What's New
Site Support

9.2 Space
9.3 Motors & Generators
9.4 Ideas-Implementation
9.7 Astrophysics
9.8 Quanta to Quarks

9.2 Space Continued
9.2 Practicals

NOTE: The notes and worksheets for this topic are divided into two separate pages in order to keep download times within acceptable limits.  To access the next page on this topic click on the button at the left marked 9.2 Continued.




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 Field

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 apart. 

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 Law as:


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 NOT squared!

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 Accelerated Motion

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 vectors.

Recall also the definitions of velocity and acceleration and the equation:

vav  = (u + v)/2

for average velocity, which applies only when the acceleration is constant.




Projectile Motion

A projectile can be considered to be any object that has been launched or thrown in a particular direction from some point.  If a projectile moves in a gravitational field and gravity is the only force that acts on it, its path or trajectory is that of a parabola.  Since such a path is two-dimensional, it is convenient to resolve the motion into vector components in the horizontal and vertical directions.  The characteristics of these components are as follows: 

u    Horizontal Motion – a constant velocity motion, the velocity at all times being the same as the horizontal component of the initial velocity.

u    Vertical Motion – a uniformly accelerated motion in which the projectile experiences a constant downward acceleration of magnitude g (on earth). 

The advantage of resolving the motion into these vector components is that the horizontal and vertical components are completely independent of each other.  Thus, we may apply the equations of uniformly accelerated motion to each component separately and then add the two components together as vectors to obtain the actual motion of the projectile. 

The following diagram illustrates this idea.



Clearly, the horizontal velocity remains constant throughout the motion, as indicated by the horizontal velocity vector staying the same length.  The vertical velocity decreases as the projectile moves upwards, is zero at the maximum height attained by the projectile and then increases again as the projectile returns to the ground.  This is due to the constant downwards acceleration due to gravity.  Note that at any time we can calculate the total velocity by adding the vertical and horizontal vector components together.  The total velocity vectors are tangents to the trajectory.

If we assume the projectile whose trajectory is shown in the diagram above, was launched at velocity v0 at an angle q to the horizontal, we can construct the following vector diagram representing these initial conditions.


The zero subscripts attached to the velocity vectors denote that this is the initial velocity of the projectile, that is the velocity at time t = 0.  Using simple trigonometry we see that:

                                    vyo = vosinq 

                          vxo = vocosq

Note that these velocities are the initial vertical and horizontal velocities of the projectile.  The velocities at any time t after t = 0 are found by applying the appropriate uniformly accelerated motion equation v = u + at.

                                    vy = vosinq - gt 

                          vx = vocosq    (since vx remains constant)

So the magnitude and direction of the total velocity at any time t is given by:

                     V = Ö (vx2 + vy2)  and  a = tan-1(vy / vx)


where a is the angle made by the total velocity vector and the horizontal.



The displacement of the projectile at any time t is found from s = ut + 0.5at2: 

                                     y = vosinq.t - 0.5gt2 

                           x = vocosq.t

So the magnitude and direction of the total displacement at any time t is given by:


                         S = Ö (x2 + y2)  and  b = tan-1(y/x) 


where b is the angle made by the total displacement vector and the horizontal.

The fact that the trajectory of the projectile is a parabola can be shown by using the displacement equations (for y and x) given above and eliminating t from the equations.  This gives the standard equation of a parabola.  Try this as an exercise.




Galileo’s Analysis 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:





Escape Velocity 

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 velocity.

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:

Newton's Cannon





A Brief Look At Circular Motion 

Satellites in low earth orbit or geosynchronous orbit move with uniform circular motionThis 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.

Example: 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 m.)

From our centripetal acceleration formula above:

             ac = (7.46 x 103)2 / (6.4 x 106 + 800 x 103)

             ac =  7.73 m/s2

So,         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 that:



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.




Important Features 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 used.

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 weightless.

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.




Satellite Orbits

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.


Altitude (km)




100 – 1000

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


Earth Observation

Weather Monitoring

Shuttle Missions


35 800 km

Allows tracking of stationary point on earth

Largest field of view



Weather Monitoring


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 satellite.


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 Orbital Decay.

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) 




Factors Affecting Re-Entry

The term “re-entry” applied to rockets refers to the return of a spacecraft into the earth’s atmosphere and subsequent descent to earth.  Perhaps the most important issues associated with safe re-entry are: 

u    How to handle the intense heat generated as the spacecraft enters the earth’s atmosphere; and

u    How to keep the g-forces of deceleration within safe limits. 

On re-entry, friction between the spacecraft and the earth’s atmosphere generates a great deal of heat.  Early spacecraft such as NASA’s Mercury, Gemini & Apollo capsules, used heat shields made from what was called an ablative material that would burn up on re-entry and protect the crew from the high temperatures.  The Space Shuttle uses an assortment of materials to protect its crew from the intense heat.  Reinforced carbon-carbon composite, low and high temperature ceramic tiles and flexible surface insulation material all play important protective roles in appropriate positions on the Shuttle. 

It would be wonderful if a spacecraft could re-enter the atmosphere vertically.  Unfortunately, the thick, bottom section of our atmosphere that is used to effectively slow the spacecraft to safe landing speed is not sufficiently thick (~ 100 km) to allow for vertical re-entry.  So, the spacecraft is forced to re-enter at an angle to the horizontal of between 5.2o and 7.2o.  This small angular corridor is called the re-entry window.  If the astronauts re-enter at too shallow an angle, the spacecraft will bounce off the atmosphere back into space.  If the astronauts re-enter at too steep an angle, both the g-forces and the heat generated will be too great for the crew to survive.




Gravity-Assisted Trajectories (Slingshot Effect)

Sometimes the gravitational fields of planets can be used to increase the speed of spacecraft relative to the Sun and thus reduce travel times and minimize fuel and energy demands.  Such spacecraft are said to be on “gravity-assist trajectories” and the whole process is often referred to as the “slingshot” effect.  Essentially, the spacecraft moves behind the planet as viewed from the Sun, and is accelerated by this moving gravity field, much as a surfer is pushed forward by a wave.  The energy gained by the spacecraft does not actually come from the gravitational field but from the kinetic energy of the moving planet, which is slowed by a tiny amount in its orbit, causing it to drop very slightly closer to the Sun. 

Let us consider the example of the Cassini spacecraft that reached Saturn in 2004.  On its long voyage, Cassini boosted its speed by passing close to Venus, Earth and Jupiter.  With each flyby its orbit was adjusted until it gained sufficient speed relative to the Sun to reach Saturn.  At that point NASA adjusted its velocity to put it into orbit.

A more recent example of a gravity assist maneuver in February 2007 saw the New Horizons spacecraft on its way to Pluto gain 4 km/s (14 400 km/h) as it was dragged along by Jupiter’s gravitational field on its approach to the massive planet.  Read more about this at: http://www.armaghplanet.com/blog/how-did-new-horizons-get-to-pluto-so-quickly.html .

Have a look at this Flash animation that allows you to simulate various slingshot trajectories around Jupiter.


Extra 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 the Sun.

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.





It is one of the great dreams of science that human beings will one day travel to not only the planets of our own Solar System but indeed to the other Stars in our Milky Way Galaxy and eventually even to other Galaxies.  One of the major factors preventing such exciting voyages at present is that current maximum velocities for spacecraft are far too slow to make such trips possible.  The Space Shuttle can travel at around 11 km/s which is about 40 000 km/h.  On their trips to the moon the Apollo spacecraft reached velocities of around the same magnitude.  Among the fastest man-made objects is Voyager 2 now hurtling towards the stars at a speed of around 1.08 x 105 km/h (ie 108 000 km/h).  A quick calculation shows that if Voyager 2 is as fast as we can go at the moment, then a trip to Mars will still take about 30 days.  A trip to Jupiter will take around a year.  A trip to a-Centauri, our nearest star outside the Sun, will take around 43 300 years.  Clearly, even if it were possible to give all spacecraft the same high speed that Voyager 2 has reached, it would only make a very small difference to our ability to travel to distant places. 

New technology is being researched all the time.  Nuclear fusion may one day be able to supply a steady thrust to spacecraft.  Ion engines have been investigated in which thrust is produced by the expulsion of ions from a spacecraft.  On the drawing board side, there is a proposal to use huge mirrors to catch the momentum of photons of light from the Sun as they are reflected from the mirrors.  This is called “Solar Sailing”.  There are even theoretical suggestions that a process called Antimatter Initiated Microfusion (AIM) may produce an efficient hybrid fission/fusion drive (see New Scientist No.2260, 14 October 2000, p.6, for more details). 

Another aspect of concern for future space travel is that of communication.  The difficulties associated with effective and reliable communications between spacecraft and earth can be attributed to: 

u    Distance: Very specialised receiving equipment is needed to enable the detection of the weak microwave signals sent back to earth from distant spacecraft.  This is due to the fact that electromagnetic radiation obeys the inverse square law and so the intensity of the EM radiation decreases as the square of the distance from the source. 

Due to size and energy limitations, most spacecraft, especially unmanned probes, carry only low power radio transmitters (~ 20W).  Microwave signals are therefore normally used for communication with spacecraft since they can be sent in a particular direction in a single beam.  This is very useful when sending signals to a single point in space, a huge distance away.  Normally one frequency is used for the signal sent from earth to the satellite (the uplink) and a slightly different frequency in the same band is used for the signal from the satellite to earth (the downlink).

u    The Van Allen Radiation Belts: As you will recall from the Cosmic Engine topic, the Van Allen radiation belts are two regions of energetic charged particles, mainly electrons and protons, surrounding the earth.  The ring current formed by the motion of the charged particles in the outer belt increases once or twice a month.  This causes an increase in the magnitude of the associated magnetic field, which can lead to communication disruptions such as interference in short wave radio transmissions, and data transmission errors in communications satellites.

u    Sunspot Activity: Again, as was seen in the Cosmic Engine topic, sunspots are relatively cool areas of the sun’s photosphere with associated strong magnetic fields.  As sunspot activity increases during the eleven-year sunspot cycle, so too does the number of particles in the Solar Wind.  This increased concentration of charged particles in the Solar Wind can disrupt the earth’s magnetic field and cause similar disruption to communications as mentioned above.





1        A football is kicked with a speed of 25 m/s at an angle of 30o to the horizontal.  Neglecting air resistance determine:

(a)   The time of flight for the football.  (2.55 s) 

(b)   The maximum height reached by the football.  (7.97 m)

(c)    The horizontal range of the football.  (55.2 m)

2        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 m)

3        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:

(a)   The horizontal distance travelled by the stone from the foot of the cliff.  (60 m)

(b)   The height of the cliff.  (44.1 m)

(c)    The velocity of the stone just before it hits the ground.  (35.6 m/s at an angle of 55.8o below the horizontal)


4        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(8 s)

Click on this link for a solution to question 4.





1        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)

2        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 each planet.)

3        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)

4        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 the astronauts.

5        Identify data sources, gather, analyse and present information on the contribution to the development of space exploration of ONE of the following:

Tsiolkovsky, Oberth, Goddard, Esnault-Pelterie, O’Neill or von Braun.

6        The OPTUS satellites occupy geostationary orbits around the earth.

(a)   Explain the meaning of the term “geostationary orbit”.

(b)   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 104 km)

(c)    If the earth’s radius is 6370 km, determine the altitude of an OPTUS satellite orbit in kilometres.  (3.5877 x 104 km)

(d)   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.  (15.6 N)



1.      Hawking, S.W. & Israel, W. (1979).  General Relativity – An Einstein Centenary Survey, Cambridge, Cambridge University Press

2.      Hawking, S.W. (1988).  A Brief History of Time, London, Bantam Press

3.      Whitrow, G.J. (1984). The Natural Philosophy of Time, Oxford, Oxford University Press

4.      Born, M. (1965). Einstein’s Theory of Relativity, New York, Dover Publications Inc.

5.      Resnick, R. (1968). Introduction To Special Relativity, New York, Wiley

6.      Bunn, D.J. (1990). Physics for a Modern World, Sydney, The Jacaranda Press

7.      Halliday, D. & Resnick, R. (1966). Physics Parts I & II Combined Edition, New York, Wiley


Go to the rest of the notes on Topic 9.2 Space


Last updated:

© Robert Emery 2002 - view the Terms of Use of this site.