for 10 complete horizontal circles is
taken using a stopwatch. Be sure that the paper clip does not move up and down or touch the
end of the glass tube.

7. The period T, time for one complete revolution is determined. Then the velocity v of the rubber
stopper is calculated using the formula v = 2Tpr.

8. The weight of the 50 g mass is calculated. This weight is the gravitational force, Fg and this force
of gravity is equal to the tension of the nylon string holding the rubber stopper.

9. The centripetal acceleration is determined using the formula ac = v2 , and then the centripetal
Fthc e=gmraavcit=atimornv2alwfohrecree, mFg.is the r rubber stopper. The
force is calculated using the formula, mass of
centripetal force, F is compared with

10. The activity is repeated by adjusting the radius of circle to 75 cm. Then add another 50 g mass

of slotted weight for a total of 100 g with radius of circle 75 cm.

76

Physics SPM  Chapter 3 Gravitation  

Results:

Mass of Weight Radius of Time for 10 Period, Velocity Centripetal Centripetal
slotted of slotted circle revolutions time for 1 v / m s–1
weight r/m revolution acceleration force
m/g weight t10 / s 2.18
Fg / N 0.50 T/s ac / m s–2 Fc / N
50 14.4
0.5 1.44 9.50 0.48

50 0.5 0.75 16.8 1.68 2.80 10.4 0.52Penerbitan Pelangi Sdn Bhd. All Rights Reserved.

100 1.0 0.75 17.2 1.72 2.74 10.0 1.00

Discussion: Form

1. When the rubber stopper is whirled in horizontal circles with constant speed, the direction of the 4
speed is always changing. The rubber stopper moving with changing speed in the same direction
is undergoing acceleration. This acceleration is also known as centripetal acceleration and is
always directed towards the centre of the circular path.

2. The force that is used to maintain the rubber stopper in a circular motion is called the centripetal
force. This force is provided by the tension of the string and is always directed towards the centre
of the circular path.

3. The above activity shows that the centripetal force acting on the rubber stopper is equal to the
weight of the slotted weight.

4. From the formula of centripetal force, Fc = mrv2, the factors that can affect the centripetal force are
the mass of the object, the velocity of the object and the radius of the circular path.

Centripetal Acceleration and Centripetal 3. Swinging an object on a string requires string
Force tension to provide the centripetal force to hold
the object in a circular motion. If the string
1. In Figure 3.14, the force needed by a body is released, the object will continue to move
of mass  m, to keep in a circular motion along a straight line tangentially to the circle
at a distance r, from the centre of a circle as shown in Figure 3.15.

with velocity v, is the centripetal force,  Fc = v
cmervn2t.reTohfethdeireccirtciounlaorfmthoetiofonr.ce is towards the
ac
m
Centripetal v
force, Fc

r

Figure 3.14 Figure 3.15

2. The centripetal acceleration is given as 4. Gravitational force is a universal force of
attraction between masses. It provides the
caec n=trevr2o,f and is always directed towards the centripetal force needed to keep a satellite
the circular motion. in orbit around a planet, or a planet
in orbit around a star such as the Sun.

77

  Physics SPM  Chapter 3 Gravitation

Satellite To Determine the Mass of Earth and Sun
To determine the mass of Earth
Instantaneous Gravitatinal force = 1. Consider a small body on Earth’s surface as
velocity centripetal force
shown in Figure 3.18.
Free fall Earth
Small body on
Earth’s surface

m

MR
Free fallPenerbitan Pelangi Sdn Bhd. All Rights Reserved.
Form
Free fall
Figure 3.16 Earth

4 Without gravity Figure 3.18

With gravity The gravitational force between the small body

and the Earth is given as:

F = G Mm
R2

This gravitational force is the same as the

weight, W = mg, of the body due to gravity:

W = mg

Therefore,

G Mm = mg → M = gR2
R2 G

where,

Figure 3.17 g = 9.81 m s–2 (acceleration due to gravity)

5. Figure 3.16 shows that the satellite that orbits G = 6.67 × 10–11 N m2 kg–2 (Universal
the Earth always experience free fall towards
the centre of the Earth. The Earth’s gravity gravitational constant)
pulls on the satellite to keep the satellite
moving round the Earth in its orbit. R = 6.37 × 106 m (Radius of the Earth)

6. In the absence of this gravitational force, the Then, the mass of Earth,
satellite will move in a straight line tangent to
its orbit as shown in Figure 3.17. M = 9.81 × (6.37 × 106)2
6.67 × 10–11
7. The movement of the satellite system and
the planet is a circular motion that always = 5.97 × 1024 kg
experiences a centripetal acceleration, ac or a
centripetal force, Fc where: Moon

v m Moon’s
and orbit

r F = m –vr–2

Earth

M

Figure 3.19

2. Alternatively, consider the Moon orbiting the
Earth as shown in Figure 3.19.

78

Physics SPM  Chapter 3 Gravitation  

3. The gravitational force between the Moon and T = 3.15 × 107 s (Earth’s orbitual period = 365
days)
the Earth is given as,

F = G Mm Then, mass of the Sun,
r2
4(p2)(1.50 × 1011)3
and the centripetal force acts on the Moon, M = (6.67 × 10–11)(3.15 × 107)2

F = mv2 = 2.01 × 1030 kg
r

These two forces are equal, therefore, Checkpoint 3.1

Q1 The diagram shows two bowling balls, A and B,
each having a mass of 7 kg and placed 2 m apart
from each other.

AB
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. GMm=mv2→GM=v2
r2 r r

The velocity of Moon orbiting the Earth is Form

given as, v = 2pr
T
Then,
7 kg 7 kg
M  2pr 2 4
G r = T

And mass of Earth, 2m

M= 4(p2)(r3) = 4p2­ r3 What is the magnitude of the gravitational force
GT2 G T2 applied by ball A on ball B?

where, Q2 An space probe was launched into space from
the surface of the Earth.
r = 3.85 × 108 m (Moon’s orbital radius) (a) State the relationship between the magnitude
of the gravitational force, F, imposed on
G = 6.67 × 10–11 N m2 kg–2 (Universal Earth by the space probe and the distance,
r between the space probe and the centre of
gravitational constant) the Earth.
T = 2.36 × 106 (Moon orbital period = 27.3 (b) Sketch a graph of F versus r to represent the
relationship.
days)

Therefore, mass of Earth, Q3 A 60 kg student weighs 1560 N on the surface
of planet X. What is the magnitude of the
M = 4(p2)(3.85 × 108)3 acceleration due to the gravity on the surface of
(6.67 × 10–11)(2.36 × 106)2 planet X?

= 6.06 × 1024 kg

To determine the mass of the Sun Q4 The mass of the Earth is approximately 81 times
the mass of the Moon. If the Earth exerts a
4. Consider the Earth of mass m moving with gravitational force of magnitude F over the Moon,

velocity v around the Sun of mass M in a determine the magnitude of the gravitational force
of the Moon acting on the Earth in terms of F.

circular orbit of radius R as shown in Figure Q5 The diagram shows an experiment to study
centripetal acceleration.
3.20. v

R R m
Sun Earth Glass tube Rubber stopper
Mm

Figure 3.20

Mass of the Sun is given as, Nylon string

4p2 R3 Load M
G T2
M= If the radius of the circle, R = 0.6 m, the mass of
where load, M = 600 g and the mass of rubber stopper,
m = 50 g, what is the centripetal acceleration of
R = 1.50 × 1011 m (Earth’s orbital radius around the rubber stopper? Next, determine the velocity
of the rubber stopper in a circular path.
the Sun)

G = 6.67 × 10–11 N m2 kg–2 (Universal

gravitational constant)

79

  Physics SPM  Chapter 3 Gravitation

3.2 Kepler’s Laws Sun Planet

1. Kepler’s laws were discovered by an Focus
astronomer named Johannes Kepler. Figure 3.21

2. He discovered three laws to explain the 3. The orbit or path of the planet around the Sun
movement of celestial objects (planets). is an ellipse, oval in shape and not a perfect
circle.
Kepler’s First Law
1. Kepler’s first law is also known as the law of 4. The elliptical path has two foci. The sun is
at one of the two foci of the elliptical path as
ellipses. shown in Figure 3.21.
2. Kepler’s first law states that the orbits of the
5. The shape of the ellipse is based on dual focus.
4 planets are ellipses with the Sun in one of This ellipse can be drawn by using threads and
its focus. pencils as shown in the activity below.

ACTIVITY 3.2
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Aim: To draw an ellipse base on the concept of dual-focus ellipse using string and pencil
Apparatus and Materials: Drawing paper, compass, string and pencil.
Instructions:

1 The horizontal axis (major) and the vertical Vertical axis
axis (minor) of the ellipse is drawn. (minor)

Horizontal axis
(major)

2 The compass width is set to half the width of
the major axis.

3 The compass point is moved to one end of F1 F2
the minor axis and two arcs across the major
axis are drawn.

4 Where these arcs cross the major axis are the
foci of the ellipse. The foci are labeled as F1
and F2.

80

5 In each end of the major axis is put a pin, and Physics SPM  Chapter 3 Gravitation  
a string is tied to them so that the string is taut FF
between them.
12
6 Leaving the string attached, the pins are moved
to the focus point F1 and F2. A pencil is put
against the string and the string is pulled taut
with the pencil.

7 Keeping the string taut, the pencil is moved in
a large arc. The pencil draws out the desired
ellipse.

8 Done, the ellipse is drawn through the four
initial points which define the ends of the
major and minor axes.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. F1 F2 Form
F1 F2
F1 F2 4

Kepler’s Second Law   Kepler’s Third Law
1. Kepler’s third law is also known as the law of
1. Kepler’s second law is also known as the law
of areas. periods.
2. Kepler’s third law states that the square of
2. Kepler’s second law states that a line joining
the planet with the Sun sweeps over equal the period is directly proportional to the
areas in equal intervals of time as the planet cube of the orbital’s radius.
moves in orbit. 3. Kepler’s third law can be written as T 2 ∝
r3, where T is the orbital period and r is the
Distance Planet’s orbit average radius of the orbit.
x around the
sun Derivation of Kepler’s Third Law
Area x
v
Sun
Area y m
Distance y
r
Figure 3.22
M
3. Figure 3.22 shows that the planet moves a
distance x and distance y in same time interval. Figure 3.23
Area x equals area y. This means that the planet
moves faster when it is closer to the sun and 1. When a planet moves in an orbit round the Sun
moves slower when it is further from the Sun. with velocity v, the centripetal force is caused
by gravitation force.

81

  Physics SPM  Chapter 3 Gravitation

Then, centripetal force = gravitational force EXAMPLE 3.6

mv2 = G Mm …… Mars takes about 1.88 Earth-years to complete one
r r2 orbit around the Sun. What is the average distance
of Mars from the Sun in terms of Earth distance
The velocity of the planet is given as, from the Sun?

v = 2pr ……
T

where T is the orbital period. Solution
Substituting  into ,
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. Using,
2pr
 m T 2 G Mm T21 = T22
r2 r­31 r­32
r =

Form 4p2 = GM 1.R83­82 = 12 → R3 = 1.882
T2 r3 r­3 r­3
4 4p2
T2 = GM × r3 R = 31.882
r­
T2 = kr3
R = 1.52 r

k = 4p2 is a constant. The average distance of Mars from the Sun is 1.52
GM times the average distance of the Earth from the
where, Sun.

Therefore,

EXAMPLE 3.7
2. For a planet system orbiting the Sun, M is the
The average distance from the centre of Earth to
mass of the sun. the centre of Moon is 3.82 × 108 m. Determine the

3. For a satellite system orbiting the Earth, M is period of the Moon round the Earth.
the mass of the Earth. [Mass of Earth = 5.98 × 1024 kg, G = 6.67 × 10–11

Solving Problems using Kepler’s Third Law N m2 kg2]

1. To solve problems involving Kepler’s third Solution

law, we can use the proportion method. Since Using,

T2 = k, we can use the proportion T2 = 4p2 × r3
r­3 GM
4p2
= 6.67 × 10–11 × 5.98 × 1024 × (3.82 × 108)3
where, T is the orbital period and r is the
T2 = 5.517 × 1012
orbital radius.
T = 2.35 × 106 s

EXAMPLE 3.5 Checkpoint 3.2

Saturn is about 9 times further from the Sun than Q1 The distance of Earth from the Sun is 1.5 × 108 km
the Earth, how long is its orbital period in terms of CanadlctuhlaeteperTiro32d. of Earth around the Sun is 1 year.
earth years?
Q2 The average distance of the Earth from the Sun is
Solution 1.50 × 1011 m and the average distance of Mercury
from the Sun is 5.79 × 1010 m. The period of the
Using, Earth moving in orbit round the Sun is 1 year.
Determine the period of Mercury moving in orbit
T 2 = T 2 round the Sun.
1 2
Q3 Determine the height of a geostationary satellite
r­31 r­32 above the surface of the Earth.

T 2 = TE2         [Mass of the Earth, M = 5.98 × 1024 kg, G = 6.67
S (rE)­ 3 × 10–11 N m2 kg–2, radius of Earth, R = 6.37 × 106
m and period, T = 24 hours]
(9rE­)3

T 2S = (9rE­)3 TE2
r­3E

Therefore, TS = 27 Earth-years.

82

Physics SPM  Chapter 3 Gravitation  

3.3 Man-made Satellites 8. In Figure 3.24, for a satellite of mass m moving
with speed v in a circular orbit of radius r
1. A man-made satellites is a satellite created and (measured from centre of the Earth), we have
built by humans. It is launched into space using
rockets and is placed in specific orbits of Earth Centripetal force = Gravitational force
to collect information or communications.
mv2 = G Mm
2. The moon is a natural satellite that orbits the r r2
Earth. Most man-made satellites orbit the
Earth, but some orbit other planets, such as where M is the mass of the Earth.
Saturn, Venus or Mars.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. Then, the orbital velocity of the satellite is:
3. A communication satellite is a man-made
satellite in space used for receiving and sending Form
telecommunication, radio and television where,
signals. M is the mass of Earth, and, 4
r = R + h (R = Earth’s radius and h = height
4. The world’s first man-made satellite, Sputnik I
was launched on 4 October 1957 by the Soviet of satellite from Earth)
Union.
9. The equation shows that satellites that orbit at
5. When a satellite is launched by rocket, it is different altitudes have different orbital speeds.
placed in orbit around the Earth. The satellite Satellites that are further away actually travel
must be raised to a desired height and given slower.
the correct speed and the direction by the
launching rocket. EXAMPLE 3.8

6. The only force that acts on the satellite when The International Space Station (ISS) has a Low
it is in orbit is the gravitational force exerted Earth Orbit (LEO), about 400 km above the Earth’s
on the satellite by the Earth. This force is surface. Calculate its orbital speed v.
constantly pulling the satellite towards the [Mass of Earth, M = 5.98 × 1024 kg, G = 6.67 ×
centre of the Earth. 10–11 N m2 kg-2, Earth’s radius, R = 6.37 × 106 m
and height of satellite, h = 400 km]
7. Satellites in orbit will not fall straight to
the Earth because of its velocity which is Solution
always perpendicular to the force of gravity.
Therefore, there is a perfect balance between v = GrM = RG+Mh
the gravitational force of the Earth and the  =
centripetal force necessary to keep the satellite 6.67 × 10–11 × 5.98 × 1024
in the orbit. 6.37 × 106 + 400 000

= 7 676 m s–1 = 7.68 km s–1

Satellite EXAMPLE 3.9
m
MEASAT-3 is a Malaysian communications satellite
h launched on 11 December 2006 with an orbital
v height of 3.58 × 104 km. Calculate its orbital speed.

rR Earth Solution
O
M  v =
 =
GM

R+h

6.67 × 10–11 × 5.98 × 1024
6.37 × 106 + 3.58 × 107

Figure 3.24 = 3 075 m s–1 = 3.08 km s–1

83

Form   Physics SPM  Chapter 3 Gravitation Penerbitan Pelangi Sdn Bhd. All Rights Reserved. 4. Since a geostationary satellite has the same
orbital period as Earth, and it also travels
v = Orbital speed from west to east (the direction in which
vA < v A B vB < v Earth rotates on its axis), it therefore appears
to hover at a single point in the sky when
C vC = v observed from a given point on the ground. A
Figure 3.25 satellite in geosynchronous orbit has the same
orbital period, i.e., 24 hours or one day, as that
10. Refer to Figure 3.25, a satellite will orbit of a satellite in a geostationary orbit.
around the Earth along the trajectory C with
5. The only difference between a geosynchronous
4 orbital speed. satellite and a geostationary satellite is that a
11. If the speed of the satellite is lower than the geosynchronous satellite may or may not be
orbital speed, this will cause the satellite to fall following an inclined orbit (with respect to
back to Earth, for example, trajectory A and B. the equatorial plane), while a geostationary
This is because the centripetal force is smaller satellite must follow a non-inclined orbit.
than the gravitational force.
6. In other words, a geostationary satellite always
Geostationary and Non-geostationary remains exactly above the Earth’s equator.
Satellites
1. A geostationary satellite is an earth-orbiting 7. Geosynchronous satellites are typically used
for various purposes, such as communicating
satellite. It is a type of geosynchronous back and forth with spacecraft (like the Hubble
satellite. Space Telescope and space shuttles), voice
2. A satellite in geostationary orbit is known as a communication, Internet, broadcasting cable
geostationary satellite. It has an orbital period TV and radio signals, as well as weather
of 24 hours or one day, which means that it forecasting.
completes one revolution around Earth in the
same time as Earth completes one rotation on 8. Geostationary satellites can give detailed
its axis. terrestrial and weather-related information
3. A geostationary orbit (also known as a about a geographical region, making them an
Geostationary Earth Orbit, or Geosynchronous ideal choice to predict climate trends in that
Equatorial Orbit, or simply GEO) is a circular region.
orbit located at an altitude of 35 786 km above
the surface of Earth with zero inclination to the 9. Table 3.4 shows some examples of
equatorial plane. communication satellites launched by Malaysia.

Axis of rotation Table 3.4

Geosynchronous orbit Satellites Launched Mission
date
Geostationary orbit
Measat 1 12 Jan 1996 Communications
Figure 3.26
Measat 2 13 Nov 1996 Communications

Tiungsat-1 26 Sep 2000 Earth imaging

Measat-3 11 Dis 2006 Communications

Measat-3a 21 Jun 2009 Communications

Razaksat 14 Jul 2009 Earth Remote
Sensing

10. Satellites that are not in geostationary orbit are
known as non-geostationary satellites.

84

Physics SPM  Chapter 3 Gravitation  

11. Nongeostationary Satellite Orbit (NGSO) satellites almost always orbit closer to the Earth than
Geostationary Satellite Orbit (GSO) satellites.

12. NGSO satellites orbit with varying velocities dependening on the orbit’s height.

13. Table 3.5 shows the types of satellite orbit.

Table 3.5

Type of satellite orbit Height Period
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
LEO – Low Earth Orbit 500 – 1 000 km 1.6 to 1.8 hours

NGSO MEO – Medium Earth Orbit 8 000 – 12 000 km Approximate to 6 hours
(at 10 000 km)
Form
HEO – High Earth and Geosynchronous Orbit ≥ 36 000 km

GSO GEO – Geostationary Earth Orbit Approximate to 36 000 km Approximate to 24 hours

4
5. Gravitational potential energy of a body with

Axis of the Earth’s Geostationary mass m on the surface of the Earth is given by
earth’s rotation orbit
Low Earth Orbit U = –G Mm
R

and kinetic energy is given by

Medium Earth Orbit K = 1 mv2
2

High Earth Orbit From equation,

Gravitational Minimum

potential + kinetic energy, = 0

Figure 3.27 energy, U KE

14. The average life span of a LEO satellite is –G Mm + 1 mv2 = 0
approximately 5 years, but the average life R 2

Therefore, escape velocity,

span for a GEO satellite is approximately 8

years.



Escape Velocity where,

1. The escape velocity is the minimum velocity G = the gravitational constant
an object must have in order to overcome the M = the mass of Earth
gravitational pull of the Earth and escape into
r = distance from centre of Earth

outer space without ever falling back. EXAMPLE 3.10

2. An object which has this velocity at the surface Given that the mass of Earth, M = 5.97 × 1024 kg,
of the Earth, will totally escape the Earth’s universal gravitational constant, G = 6.67 × 10–11
gravitational field, ignoring the losses due to N m2 kg–2 and the Earth’s radius, r = 6.37 × 106 m.
the atmosphere. Calculate the escape velocity of a body from the
surface of the Earth.
3. The escape velocity from Earth is about
11.2 km s–1 at the surface.  Solution

4. The escape velocity will be achieved when The escape velocity
the minimum kinetic energy supplied to the  v =
object has overcome the gravitational potential 2 × 6.67 × 10–11 × 5.97 × 1024
energy. 6.37 × 106

= 11 181 m s–1 = 11.2 km s–1

85

  Physics SPM  Chapter 3 Gravitation 8. Aeroplanes can never escape from Earth
because:
6. Earth’s atmosphere is a thin layer of air (a) The air is too thin above that altitude to
surrounding our planet. How does Earth hold hold the plane up.
on to this thin layer of atmosphere? The (b) The speed of aeroplanes can never reach
answer is the force of gravity, the same force the escape velocity.
that keeps us on Earth.
Solving Problems involving Escape
7. The atmosphere is made up of air particles or Velocity of Rockets from the Surface of
molecules. The molecules in the atmosphere the Earth, Moon, Mars and the Sun
are constantly moving at high velocity due
to the hot sunlight. Some light molecules, 1. The escape velocity formula can be used to
like hydrogen and helium, move faster than calculate the escape velocity for each planet
heavier ones, like oxygen and nitrogen. The in the Solar System.
light molecules are more likely to reach escape
velocity and escape to space. Most of the 2. The following table shows the escape velocity
for the Moon, the Sun, and the planets in the
4 heavier ones will remain in the atmosphere Solar System.
due to gravity.
Form Penerbitan Pelangi Sdn Bhd. All Rights Reserved.

Planet Mass, M Radius, R Acceleration due to Escape velocity, v
graviti, g
Mercury 3.29 × 1023 kg 2.44 × 106 m 3.69 m s–2 4.24 km s–1
Venus 4.87 × 1024 kg 6.05 × 106 m 8.87 m s–2 10.36 km s–1
Earth 5.97 × 1024 kg 6.37 × 106 m 9.81 m s–2 11.18 km s–1
Moon 7.35 ×1022 kg 1.74 × 106 m 1.62 m s–2 2.37 km s–1
Mars 6.39 × 1023 kg 3.39 × 106 m 3.71 m s–2 5.01 km s–1
Jupiter 1.90 ×1027 kg 6.99 × 107 m 25.94 m s–2 60.22 km s–1
Saturn 5.68 × 1026 kg 5.82 × 107 m 11.18 m s–2 36.08 km s–1
Uranus 8.68 × 1025 kg 2.54 × 107 m 8.97 m s–2 21.35 km s–1
Neptune 1.02 × 1026 kg 2.46 × 107 m 11.24 m s–2 23.52 km s–1
1.99 × 1030 kg 6.96 × 108 m 274 m s–2 617.59 km s–1
Sun

3. The formula for gravitational acceleration on Checkpoint 3.3

the surface of a planet is given as Q1 What are man-made satellites and natural
satellites? Give one example in each case.
g = GM or gR2 = GM
R2 Q2 A satellite moves with velocity v in its own orbit.
The height of the satellite from Earth’s surface is
The formula for escape velocity on the surface 500 km. What is the orbital velocity of the satellite
v?
of a planet is given as
v = 2GRM [G = 6.67 × 10–11 N m2 kg–2, mass of Earth, M =
5.97 × 1024 kg, Earth’s radius, R = 6.37 × 106 m]

The formula for gravitational acceleration and Q3 The escape velocity of planet Y is 1791 m s_1.
Calculate the escape velocity of planet X if the
escape velocity can be combined as mass of planet X is 1.41 times the mass of planet
v = Y, and its radius is 0.919 times the radius of
2gR2 = 2gR planet Y.
R

Therefore, escape velocity from the surface

of a planet can be calculated from the value

of gravitational acceleration and radius of the

planet.

86

CONCEPTPMenAeP GRAVITATION
rbitan Kepler’s Law
PelanF Man-made Satellite

G m1m2 Solving problems for: Kepler’s first law
r2 (i) two bodies at rest
Sun
on Earth
(ii) a body above the

surface of the Earth
(iii) Earth and satellite
(iv) The Earth and the

Sun
gF = gravitational force (N)=
im1, m2 = mass of respective object
S(kg)
d r = distance between two objects
Satellite orbital Geostationary and
not geostationary
velocity and its effect satellites.

on satellite motion

n(m) Focus Ellipse v= GM
r
G = universal gavitational constant
Bhd. Relationship of gravitational
Kepler’s second law Conceptualising Solving problems
involving the escape
Determine the mass of S2 escape velocity velocity of rockets
S1 on Earth, the Moon,
A acceleration, g on Earth with 2GM Mars and the Sun
the Earth and the Sun v= r

luniversal gravitational constant, G.
using the formula
l Rg = gR2
GM M = G
r2
ights ReThe importance of knowing theor Same area at same time
sgravitational acceleration of planets interval
erved.in the Solar SystemM=4p2R2
G T2 Area S1 = area S2

Physics SPM  Chapter 3 Gravitation  Centripetal force inKepler’s third lawSolving

4satellite and planetaryrFormulatingproblems using
Kepler’s third
87 motion systems T : period law Kepler’s third
v2 r : orbit radius
F=m r T2 T2 ∝ r3­ law

T21 = T22
r21 r22

r3 = constant

Form

  Physics SPM  Chapter 3 Gravitation

SPM Practice 3

Objective Questions 5. Which of the following best By comparing the magnitude
of the gravitational force
1. The graph below shows the describes the force of gravity, between satellite A and the
relationship between the Fg, acting between satellites planet with the magnitude
force of gravity and mass of S and the Earth? of the gravitational force
an object near the surface of between satellite B and the
the Earth. A S C planet, which statement is
correct?
Fg A The gravitational force
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. S between satellite A and
Fg the planet is half as large.
B The gravitational force
Form Force Earth Fg between satellite A and
of Fg Earth the planet is twice as
gravity large.
C The gravitational force
4 B S Fg D S Fg between satellite A and
0 Mass the planet is one quarter
as large.
The gradient of the graph Earth Fg D The gravitational force
represents Fg Earth between satellite A and
the planet is four times as
A the acceleration due to 6. What is the gravitational larger.
gravity force between two planets if
their mass is 1.23 × 1026 kg 9. An object of mass 2.00 kg
B the universal gravitational and 5.21 × 1022 kg, and the weighs 19.6 N on Earth.
constant distance between them is If the acceleration due to
2.30 × 1011 m? gravity on Mars is 3.71 m s–2,
C the weight of the object what is the mass of the
D the momentum of the [G = 6.67 × 10-11 N m2 kg–2] object on Mars?
A 1.83 × 1015 N A 2.64 kg
object B 1.83 × 1027 N B 2.00 kg
C 8.08 × 1015 N C 7.42 kg
2. How does the gravitational D 8.08 × 1027 N D 19.6 kg
force, F between two planets
7. An astronaut weighs 800 N 10. The orbit of the Hubble
change with the distance on the surface of the Earth. telescope is 5.6 × 105 m
apart, r? What is the weight of the above the surface of the
A F ∝ 1r C F ∝ r astronaut at a height of R m Earth. The telescope has a
1 mass of 1.1 × 104 kg. The
B F ∝ r2 D F ∝ r2 above the surface of the earth exerts a gravitational
Earth? The radius of the force of 9.1 × 104 N on
3. The Earth attracts the Earth is R m. the telescope. What is
Moon with a gravitational A 0.00 N C 1 600 N the strength of the Earth’s
force of 1020 N. What is the B 200 N D 3 200 N gravitational field at the the
gravitational force of the position of Hubble telescope?
Moon attracting the Earth? 8. The diagram shows two A 1.5 × 10–2 N/kg
A 10–20 N C 1010 N satellites A and B, equal in B 0.12 N/kg
B 102 N D 1020 N mass and orbiting around a C 8.3 N/kg
planet. D 9.8 N/kg
4. Two bodies are at a certain
distance apart. If the mass 2R A
of the two bodies is doubled, R
and the distance between
them does not change, how Planet
much does the gravitational
force between them change? B
A ¼ times
B ½ times
C 4 times
D No change

88

Physics SPM  Chapter 3 Gravitation  

11. A rocket is launched into When the rock reaches O 18. Which of the following is
space from the Earth. Which and released, which way will
of the following graphs the stone keep moving? the formula for the escape
represents the force of A OP
gravity F acting on the rocket B OQ velocity of an object from the
as the distance r from the C OR
Earth’s surface increases? D OS surface of the Earth?
A F 2GM
A v = M

14. A car is turning a bend at B v = GM
R
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. C v = 2GMR
a speed of 20 km h–1. The
2GM
centripetal force experienced D v = R

r by the car is 100 N. If the car

is now turning the same bend Form

B F at 40 km h–1, what is the new

centripetal force acting on it? 19. The diagram shows a rocket
launched from O at an
A 50 N C 200 N altitude from the surface of 4
the Earth with a horizontal
B 100 N D 400 N velocity u. The rocket can
only move around the Earth
r 15. An object of mass 500 g in a circular orbit.
is fastened to a thread of
C F
length 100 cm and rotated
r Rocket S R
in a horizontal circle with a u Q
uniform velocity of 5.0 m s–1.
O
What is the tension of the P

D F thread?

A 2.5 N Earth
B 12.5 N
C 15.0 N
D 125 N

r 16. Which of the following
statements is true about
geostationary satellites? Circular orbit

12. A satellite is placed in a fixed I It orbits the Earth in If the rocket launches with a
orbit around the Earth. The the direction follow 3
velocity 4 u, which trajectory,
satellite can remain in its the direction of Earth’s
P, Q, R or S, the rocket will
orbit without escape to outer rotation.
move? C R
space because II Its orbital period is equal A P D S
to the period of Earth’s B Q
A it is experiencing
weightlessness. rotation. 20. A geostationary satellite

B it is moving in a vacuum. III Its orbit is concentric to orbits the Earth in a circular
C it is always moving at the equator.
path approximately 36 000 km.
high speed. A I and II only
B II and III only
D it is always pulled by the C I, II and III
Earth’s gravitational force.

13. A rock is fastened to one 17. A satellite is always visible The radius of the Earth is
end of a rope and rotated to on the surface of the Earth
form a circle as shown in the approximately 6400 km and
diagram. at the same place throughout
the period of rotation of the
the year. How much time
Earth is 24 hours. What is
does it take for the satellite to
the period of a spy satellite
move in orbit?
Q R orbiting the Earth at a height
P S A 1 hour
B 12 hours of 400 km above the Earth’s
C 24 hours
O D 365 days surface?

Stone A 1 hour C 2 hours
2

B 1 hour D 4 hours

89

  Physics SPM  Chapter 3 Gravitation

21. The diagram shows two 23. The diagram shows a narrow 24. A satellite is in an orbit at a
satellites, P and Q orbiting tube T in which a thread height h above the surface
around the Earth. passes through, the mass of a planet of mass M with
m and M is fastened at both radius R. What is the velocity
2R P ends. Initially the mass m
R is rotated so that it moves of the satellite in its orbit?
in a horizontal circle with a GM(R + h)
Earth constant radius r, at uniform A R
Q velocity, v.
B GM(T + h)
m R
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. GM
C R+h

Form T D GM
R+h
The orbital velocity of the
satellite P is v. What is the 25. The Moon located at a
distance R from the centre
4 orbital velocity of the satellite
Q in terms of v? M of the Earth has a period of
1 C 2 v
A 2 v 27 days in orbit around the

1 Which of the following Earth. What is the period of
B v 2 v expressions is equivalent to 1
D a satellite with radius 2 R

22. If vL, is the escape velocity Am asms2vMr2? orbit round the Earth?
and vO is the orbital velocity
A 3.4 days

of a satellite close to the B mv2rg B 7 days

surface of the Earth, then the mv2g C 21 days
relationship between vL and mvr2
vo C rg D 108 days
A D
is = 2vO
vL

B vL = 2vO

C vL = vO

 D vL = 1 vO
2

Subjective Questions

Section A

1. (a) Complete the following sentence by ticking [✓] the box for the correct definition. [2 marks]
Newton’s law of universal gravitation states that:

[ ] the force is directly proportional to the product of their masses.

[ ] The force is directly proportional to the square of the distance between two bodies.

[ ] force is directly proportional to the inverse square of the distance between two bodies.

(b) Two satellites P and Q orbiting round the Earth in the same orbit. The mass of satellite P is twice the
mass of satellite Q. Compare

(i) the linear speed of satellite P and satellite Q round the Earth. [1 mark]
(ii) the centripetal force acting on the satellite P and satellite Q. [1 mark]

90

Physics SPM  Chapter 3 Gravitation  

2. In a laboratory experiment, a rubber stopper is fastened to one end of a thread. The other end of the thread
is inserted through a glass tubing and a load is then fastened to the end. The rubber stopper is whirled in a

horizontal circle as shown in Figure 2.

1.0 m Rubber stopper
T 100 g

Glass
tubing

Penerbitan Pelangi Sdn Bhd. All Rights Reserved. Load 200 g Form
Figure 2

(a) The speed of the rubber stopper is adjusted so that it moves in a horizontal circle of radius of 1.0 m and

is balanced by a load of mass 200 g. [1 mark] 4
(i) What is the tension T experienced by the thread? [1 mark]
(ii) What is the centripetal force acting on the rubber stopper?

(b) If the mass of the rubber stopper is 100 g, determine the speed of the rubber stopper. [2 marks]

(c) If the speed of the rubber stopper is made constant, what will happen to the radius of the circular orbit
[1 mark]
if the mass of the load is increased?

(d) While the stopper is moving in a constant horizontal orbit, though there is a force acting on it, but no
[2 marks]
work is done. Explain why?

3. Figure 3(i) shows a rocket of mass m on the surface of the Earth. Figure 3(ii) shows the rocket moving in its
orbit at a height h above the Earth’s surface. The mass of the Earth is M and the radius of the Earth is R.

m

h

m R

R Earth O
M

Earth O
M

Figure 3(i)  Figure 3(ii)

(a) (i) On the Earth’s surface, the strength of the gravitational field is g N kg–1. Write an expression to show
the gravitational force, F acting on the rocket. [1 mark]
(ii) From Newton’s Law of Universal Gravitation, write an expression to show the gravitational force, F,
[1 mark]
acting between the Earth and the rocket on the surface of the Earth.
g,
(iii) Then, show t=heGRrMe2 la. tionship between the force of gravity, and the universal gravitational constant,
G, that is, g [1 mark]

(b) (i) Write an expression to show the force of gravity, g, at the height h above the surface Earth. [1 mark]
(ii) Then, write an expression to show the relationship between g and gh. [1 mark]

(iii) If the weight of an astronaut on the surface of the Earth is 810 N, what is the weight of the astronaut

in a spacecraft located at an altitude of 12 800 km above the surface of the Earth? [3 marks]
[Earth’s radius = 6 400 km]

91

  Physics SPM  Chapter 3 Gravitation

Section C

4. (a) To broadcast live telecast on sports round the world, a network of communications satellites is needed
to cover the entire Earth. For this purpose, geostationary satellites need to be placed above the Earth in
certain fixed places.

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Form

4


(i) What is a geostationary satellite? [2 marks]

(ii) What is the minimum number of geostationary satellites required for worldwide live TV
broadcasts? [1 mark]

(iii) In the space given, draw the position of the geostationary satellites to show how this worldwide
[2 marks]
communications network can be made.

(b) The table below shows the orbital radius and the orbital period of the planets around the Sun.

Planet Average orbital radius of Period (T) / day R3 / km3 T2 / day2
the planet (R) / km

Mercury 5.8 × 107 88 1.95 × 1023 7744

Saturn 1.08 × 108 225 1.26 × 1024 50625

Mars 2.28 × 108 687 1.26 × 1024 471969

(i) Work odthuoetytKoheuepnvleaorlt’uisceeLaaRTwb23otuhfotartthtcheoenvaftihrlumreeseotpfhleaRTne23extpfMoreresarscliluoprnylaanRTne23dts. ?M ars. [2[[11mmmaaarkrrkks]]]
(ii) What
(iii) State

(iv) Can the Kepler Law’s mentioned in b(ii) apply to man-made satellites orbiting the Earth? [1 mark]

(c) Sir Isaac Newton in one of his paper recorded that he observed that an apple fall from its tree to the
Earth and developed the idea that the same force that caused the fall of the apple keep the Moon in its
orbit. This means that the Moon is also falling towards the Earth.

Give explanation of the phenomena mentioned in the three statements in italics. [10 marks]
Why did the Moon cannot reach the Earth?

92

5Chapter Form 5

Electronics

CHAPTER FOCUS
5.1 Electron
5.2 Semiconductor Diode
5.3 Transistor
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Why are electrons important in the electronics field?
What is the function of a semiconductor diode and its uses in electronic circuits?
Why are transistors important in the electronics industry?

397

  Physics SPM  Chapter 5 Electronics Some electrons are free from the surface

5.1 Electron

1. An atom is made up of protons, neutrons, and
electrons. From these three particles, electron
has the smallest mass.

Electron
(<10–16 cm)
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Atom Nucleus Proton Heat is supplied
(~10– 8 cm) (~10– 12 cm) +
Figure 5.3
Figure 5.1 Neutron
(~10– 13 cm) 4. If the metal is heated to a high temperature as
shown in Figure 5.3, some of the free electrons
2. The electron is a stable subatomic particle with may gain sufficient energy to escape from the
a negative electric charge. metal surface.

3. Each electron carries one unit negative charge 5. The electrons are more easily emitted if the
(1.6 × 10–19 C) and has a very small mass metal is coated with an oxide of barium or
strontium.
compared to neutron or proton.
6. The electrons that escaped from the metal
4. The mass of an electron is 9.1 × 10–31 kg. This surface are known as thermion.
1
is about 1836 of the mass of a proton. 7. The process of emission of electrons from
the surface of a heated metal is known as
Thermionic EmTheirsmsioioninc emission thermionic emission.

Metal surface Electrons

Thermionic
emission

VIDEO

Form 8. The factors affecting the rate of thermionic
emission are:
5 Figure 5.2 • surface area of the metal
• temperature of the metal
1. Figure 5.2 shows the surface of a metal • type of the metal
containing many electrons move freely at • nature of the metal surface
random.
Cathode Rays
2. The electrons are free to move at the surface
but cannot escape from the metal surface. 1. Cathode rays are streams of electrons moving
with high speed in a vacuum tube.
3. The electrons at the surface cannot escape
because they are held back by the attractive 2. A vacuum tube is an evacuated glass tube
forces of the atomic nuclei near the surface. where streams of electrons can be moved
without any obstruction by air molecules.

398

Physics SPM  Chapter 5 Electronics  

3. The production of cathode rays in a vacuum tube using extra high tension (E.H.T.) is shown in Figure
5.4.

Thermionic emission from Vaccum tube allows the
a hot cathode produces a streams of electrons to
continuous supply of move without obstruction
electrons. by air molecules.

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Cathode Cathode rays Anode

E.H.T. producing a high potential To vaccum pump
difference between cathode and
anode allows the streams of –+
electrons to move at high speed
in the vaccum tube. Extra High Tension
(E.H.T)

Figure 5.4

4. Inside the vacuum tube, there is one positive electrode called anode and one negative electrode called
cathode.

5. When a high voltage is applied between SPM Highlights
the cathode and the anode, the electrons are
accelerated at high speed from the cathode to Figure 5.5 shows a cathode ray tube.
the anode. The stream of electrons that moves
at high speed is called cathode rays. Cathode (–) Anode (+)

AC ~ Cathode rays
supply
Cathode Heating filament
rays –+ Vaccum
tube
High voltage supply
VIDEO
Figure 5.5
Form

When the heating filament heats the cathode, which 5

particle is emitted from the cathode?

A Proton C Electron

B Neutron D Ion helium

Examiner’s Tips

During thermionic emission, electrons are emitted
from the metal surface when the metal is heated to
a high temperature.

Answer: C

399

  Physics SPM  Chapter 5 Electronics

Effects of Electric Field and Magnetic Field on Cathode Rays

ACTIVITY 5.1

Aim: (I) To observe the effect of electric field on a cathode rays using a deflection tube.
(II) To observe the effect of magnetic field on cathode rays using a Maltese Cross Tube.

(I) To observe the effect of electric field on cathode rays using a deflection tube.
Apparatus and materials: Deflection tube, E.H.T. supply, 6 V AC supply, connecting wires
Instructions:
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Figure 5.6 Deflection tube

1. The deflection cathode ray tube as shown in Figure 5.6 is set up and teacher will do the
demonstration.

2. The 6 V AC supply is turned on to heat the heating filament to produce electrons. The high voltage
supply (E.H.T.) is switched on and the voltage is gradually increased to 1000 V. The cathode rays
that pass through the deflection plate are observed.

Deflection plate Positive plate E.H.T.
+–

Form Vaccum tube +
Electron beam
– 6V
5 3A
Negative plate

Figure 5.7 Electron
deflection
3. Watch the video on the effect of electric field on cathode rays.

VIDEO

400

Physics SPM  Chapter 5 Electronics  

Observation:

1. The cathode ray is observed to deflect toward the positive plate and away from the negative plate
as shown in Figure 5.7.

2. The path of electron beam (cathode ray) in the electric field is in the form of a parabola as shown
in Figure 5.8.

Straight line +
path
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Parabola
path

Cathode ray

–

Electric field

Figure 5.8

Discussion:
1. The cathode ray is moving in straight line and is negatively charged.
2. The cathode ray can be deflected by electric field.

(II) To observe the effect of magnetic field on cathode rays using a Maltese cross tube.
Apparatus and materials: Maltese cross tube, E.H.T. supply, 6 V AC supply, connecting wires, a strong

magnet
Instructions:

+ E.H.T.
0 – 5 kV

+ Maltese ~
cross
– 6V

3A
a.u.

+–

Image of
Maltese cross

Anode Heating filament Form
and cathode
Fluorescent Vacuum
      screen

Maltese Cross Tube 5

Figure 5.9

1. The Maltese cross tube as shown in Figure 5.9 is set up and teacher will do the demonstration.

2. The 6 V AC supply is switched on and the dark shadow of the Maltese cross formed on the screen
is observed.

3. The E.H.T. supply is switched on and the voltage is increased slowly from zero to above 2 kV.
The shadow formed and the colour on the screen is observed.

4. The north pole of a strong magnet is directed to the side of the Maltese cross tube as shown in
Figure 5.10. What happened to the image of Maltese cross formed on the screen is observed.

401

  Physics SPM  Chapter 5 Electronics

Strong Maltese cross
magnet
VIDEO
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Image of the
maltses cross

Figure 5.10

5. Watch the video on the effect of magnetic field on cathode rays.

Observation:

1. When the 6 V AC power supply is turned on, a dark shadow of the Maltese cross is formed on
the screen.

2. When the E.H.T. power supply is switch on, a darker shadow of the Maltese cross is seen on the
screen. The shadow of the Maltese cross is surrounded by a green light.

3. When the north pole of a strong magnet is placed at one side of the Maltese cross tube, two
shadows are seen on the screen. The lighter shadow remains at the centre of the screen, the darker
one moved downward, and the shape is distorted slightly.

Discussion

1. The dark shadow of the Maltese cross is due to the light from the filament when it is turned on.

2. When the E.H.T. power supply is turned on, a darker shadow is formed on the screen. This is due
to the cathode ray flow being obstructed by the Maltese Cross. The unobstructed flow of electrons
continues to move and hit the fluorescent screen producing a green light around the Maltese
shadow.

3. When the electrons hit the fluorescent screen, the kinetic energy of the electron is converted into
light energy.

4. When the north pole of a strong magnet is placed at the side of the Maltese cross, the shadow
cast by the cathode rays is moved and distorted. The direction of deflection of the shadow can be
determined by Fleming's left-hand rule. The shadow cast by light which is not affected by magnet
will stay at the centre of the screen.

Form
SPM Highlights
Which of the following rule can be used to determine the
5 Figure 5.11 shows a shadow formed on the screen of direction of deflection of the shadow?
a Maltese cross tube. The shadow is deflected by the A The right-hand grip rule
magnet. B The Fleming’s right-hand rule
C The Fleming’s left-hand rule
S Magnet D The cork’s screw rule

6.3 V a.u. Examiner’s Tips
The deflection of moving charges in magnetic field is
Shadow on determined by Fleming’s left-hand rule.
screen
Answer: C
–+ U Magnet
VLT

Figure 5.11

402

Physics SPM  Chapter 5 Electronics  

SPM Highlights The characteristics of cathode rays are
summarised as follows:
The circuit of a Maltese cross tube is connected as
shown in Figure 5.12. • Cathode ray is a negatively charged electron
beam.
6 V AC Heating filament Vaccum
supply (Cathode) Anode • Cathode ray can be deflected by electric field
Maltese and magnetic field.
S1 cross
shadow • Cathode ray moves in a straight line and form
Cathode ray a sharp shadow.
S2 V.L.T Fluoresent
screen • Cathode ray can move at high speed and has
kinetic energy and momentum.

• Cathode ray can do work and therefore can
produce fluorescent when the electron beam
hits the fluorescent screen.

• Cathode ray can be stopped by a thin sheet of
metal.
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.Figure 5.12

What will happen when switch S1 is opened? To Determine the Velocity of an Electron
A Cathode rays move with lower speed. in Cathode Ray Tube
B Cathode rays move with higher speed.
C Shadow of Maltese cross becomes brighter. 1. In cathode ray tube, the electron gun (or
D Nothing can be seen on the screen. electron emitter) is composed of a heating
filament, a cathode and an accelerating anode
Examiner’s Tips as shown in Figure 5.14.

When switch S1 is opened, no current passed
through the filament. Therefore, no light and
electrons are produced and thus nothing can be
seen on the screen.

Answer: D

SPM Highlights Heating Cathode Accelerating anode
filament
Figure 5.13 shows a Maltese cross tube.
Heater Voltage Electron beam
Cathode rays (0.6, 3 V)

Maltese –+ Cathode ray tube
cross (Evacuated glass tube)
E.H.T.
Shadow Acceleration Voltage V

(100 V – 5000 V)

Figure 5.14 Cathode ray tube

Vaccum 2. Recall that the potential difference, V is defined Form
as the work done, E per unit of charge, q, that
Figure 5.13 is,

Which of the following explained the shadow formed on 5
the fluorescent screen?
A Cathode rays move from cathode to anode. Therefore, the work done of moving an electron
B Cathode rays can be blocked by a thin sheet of or electric potential energy is given as,

metal.
C Cathode rays move with very high speed.
D Cathode rays are positively charged particles. where e is the charge of one electron (1.6 × 10-19 C)
and V is the potential difference between the
Examiner’s Tips cathode and the anode.

Cathode rays cannot pass through Maltese
cross, part of the rays that are not blocked hit the
fluorescent screen to produce green light.

Answer: B

403

  Physics SPM  Chapter 5 Electronics

3. Based on the principle of conservation of Checkpoint! 5.1
energy, the electrical potential energy is
converted into kinetic energy. The kinetic Q1 (a) Explain thermionic emission and cathode ray.
energy of electron is given as, (b) State three characteristics of cathode ray.

KEe = 1 mev2 Q2 Figure 5.15 shows a cathode ray beam moving
2 into an electric field produced by two metal plates
connected to a E.H.T. supply.

where me is the mass of an electronPenerbitan Pelangi Sdn Bhd. All Rights Reserved.Cathode ray+
(9.1 × 10–31 kg) and v is the velocity of an V.L.T
electron. –

4. Therefore, the velocity of the electron in the
cathode ray tube can be calculated using the
following formula:

Electrical potential energy,

E = Kinetic Energy, KEe Figure 5.15
1
2 mev2 (a) On Figure 5.15, complete the path of cathode
 eV = ray through the electric field.
v = 2eV
me (b) Give one reason for your answer.

where e = the charge of one electron Q3 An electron is emitted from cathode and
accelerated through a potential difference of 4 kV
(1.6 × 10–19 C) in the cathode ray tube.
(a) How much energy acquired by the electron?
me = the mass of an electron (9.1 × 10–31 kg) (b) What is the maximum velocity of electron?
V = the potential difference between the

cathode and the anode

v = the velocity of an electron

EXAMPLE 5.1 5.2 Semiconductor Diode

A cathode ray tube is supplied with a high voltage 1. Figure 5.16 shows different types of diodes in
of 5 000 V. Calculate the velocity of the electron the market.
emitted from the electron gun inside the cathode
ray tube. Various
Given the mass of the electron, me = 9.1 × 10−31 kg types
and the charge of one electron, e = 1.6×10−19 C. of diode

Form Solution Figure 5.16

Given 2. Diodes are devices made of semiconductor
V = 5 000 V, components, usually silicon.

5 me = 9.1 × 10−31 kg,
e = 1.6 × 10−19 C

Using, eV = 1 mev2
2

Rearrange,
 v =
2eV = 2 × 1.6 × 10–19 × 5000
me 9.1 × 10–31

= 4.2 × 107 m s–1

Therefore, the velocity of the electron

= 4.2 × 107 m s–1

404

3. When silicon is doped with trivalent atom such Physics SPM  Chapter 5 Electronics  
as boron, a p-type semiconductor is produced
as shown in Figure 5.17. 6. The cathode terminal of diode is negative and
from n-type semiconductor which carry excess
p-type semiconductor electrons.

Acceptor of impurity 7. The anode terminal of diode is positive and
Si creates a hole from p-type semiconductor which carry excess
holes.

SPM Highlights

What happens to a pure semiconductor when it is
doped with pentavalent element?
A Semiconductor is negatively charged.
B Semiconductor becomes p-type.
C The resistance of the semiconductor increases.
D The free electrons are the majority carriers.

Examiner’s Tips
The pentavalent element has 5 valence electrons.
Doping of semiconductor with pentavalent element
will result in n-type semiconductor, where the
majority carriers are electrons.

Answer: D
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Boron is added Si
as impurity

Figure 5.17

In p-type semiconductor, the majority carriers
are holes (positive charges).

4. When silicon is doped with pentavalent atom
such as arsenic, a n-type semiconductor is
produced as shown in Figure 5.18.

n-type semiconductor

Donor of impurity
contributes free

Si electrons

Si As Si SPM Highlights

Arsenic is added Si Which statement is correct about p-type semiconductor?
as impurity A Doped with pentavalent element.
B Doped with trivalent element.
Figure 5.18 C More free electrons than holes. Form
D The free electrons are the majority charge carrier.
In n-type semiconductor, the majority carriers 5
are electrons (negative charges). Examiner’s Tips
The p-type semiconductor has an excess hole
5. The combination of p-type semiconductor and (positive charge) results from the doping of trivalent
n-type semiconductor forms a two terminal atoms into a pure semiconductor.
diode as shown in Figure 5.19.
Answer: B

p-type n-type
semicondutor semicondutor

Anode P Cathode
N

Figure 5.19 Semiconductor diode or p-n junction

405

  Physics SPM  Chapter 5 Electronics

The Function of a Semiconductor Diode

ACTIVITY 5.2

Aim: To observe the effect on diode in (i) forward biased circuit, and (ii) in reverse biased circuit.
Apparatus and materials: Dry cell, diode, switch, bulb and connecting wires
Instructions:
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+– Bulb –+ Bulb
Diode Diode

Switch Switch

Dry cell Dry cell

+– +–

(i) Diode in forward biased circuit (ii) Diode in reverse biased circuit

Figure 5.20

1. For diode connected in forward biased circuit, the anode is connected to positive terminal and
cathode is connected to the negative terminal of dry cell as shown in Figure 5.20(i). The state of
the bulb in the circuit is observed.

2. For diode connected in reverse biased circuit, the anode is connected to negative terminal and
cathode is connected to the positive terminal of dry cell as shown in Figure 5.20(ii). The state of
the bulb in the circuit is observed.

Observation:
1. The bulb is lit in a circuit where the diode is forward biased.
2. When the diode is reverse biased, the bulb is not lit.

Discussion:

1. The current flows in the circuit when the diode is forward biased and no current flows in the circuit
when the diode is reverse biased.

2. The function of the diode is to allow the current to flow in one direction only.

Draw a Simple Circuit for Forward and Reverse Biased Diode

Form 1. The symbol for diode is shown in Figure 5.21. The forward biased and reverse biased circuits are shown
in Figure 5.22.

Diode Symbol

5+ – Anode + – Cathode

Figure 5.21

Diode in forward biased circuit Diode in reverse biased circuit
+– –+

Current flows No current flows

++
V– V–

Bulb lit Bulb
not lit

Figure 5.22

406

2. The circuit in Figure 5.22 shows current flows Physics SPM  Chapter 5 Electronics  
through diode in forward biased position and
no current flows in the reverse biased position. SPM Highlights

3. The function of diode can be concluded that: Figure 5.24 shows an electric circuit.

Diode allows current to flow in one direction only. L



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and reverse biased circuit

VIDEO M

SPM Highlights Figure 5.24
Which of the following statements is true?
Figure 5.23 shows an electric circuit. A Only bulb L is lit
B Only bulb M is lit
DPC C Both bulbs L and M are lit
D Both bulbs L and M bulbs are not lit.
Figure 5.23
In the circuit above, it is found that the light-emitting Examiner’s Tips
diode (DPC) is not lit. DPC lights up when The diode in series with bulb M is forward biased
A the polarity of diode is reversed whereas the diode in series with bulb L is reverse
B the polarity of DPC is reversed biased.
C the polarity of battery is reversed
D the polarity of battery and diodes are reversed. Answer: B
Examiner’s Tips
The DPC will turn on when the diode and the DPC are The Use of Semiconductor Diode Form
in forward biased position. Therefore, the polarity of and Capacitor in the Rectification of
battery needs to be reversed. Alternating Current
Answer: C
1. Since the diode has the characteristic of
ACTIVITY 5.3 allowing current to flow only in one direction,
the diode can be used as a rectifier.

2. Rectifier is a device that can turn an alternating
current into a direct current.

3. Rectification is a process of converting the
alternating current into a direct current using
a diode.

5

Aim: To build a rectification circuit and observe the current on the display screen on a cathode ray
oscilloscope (C.R.O.) for:

(I) half-wave rectification using one diode.

(II) full-wave rectification using four diodes.

Apparatus and materials: Diodes, resistor, connecting wires, a.c. power supply and C.R.O.

407

  Physics SPM  Chapter 5 Electronics

Instructions:
(I) Half-wave rectification using one diode
1. The circuit is set up as shown in Figure 5.25. The setting on C.R.O. is adjusted to display at least

three or four full-wave a.c. voltage on the screen as shown in Figure 5.26.

Power supply C.R.O

Penerbitan Pelangi Sdn Bhd. All Rights Reserved.To a.c. voltageResistorTo
power supply C.R.O.

Figure 5.25 Figure 5.26

2. A half-wave rectifier circuit is made by connecting a diode in series with the resistor as shown in
Figure 5.27. The trace displayed on screen are observed and sketched.

Power supply

C.R.O

Diode
Resistor

Form Diode

5 To a.c. power Resistor To
supply C.R.O

Figure 5.27

(II) Full-wave rectification using four diodes
3. A full-wave rectifier circuit is made by connecting four diodes as shown in Figure 5.28. This rectifier

is also known as bridge rectifier. The trace displayed on screen is observed and sketched.

408

Power supply Physics SPM  Chapter 5 Electronics  
C.R.O.
Bridge
rectifier

Resistor
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Bridge rectifier

To a.c. power
supply

Resistor To C.R.O.

Figure 5.28

Observation:

1. Figure 5.29 shows the observations for cases of no rectification, half-wave rectification and full-
wave rectification.

(i) No rectification (ii) Half-wave rectifications (iii) Full-wave rectifications
Figure 5.29

Discussion:

1. Half-wave rectification: Form
(a) Achieved with only one diode.
(b) A pulsating current is produced, where the diode only lets the current to flow through the
resistor for every half-cycle.

2. Full-wave rectification: 5
(a) Achieved with four diodes.
(b) A continuous current is produced, where one pair of diodes is conducting in the first half
cycle and the other pair of diodes is conducting in the second half cycle.

The Flow of Current in Half-Wave Rectification Circuit
1. The process where only half of every cycle of an alternating current (a.c.) is made to flow in one

direction only is called a half-wave rectification.
2. In half-wave rectification, the rectifier (single diode) allows half the positive cycle of the alternating

current (a.c.) and blocking the other half of the negative cycle to pass through. This process is as shown
in Figure 5.30.

409

  Physics SPM  Chapter 5 Electronics

At each half positive cycle of a.c. input, the
diode is forward biased causing the current
to flow through the diode.

+Vmax I Diode + +V
0V A.c. input D.c. output
–Vmax R 0V
– waveform
A.c. input waveform
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In the negative halft cycle of a.c. input, the diode
is reserve biased and no current flows through the
diode. Thus, the current flows through the load
R in one direction only.

Figure 5.30

The Flow of Current in Full-Wave Rectification Circuit

1. In full-wave rectification, the bridge rectifier will reverse the negative cycle of the alternating current
(a.c.) to the positive cycle. Therefore, both the positive and negative cycles of the alternating current
will appear on the output in the same direction.

2. This process is described as shown in Figure 5.31 and Figure 5.32.

In the first half cycle, diodes D1
aDn2 danDd3
are forward biased, diodes
D4 are reverse biased.

The current follows

+Vmax D4 D1 the direction of the
0V arrows.
–Vmax
A.c. +
5 input
+Vmax
+Vmax D3 D2
0V 0V
–Vmax D.c. output

Form 410 –

Figure 5.31 First half cycle (positive cycle)

IanntdheD4saerceonfodrwhaalrfdcbyicalese, dd,ioddioedseDs2
D1 and D3 are reverse biased.

The current follows

D4 D1 the direction of the
arrows.

A.c. +
input +V

D3 D2 0V
D.c. output

–

Figure 5.32 Second half cycle (negative cycle)

Physics SPM  Chapter 5 Electronics  

3. Note that the current flowing through the resistor is always in the same direction for both half cycles
of the a.c. input.

4. The output of the half-wave and full-wave rectification can be summarised as shown in Figure 5.33.

Half
wave

The second half
cycle (negative)
is cut off
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.A.c. input D.c. output

Full The second half
Wave cycle (negative)
revert change to
positive d.c.

Figure 5.33

5. The full-wave output voltage is smoother than half-wave output voltage.

The Function of Capacitor as a Current Smoother in a Rectification Circuit

1. The process of rectification can transform alternating current (a.c. voltage) into direct current
(d.c. voltage). However, when we look at the output waveform for both rectifications, the size of the
amplitude produced is not uniform as the battery.

2. Capacitor is an electronic component that can store electrical charges, as well as discharge electricity.

3. When the capacitor is connected in parallel to the resistor or output load of the rectifier circuit, it can
smooth the half-wave and full-wave rectification output voltage as shown in Figure 5.34 and Figure
5.35.

Diode Capacitor Charging
capacitor
+Vmax A.c. Discharging capacitor
0V Input
–Vmax RL Output
C

0V

A.c. input waveform Output waveform

Figure 5.34 Capacitor smoothen the output voltage of half-wave rectification Form

+Vmax D4 D1 Charging Waveform 5
Capacitor capacitor with capacitor
0V
–Vmax D3 D2 C RL Output Vdc

A.c. input waveform

Waveform

0V without
capacitor

Discharging capacitor

Figure 5.35 Capacitor smoothen the output voltage of full-wave rectification

411

  Physics SPM  Chapter 5 Electronics

4. Capacitor of large capacitance (farad) is able to store large amount of charges when charging, and also
take longer time to discharge. Therefore, larger capacitor can smoothen the rectification output more
efficiently.

SPM Highlights Which waveform is seen on C.R.O. screen?

Figure 5.36 shows a rectification circuit for alternating A C
current (AC).

CRO
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B D

Input-Y Examiner’s Tips
The circuit shows a half-wave rectification circuit.
Figure 5.36 Capacitor is used to smoothen the output of the
Which of the following waveforms is showed on direct current that is being rectified.
C.R.O. screen?
A C Answer: B

B D SPM Highlights

Figure 5.38 shows a rectification circuit.

A.c. power To cathode ray
supply oscilloscope

Examiner’s Tips Figure 5.38
For full-wave rectification, the output waveform
without the presence of capacitor is C. Which of the following waveforms is showed on the
screen of cathode ray oscilloscope?
Answer: C A C

Form SPM Highlights

5 Figure 5.37 shows a diode, resistor R and capacitor B D
are connected to form a rectifier.

Diode Capacitor

A.c. R To
input C C.R.O.

Figure 5.37 Examiner’s Tips
The circuit shown is a half-wave rectification circuit.

Answer: C

412

Checkpoint! 5.2 Physics SPM  Chapter 5 Electronics  

1. (a) State the function of semiconductor diode. Various types
of
(b) Figure 5.39 shows an electric circuit consisting
twXoadnidodYesarDe1 and D2 and P and transistor
of two battery two bulbs When
Q. terminals.
a battery is connected across XY, the P bulb

lights up.

Penerbitan Pelangi Sdn Bhd. All Rights Reserved.X D1 D2 Figure 5.41
Y PQ
The Structure of npn and pnp Transistor
Figure 5.39
(i) Is bulb Q lights up too? Give a reason. 1. Like diode, transistor is made up from a
(ii) Which terminal is positive, X or Y? combination of semiconductors. However,
(iii) What is the function of diodes D1 and D2 the significant difference between diode and
transistor is the number of terminals. The
in this circuit? transistor has three terminals, while the diode
2. (a) What is rectifier? has only two terminals.

(b) Figure 5.40 shows a rectification circuit. 2. Figure 5.42 shows a transistor consisting of
two p-n diodes coupled end to end.
D4 D1
p-n junction p-n junction
D2 D3 P R To C.R.O. diode diode

Figure 5.40 N PN
(i) State the type of rectification circuit
Collector (C) Emitter (E)
shown. Base(B)
(ii) Name the device consisting of four
Figure 5.42 The structure of a transistor
diodes.
(iii) What is the function of the device P 3. Transistor has three terminals, namely emitter Form
(E), base (B) and collector (C). The base is a
connected parallel with resistor R? thin layer in the middle part. The right side of 5
(iv) If the input is a.c. voltage, sketch the the diode is called the emitter diode and the
left side is called the collector diode.
output voltage shown on C.R.O.
4. There are two types of transistor depending on
5.3 Transistor the position of the n-type semiconductor and
the p-type semiconductor. Figure 5.43 shows
1. Transistor is a semiconductor device used to the two types of transistor, the npn transistor
transfer a weak signal from a low-resistance and pnp transistor and their respective symbol.
circuit to a high-resistance circuit.

2. It is also a switching device that controls and
amplifies electrical signals such as voltage or
current.

413

  Physics SPM  Chapter 5 Electronics

Collector npn transistor C
C n
NP N Emitter Bp
E n
E
B      
Base C
Penerbitan Pelangi Sdn Bhd. All Rights Reserved. p
Collector pnp transistor P Emitter Bn
C PN E p
E
B      
Base

Figure 5.43 Two types of transistor

5. Note that the direction of the arrow shows on SPM Highlights
the transistor symbol is always from p → n,
this is the direction of current flow when the Which is the symbol for pnp transistor?
transistor is turned on in the circuit.
A Emitter C Collector
SPM Highlights
Base Base
Which statement about a transistor is true?
A It has two terminals. Collector Emitter
B It acts as a half-wave rectifier.
C It has its own internal energy supply. B Emitter D Emitter
D It functions as a current amplifier.
Base Base
Examiner’s Tips
Form A transistor has three terminals, it cannot act as Collector Collector
rectifier and does not has its own internal energy.
Transistor is a current amplifier. Examiner’s Tips
The direction of arrow in pnp transistor is from emitter
5 Answer: D to base.

Answer: C

414

The npn and pnp Transistor Circuits Physics SPM  Chapter 5 Electronics  
pnp transistor circuit
npn transistor circuit

Load RL + Load RL –

Penerbitan Pelangi Sdn Bhd. All Rights Reserved.IC (big)VC IC (big) VC
– VS +
C C

RB B RB B
IB VCE VCE
IB
VS (small) E (small) E

Base curcuit Emitter curcuit Base curcuit Emitter curcuit

Similarity

• Transistor circuit consists of base circuit and emitter circuit.
• The base circuit needs to reach a minimum voltage to turn the circuit on. The characteristic of this transistor

circuit is known as switch.
• A small current, IB on the base circuit can control a larger current, IC on the emitter (output) circuit. Thus, this

transistor circuit is known as amplifier circuit.
• The resistance, RB connected to base should be large enough to limit the base current.

Differences

• The collector terminal is connected to the positive • The collector terminal is connected to the negative
terminal of the supply voltage, +VC. terminal of the supply voltage, –VC.

• The emitter terminal is connected to negative terminal • The emitter terminal is connected to the positive
of voltage supply, –VC. terminal of the supply voltage +VC.

• The base terminal is connected to the positive of source • The base terminal is connected to the negative source
voltage, +VS. voltage, –VS.

Figure 5.44

Note that: The minimum base voltage for turning on the silicon transistor is 0.7 V and 0.3 V for Germanium
transistor.

Potential Divider X Form

1. The npn and pnp transistor circuits shown in R1 Load RL + 5
Figure 5.44 have two voltage supplies, a small VC
voltage on the base circuit, VS and a larger Y IC (big)
voltage on the output circuit, VC. R2
C
2. These two sources can be combined to become
a single source that form a voltage divider RB B VCE
circuits to forward bias the base circuit of the
transistor as shown in Figure 5.45. IB
(small)
E

Z Emitter circuit
Base circuit

Figure 5.45

415

  Physics SPM  Chapter 5 Electronics

3. The advantage of a voltage divider is that the SPM Highlights
value of the base voltage, VYZ can be controlled
by selecting the appropriate values for resistors Figure 5.46 shows a transistor circuit.
R1 and R2. To turn on the transistor, the voltage
across R2 or VYZ needs to exceed a minimum V1 R
voltage value. This is because VYZ is the base V2 LDR
voltage.

Penerbitan Pelangi Sdn Bhd. All Rights Reserved. 4. The voltage divider formula can be derived as V

follows:

The ratio between voltages across YZ to voltage

across XZ, VR2
VR2 + R1
VVXYZZ = Figure 5.46
What is voltage V when V1 = 1.5 V and V2 = 4.5 V?
VVYCZ = IR2 A 1.5 V
I(R2 + R1) B 4.5 V
C 6.0 V
 VYZ = D 7.5 V
R2 × VC
R2 + R1 Examiner’s Tips
Based on voltage divider,
 VXY =
R1 × VC V = 1.5 V + 4.5 V = 6 V
R1 + R2
Answer: C

SPM Highlights

A transistor circuit functions as an automatic switch. SPM Highlights
This circuit turns on the LED bulb during day time.
Figure 5.47 shows a transistor circuit.
Photoresistor

LED npn Battery V RT Diode Relay switch
Resistor R2 transistor RB = 1 k⍀ 9V

Resistor R1 6 V R = 10 k⍀ Alarm

What action should be taken to light the LED bulb

Form at night?

A Replace the npn transistor with a pnp transistor
RR12
B Interchange the position of photoresistor and Figure 5.47
C Interchange the position of photoresistor and

5 D Increase the number of battery What is the potential difference across the thermistor
RT?
A 2 V C 3 V
Examiner’s Tips
B 5 V D 8 V
Photoresistor and R form a voltage divider.
1 of photoresistor and R1 will
Exchanging the position
Examiner’s Tips
turn on the LED bulb at night. Based on voltage divider, 6 + V = 9
Thus, V = 9 – 6 = 3 V
Answer: B
Answer: B

416

Physics SPM  Chapter 5 Electronics  

Function and Uses of Transistor as Current Amplifier

ACTIVITY 5.4

Aim: To study the function of transistor as current amplifier by using a transistor circuit kits and to
calculate the amplification factor.

Apparatus and materials: Battery (6 × 1.5 V), electronic kit with transistor as amplifier, milliammeter
(0 - 1 mA) and milliammeter (0 – 100 mA). [The electronic kit is supplied
with a circuit board set up with electronic components with transistor as an
amplifier]

Instructions:
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Voltage RC 0 – 100 mA
divider A2

+ RB 0 – 1 mA +
3.0 V – A1 6V

V input –

Figure 5.48

1. A circuit board equipped with electronic components with transistor as amplifiers as shown in
Figure 5.48 is provided.

2. The electronic components set up in the circuit board are studied carefully. The type of transistor
used is named.

3. mT=hiVellmivainomimlutmamgetetoedrtuiAvri2ndaeosrnIicst.haedjturasntesdistuonr.tiRl ethaedimngilloianmmmilelitaemr Am2estherowA1s a minimum reading, that is,Vinput
is recorded as Ib and reading on

4. Step 3 is repeated by increasing the bfoarsme .voGltraagpehVoinpfutcsoolltehcatto5r cseutrsreonf tr,eIacdainggasinIbstanbdasIec are obtained.
All readings are recorded in table current, Ib is
plotted.

5. From the graph plotted, the gradient of the graph is determined. The gradient of the graph represents
Ic
the amplification factor of the transistor amplifier, b = Ib . Form
Results:

Ib / mA Ic / mA 5
0.1 5.0

0.3 16.0

0.5 24.0

0.6 30.0

0.8 40.0

417

  Physics SPM  Chapter 5 Electronics

Graph of collector current, Ic against base current, Ib.

IC / mA
40

30
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20

10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 IB / mA

Gradient of graph, b = IIbc = 40 = 50
0.8

Therefore, amplification factor is 50.

Discussion:

1. The transistor circuit used in this activity is known as the current amplifier circuit.

2. The type of transistor used in the circuit is the npn transistor.

3. The results of the activity show that a small change in base current, Ib will cause a large change
in collector current, Ic.
4. The current gain or amplification can be calculated from the gradient of graph Ic against Ib.
5. This activity shows that the transistor acts as a current amplifier.

Form A Transistor Functions as a Current Amplifier

1. The amplifier circuit is the circuit used to amplify a signal.

2. The input of amplifier circuit is a voltage or current, while the output of the amplifier circuit is an
amplified input signal.

5 3. A small input signal sent through the transistor base circuit will produce a large signal through the
output circuit.
IC
IC

IB Input IB Output
signal signal

RC Output

0t IE

VB 0 IC
VC t

Figure 5.49

418

Physics SPM  Chapter 5 Electronics  

The Use of a Transistor as an Automatic Switch

ACTIVITY 5.5

Aim: To show the function of transistor as an automatic switch.

Apparatus and materials: 4 x 1.5 V, battery, an electronic circuit used transistor as an automatic switch
controlled by light (Light Dependent Resistor, LDR).

Instructions:

1. An electronic circuit set up as shown in Figure 5.50 is given.
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LED

10 k⍀ Switch
6V
1 k⍀

Light
Dependent
Resistor (LDR)

Figure 5.50 Form

2. The circuit is turned on and the LDR is covered with a piece of black paper. The state of LED is
observed.

3. The black paper is removed so that LDR is exposed to light. The state of LED is observed.

4. Steps 2 and 3 are repeated by interchanging the position of the LDR and resistor of 10 kΩ.

5. The role of LDR and the function of transistor in the circuit as an automatic switch are explained.

Observation:

1. When the LDR is closed with a black paper, the LED lighted up.

2. When the black paper is removed, the LDR is exposed to light and the LED is cut off.

3. When position of LDR and resistor of 10 kΩ are interchanged, the LED lighted up when the LDR
is exposed to light and cut off when covered with black paper.

Discussion:

1. Light Dependent Resistor (LDR), or photoresistor, is a light-sensitive resistor. In a dark condition,
LDR has a resistance of approximately 1 million Ohm (1 MΩ). When exposed to light, the LDR
has a resistance of only a few hundred Ohm.

2. The table below how the resistance of LDR affects the base voltage, base current and collector
current in bright and dark conditions.

Situation 1 Situation 2 5

10 k⍀ LED Light LED
1 k⍀ Dependent
Resistor (LDR) 1 k⍀ 6V

6V

Light 10 k⍀
Dependent
Resistor (LDR)

419