Physics HSC
Space
Notes and Questions by Dot Point
2.
Many factors have to be taken into account to achieve a successful rocket launch, maintain a stable
orbit and return to Earth

describe the trajectory of an object undergoing projectile motion within the Earth’s gravitational field in
terms of horizontal and vertical components

describe Galileo’s analysis of projectile motion
o
solve problems and analyse information to calculate the actual velocity of a projectile from its
horizontal and vertical components using :
v2x  u2x
v  u  at
2
2
v y  u y  2a y y
x  u x t
1
y  u yt  a y t 2
2
o
perform a first-hand investigation, gather information and analyse data to calculate initial and
final velocity, maximum height reached, range and time of flight of a projectile for a range of
situations by using simulations, data loggers and computer analysis
In the preliminary course we investigated accelerated motion in a straight line. We derived a series of equations
(they are on the formula sheet for the exams) which describe mathematically this type of motion.
Projectiles are launched at an angle other than 90o to the Earth’s surface. Their subsequent motions involve both a
horizontal component and a vertical component. The horizontal component is unaffected by any forces (if we
ignore air resistance) and so remains constant (recall Newton’s first law of motion). On the other hand the vertical
component is subjected to the weight force of the projectile. This causes an acceleration of 9.8 ms-2 towards the
centre of the Earth. Thus the vertical component of the motion is undergoing a constant acceleration.
The resultant path of the projectile is therefore parabolic. You do not need to prove this for the purposes of the
physics course. However, you will need to be able to analyse the motion of the projectile in a mathematical sense.
To this end you should understand the following:

The initial vertical velocity is vsin where  is the angle of projection and v is the speed of
projection.

The initial horizontal velocity is vcos.

The horizontal velocity remains constant throughout the motion.
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 1

The vertical velocity changes by 9.8 ms-1 every second. As such the projectile will stop rising and
then accelerate back to Earth, moving sideways the whole time.

It is possible to apply the equations of uniformly accelerated motion to the vertical component of
the projectile’s motion. In each equation a = g = 9.8 ms-2.

The application of these equations results in a formula for the maximum height, the time of flight
and the range of the projectile. These formulas are limited in their application as very few
projectiles are projected from, and return to, a perfectly horizontal surface.
The following questions will help you to understand the concept of projectile motion.
1.
An object is projected at an angle of 30o from the top of a building 53.9 m tall. Its initial speed is
98 ms-1. Determine:
(a)
the initial vertical velocity of the projectile
(b)
the initial horizontal velocity of the projectile
(c)
the time it will take for the projectile to stop rising
(d)
the height of the projectile at this time
(e)
its horizontal displacement at this time
(f)
the time it takes to fall to the ground from this position
(g)
its horizontal displacement on reaching the ground
(h)
the final vertical velocity
(i)
the final horizontal velocity
(j)
the final velocity of the projectile at the instant it hits the ground
2.
Sketch the trajectory of the projectile.
3.
Is it possible to use the range, time of flight and maximum height equations for this type of problem?
Comment.
4.
Describe how the horizontal distance traveled by the projectile would be affected if we took air resistance
into account.
Please attempt the assignment containing questions from past examination papers. The assignment is called
Space Syllabus Section 2 Assignment A.
The assignment can be downloaded from the school’s Moodle site.
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 2

explain the concept of escape velocity in terms of the:
o gravitational constant
o mass and radius of the planet

outline Newton’s concept of escape velocity
o
identify data sources, gather, analyse and present information on the contribution of one of the
following to the development of space exploration: Tsiolkovsky, Oberth, Goddard, EsnaultPelterie, O’Neill or von Braun
When studying the work of Galileo it must be remembered that he worked under huge constraints. The “scientific
truths” of great thinkers like Aristotle had held sway for many centuries, primarily because they fitted neatly with
the teachings of the church. To questions such “truths” was to question the wisdom of the church, which was
deemed to be heretical (and possibly punished by death). Galileo was not only a thinker but also an experimenter.
He, like all good scientists, was not prepared to accept someone else’s word for things that could be tested.
However, since experimentation was frowned upon, he sometimes had to perform gendaken, or thought
experiments. Amazingly, with no algebra and less mathematics than a year 8 student of today, he was able to
arrive at conclusions which have since been verified by experimental and mathematical means.
Galileo postulated that all objects, regardless of mass, would accelerate uniformly to Earth. This is of course true,
providing air resistance is removed (i.e. in a vacuum). It was very difficult to prove it at the time however.
Galileo also analysed projectile motion and in doing so introduced the concept of frames of reference. To do this
he considered the example of a ball dropped from the crow’s nest of a sailing ship that was itself moving. To an
observer on the ship, the ball falls directly to the deck of the ship i.e. in a straight line. To a stationary observer
(perhaps on shore) the ball follows a parabolic path as it falls under the influence of its own weight force. (Note
that Aristotle had argued that the ball would be left behind by the ship!)
Galileo argued (as we would today) that the ball had both a vertical velocity, which was increasing, and a
horizontal velocity, which effectively remained constant. This is the basis of our understanding of projectile
motion today.
Soon after Galileo Sir Isaac Newton proposed an analysis of projecting an object at such a speed that it would not
return to the Earth. His argument went along these lines. If you could place a cannon on top of a high mountain
and fire a cannon ball from it, the horizontal distance travelled by the cannon ball would be dependent upon its
initial velocity. We have seen that this true from our equations relating to projectile motion.
Newton reasoned that if you fired the cannon ball fast enough it would enter into a circular orbit around the Earth!
The speed would have to be about 8 kms-1. The reason for this is that the Earth’s curvature is approximately 5 m
vertical drop for every 8 km horizontally travelled. This would be exactly what the cannon ball would do as it
drops approximately 5 m vertically in 1 second. Thus the ball would effectively remain parallel with the Earth’s
surface and thus enter into a circular orbit, returning to its starting point again and again. The speed required has
since been shown to be approx. 7.8 kms-1.
Newton reasoned that for speeds greater than 7.8 kms-1 the projectile would enter into an elliptical orbit. As the
speed of projection increases, the eccentricity of the ellipse becomes greater i.e. more elliptical, less circular.
At a speed of 11.2 kms-1 (or greater), the cannon ball will continue on into space. This speed is referred to as the
escape velocity of the Earth, and is defined as the minimum velocity required for a projectile to escape the Earth’s
gravitational field. (Note that some texts refer to this as going into an open orbit, but this seems like a
contradiction in terms.)
The following questions are designed to help with your understanding of the concept of Escape Velocity.
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 3
1.
Draw a series of diagrams to contrast the views of Aristotle and Galileo with regard to objects falling from
a building or tower towards a rotating Earth.
2.
Explain how Galileo accounted for the fact that the dropped object does not get “left behind” by the
rotating Earth.
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3.
Bearing in mind that Galileo had no access to algebra, draw a diagram (similar to the one he might have
drawn) to show that a body that has a uniform horizontal velocity and an accelerated vertical velocity
traces out a parabola.
4.
Newton argued that a body projected from the Earth’s surface at an angle would always crash back to
Earth. Outline his argument for this point of view.
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5.
Some scientists use the term open orbit (as mentioned in the notes for Galileo, Newton and Escape
Velocity). Define what is meant by an open orbit. Explain how this might be considered to be a
contradiction in terms.
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For the mathematical derivation of a value for the escape velocity of any massive body, see the following notes.
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 4
To follow this argument you need to remind yourself of the concepts of gravitational potential energy and kinetic
energy. Yu will remember that gravitational potential energy increases as you move away from the surface of the
Earth i.e. the further you go, the greater the GPE becomes. Kinetic energy is independent of height and is
dependent upon mass and speed.
KE is determined by the formula ½ mv2.
(Preliminary course)
GPE is determined by the formula Gm1m2 / r
(See earlier notes on GPE)
Rockets that are projected from the Earth’s surface are subjected to enormous retarding frictional forces as they
attempt to move through the Earth’s atmosphere at great speed. However once a rocket has attained a suitable
height above the Earth (say between 200 and 300 km) the mass of the rocket may be considered to be constant (i.e.
there are no more fuel tanks to be jettisoned). Let us assume that the mass of the rocket is m2 and it has a velocity
(speed) of v.
The rocket continues to move away from the Earth and is thus being decelerated by the Earth’s gravitational field
(v is decreasing). Thus KE is decreasing. However, as the altitude of the rocket increases, so too the GPE
increases. The law of conservation of energy requires no overall loss of energy so the loss in KE is equal to the
gain in GPE i.e.
1/2m2v2
=
Gm1m2 / r
where m1 is the mass of the Earth, and
r is the distance from the centre of the Earth
The mass of the rocket (m2) cancels, and with a little fine-tuning, the equation can be written
v
=
(2Gm1/r)
Since G is the universal gravitational constant it can be seen that the escape velocity for the Earth depends only on
the mass of the Earth and the Earth’s radius. Substituting these values into the equation gives
v
=
11 000 ms-1, or 11 kms-1 (39 600 kmph)
Also worthy of note is the fact that the mass of the rocket does not appear anywhere in the equation. This means
that the escape velocity for any object is the same, from the smallest molecule to the largest mass. Any textbook
will have a table of values of escape velocities for various planets and moons. A close analysis shows that an
increase in mass will cause an increase in escape velocity but the opposite is true for the radius. Neither of these
relationships is linear, as both are dependent upon the square root of the quantity.
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 5
1.
Verify that the escape velocity of a body at the Earth’s surface is 11.2 kms-1.
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2.
Explain why this value is greater than the value quoted in the notes i.e. 11.0 kms-1.
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___________________________________________________________________________________
3.
Given that the mass of Mercury is 0.06 times the mass of the Earth and the radius of Mercury is 0.38 times
the radius of the Earth, calculate the escape velocity on Mercury.
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___________________________________________________________________________________
4.
Consider a fictional planet whose radius is 0.5 times the Earth’s radius and mass is 8 times the mass of the
Earth. Calculate the escape velocity on this planet.
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___________________________________________________________________________________
5.
A student suggests to his colleagues that the more massive planets have the greatest escape velocities,
principally because of their greater mass. By means of research, determine if the student is correct.
Discuss the student’s reasoning making reference to the formula for escape velocity.
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Please attempt the assignment containing questions from past examination papers. The assignment is called
Space Syllabus Section 2 Assignment B.
The assignment can be downloaded from the school’s Moodle site.
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 6

identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch

discuss the effect of the Earth‘s orbital motion and its rotational motion on the launch of a rocket

analyse the changing acceleration of a rocket during launch in terms of the:
o Law of Conservation of Momentum
o forces experienced by astronauts
We have already seen that Newton, long before the reality of the space age, had contemplated propelling an object
into space. Recall his thought experiment with the “super cannon”. Such “gun launches” are doomed to failure
because the projectile will attempt to follow an elliptical orbit, with one of the foci of the ellipse as the Earth’s
centre. Thus the projectile would have to carry a means of secondary propulsion which would be able to be used
to alter the trajectory at an appropriate altitude and speed. So GTLS (gun launch to space) is theoretically
possible, but not a reality at this time.
The current conventional method is to use a rocket-powered launch vehicle. Such vehicles accelerate over
minutes, rather than seconds as in the GLTS) and thus create g forces which are acceptable for human on board the
vehicle. This would not be the case in GLTS!
The engines of the rocket require fuel and an oxidiser (often oxygen). In order to provide the necessary thrust a
great deal of fuel and oxidiser is required, contributing significantly to the total mass of the rocket and payload.
However all launch vehicles have several sections, usually called stages. As the fuel in each stage is used so the
containers (now empty) may be jettisoned. This means that the mass of the launch vehicle decreases during the
launch and the large accelerations necessary can be achieved.
Vertical take-offs are favoured in order to get the vehicle as far from the Earth’s surface as quickly as possible.
This is because in the first stage of launch, the gravitational deceleration is greatest, the mass of the launch vehicle
is greatest and air resistance is greatest. For the rocket which launches the Space Shuttle, the following
information applies.
After approximately 2 minutes, at an altitude of 45 km and a speed of about 1.4 kms-1, the solid fuel rocket
boosters are released. They are recovered after parachuting to Earth. After a further 6 minutes the very large
external fuel tank is jettisoned and burns up in the atmosphere. At this point the shuttle is travelling at a
sufficiently large speed to maintain a low-Earth orbit. If necessary, further manoeuvring can be achieved by
means of relatively smaller on-board boosters.
Most launches occur as close to the equator as possible. This is done so as to make maximum use of the Earth’s
orbital speed, which is greater at points on the equator than at other latitudes. Recall that the Earth is rotating on
its axis from west to east at about 464 ms-1. This velocity is shared by the launch vehicle as it waits to lift off. It
contributes to the attainment of the necessary speed without the consumption of any fuel. Of course a vehicle
launched in a westerly direction has to carry more fuel to overcome this easterly component.
1.
Show that velocity of a point on the ……………………………………………………………………
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 7
2.
Earth’s surface at the equator is
……………………………………………………………………
approximately 464 ms-1.
……………………………………………………………………
Calculate the radius of the Earth at
……………………………………………………………………
a latitude of 22o N. Use this answer ……………………………………………………………………
3.
to calculate the velocity of a point
……………………………………………………………………
on the Earth’s surface at this
……………………………………………………………………
latitude.
……………………………………………………………………
Using your answers to questions 1 and 2, discuss the reason(s) why equatorial launch sites are popular for
launching rockets.
……………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
4.
Discuss the method of propulsion of the launch vehicle, with particular reference to the law of
conservation of momentum. ……………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
……………………………………………………………………………………………………………
Please attempt the assignment containing questions from past examination papers. The assignment is called
Space Syllabus Section 2 Assignment C.
The assignment can be downloaded from the school’s Moodle site.

analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting
the Earth
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 8

compare qualitatively low Earth and geo-stationary orbits

define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity,
the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using
Kepler’s Law of Periods

account for the orbital decay of satellites in low Earth orbit
o
solve problems and analyse information to calculate the centripetal force acting on a satellite
undergoing uniform circular motion about the Earth using:
2
mv
r
solve problems and analyse information using:
F
o
3
r
GM
2 
T
4 2
Communications satellites orbit the Earth at a range of altitudes and speeds. The purpose of the satellite will
determine exactly how it orbits. The altitude will also determine the speed at which the satellite orbits.
Recall that the force required to keep a body moving in a circular path is called a centripetal force. This is equal to
mv2 / r. This force is provided by the body’s weight force i.e. mg. Equating these two expressions gives:
g = v2 / r
Or
v2 = gr
where r is the radius of the orbit (including the radius of the Earth); and g is
the acceleration due to gravity at this altitude.
Recall that the value of g decreases with altitude. It can be calculated using g = GM / r2.
In order to remain in orbit at a given altitude a certain speed must be maintained. If the speed becomes less than
the required speed the satellite will fall back to Earth. If the speed becomes too large the satellite will move away
from the Earth and eventually return following an elliptical path. If the speed is greater than the escape velocity,
the craft would not return.
Low-Earth orbits are those for which the altitude of the satellite is around several hundred kilometres above the
Earth’s surface. At this altitude they circle the Earth several times a day. There are exceptions: Telstar, a
communications satellite, launched in 1962, orbited at 3 300 km above the Earth’s surface. At this altitude it had a
period of 2 hours 38 minutes, so it orbited the Earth many times a day. This, of course, illustrates the problem of
using such low altitudes for communications satellites. They are only visible (available) to a sending or receiving
station on the Earth’s surface for short periods of time (in Telstar’s case about 20 minutes). However this can be
of some use. Such a satellite can survey the entire surface of the Earth over a period of several days. Thus they
make excellent spy satellites. Their low altitudes do pose other problems though. The major one is friction with
the atmosphere (thin though it is it still has to be taken into consideration). Friction provides a retarding force and
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 9
causes a slowing of the satellite. If this is not corrected the satellite will spiral towards Earth and burn up in the
lower atmosphere.
Geostationary and geosynchronous orbits both have a period of 24 hours. This of course matches the period of the
Earth’s rotation. Thus if the satellite is placed in an equatorial orbit it will remain above the same point on the
Earth’s surface i.e. their relative velocities will be zero. It is possible to show that the altitude of such a satellite
needs to be 35 880 km (see Satellite Orbits Worksheet 1). Three such satellites can form a communications
network which can cover almost all of the Earth’s surface. If the satellite is placed in a non-equatorial orbit, but its
period is still 24 hours, then its orbit is referred to as geosynchronous. Such satellites do not remain above the
same place on the Earth’s surface. Because their orbits have to have a north-south component, their apparent
paths (as seen by an observer on Earth) take on shapes like figure eights, for example.
1.
2.
Calculate the gravitational acceleration
……………………………………………………………..
due to the Earth (g) at an altitude of
……………………………………………………………..
(a)
300 km
……………………………………………………………..
(b)
3 300 km
……………………………………………………………..
Explain why the gravitational acceleration decreases with altitude. ……………………………………..
……………………………………………………………………………………………………………..
……………………………………………………………………………………………………………..
3.
4.
Using Kepler’s law of periods
……………………………………………………………..
(i.e. r3 / T2 = constant), show that
……………………………………………………………..
the altitude required for a satellite to
……………………………………………………………..
have a period of 24 hours is 35 880 km.
……………………………………………………………..
(Hint: SI units first, then convert.)
……………………………………………………………..
The Telstar satellite launched in 1962
……………………………………………………………..
had a period of 2 hours and 38 minutes.
……………………………………………………………..
By using appropriate calculations verify
……………………………………………………………..
that the altitude of its orbit was 3 300 km.
……………………………………………………………..
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 10
5.
Imagine a satellite in orbit over the geographical poles of the Earth i.e. tracing a “vertical” circle rather
than a horizontal one). Assume that the period of its orbit is 6 hours.
(a)
Describe the path of the satellite as viewed from an observer on the ground. Assume that the
satellite first appears low in the southern sky.
………………………………………………
……………………………………………………………………………………………………..
……………………………………………………………………………………………………..
……………………………………………………………………………………………………..
……………………………………………………………………………………………………..
(b)
Calculate (estimate?) the number of days it would take for the satellite to “see” the entire surface
of the Earth. ……………………………………………………………………………………..
Objects moving in a circular path with a constant radius and at a constant speed are said to be undergoing uniform
circular motion. This type of motion is created by a centripetal force. A centripetal force is always directed at
right angles to the direction of the object’s velocity. For this reason the force neither enhances nor reduces the
object’s speed, but does effect a change in the direction of the object’s velocity. This creates an apparent anomaly
– an object subjected to an unbalanced force must accelerate, but here we have no change in speed i.e. the object is
neither speeding up nor slowing down but it is still accelerating! The answer lies in the fact that velocity is a
vector quantity – so the size of the velocity (speed) remains constant but the direction of the velocity is constantly
changing. The acceleration is directed towards the centre of the circle and so is called a centripetal acceleration.
In order to analyse the motion mathematically we must first introduce a new unit of angular measure. This is the
radian. One radian is the size of the angle created at the centre of the circle if the arc length is equal to the radius
of the circle. Thus there are 2 radians in one revolution. (This is equivalent to 360o so 1 radian is approximately
57 o.)
The time taken for one revolution is called the period, T. This can be measured in any unit from seconds to years.
If the radius of the circle is r metres, and the period of the motion is T seconds, the following applies:
linear speed:
v
=
distance / time
angular velocity:

=
2 / T
Linking these two equations gives:
=
2r / T
in ms-1
in rad.s-1
v
=
r
Note that linear speed is sometimes called orbital speed, as the object does not move in a straight line.
We have noted above that the direction of the acceleration is centripetal i.e. directed towards the centre of the
circle. The size of the acceleration is given by:
acceleration:
a
=
v2 / r
in ms-2
Since it is the force that produces the acceleration then Newton’s second law applies. So the size of the force is
given by:
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 11
force:
F
=
ma
=
mv2 / r
in newton
where m is the mass of the object
Masses which orbit heavenly bodies e.g. man-made satellites, or natural satellites like our moon, or the planets in
our solar system orbiting the Sun, obey these equations (although technically these masses may move in elliptical
orbits, so some modification is required!). Newton’s law of universal gravitation also applies. So now we have
two expressions for the size of the force keeping a mass in orbit i.e.
F
=
mv2 / r
and
F
=
Gm1m2 / r2
Equating these two equations verifies Kepler’s law of periods.
1.
Draw a diagram of a circle which
illustrates the definition of the radian.
(Hint: arc length has to equal radius.)
2.
3.
4.
Calculate the number of radians in the following:
(a)
one revolution
………
(b)
10 revolutions
………
(c)
one quarter of a rev. ………
(d)
60o
………
The period of the moon’s orbit is
…………………………………………………………
approximately 28 days. Calculate
…………………………………………………………
the angular velocity of the moon
…………………………………………………………
in rad.s-1.
…………………………………………………………
The radius of the Earth’s orbit averages
…………………………………………………………
1.496 x 1011 m. If we approximate the
…………………………………………………………
Earth’s orbit to be circular (a very round
…………………………………………………………
ellipse!), calculate the average linear
…………………………………………………………
speed of the Earth in km.s-1.
…………………………………………………………
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 12
5.
Two ice skaters skate in two perfect concentric circles (same centre). Skater A travels in a circle whose
radius is exactly four times as large as the radius of the circle traced out by skater B. If skater A takes
twice as long to travel around her circle, calculate:
(a)
the angular velocity of skater B compared to skater A
(b)
the linear speed of skater A compared to skater B
………………………………………………………………………………………………………………
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6.
Calculate the size of the centripetal force
………………………………………………………………
acting on a satellite of mass 2 000 kg
………………………………………………………………
moving in a circle of radius 6 680 km
………………………………………………………………
with a period of 24 hours.
………………………………………………………………
Please attempt the assignment containing questions from past examination papers. The assignment is called
Space Syllabus Section 2 Assignment D.
The assignment can be downloaded from the school’s Moodle site.
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 13

discuss issues associated with safe re-entry into the Earth’s atmosphere and landing on the Earth’s surface

identify that there is an optimum angle for safe re-entry for a manned spacecraft into the Earth’s
atmosphere and the consequences of failing to achieve this angle
Please attempt the assignment containing questions from past examination papers. The assignment is called
Space Syllabus Section 2 Assignment E.
The assignment can be downloaded from the school’s Moodle site.
Physics HSC : Space Syllabus Section 2 : Notes by Dot Point : page 14