Physics
Home > Physics > Core > Space > Space: 2. A successful rocket launch
9.2 Space: 2. A successful rocket launch
Syllabus reference (October 2002
version) 
2. Many factors have to be taken into account to
achieve a successful rocket launch, maintain a stable
orbit and return to Earth

Students learn to:

Students:
 solve problems and analyse information
to calculate the actual velocity of a projectile from
its horizontal and vertical components using:
 perform a firsthand investigation,
gather information and analyse data to calculate
initial and final velocity, maximum height reached,
range, time of flight of a projectile, for a range of
situations by using simulations, data loggers and
computer analysis
 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
 solve problems and analyse information
to calculate centripetal force acting on a satellite
undergoing uniform circular motion about the Earth
using:
 solve problems and analyse information
using:

Extract from Physics Stage 6 Syllabus (Amended
October 2002). © Board of Studies, NSW.
[Edit: 11 June 10]
Prior learning: Preliminary modules 8.4
(subsection 1), 8.5 (subsection 4)
Background: This section is concerned
with the context of launching a spacecraft, orbiting it
around the Earth and the issues of reentry. It begins by
considering projectile motion, within this context, and then
proceeds to rocket launches.
solve
problems and analyse
information to calculate
the actual velocity of a projectile from its horizontal and
vertical components using:

The actual velocity of a projectile can be
analysed into two components,
perpendicular to each other. In a convenient Earth frame
of reference, they are usually the vertical and
horizontal components. The vector sum of those components
is the actual velocity. Refer to the figure below.
Similarly, the velocity of a projectile at any time can
be resolved into components as follows:
Here are two sample problems with solutions:

Five seconds after launch, a projectile is tracked on
radar moving upwards at 26 m s^{1} at an angle
of 30^{0} to the horizon. What are the vertical
and horizontal components of motion?
Solution: The vertical component is
v_{y}; the horizontal component in
this case is v_{x}

Five seconds after launch from Earth, a projectile has
a horizontal velocity of 23.0 m s^{1} and a
vertical velocity of 11.5 m s^{1} up.
Determine its velocity.
Hint: when calculating velocity, which is a vector
quantity, it is usual to show both the magnitude and
direction of motion. This is a general rule to use when
calculating all vector quantities when no other
direction is given in the question.
Solution:
Here are two other problems for you to
solve:
 A test rocket flight is tracked on radar, which
displays its horizontal component of velocity at 7.5 m
s^{1} and its vertical component of velocity at
1.4 m s^{1} down. Calculate its actual velocity.
 A piece of space junk is being tracked by radar on a
ship in the Pacific as it plunges into the ocean. The
vertical and horizontal components of the object’s
velocity are 7.5 m s^{1} and 16 m s^{1}
respectively. Calculate the actual velocity of the
falling object.
Answers
describe
the trajectory of an object undergoing projectile motion
within the Earth’s gravitational field in terms of
horizontal and vertical components

The trajectory of a projectile in the Earth’s
gravitational field is parabolic, provided that air
resistance is ignored and the acceleration due to gravity
is uniform.
 This complex motion can be analysed by considering
its horizontal and vertical components at particular
instances during the flight. The horizontal motion of
the projectile is a constant velocity (air resistance
is assumed negligible). Its vertical motion is changing
all the time due to the effect of gravity, which causes
the projectile to accelerate at 9.8 m s^{2}
downwards.
What's the big idea in Physics? NSW CLIC
describe
Galileo’s analysis of projectile motion

Galileo postulated that all objects, regardless of their
mass, fall at the same rate. In other words, the
acceleration due to gravity is the same for all objects.
This is only true if air resistance can be ignored.
 Experiments based on dropping different objects
proved inconclusive so Galileo developed a method of
rolling balls down highly polished ramps in order to
make his comparisons. This enabled him to slow down the
motion enough to make more accurate observations of the
motion.
perform a
firsthand investigation, gather
secondary information and analyse data
to calculate
initial and final velocity, maximum height reached, range,
time of flight of a projectile for a range of situations by
using simulations, data loggers and computer analysis
 There are many suitable practical experiences that can
be performed as firsthand investigations
to achieve these measurements and
analyses. When gathering
secondary information, a variety of sources may be
suitable, including the Internet.
An online investigation can be found at Projectile Motion The University of Tennessee, Department of Physics and Astronomy.
Another investigation
Projectile Motion is
available at the web site of the Physics Education Technology,
University of Colorado at Boulder, Colorado, USA.
explain
the concept of escape velocity in terms of the:
 gravitational constant
 mass and radius of the planet
 Escape velocity is the initial velocity required by a
projectile to rise vertically and just escape the
gravitational field of a planet.
During this rise, the projectile’s kinetic energy
transforms into gravitational potential energy, so that:
E_{k} initially = E_{p} finally
Background
A useful equation
From the relationship described above, the following
expression for escape velocity of a planet can be
deduced.
The physics syllabus does NOT require
students to know or manipulate this
equation.

From the expression provided in Background
above, it can be seen that the escape velocity depends
upon the universal gravitational constant, G, the mass of
the planet and the radius of the planet. It does not
depend on any intrinsic property of the projectile.The
escape velocity for objects leaving the Earth works out
to be approximately 40 000 km h^{1}.
It is worth noting that this is a hypothetical concept
only. It would not be possible to successfully perform
such a launch for two reasons.
 Trying to travel through the Earth’s dense
lower atmosphere at this speed would produce an
enormous amount of heat, sufficient to vaporise the
projectile.
 Any living thing or delicate equipment would be
crushed by the enormous g forces created by
the process of suddenly being accelerated to 40 000 km
h^{1}.
outline
Newton’s concept of escape velocity

Newton created a hypothetical scenario as follows. A
person climbed a very tall mountain and launched a
projectile horizontally from the peak. The projectile
follows a parabolic path (see the above discussion
relating to projectile motion) before striking the
ground. If another projectile were launched faster than
the first, then it would travel further before striking
the ground. If yet another projectile were launched fast
enough, then it should be able to travel right around the
Earth because, as it falls, the surface of the Earth
curves away from it. The curve of the projectile’s
motion would follow that of the earth’s surface and
thus not hit it. This projectile would then be in a
circular orbit at a fixed height above the earth’s
surface. (For an analysis of the forces operating to
establish this circular motion, see below.)
 If a projectile is launched still faster, its orbit
will stretch out into an elliptical shape. Even faster
launch velocities result in the projectile following a
parabolic or hyperbolic path away from the Earth, escaping
it entirely.
identify why
the term ‘g forces’ is used to explain the forces
acting on an astronaut during launch
 The term ‘g
force’ is used to express apparent
weight as a proportion of true weight. True weight cannot
be felt as it is a gravitational forceatadistance that
acts on each atom of a person’s body. Apparent weight
is what a person experiences or feels when an external
force acts on them to cause a change in their motion
(either magnitude or direction or both). These forces are
also called inertial forces because they arise from the
body’s inertia or resistance to having its motion
changed. They can be calculated.
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
 Some useful data sources on rocket
pioneers that are available from the Internet are provided
below. Decide on the type of information you will collect.
Presenting information in chronological
sequence would be appropriate.
The evolution of the rocket NASA
Robert Goddard and his rockets NASA
The pioneers of rocketry & space
travel The ThinkQuest web site
Note that there are several other important rocket
pioneers not mentioned in this syllabus point for which
information is readily available. Conversely, not all of
the pioneers mentioned have a plentiful supply of
information available. Nevertheless, there is a great deal
of information available on Tsiolkovsky and Goddard, in
particular.
 Gather information from a range of
sources. Analyse the information by
identifying trends and relationships as well as
contradictions in data and information.
 Select and use an appropriate media to
present your data and information. The use
of an annotated timeline would be appropriate.
discuss
the effect of the Earth's orbital motion and its
rotational motion on the launch of a rocket
 The rotation of the Earth on its axis, and the orbital
motion of the Earth around the Sun, can be used to provide
a launched rocket with a velocity boost. This allows the
operators of the rocket to save fuel in achieving the
target velocity.
 Rockets launching into orbit are launched to the east,
in the direction of the Earth’s rotation. They
therefore receive a 1700 km h^{1} boost toward
their target velocity of approximately 30 000 km
h^{1} for a low Earth orbit.
 Rockets that are heading away from the Earth and
further into space are not launched until the direction of
the Earth in its orbit around the Sun corresponds with the
desired direction. The rocket is then launched up to a low
Earth orbit, before firing its rockets again to accelerate
ahead of the Earth. The velocity boost received from the
Earth’s orbital motion in this case is approximately
107 000 km h^{1}.
analyse
the changing acceleration of a rocket during launch in terms
of the:
 Law of Conservation of Momentum
 forces experienced by
astronauts
 Law of Conservation of Momentum
 Rocket propulsion is derived from a force pair (as
described in Newton’s third law).
 This is an example of how the Law of Conservation
of Momentum can be used to analyse rocket motion. The
initial momentum of the rocket and its fuel is zero. This
sum must be preserved (the Law of Conservation of
Momentum). Note that while the mass of fuel burned in
a second is much less than the mass of the rocket, the
velocity of the hot exhaust gases is much greater than the
velocity of the rocket.
Note also that while the left side of the equation remains
quite constant during a burn, the terms on the right side
are changing. The mass of the rocket is decreasing
significantly as the fuel is burned (typically, 90% of a
rocket’s mass is fuel). This means that the velocity
of the rocket must increase significantly.

Forces experienced by astronauts
Note that two forces act upon an astronaut during launch:
the upward thrust (T) as well as the downward weight (W
or mg). Newton’s second law can be used to derive a
simple expression for acceleration of a rocket that is
launched directly up (using the diagram above):
 As described above, if the mass of the rocket decreases
during flight and the thrust remains constant, the
acceleration of the rocket (and astronauts) increases. Thus
the force experienced by the astronaut increases. Refer to
the graph above to see how the forces change at different
times during the flight into orbit around the earth.
analyse
the forces involved in uniform circular motion for a range of
objects, including satellites orbiting the Earth
 Objects do not perform uniform circular motion unless
they are subject to a centripetal force. This is a force
that is always perpendicular to the velocity of the object.
That force causes the moving object to continually change
direction so that it follows a circular path. The
centripetal force is always directed toward the centre of
the circular motion.
 The source of the centripetal force for a range of
circular motions is listed here.
Circular motion 
Source of centripetal force 
Ball on a string whirled in a circle 
Tension in the string 
Car driving around a corner 
Friction between the tyres and the road 
Satellite orbiting the Earth 
Gravitational attraction between the Earth and the
satellite 
compare
qualitatively low Earth and geostationary orbits
 A low Earth orbit is an orbit that lies above the
Earth’s atmosphere but below the van Allen radiation
belts. This means that its altitude is from approximately
250 km to 1 000 km above the surface of the Earth. A
satellite in low Earth orbit will need an orbital velocity
of approximately 28 000 km h^{1} to maintain the
orbit, which gives it an orbital period of approximately 90
minutes. Examples are the space shuttle (altitude
250–400 km) and the Hubble telescope (altitude 600
km).
 A satellite in a geostationary orbit has an orbital
period equal to the Earth’s, that is, it takes one
sidereal day (23 hours 56 minutes) to complete an orbit. To
achieve this, the satellite must have an altitude of
approximately 35 800 km, which places it near the upper
edge of the outer van Allen belt. Its orbital velocity is
approximately 11 000 km h^{1}. From the ground, a
satellite in this type of orbit appears fixed in the sky,
which makes it especially useful for communications and
weather satellites.
Background information outside the
syllabus
If you want to work through these figures for yourself,
the formulae you need are provided below. They may assist
you to understand what is happening in the next activity as
well.
Note: the formulae and calculations here are beyond the
scope of the syllabus at this point.
The centripetal force to maintain an orbit is provided
by gravity.
F_{gravity} = F_{centripetal}
GMm_{s}/r^{2} =
m_{s}v^{2}/r
M = mass of earth (kg); m_{s} = mass of satellite
(kg); G = universal gravitational constant; v = orbital
speed in linear terms (ms^{1}) & r = the
orbital radius (m)
The orbital speed (v) can also be calculated from the
orbit path (2πr) divided by the period (T) of motion
measured in seconds.
v = 2πr/T
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

Orbital velocity is the instantaneous speed (magnitude)
in the direction indicated by an arrow (directional)
drawn as a tangent to the point of interest on the
orbital path.
Quantitative relationship: (where
d = average distance between centres of the two
masses; v = orbital velocity; M =
larger mass; ms = smaller mass; G =
gravitational constant; T = period of orbit)
Fg = Fc (gravity is the centripetal
force)
Using Kepler’s Law of Periods,
:
solve
problems and analyse
information to calculate
centripetal force acting on a satellite undergoing uniform
circular motion about the Earth using:
F=mv^{2}/r

Recall that the centripetal force for a satellite’s
orbital motion is provided by the gravitational force of
attraction between the satellite and the Earth. Hence, by
using Newton’s Law of Universal Gravitation to
calculate this gravitational force, the centripetal force
can be determined.
Sample problem: The Hubble telescope
has a mass of 1 200 kg and orbits at an altitude of 600
km. Calculate the centripetal force acting on this
satellite as it performs circular motion around the
Earth and then use his calculated value to determine
the tangential velocity of the Hubble telescope.
Solution: The centripetal force can be determined by
calculating the gravitational force of attraction between
the Hubble telescope (m_{ht}) and the Earth (M)
and then the centripetal force can be equated to the
gravitational force to calculate the tangential velocity:

Problems could take the form of absolute calculations or
satellite comparisons. The expression below is a more
useful form of Kepler’s third law; the law of
periods. It can be applied to any orbiting object, e.g. a
satellite orbiting a planet or a planet orbiting the Sun.
If comparing two objects orbiting the same central body,
then the right hand side of the expression has a fixed
value and hence:
This was the form of the equation used by Kepler when
comparing planets to the Earth.
Sample problem 1: Determine the period
of the orbit of the Hubble space telescope, given that its
altitude is 600 km. Mass of the Earth = 6 ×
10^{24} kg. Radius of Earth = 6 380 km.
Solution: Radius of the orbit = 6380 km +
600 km = 6.98× 10^{6} m
Sample problem 2 (taken from specimen
paper): A planet in another solar system has three moons,
all of which travel in circular orbits. Some information
about these moons is given in the table.
Moon 
Radius of orbit (orbs) 
Period of revolution (reps) 
Alpha 
4.0 
16 
Beta 
9.0 
54 
Gamma 
2.5 

The radius of orbit and period of revolution are
measured in orbs and reps respectively, which are not
metric units.
 Use the data to show that Kepler’s third law is
obeyed for the moons Alpha and Beta.
 Calculate the speed of moon Gamma in orbs/rep.
Solution:
The first step is to calculate the period of
Gamma’s orbit using Kepler’s third law:
The orbital speed of Gamma can now be found as the
circumference of its orbit divided by the period
(remembering from Module 8.4, that the speed is related to
distance and time where the distance here is the
circumference and the time is the period).
account
for the orbital decay of satellites in low Earth orbit
 A satellite in a stable orbit around the Earth
possesses a certain amount of mechanical energy, which is
the sum of its kinetic energy (due to its high speed) and
its gravitational energy (due to its altitude). The lower
the altitude of the orbit, the lower the total mechanical
energy is.
 Satellites in low Earth orbit are subject to friction
with the sparse outer fringes of the atmosphere. This
friction results in a loss of energy. The loss of energy
means that this orbit is no longer viable and the satellite
drops down to an altitude that corresponds with its new,
lower energy. Ironically, the satellite will be moving
faster than before (recall that lower orbits require faster
orbital velocities) however the extra kinetic energy is
derived from the lost potential energy.
 This is the process of orbital decay, and it is cyclic,
as the satellites new lower orbit resides in slightly
denser atmosphere, which leads to further friction and loss
of energy. The process is not only continuous but speeds up
as time goes on.
discuss
issues associated with safe reentry for a manned spacecraft
into the Earth’s atmosphere and landing on the
Earth’s surface
 Heat: The considerable kinetic and potential energy
possessed by an orbiting spacecraft must be lost during
reentry. As the atmosphere decelerates the spacecraft, the
energy is converted into a great deal of heat. This heat
must be tolerated and/or minimised. The heat can be
tolerated by using heat shields that use ablating surfaces
(as used on Apollo capsules) or insulating surfaces (as
used on the space shuttle). The heat can be minimised by
taking longer to reenter, thereby lengthening the time
over which the energy is converted to heat. The space
shuttle uses this technique.
 g forces: The deceleration of a reentering
spacecraft also produces g forces, typically greater than
those experienced during launch. High g forces can
be better tolerated by reclining the astronaut, so that
blood is not forced away from the brain, and by fully
supporting the body. The g forces can be minimised
by extending the reentry, slowing the rate of descent.
This strategy is employed by the space shuttle.
 For a time during reentry, there is a radio blackout
caused by overheated air particles ionising as they collide
with the spacecraft. This may be a safety issue if contact
is needed between the spacecraft and earth at this phase of
its flight.
 Reaching the surface: Even after surviving the issues
listed above, the spacecraft must touch down softly onto
the surface of the Earth. Several solutions to this problem
have been employed, such as first using parachutes and then
splashing into ocean, or using many parachutes before
crunching onto the ground, or by landing on an air strip
(as performed by the space shuttle).
identify
that there is an optimum angle for reentry into the
Earth’s atmosphere and the consequences of failing to
achieve this angle
 For any given spacecraft wishing to reenter safely, an
optimum angle of reentry exists. For Apollo capsules this
angle was between 5.2° and 7.2°, although this
would differ for other spacecraft.
 If the angle is too shallow then the spacecraft will
rebound, due to compression of the atmosphere beneath it.
 If the angle is too steep then the spacecraft will
decelerate too quickly, creating too much heat and burning
up the spacecraft.