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Scientists have drawn on advances in areas such as aeronautics,
materials science, robotics, electronics, medicine and energy production
to develop viable spacecraft. Perhaps the most dangerous parts
of any space mission are the launch, re-entry and landing.
A huge force is required to propel the rocket a sufficient distance
from the earth so that it is able to either escape the earth’s gravitational
pull or maintain an orbit. Following a successful mission,
re-entry through the earth’s atmosphere provides further challenges
to scientists if astronauts are to return to earth safely.
Rapid advances in technologies over the past fifty years have
allowed the exploration of not only the Moon, but also the Solar System and, to an increasing extent, the Universe. Space exploration
is becoming more viable. Information from research undertaken
in space programs has impacted on society through the development
of devices such as personal computers, advanced medical equipment,
communication satellites and the accurate mapping of natural resources.
Space research and exploration of space increases our understanding
of the earth’s own environment, the Solar System and the Universe.
This module increases students’ understanding of the history,
nature and practice of physics and the implications for society
and the environment.
Note: Some internet browsers (eg Firefox) do not
accurately display text symbols such as Greek letters used to represent
quantities in Physics. For example, capital delta is displayed as D in
Firefox. This is just something to be
aware of in case you do come across such issues. The square root sign is
another one not displayed properly by some browsers. Any symbols used in
equations produced by equation editors will of course display properly.
Gravity and the Gravitational
In 1687 Isaac Newton published his Principia Mathematica,
in which he explained his three Laws of Motion
and his Law of Universal Gravitation. The
Law of Universal Gravitation states:
where F = the force of
gravitational attraction between two masses, m1 and m2
a distance d apart and G = the Universal Gravitational Constant =
6.673 x 10-11 SI Units.
In words this law can be expressed as: The force of attraction between any
two bodies in the universe is proportional to the product of their
masses and inversely proportional to the square of their distance
Newton’s law suggests that every mass in the universe, no
matter how small, has its own gravitational field surrounding it.
The larger the mass, the stronger the gravitational field
around it. This field
is a region of influence in which another mass would experience a
force due to the presence of the first mass.
Definitions of Mass and Weight:
The mass of an object is a measure of the amount of matter
contained in the object. Mass is a scalar quantity.
The weight of an object is the force due to a gravitational
field acting on the object. Weight is a vector quantity.
The weight, W, of an object is given by Newton’s 2nd
where m is the mass of the object and g is the acceleration
due to gravity (9.8 ms-2 close to the earth’s surface).
Exercise: Use Newton’s Law of Universal
Gravitation equation to determine the acceleration due to gravity
on other planets. Hint – Consider a mass m at rest on the
surface of the planet in question and express the weight of the
mass m both in terms of the universal gravitation equation and the
Newton’s second Law equation F = ma. Compare the values obtained
with those from an appropriate reference (ref. 6 p.96).
Gravitational Potential Energy:
When we lift an object from the ground to a particular height
above the ground we must do work against the gravitational field
of the earth. This work goes into increasing the gravitational
potential energy of the body. The amount of work done is equal
to the change in gravitational potential energy (GPE) of the body.
For bodies near the earth’s surface, where the acceleration due
to gravity is fairly constant, the GPE of a body is given by:
Ep = mgh
where m is the mass of the body, g is the acceleration due to
gravity and h the height. In this situation we define the
ground level to be the zero potential energy level.
For objects a long way from the Earth’s surface, we must use a
more general definition: “The gravitational potential energy
of an object is the work done in moving the object from a very large
distance away to a point in the gravitational field of the Earth”.
This definition sets the zero of potential energy at an infinite
distance from the centre of the Earth. It leads to the following
expression for the gravitational potential energy of a mass m1
at a distance r from a mass m2 in the gravitational field
due to m2:
Note that the r in the denominator is
Exercise: Determine the GPE for a satellite
of mass 200 kg launched from the surface of Mars into an orbit 650
km above the planet’s surface. (Data: radius of Mars
= 3.4 x 106 m, mass of Mars = 6.5 x 1023 kg,
G = 6.67 x 10-11 SI Units. Answer: EP
= - 2.14 x 109 J)
Equations of Uniformly
By starting with some of our basic definitions given in the Moving
About topic it can easily be shown that:
where u = initial velocity, v
= final velocity, a = acceleration, s = displacement
and t = time. These three equations are referred
to as the equations of uniformly accelerated motion. They
may be used whenever the acceleration is uniform (constant or zero)
and the motion is considered in one dimension. The correct
sign must accompany each value as the quantities (except time) are
Recall also the definitions of velocity and acceleration
and the equation:
for average velocity, which applies only when the acceleration
the angle made by the total velocity vector and the horizontal.
is the angle made by the total
displacement vector and the horizontal.
of Projectile Motion
Our understanding of projectile motion owes a great debt to Galileo,
who in his work entitled “Dialogues Concerning Two New Sciences”,
presented his classic analysis of such motion. Galileo argued
that projectile motion was a compound motion made up of a
horizontal and a vertical motion. The horizontal
motion had a steady speed in a fixed direction, while the vertical
motion was one of downwards acceleration. Using a geometric
argument, Galileo went on to show that the path of a particle
undergoing such motion was a parabola.
In his work Galileo admits that his assumptions and results
are only approximations to the real world. He admits that
due to resistance of the medium, for instance, a projectile’s horizontal
motion cannot be truly constant in speed. He states quite
clearly that in reality the path of the projectile will not be exactly
parabolic. He argues, however, that his approximations
can be shown by experiment to be close enough to the real world
to be of very real use in the analysis of such motion.
In doing this, he became perhaps the first scientist to demonstrate
this modern scientific attitude. His approach was certainly
very different from that of the ancient Greek geometers, who were
only interested in exact results. A translation of Galileo’s
analysis of projectile motion can be read at:
Isaac Newton was the
first to propose that objects could be projected from the earth
and placed into orbit around the planet. He suggested that
such a projectile would have to be launched horizontally from
the top of a very high mountain. He argued that as the launch
velocity was increased, the distance that the projectile would fall
before hitting the earth would increase, until eventually, the downward
fall of the projectile would be just matched by the earth’s surface
falling away. At this point the projectile would never hit
the ground. This is just the motion, familiar to us now,
of a satellite in low earth orbit, which travels at about 8000 m/s.
Newton proposed a horizontal launch because he realized that any
projectile launched at an angle to the horizontal would attempt
to follow an elliptical path and would therefore eventually crash
back to earth. He also envisaged that if the launch velocity
became too great, the projectile would proceed away from the earth
and not return. Such a launch velocity became known as escape
Today we define escape velocity as the velocity at which an
object on the surface of a body must be propelled in order not to
return to that body under the influence of their mutual
gravitational attraction. There are many equivalent
definitions. I like this one in particular because it avoids
implying that the object being launched has some how magically
escaped completely from the body's gravitational field. In the
Newtonian theory of gravity, the object being launched will never
completely escape the gravitational field of the body from which it
was launched because the gravitational fields of all bodies are
infinite in range. However, it is certainly true to say that
if an object is launched with a sufficiently high velocity, it will
escape from any effective influence of the gravitational
field of the body from which it was launched.
By considering the mechanical energy (KE & PE) of an object
trying to escape from earth’s gravitational field, it can be shown
that the escape velocity, ve, for earth is:
where G = the gravitational constant, ME = mass of
earth and RE = radius of earth.
Clearly, escape velocity depends on the gravitational constant
and the mass and radius of the planet from which the object is to
escape. Earth’s escape velocity works out to be 11.2 km/s.
For an applet on Newton's Cannon click
on the link that follows:
A Brief Look At Circular
Satellites in low earth orbit or geosynchronous orbit move with
uniform circular motion. This is motion in a circular
path at a constant speed. Obviously, although the speed
is constant, the velocity is not, since the direction of the motion
is always changing. It can be shown that for an object executing
uniform circular motion (UCM), the acceleration keeping the object
in its circular path is given by:
ac = v2/r
where ac is called the centripetal
("centre-seeking") acceleration, v = velocity of the object and r = radius of the circular path.
As the name implies, centripetal acceleration is directed towards
the centre of the circle.
Clearly, the centripetal force, Fc, acting on an object
undergoing UCM is given by:
where m = mass of the object. This force is also directed
towards the centre of the circle.
The Australian satellite FedSat has a mass of 58 kg and executes UCM with a speed of 7.46 x 103 m/s at an altitude
of 800 km. Find the centripetal acceleration and the centripetal
force acting on this satellite. (Radius of earth = 6.4 x 106
From our centripetal acceleration formula above:
ac = (7.46
x 103)2 / (6.4 x 106 + 800 x 103)
ac = 7.73 m/s2
Fc = m . ac = 58 x 7.73 = 448 N
Note that for a satellite, the centripetal force is supplied
by the gravitational attraction between the earth and the satellite.
Thus, using our knowledge of circular motion theory and an understanding
of Newton’s Law of Universal Gravitation, we can calculate much
about the motion of satellites. For instance we can write
where G is the gravitational constant, M = mass of the earth,
m = mass of satellite, v = speed of satellite and r = distance of
satellite from centre of earth. This can be used to calculate
the velocity of the satellite at any value of r.
Also, the fact that the centripetal acceleration of a satellite
is supplied by the gravitational attraction between the earth and
the satellite, means that we can write:
ac = g = v2/r
where g is the acceleration due to gravity
at a distance r from the centre of the earth.
of Rocket Launches
A rocket is a system that undergoes a kind of continuous explosion.
The launch begins with the rocket producing thrust by burning fuel
and expelling the resulting hot gases out one end. These hot
gases have a momentum in one direction, and since the total momentum
of the rocket-fuel system is zero, the rocket itself has an equal
momentum in the opposite direction. Thus,
the rocket moves off in the opposite direction to the expelled gases,
in accordance with the Law of Conservation of Momentum.
As the launch proceeds, fuel is burnt, gases expelled and the
mass of the rocket decreases. This produces an increase in
acceleration, since acceleration is proportional to the applied
force (the thrust) and inversely proportional to the mass.
The initial acceleration is small, around 1 m/s2 but
continues to build as the mass of the rocket decreases and the atmosphere
becomes thinner. Clearly, as the acceleration of the rocket
increases, so too do the forces experienced by the astronauts.
The term “g-forces” is often used to describe
the forces acting on pilots and astronauts. This term describes
the size of the force in terms of the acceleration produced.
A g-force of 2g, for instance, is a force that produces acceleration
equivalent to 2 times the acceleration due to gravity at the surface
of the earth (ie 19.6 m/s2). Accelerations
during a launch can go as high as 10g depending on the type of rocket
The Saturn V rockets that were used to launch the Apollo spacecraft
stood 110.6 m tall (just less than the height of The Giant
Drop at Dreamworld), had a mass of 2.7 million kg and developed
a thrust at lift off of about 33.7 million N. Such rockets
produce g-forces as high as 10g on astronauts for several seconds
during launch. The Space Shuttle has an initial mass of about
2 million kg and produces about 31 million N of thrust. The
design of the shuttle engines enables much lower accelerations during
the final stages of launch than previous generations of rocket.
During launch an astronaut is seated
horizontally in a specially contoured chair. That
is, the astronaut has his/her back and legs below the knees in the
horizontal plane. In this way most of the accelerations
the astronaut experiences during launch are directed vertically
upwards through his/her well-supported, horizontal back.
Such, so-called transverse accelerations cause much less
stress to the body than accelerations applied along the long axis
of the body. Humans can tolerate up to 12g of transverse acceleration
without undue discomfort or visual disorders. Accelerations
acting along the long axis of the body can cause serious stress
for fairly low g-forces.
A pilot can experience acceleration applied in the direction
from the feet to the head. This results when the pilot pulls
out of a steep dive. The pilot feels heavier and is therefore
said to be experiencing positive g-forces. In this
case, the blood is left behind in the feet so to speak. At
3g to 4g this loss of blood from the head may cause “grey outs”
(blurring of vision caused by lack of blood flow in the eyes) or
“black outs” (loss of vision) and from 3g to 5g loss of consciousness.
A pilot can also experience acceleration applied in the direction
from the head to the feet. This results when the pilot executes
an outside loop or pushover at the start of a dive. The pilot feels lighter or weightless
and is therefore said to be experiencing negative g-forces.
In this case, the blood is left behind in the upper body.
At 3g to 4.5g this accumulation of blood in the upper body can cause
red outs (all objects appear red) or loss of vision or unconsciousness.
The average endurable times for tolerating negative g-forces are:
15 seconds at 4.5g and 30 seconds at 3g.
You do not need to be an astronaut or pilot to have experienced
positive and negative g-forces. Any person who has ridden a roller coaster
has experienced these effects. As you pull out
of a steep drop or negotiate an inside loop, you experience positive
g-forces and feel heavier. This is a similar feeling to
that experienced by astronauts at launch. As you fall
from a height, you experience negative g-forces and feel somewhat
The final aspect of launching a rocket that we will deal with
is the impact that the earth’s orbital motion
around the Sun and its rotational motion around its axis has on
the launch. The earth rotates on its own axis in
an easterly direction. At the latitude of NASA’s Cape
Canaveral launch site, the speed of rotation is about 400 m/s.
So, it makes good sense to launch a rocket in the easterly direction.
That way, it already has 400 m/s worth of speed in the direction
it wishes to go.
To penetrate the dense lower portion of the atmosphere by the
shortest possible route, rockets are initially launched vertically
from the launch pad. As the rocket climbs, its trajectory
is tilted in the easterly direction by the guidance system to take
advantage of the earth’s rotational motion. Eventually,
the rocket is travelling parallel to the earth’s surface immediately
below and can then be manoeuvred into earth orbit.
For a spacecraft to go on a mission to another planet, it is first
necessary for the spacecraft to achieve escape velocity and to
go into its own elliptical orbit around the Sun. The earth
orbits the Sun at about 30 km/s. Again it makes good sense
to use this speed to help a spacecraft achieve escape velocity
for trips to other planets. So, if the spacecraft is to go
on a mission to planets beyond the earth’s orbit, it is launched
in the direction of earth’s orbital motion and achieves a velocity
around the Sun greater than the earth’s 30 km/s. Thus, the
spacecraft’s orbit is larger than that of the earth and is arranged
to intersect with the orbit of the planet to which it is heading
at a time when the planet will be at that point. Similarly,
if the target is Mercury or Venus, the spacecraft is launched in
the opposite direction to the earth’s motion through space.
Then, the spacecraft achieves an escape velocity less than 30 km/s,
enters an elliptical orbit around the Sun that is smaller than the
earth’s and can thus intercept either planet.
It should be noted that the calculations involved in ensuring
that spacecraft are launched at the best possible time to intercept
their target are extremely complicated. One important equation
is the familiar Kepler’s
Third Law equation:
where T = period of satellite around central body, r = distance
from centre of central body to satellite, M = mass of central body
and G = gravitational constant. This equation can be used
to calculate the period of the spacecraft’s orbit around the Sun,
which is one of the important pieces of data in arranging the interception
with the target planet.
Satellites are placed in one of several different types of orbit
depending on the nature of their mission. Two common orbit
types are “Low Earth Orbit” and “Geostationary Orbit”.
Low Earth Orbit (LEO) is defined to be from 100 to 1000 km above
the earth’s surface. The Space Shuttle uses this type
of orbit (200-250 km). Geostationary Orbit (GEO) is defined
as one in which the satellite has a circular orbit in the earth’s
equatorial plane with a period of 24 hours. A satellite in such
an orbit remains above the same point on the equator at all times.
The following table shows a brief comparison of the features of
these two orbits.
60 – 90 minute orbit time
Smallest field of view
Frequent coverage of specific or varied locations
Orbits at less than 400 km difficult to maintain due
to satellite drag
Allows tracking of stationary point on
Largest field of view
To achieve and maintain a stable orbit around a
planet, a satellite must have a certain velocity. In general, we define the term orbital
velocity to be the velocity required by a satellite to enter
and maintain a particular orbit around a celestial object.
If we assume the orbit of the satellite around the celestial object
is circular, we can use Kepler’s Third Law (The Law of Periods)
to obtain an equation for the orbital velocity of the satellite.
Starting with the Law of Periods equation:
where T = period of satellite around the central
body, r = distance from centre of the central body to the satellite,
M = mass of central body and G = gravitational constant, and substituting
T = (2pr
/ v) for T, we obtain:
where v = orbital velocity of the
Satellites in Low Earth Orbit experience friction as they
move through or skim across the top of the very thin atmosphere
at LEO altitudes. This frictional effect is called atmospheric
(or satellite) drag. This causes a satellite to slow down
and lose altitude. This loss of altitude is referred to as
EXERCISE: Determine the orbital speed of the Earth around
the Sun, given that the mass of the Sun is 1.989 x 1030
kg, the radius of Earth’s orbit around the Sun is 149.6 x 106
km and the gravitational constant is 6.67 x 10-11 SI
Units. (v = 29.8 km/s)
Trajectories (Slingshot Effect)
Material beyond what the Syllabus requires: Let us
examine the basic physics of gravity-assists with reference to the
Cassini flyby of Venus in 1999.
Gravity causes the spacecraft to accelerate toward Venus,
reaching maximum speed at closest approach.
As it climbs up out of the gravitational field, the
spacecraft decelerates. Its
departure speed, relative to Venus is the same as its approach speed
but its direction of motion has changed.
Effectively, the spacecraft has picked up some of Venus’
angular momentum (mvr) and in so doing has gained speed relative to
Due to the large difference
in mass, the effect on the spacecraft is much greater than the
effect on the planet. When
the encounter is over, Venus is moving a little bit slower and the
spacecraft is moving a good deal faster.
As an aside, as Venus slows it falls ever so slightly closer
to the Sun, since it has lost some of its angular momentum.
However, as it falls it is accelerated and ends up moving
faster than it did before, by a very tiny amount.
It is also worth noting that gravity-assists can be used
to slow a spacecraft relative to the Sun.
To do this the spacecraft needs to pass in front of the
planet as viewed from the Sun.
FUTURE SPACE TRAVEL
A football is kicked with a speed of 25 m/s at an angle of 30o
to the horizontal. Neglecting
air resistance determine:
The time of flight for the football.
The maximum height reached by the football.
The horizontal range of the football.
A crate of supplies for a scientific expedition to Greenland
is being dropped by plane. When
the supplies are dropped, the plane is travelling at 40 m/s
horizontally at a height of 50 m.
Sadly the parachute fails to open and the package falls to
the ground at 9.8 m/s2.
Find the horizontal distance travelled by the package as it
falls given that it hits the ground right at the feet of the
scientific party. (127.8
A stone is thrown horizontally from the top of a vertical
cliff. Given that the
initial velocity of the stone is 20 m/s and that it hits the
horizontal ground below the cliff 3 seconds later, calculate:
The horizontal distance travelled by the stone from the foot
of the cliff. (60 m)
The height of the cliff.
The velocity of the stone just before it hits the ground. (35.6 m/s at an angle of 55.8o below the
A stone is thrown vertically upwards with a velocity of 29.4
m/s from the edge of a cliff 78.4 m high.
The stone falls so that it just misses the edge of the cliff
and falls to the ground at the foot of the cliff.
Determine the time taken by the stone to reach the ground.
Assume that the acceleration due to gravity is 9.8 ms-2.
Click on this link for a
solution to question 4.
TOPIC PROBLEMS I
Use Newton’s Law of Universal Gravitation equation to
determine the acceleration due to gravity on Mercury, Saturn &
Pluto to one decimal place. Hint
– Consider a mass m at rest on the surface of the planet in
question and express the weight of the mass m both in terms of the
universal gravitation equation and the Newton’s Second Law
equation F = ma. Necessary
data can be obtained from Ref 6 p.96.
Compare the values of g obtained with those in Ref 6 p.96.
(3.9 m/s2, 10.2 m/s2, 2.7 m/s2)
Using the value of g calculated for Mercury in question 1,
determine the weight force for a body of 65 kg on Mercury and
compare this to the weight force for the same body on Earth. What is the mass of the body on each planet?
(W = 253.5 N on Mercury & 637 N on Earth; Mass = 65 kg on
A projectile is launched at an angle of 45o to the
horizontal and just clears a fence 6 metres high at a horizontal
distance of 100 metres from the launch site.
Determine the velocity at which the projectile was launched. (32.3 m/s)
Explain why the acceleration of the Space Shuttle increases
during the initial periods of launch. Describe any effect this increase in acceleration may have on
Identify data sources, gather, analyse and present
information on the contribution to the development of space
exploration of ONE of the following:
The OPTUS satellites occupy geostationary orbits around the
Explain the meaning of the term “geostationary orbit”.
Given that the mass of the earth is 5.976 x 1024
kg and that the universal gravitational constant, G = 6.673 x 10-11
SI units, calculate the radius of the orbit of an OPTUS satellite in
kilometres. (4.2247 x
If the earth’s radius is 6370 km, determine the altitude of
an OPTUS satellite orbit in kilometres.
(3.5877 x 104 km)
If an OPTUS satellite has a mass of 70 kg determine the
centripetal force acting on the satellite as it undergoes circular
motion around the earth with an orbital velocity of 3067 m/s.
Hawking, S.W. & Israel, W. (1979).
General Relativity – An Einstein Centenary
Survey, Cambridge, Cambridge University Press
Hawking, S.W. (1988).
A Brief History of Time, London, Bantam
Whitrow, G.J. (1984). The Natural Philosophy
of Time, Oxford, Oxford University Press
Born, M. (1965). Einstein’s Theory of
Relativity, New York, Dover Publications Inc.
Resnick, R. (1968). Introduction To Special
Relativity, New York, Wiley
Bunn, D.J. (1990). Physics
for a Modern World, Sydney, The
Halliday, D. & Resnick, R. (1966). Physics
Parts I & II Combined Edition, New York, Wiley
to the rest of the notes on Topic 9.2 Space