# Category Archives: HSC Physics – Space

HSC Physics – Space

# Outline the features of the aether model for the transmission of light

The concept that the aether is a stationary or absolute rest frame requires an understanding of frames of reference and relative motion. Scientists today agree that there is no absolute reference frame and the motion of objects can only be measured relative to other objects. In turn these other objects may be moving relative to still other objects. For example, a person on a train throws a ball. Relative to the train, the ball is travelling north at 5m/s. However, the train is travelling south at 20m/s, and so relative to a person on the Earth’s surface next to the train the ball is travelling south at 15m/s. A person on an aircraft travelling north at 40m/s observes this same event, and sees that the ball is travelling south at 55m/s relative to him. An observer outside the solar system will see the ball’s motion in light of the orbital motion of the Earth, and an observer outside the galaxy will see the ball’s motion in light not only of the orbital motion of the Earth, but the motion of the Sun as it orbits around the centre of the galaxy. In this way it is impossible to “truly” determine an object’s velocity in absolute terms- there is no one “correct” answer for the ball’s velocity, and each of the observations made (in the train, outside the train, in the aircraft etc.) is equally valid. Previously, scientists thought that motion could be determined in absolute terms by measuring motion relative to the aether. Under such a model, the ball may be travelling west at 30m/s relative to the aether (an arbitrary figure) and this would be its true velocity. This is what is meant by the aether being a stationary frame, with all objects moving relative to it. This explanation is not part of the dotpoint and so is not necessary for an exam response. It exists only to clarify the meaning of “absolute rest frame”.

According to the aether model for transmission of light, light was a wave that propagated through a material called the “aether”. According to the model, aether had no mass, could not be seen, heard or felt, and was distributed evenly throughout the universe residing between the particles that make up matter. Further, it was considered to be an absolute rest frame, meaning that the absolute motion of all objects in the universe could be measured relative to the aether.

Remember- The aether was invisible, without mass, existed at all points in the universe, is an abso- lute rest frame, and was the medium for light.

# Describe and evaluate the Michelson-Morley attempt to measure the relative velocity through the aether

Be aware that the failure of the Michelson-Morley experiment to observe a changing interference pattern does not disprove the existence of the aether. All it does is question the theory and prove that either the theory or the experiment is flawed. Einstein subsequently interpreted this experiment as disproving the aether, but the experiment itself did not disprove the aether.

If the aether is stationary and the Earth is moving through the aether, then it follows that there is an aether “wind” that will affect the apparent speed of light to an observer on the Earth. The Michelson-Morley experiment was designed to analyse the aether wind, and thus calculate the velocity of Earth through space. A beam of light was split and sent into two directions at 90 degrees to each other horizontally by a half-silvered mirror. They were then reflected back and combined, such that both rays had travelled the same distance. This recombining process resulted in an interference pattern. The device was floated on liquid mercury, which enabled smooth rotation of the entire experiment. As the device was rotated, the aether wind was expected to cause the light to travel at different speeds in each direction, thus causing the interference pattern to change. The velocity of the Earth would be calculated by analysing the changing interference pattern. However, despite extensive testing, no change in the interference pattern was observed. This led to the conclusion that the aether model was flawed, which subsequently led to the conclusion that the aether did not exist. In terms of calculating the velocity of the Earth, the Michelson-Morley experiment was a failure, but its conclusion, based on results that were both valid and reliable changed scientific theory dramatically, making it one of history’s most important experiments.

Remember- The Michelson-Morley experiment failed in its goal to determine the speed of the Earth through the aether.

# Gather and process information to interpret the results of the Michelson- Morley experiment

The Michelson-Morley experiment was designed to calculate the velocity of the Earth through the aether, on the grounds that light would travel faster in certain directions and slower in others, due to the relative motion between the Earth and the aether. The Michelson-Morley experiment split a light beam, creating two beams at right-angles to each other, and after letting them travel for a short distance, recombined them. As the differences between the speed of light change when the device rotates, the interference pattern formed also changes as the phase difference between the two beams change. However, despite much repetition the experiment showed that light seemed to travel at the same speed in all directions, because the interference pattern formed never changed even when the orientation of the experiment was changed by rotating the apparatus. The experiment therefore provided a null result, neither disproving nor proving the existence of the aether. However, the results of the experiment could be taken in two ways- that the Earth wasn’t moving through the aether, or that the aether model was flawed. Since the Earth was known to move, the Michelson- Morley experiment provided the final evidence that debunked the aether model for light transmission. Einstein interpreted the results of the experiment as confirming his theories as to the constancy of the speed of light, as well as the non-existence of the aether.

Remember- The Michelson-Morley experiment demonstrated that the speed of light on Earth was constant in all directions, significant evidence towards disproving the aether model.

# Discuss the role of the Michelson-Morley experiments in making determina- tions about competing theories

The Michelson-Morley experiment produced startling results that in the end disproved the aether model for transmission of light and instead supported Einstein’s model of light. At the time of the experiment there were two competing theories- the aether model in which light propagated through a stationary aether through which the Earth moved, and Einstein’s model, part of which specifying that light travelled at a constant speed under all circumstances. The Michelson-Morley experiment showed that light travelled at a constant speed in all directions, and challenged the aether theory by showing that there was no aether wind. So the Michelson-Morley experiment provided pivotal evidence that determined the survival of competing theories as to the transmission of light.

Remember- The Michelson-Morley experiment helped prove Einstein’s theory while debunking the aether theory.

# Outline the nature of inertial frames of reference

In terms of Newton’s laws holding true, an inertial reference frame is one in which fictitious forces are not required to account for motion. For example, consider the rotating ride “Rotor” at Luna Park (Sydney), a ride where people are placed inside a rapidly spinning cylinder so that they are pinned to the sides of the cylinder. To an observer on the deck above, it is quite clear that the people inside the ride travel in a circular path because the walls of the ride exert centripetal force. However, an observer in the ride feels a force pressing them into the walls of the ride. To the person outside, this is simply their inertia pushing them against the wall. But to the observer inside, they may not even be moving- all the objects inside “Rotor” are stationary relative to them (as they are spinning along with the ride). Therefore, the fictitious force centrifugal force is pressing them against the wall of the ride. This force is fictitious because it does not exist as an “action” force in all inertial frames of reference- it exists in the frame inside Rotor but in the frame outside it is observed as a reaction force. Fictitious forces only exist in non-inertial reference frames, and so it can be concluded from this that the rotating cylinder in “Rotor” is non-inertial (which is true, as it is constantly accelerating because it rotates). Further, if the rider threw a ball straight into the middle of the ride, they would find that the ball would not travel in a straight line, disobeying Newton’s laws, again showing that the laws only hold directly true in an inertial frame. This needn’t be detailed in an answer- however, it is an important concept to understand. This answer has focussed on the use of the fictitious centrifugal force to show the difference between inertial and non-inertial frames of reference. For a more in-depth examination of centrifugal force itself, see the Extra Content chapter at the end of the Guide.

A frame of reference is essentially the environment from which measurements are taken by an ob- server. It can be a stationary room or a moving train. An inertial frame of reference is one in which no net force is acting, and in which all of Newton’s laws hold true. No mechanical experiment or observation from within the frame can reveal if the frame is moving with constant velocity or at rest.

Remember- An inertial frame of reference is any frame that isn’t accelerating.

# Perform an investigation to help distinguish between non-inertial and inertial frames of reference

The experiment we carried out distinguished between non-inertial and inertial frames of reference by considering the definition of an inertial frame- one where all the laws of physics hold directly true and one which is indistinguishable from another inertial frame. In our experiment, we had a pulley with a string attached to a spring balance, holding a 100g weight. We took the apparatus as being an inertial reference frame when stationary- at that point the spring balance registered 100g. When we pulled the rope to cause the balance and weight to rise at a constant velocity, the spring balance still indicated 100g, showing that the constant-velocity frame was inertial. However, when we pulled the rope increasingly faster to cause the spring to accelerate upwards, it registered more than 100g, because according to F = ma, it was exerting extra force on the weight to cause it to accelerate upwards. Because this accelerating frame indicated a different value from the stationary 100g, we identified it as a non-inertial frame where the laws of physics do not directly hold true (in this case, because the 100g weight was indicated as weighing more by the spring balance while accelerating).

Remember- Spring balance with a pulley experiment, pulling the rope changed the reading on the balance.

# Discuss the principle of relativity

Although some interpret this dotpoint as only covering classical Galilean relativity, it is useful at this point to consider Einstein’s special relativity as well.

The classical principle of relativity was first explored by Galileo, and then developed upon by New- ton, and states that no measurement made from within an inertial reference frame can be used to determine the velocity of that frame. This means that when within an inertial frame of reference, it is impossible to determine whether the frame is moving or not, unless measurements are taken involving observations outside the frame. For example, consider a train that is travelling at a con- stant velocity. From within the train, there is no observation that can be made to determine whether the train is stopped at a station (with a constant velocity of 0) or travelling at a constant velocity. This is because the train is an inertial frame of reference (so long as it is travelling at a constant velocity). The only way to determine the motion of the train is to make observations of other frames from within the train- for example, looking out of the window to the frame outside the train to see whether the train is moving or not. Effectively, this means that all inertial frames of reference are equal and equally correct- there is no such thing as an absolute rest frame against which all motion can be measured since all inertial reference frames are equal.

In 1905 Einstein devised his theory of special relativity. It was based on two key postulates- firstly, that the laws of physics are the same for all inertial reference frames (and by that it is meant that all inertial frames are equal and cannot be distinguished from another- there is no absolute rest frame) and secondly that the speed of light is constant for all observers. The idea that the speed of light is constant for all observers was extremely revolutionary because of its implications. Thought experiments, and subsequently physical experiments, showed that as observed velocity increases, time dilates, length contacts, and mass increases. Essentially, the principle of relativity states that nothing in the universe is constant except for the speed of light, and everything else is dependant on the relative movement between frames of reference. Although it was able to explain evidence (such as the Michelson-Morley experiment) and make predictions about the behaviour of light, this extremely revolutionary idea had little evidence to directly prove it when it was formulated. As a result, it took many years for the principle of relativity to become part of mainstream science.

Remember- No measurement from within an inertial reference frame can determine anything about the movement of that frame, all motion occurs relative to something else, and the speed of light is constant for all observers.

# Describe the significance of Einstein’s assumption of the constancy of the speed of light

Einstein’s key postulate was that the speed of light is constant for all observers. This means that whenever an observer takes measurements to determine the speed of light, the value calculated is always the same. However, in many cases Newtonian vector addition will increase the distance travelled by light as observed by a stationary observer. Under traditional vector addition, calculating

the velocity by dividing distance by time would break Einstein’s postulate resulting in a value greater than 3 × 108. The consequence and significance of the speed of light being constant is that mass, length and time change so that the speed of light can never be exceeded. This is extremely significant to predicting how objects behave at relativistic velocities.

Remember- The speed of light being constant is significant because it means mass, time and length all become variable.

# Analyse and interpret some of Einstein’s thought experiment involving mirrors and trains and discuss the relationship between thought and reality

Make sure you understand everything in this dotpoint, and practice writing a response to this dotpoint. If you are not clear and concise, it’s easy to not fully answer the question or to end up with an extremely long answer that wastes time in a test.

Einstein had two main thought experiments- looking at himself in a mirror on a train moving at the speed of light, and bouncing light from the roof to the floor and back in a moving train. Both these experiments showed that with conventional models such as vector addition, it would be possible for a stationary observer looking to the train to see light travelling faster than c. However, this ran against his principle of the speed of light being constant.

In the mirror thought experiment, Einstein wondered whether he would be able to see his face normally in a mirror held in front of him if the train was travelling constantly at the speed of light. He decided that he would be able to, because he was in an inertial frame and should have no way to determine he was moving at c. But with vector addition, a stationary observer would see light travelling away from Einstein’s face at c, but as the train was moving at c as well, the observer would see light travel twice the distance in the same amount of time. Einstein’s interpretation of this was that the time observed for light to travel that distance changed, so that a stationary observer would still see light travelling at c.

In the light bouncing experiment, light was seen to travel a longer path by an observer. Again, the interpretation was that time changes so that c remains constant. In terms of discussing the relationship between thought and reality, thought experiments can be valuable tools to “perform” experiments that cannot be performed in reality, such as a train moving at relativistic speeds, and to make meaningful conclusions as Einstein did. This makes them extremely useful tools. On the other hand, it is very easy to misinterpret thought experiments, either through flawed logic or failing to take account of other factors, possibly unknown to science that would affect an experiment in reality. So while they are very useful tools, they need to be used carefully when drawing conclusions.

Remember- The thought experiments were Einstein looking at a mirror on a train, and bouncing light from the roof of a train to the floor and back as observed by a stationary observer.

# Identify that if c is constant then space and time become relative

This is an identify dotpoint, and so requires very little detail. It would be better to study the previous dotpoint as it goes into more detail about the impacts of the speed of light being constant.

In traditional physics, the behaviour of light had to adapt to the motion of the observer. With the speed of light being a constant under Einstein’s theory, the dimensions involved in motion have to adapt to light. This means that space and time become relative to velocity so that c is always a constant.

Remember- When the speed of light is constant, space and time become relative.

# Discuss the concept that length standards are defined in terms of time in contrast to the original metre standard

Originally, a metre was defined as $/frac {1}{10,000,000}$th of the circumference of the Earth, and then later as the distance between two lines on a platinum-iridium bar, which provided the standard measure of a metre. However, today the metre is defined as the distance light travels in $/frac {1}{299792458}$ seconds. This means that distance is calculated based on time- a unit of distance is measured in terms of how much distance light travels in a period of time. A light-year is another distance measured by time, and it is the distance light travels in one year.

# Analyse information to discuss the relationship between theory and the evidence supporting it, using Einstein’s predictions based on relativity that were made many years before evidence was available to support it

Ensure you memorise the evidence, and also make sure you can link it back to Einstein’s theory clearly showing how the evidence supported the theory.

Einstein’s key prediction that was made before available evidence was that space and time are relative to observed movement, and that the speed of light is constant. The consequences of this were that observed time could vary, so time is not constant. In 1971, the Hafele-Keating experiment took 4 synchronised atomic clocks, placed 2 of them on commercial airline flights, and flew them in opposite directions around the world. When later compared after circumnavigating the world, both the clocks showed less time had passed than the clocks on the ground, with differences of around 50 nanoseconds in an easterly direction, and around 270 nanoseconds in a westerly direction, which almost exactly matched up with Einstein’s predictions.

Other experiments using muons found similar effects. The muon is a particle similar to an electron, but heavier. When stationary it has a half-life of around 2 microseconds, but when accelerated in a particle accelerator to speeds up to 0.9994c, it was found their observed half-life was around 60 microseconds- confirmation of Einstein’s theory. There is a distinct link between theory and evidence supporting it. No hypothesis can be considered a theory until there is evidence confirming that the hypothesis is correct. Therefore, Einstein’s conclusions were merely predictions of what would happen at relativistic speeds and nothing more at the time he devised them, and his ideas only became theory later after evidence confirmed his ideas.

Remember- Longer muon decay in accelerators, and atomic clocks in aircraft circumnavigating the Earth.

# Explain qualitatively and quantitatively the consequences of special relativity in relation to the relativity of simultaneity, the equivalence between mass and energy, length contraction, time dilation, and mass dilation

See the Formulae chapter for a comprehensive guide to quantitatively determining the effect of relative motion.

Relativity has many consequences. Among the most counter-intuitive ideas is the relativity of simultaneity- meaning that because of special relativity, events observed to be simultaneous in one frame may not be observed as simultaneous in another. Consider a train moving at a relativistic velocity (i.e. an appreciable portion of the speed of light, perhaps 0.5c or more). In the middle of a carriage is a light, and at either end of the carriage are doors with light sensors. When the light in the middle of the carriage is turned on, light travels to the doors, and the doors open as soon as their light sensors detect the light. To the person inside the train, both doors open at the same time because the distance to each door from the light source is equal. However, a person outside the train sees the doors opening as non-simultaneous. When the light turns on, the distance to each door is equal. However the observer from outside sees the train moving. This means that the light reaches the rear door faster than it reaches the front door (since the train is moving forwards, the front door is moving away from the point where the light was originally turned on). This illustrates the idea that simultaneity is dependant on the frame from which events are observed.

Mass and energy are linked by the formula E = mc2, which shows the “rest energy” of an object and also the amount of energy released if matter is destroyed and converted to pure energy. There are several equations that together govern the mathematics of simple relativistic effects:

$L_v=L_0 \sqrt {1- \frac {v^2}{c^2} }$

$T_v= \frac {T_0}{ \sqrt {1- \frac {v^2}{c^2}} }$

$M_v= \frac {M_0}{ \sqrt {1- \frac {v^2}{c^2}} }$

Length contraction means that as observed velocity increases, length appears to contract in the direction of movement according to (1.1). Time dilation means moving clocks appear to run slower as observed velocity increases, according to (1.2). Mass appears to increase as observed velocity increases according to (1.3). All of these observations are true only when the frame being observed and the frame of observation are both inertial reference frames. Note also that these changes are actual changes in the properties of space-time. Moving clocks appear to run slower because in the moving frame, time is actually elapsing at a different rate to time in the frame from which the observation is being made.

# Discuss the implications of mass increase, time dilation and length contraction for space travel

Be very careful regarding the implications of time dilation. According to the twin paradox outlined by Einstein, the paradox exists because the other twin will appear younger for each of the twins. However, according to Einstein’s theory of general relativity the non-accelerated frame takes precedence, and so the twin on the spacecraft will actually be younger.

Relativistic effects have several implications for space travel. Mass increase shows that as speed increases towards c, mass increases up to infinity. What this means is that as a spacecraft gets faster, its mass increases and its acceleration progressively decreases. While acceleration never gets to zero, because mass increases a spacecraft can never travel at the speed of light. Time dilation means that astronauts in a relativistic spacecraft will age slower than people back on Earth, which means that they can effectively live longer during relativistic flight compared to a stationary observer, who will pass away well before the astronaut. Finally, length contraction means that as a spacecraft speeds up, the apparent distance to objects ahead of it decreases. This means that trips on a relativistic spacecraft will appear to cover less distance to observers in the spacecraft.

Remember- Conventional spacecraft can never travel at the speed of light, astronauts will age more slowly, and trips will appear to cover less distance from within the spacecraft.

## Describe a gravitational field in the region surrounding a massive object in terms of its effect on other masses in it

A gravitational field provides a force on objects within it that drags objects to the centre of the field.

The strength of the field is related to the mass of the object that produces it, with larger masses resulting in stronger fields. A massive object will have a strong gravitational field that will attract other masses near it. If these masses have little or no tangential velocity, they will be dragged into the massive object. If they have some degree of tangential velocity, they will be pulled into orbit, or they will have their trajectory through space altered by the massive object with the force acting on the object pulling it towards the massive object.

Remember- A massive object has a gravitational field that drags other masses towards it.

## Define Newton’s Law of Universal Gravitation

Newton’s Law of Universal Gravitation provides a formula by which the force exerted by gravity in a field can be calculated based on the masses involved and the distance between them. Gravitational force is equal to the multiple of the masses of the two objects, divided by the distance between them squared, then multiplied by the gravitational constant. $F = \frac{Gm_1m_2}{d^2}$. This formula serves to calculate the force experienced each of the bodies- however, the body with the larger mass will be less affected, because according to, F = ma if F is constant and m is large, then acceleration must be small.

Remember – Universal gravitation calculates the force experienced by each of the objects, and is experienced by both of them equally.

## Present information and use available evidence to discuss the factors affecting the strength of gravitational force

There are numerous factors affecting the strength of gravity on Earth. Firstly, as the Earth spins it bulges at the equator, flattening at the poles. This causes the poles to be closer to the centre of the Earth than the equator. According to the formula for gravitation force, the force experienced depends on the distance from the centre of the field. This means that Earth’s gravitational field is stronger at the poles than at the Equator. Secondly, the field of the Earth varies with the density of nearby geography. Places where the lithosphere is thick, or where there are dense mineral deposits or nearby mountain experience greater gravitational force compared to places over less dense rock or water. Thirdly, as gravitational force depends on altitude, places with greater elevation such as mountain ranges experience less gravitational force, than areas at or below sea level. Finally, and more generally, gravitational force also depends on the mass of the central body, so that planets or bodies with less mass have weaker gravitational fields and therefore weaker gravitational force.

Remember- The Earth’s gravitational field is changed by distance from the equator, altitude, and lithosphere composition.

## Discuss the importance of Newton’s Law of Universal Gravitation in under- standing and calculating the motion of satellites

In order to launch a satellite, the orbital velocity required must be known. The centripetal force acting on a body in orbit must be equal to the force that gravity exerts in order to keep the body in orbit. This means

$F_c = F_g$

and therefore

$\frac{Gm_pm}{r^2}=\frac{mv^2}{r}$

where $m_p$ is the mass of the planet, and m is the mass of the satellite. Simplifying this expression yields

$v=\sqrt{\frac{Gm_p}{r}}$

Since Newton’s Law of Universal Gravitation is required to quantify the value of Fg in the derivation of orbital velocity (and indeed in any calculation involving gravitational field strength), it is therefore vital to understanding and calculating the motion of satellites. Further, Newton’s Law can be used to derive Kepler’s Law of Periods, an integral tool in understanding the motion of satellites in a given system. So although it is by no means a complete solution to understanding orbital motion, it is nonetheless an integral tool.

Remember- Newton’s Law of Universal Gravitation is vital to mathematically modelling orbits, and was used to derive Kepler’s Law of Periods.

##### 1.3.5    Identify that a slingshot effect can be provided by planets for space probes

Note that some resources have the probe approaching the planet from the front, i.e. against the planet’s orbital direction. This also provides the same slingshot effect, but it is harder to visualise and understand.

If trajectories are calculated carefully, space probes can use the motion of planets through space in order to increase the probe’s velocity. In order to take advantage of the slingshot effect, the space probe approaches a planet in the same direction as the planet’s orbital path i.e. it approaches the planet from behind. When the probe enters the field, the probe is accelerated. However, the field itself is moving at the same time, because the planet is moving. This additional momentum is also given to the probe, as the probe is effectively dragged by the planet. When the probe leaves the gravitational field, the momentum it gained simply by falling into the field is lost (since it is climbing up and out of the gravitational field). However, the momentum gained by the dragging effect is retained, boosting the velocity of the probe. This is the slingshot effect- using the motion of planets to accelerate space probes. Another application of the slingshot effect is the altering of trajectory. For a probe to travel to the outer planets, it must travel away from the sun. However, the energy required to leave the sun’s gravitational field is immense. The probe’s trajectory outwards is gradually curved into an orbital path by the sun’s gravity. Using a variation of the slingshot effect, the probe can use a planet’s gravitation field not to gain velocity, but to alter its trajectory away from the sun. Ordinarily this trajectory change would consume large amounts of fuel, but the harnessing of the motion of planets removes this need, as well as reducing the time taken for a probe to visit the outer planets.

Remember- The slingshot effect uses the movement of planets to change a space probe’s speed or direction to help it reach outer planets.

## 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 an object in projectile motion on Earth is a parabola. The motion of an object can be derived through analysing the horizontal and vertical components of its motion and then adding the vectors to produce the resulting direction and magnitude of the object’s velocity (the object’s net velocity vector). In standard projectile motion on Earth, the horizontal component is constant, and is equal to the original horizontal component at the point of release. The vertical component is constantly changing, being affected by the gravitational field. The change occurs directly towards the centre of the field, and in the Earth’s case, acts in this direction at 9.8m/s for every second in flight. At any given time, the vertical component is equal to the initial vertical component at the time of release, minus 9.8 times the time elapsed, where a negative value is downward motion.

Remember- An object in projectile motion travels in a parabola with a constant x-component and a changing y-component.

## Describe Galileo’s analysis of projectile motion

You’ll need to memorise what Galileo said and how he devised his vector analysis. This is a history lesson, but it also tests whether you understand how the component system works so make sure you explain that too.

Galileo was the first to analyse projectile motion mathematically and have his work documented. Instead of considering the motion of the object as a whole, he divided the motion into a horizontal and a vertical component, which when added provide the total motion of the object. Galileo realised that during projectile motion, only the vertical component would change (excluding air resistance) while the horizontal component would remain constant. He also realised that the motion of projectiles is parabolic in nature due to the uniform acceleration vertically with constant horizontal motion.

Remember- Galileo was the first to break a projectile’s motion into components.

## Explain the concept of escape velocity in terms of the gravitational constant and the mass and radius of the planet

This dotpoint is essentially memorising the formula, and explaining the concept of what escape velocity is.

Escape velocity is the velocity required at a planet’s surface to completely leave its gravitational field without further energy input. This means that it must have the same amount of kinetic energy as the absolute value of the gravitational potential energy it has at the point of takeoff. Assuming takeoff from the planet’s surface, this means $\frac{1}{2}mv^2 = \frac{Gmm_p}{r_p}$ where $m_p$ refers to the mass of the planet. Cancelling, $v^2 = \frac{2Gm_p}{r_p}$. This formula links escape velocity to the gravitational constant and the mass and radius of the planet. If at the surface of the planet $v^2$ is equal to the RHS, then the rocket will be able to escape the gravitational field. Thus the $v$ value at this point is the escape velocity. Escape velocity increases as the mass of the planet increases, and decreases as the radius of the planet increases.

Remember- Escape velocity is the velocity needed at the surface to exit the gravitational field. More mass and a smaller radius make it bigger.

## Outline Newton’s concept of escape velocity

Make sure you can properly explain this, it has caused people trouble before. Memorise it.

Newton envisaged a cannon firing a projectile horizontally from the Earth’s surface. Ignoring air resistance, the projectile would prescribe a parabola, eventually falling back to Earth. However, as the speed of the projectile is increased, the projectile will take progressively longer to hit the ground, because although gravity is pulling towards the centre of the field, the Earth’s surface is falling away from the projectile at the same time due to its horizontal motion. Increase the speed enough, and the projectile will never hit the ground, instead travelling in a circle around the Earth. As the velocity increases even more, the circle becomes an ellipse, and if the speed is increased enough, the trajectory becomes hyperbolic. At this point, the projectile has enough velocity to leave the gravitational field. The velocity corresponding to the time when this first occurs is then the escape velocity.

Remember- Newton used a horizontal cannon to visualise orbits and escape velocity.

## Identify why the term “g forces” is used to explain the forces acting on an astronaut during launch

This dotpoint is comparatively easy, but when considering G-forces take care to add the forces correctly. It may be easiest to visualise yourself in the scenario to get an idea as to how the forces interact.

‘G-Forces’ refers to the force experienced by an astronaut in terms of the Earth’s gravitational field strength at the Earth’s surface. 1G is equal to the force experienced by an astronaut on the surface of the Earth: w = mg where g = 9.8. If a rocket is accelerating upwards at 9.8m/s2 , then the astronaut experiences a net force equal to 2Gs (twice the force they would experience due to Earth’s gravity). If an astronaut is in freefall, they experience 0Gs. The term g forces is used because it is easy to relate to, and because it is eases calculations as to the forces which the human body can withstand during launch.

Remember- G-force measures acceleration in terms of Earth’s gravity.

## 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 this experiment, we placed a grid against a wall and then threw a ball in a parabola in front of the grid. A video camera recorded the experiment so that we could see the ball travelling in front of the grid. Using the grid, we were able to calculate the position of the ball. Times were calculated based on each video frame representing $\frac{1}{25}$th of a second. By analysing the movement of the ball between frames, we were able to use the standard motion equations in the X and Y axes to calculate the initial and final velocities, as well as the maximum height reached and the range of the projectiles, in this case, a tennis ball. There would have been errors caused by the ball not travelling in a straight line (i.e. It did not travel only vertically and horizontally, but laterally too) resulting in erroneous readings, and it is likely that the camera did not record frames at exactly $\frac{1}{25}$th of a second intervals, producing further errors.

Remember- Grid on the wall, tennis ball, video camera, analyse changes between frames.

## Analyse the changing acceleration of a rocket during launch in terms of the Law of Conservation of Momentum and the forces experienced by astronauts

The key part of this dotpoint is analysis in terms of Conservation of Momentum. To say that the thrust is constant and the weight decreases, so acceleration increases by F = ma is incomplete. Make sure you deal with Conservation of Momentum as well.

The Law of Conservation of Momentum states that in a closed system, the sum of the momenta before a change is equal to the sum of momenta after the change. In a rocket, the change is the release of exhaust gas. The momentum of the exhaust gas is the same as the rocket’s momentum, with a reversed direction, so that when added, they amount to 0. P = mv . This equation links velocity to mass and momentum. Because the sum of the momentum of the exhaust gas and the rocket is zero, |mexhaust × vexhaust| = |mrocket × vrocket| (taking absolute values because one side of this equation will be negative, since the rocket and the exhaust travel in opposite directions). As the rocket travels into space, it burns fuel and so its mass decreases. But because the momentum of the exhaust is constant, this means the rocket’s velocity must rise in order to balance the equation. This means that when the burn is completed, the rocket is travelling faster than if the rocket had maintained a constant mass (because vrocket is now larger as mrocket decreased while procket and pexhaust remained constant). This in turn implies that the acceleration of the rocket has increased during the burn in order to fulfil conservation of energy. This can be seen through F = ma, where F is the thrust of the rocket motor. Because the rocket motor provides constant thrust, F is a constant. As the rocket burns fuel, its mass decreases, and so for ma to remain constant the rocket’s acceleration must increase. This means that as the rocket takes off, its acceleration becomes progressively higher as it burns its fuel and becomes lighter. For the astronauts, this means an increasing force. So as the rocket lifts off, its thrust needs to be progressively reduced to protect its occupants.

Remember- As a rocket burns fuel, it accelerates faster.

## Discuss the effect of the Earth’s orbital motion and its rotational motion on the launch of a rocket

A question on this area will need a comprehensive answer, so make sure that you address the positives and negatives of both Earth’s rotation and its orbital motion.

The orbital motion of the Earth and the rotational motion of the Earth both have related effects, the orbital motion affecting interplanetary travel and rotational motion affecting satellites orbiting the Earth. The effect arises because when a rocket is launched, its velocity is not simply that provided by the rocket motor, but also the velocity it has because of the Earth’s movement through space.

In terms of orbital motion, space probes launched in the same direction as the Earth’s orbit carry its orbital velocity, again reducing fuel requirements, resulting in greater payloads or cheaper missions.

For rotation, the Earth rotates constantly in an anticlockwise direction as viewed from above the North Pole. Rockets launched in an easterly direction therefore carry extra momentum with them, giving them around an additional 0.5km/s towards their velocity. This means that to achieve orbit, the rocket only needs to accelerate 7.5km/s, with the additional 0.5km/s resulting from the motion of the Earth. This means that less fuel is required, and/or a greater payload can be carried.

On the other hand, the orbital and rotational motion makes it hard to launch rockets in a direction against the motion. For example, to launch a rocket in a westerly direction into orbit would take an acceleration of 8.5km/s, significantly greater. Likewise, to launch a space probe against the motion of the Earth would result in far greater fuel requirements to achieve the same trajectory.

Remember- If you launch a satellite in the direction of the Earth’s orbit or rotation, it effectively has more velocity, saving fuel.

## Analyse the forces involved in uniform circular motion for a range of objects including satellites orbiting the Earth

The forces involved for uniform circular motion for satellites are the same as uniform circular motion in any situation. There will always be tangential velocity, and there will always be a centripetal force that causes the object to travel in a circular path. The only difference is the source of the forces. For an examination of the virtual force centrifugal force, see the Extra Content section at the end of the Guide.

Uniform circular motion refers to the motion of objects that prescribe a perfect circle as they move. The key force in uniform circular motion is centripetal force. Centripetal force is a centre-seeking force that always acts in a direction towards the centre of the circle in uniform circular motion. The
formula for centripetal force is $F = \frac{mv^2}{r}$ The forces for uniform circular motion may be sourced differently, but all are centripetal in nature and all follow this formula. This is true of satellites in orbit around the Earth, cars as they turn, and a charged particle in a magnetic field.

Remember- Uniform circular motion always requires centripetal force, which can come from a variety of sources.

## Compare qualitatively low Earth and geo-stationary orbits

A low Earth orbit is one that is approximately 300km from the Earth’s surface, although technically it refers to any satellite below 1500km in altitude. LEOs (Low Earth Orbit satellites) have an orbital period of around 90 minutes, with an orbital velocity of about 8km/s. Geostationary satellites remain above a fixed position on the Earth, because their orbital period is exactly 24 hours. They are far higher up than LEOs, at around 36000km in altitude, and have a lower orbital velocity (around 3km/s). A geo-stationary orbit is a special type of geo-synchronous orbit. A geo-synchronous orbit refers to any orbit with a period of 24 hours. However, not all geo-synchronous orbits are geo- stationary, because geo-stationary orbits must be equatorial, travelling directly above the equator. A polar orbit may be geo-synchronous, but it cannot be geo-stationary.

So essentially, compared to the low Earth orbit, a geostationary orbit is higher up, has a longer orbital period and a lower orbital velocity.

Remember- A low Earth orbit is low and fast with a short period, and geo-stationary is high and slow with a 24-hour period.

## Identify data sources, gather, analyse and present information on the contri- bution of one of the following to the development of space exploration: Tsiolkovsky, Oberth, Goddard, Esnault-Pelterie, O’Neill or von Braun

Konstantin Tsiolkovsky (1857-1935), a Russian scientist, while not contributing directly to space travel during his lifetime, devised many new ideas that were almost prophetic and extremely important in space travel. The key ideas he had were firstly the principles behind rocket propulsion, secondly the use of liquid fuels, and finally multi-stage rockets. Tsiolkovsky showed how Newton’s 3rd law and how conservation of momentum can be applied to rockets. This principle underlies the functioning of all rockets, and is vital to understanding their operation. Secondly, Tsiolkovsky proposed using liquid hydrogen and liquid oxygen as rocket fuels so that the thrust produced by a rocket could be varied. These same fuels were implemented in the Saturn V rocket that powered the Apollo missions to the moon, and the use of liquid fuels has proved vital in manned spaceflight because they allow g-forces experienced by astronauts to be controlled, unlike in solid fuel engines. Also, liquid fuels are used in satellites and space probes, where intermittent firing of rockets is desired rather than a continuous burn as provided by a solid rocket motor. Finally, Tsiolkovsky visualised a 20-stage rocket train that dropped stages as each stage ran out of fuel, to cut weight and improve efficiency. Although 20 stages was rather extreme, the multistage rocket proved vital in high-energy launches for manned space missions such as Apollo as well as missions with large payloads. So while Tsiolkovsky didn’t directly impact space exploration during his lifetime, he devised many ideas that are vital to spaceflight today.

Remember- Tsiolkovsky devised concepts well before they could be practically implemented.

## Define the term orbital velocity and the quantitative and qualitative relation- ship 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

Although the dotpoint mentions the relationship between orbital velocity and the mass of the satellite, the mass of the satellite is irrelevant. Looking at the 2 equations provided here, the only 2 variables are the mass of the central body and the orbital radius. This means that there is no relationship between the mass of the satellite and orbital velocity, providing the satellite is significantly lighter than the central body (as otherwise more complicated effects would come into play).

Orbital velocity is simply the speed at which a satellite is travelling, calculated by dividing the distance it travels in its orbit (which is the circumference of the circle in a circular orbit) by its orbital period. Orbital velocity is linked to the gravitational constant, the mass of the central body and the radius of the orbit according to the formulae $\frac{r^3}{T_2} = \frac{Gm_c}{4 \pi^2}$ and $v = \frac{2 \pi r}{T}$. Essentially, orbital velocity increases when the mass of the central body increases, and decreases when the radius of the orbit is increased. The mass of the satellite has no bearing on the orbital velocity, as it cancels out when calculating orbital velocity.

## Account for the orbital decay of satellites in low Earth orbit

LEOs continually lose orbital speed and require periodic rocket boosts in order to stay in orbit, preventing them from crashing. The reason LEOs lose velocity is because the Earth’s atmosphere extends far into space. The boundary between the atmosphere and the vacuum of space isn’t clearly defined, and there are still air particles high above the Earth’s surface. As LEOs collide with these particles they slowly lose orbital velocity through friction, resulting in orbital decay. Orbital decay is where a satellite loses orbital velocity and therefore moves into a lower orbit closer to the Earth’s surface. If orbital decay continues, the satellite will eventually crash.

Remember- LEOs crash because they collide with air particles.

## Discuss issues associated with safe re-entry into the Earth’s atmo- sphere and landing on the Earth’s surface (including “Identify that there is an op- timum angle for safe re-entry for a manned spacecraft into the Earth’s atmosphere and the consequences of failing to achieve this angle”)

Re-entry is a complex procedure due to the high velocities and temperatures encountered, as well as the fine balance of trajectory required to land safely. To land a space vehicle, the vehicle must firstly slow down, and secondly travel back down through the atmosphere. These are done simultaneously

with atmospheric drag slowing the vehicle as it descends. The high velocity of the vehicle results in a great deal of friction, which heats the vehicle to up to 3000◦C depending on airflow. This necessitates highly temperature resistant shielding, usually ceramic or carbon based, that can withstand the

temperatures and protect the rest of the vehicle as it descends. Modern designs also feature blunt noses and have the spacecraft descend belly-first, which ensures the majority of the vehicle is shielded. Without appropriate shielding, the vehicle will be unable to return, as recently seen in the 2004 Columbia space shuttle accident in which its heat shielding was compromised. Secondly, the angle of re-entry is critical. If the angle is too steep, the descent rate will be too fast, and the vehicle will encounter the higher density atmosphere closer to the Earth’s surface while it retains too much of its velocity. Higher density air provides more drag, which therefore decelerates the vehicle faster and leads to higher temperatures. This will result in at the very minimum excess g-forces for the crew, and at worst, the extra heating could destroy the entire vehicle. On the other hand, if the angle is too shallow, the spacecraft will retain too much of its velocity and exit the atmosphere by effectively skimming it, returning to space. The vehicle must have an angle between 5.2 and 7.2 degrees to make a safe re-entry. During re-entry, the high temperature of the spacecraft results in the air around it becoming ionised. This results in an ionisation blackout, with the ionised air blocking radio communication with the ground during re-entry. Although not a direct hazard, it can cause complications in the event of a safety issue arising during re-entry which could endanger the spacecraft. Finally, in order to land, the descent rate must be slowed dramatically. In the Apollo missions and with non-reusable space probes, parachutes are used to slow the descent to make a gentle landing. The space shuttle uses wings to generate lift, enabling it to glide to a gentle landing.

Remember- To re-enter, you need strong heat shielding and an approach with a specific angle of descent.

## Define weight as the force on an object due to a gravitational field

Weight is the force experienced by an object due to a gravitational field. It is directly related to the strength of the gravitational field at the point where the object is located, and is equal to the force which the field is exerting on the object.

Remember- Weight is the force on an object due to a gravitational field.

## Explain that a change in gravitational potential energy is related to work done

This section will be hard to answer if you don’t fully understand how potential energy works. If this here isn’t enough, make sure you read through the various textbooks and look for other resources to make sure you understand potential energy properly.

Work done is the measure of how much energy was used to displace an object a specified distance. W = Fs where s is displacement. When an object is moved away from a gravitational field, it gains energy. This is because by raising it up from the field’s origin, work is done. If a 1kg stone was raised 100m, then work done would be 980J. However, conservation of energy states that this energy cannot be destroyed. The 980J is now 980J of gravitational potential energy, because if the stone was dropped from 100m then it would regain 980J in the form of kinetic energy due to the gravitational field. Gravitational potential energy is the potential to do work, and is related to work done.

Remember- Potential energy is the work done to raise an object in a gravitational field.

## Perform an investigation and gather information to determine a value for acceleration due to gravity using pendulum motion or computer-assisted technology and identify reasons for possible variations from the value 9.8 $m/s^2$.

This experiment will definitely give you a value that differs from 9.8 $m/s^2$, so make sure you know both experimental reasons for your error, as well as the factors affecting gravity itself.

In our investigation we used a pendulum consisting of a weight attached to a thick, non-elastic string that was tied to a clamp on a retort stand. We set the pendulum in motion by swinging it, being careful to ensure that the pendulum was deflected no more than 30◦ at maximum deflection, to minimise errors caused by tension in the string (because the string will lose tension at angles greater than 30◦). We timed the pendulum over 10 complete cycles (time taken to return to its point of origin) in order to minimise timing errors and random factors affecting individual swings. We then used the formula $T=2\pi \sqrt{\frac{l}{g}}$ where $T$ is the period (time taken for one complete cycle), $l$ is the length of the string (measured from the knot on the clamp to the centre of gravity of the weight) and $g$ is gravitational acceleration, in order to calculate a value for $g$.

There are numerous factors affecting the strength of gravity on Earth (aside from experimental errors producing a result different to 9.8 $m/s^2%$).

Firstly, as the Earth spins it bulges at the equator, flattening at the poles. This causes the poles to be closer to the centre of the Earth than the equator. According to the formula for gravitational force, the force experienced depends on the distance from the centre of the field. This means that Earth’s gravitational field is stronger at the poles than at the Equator.

Secondly, the field of the Earth varies with the density of nearby geography. Places where the lithosphere is thick, or where there are dense mineral deposits or nearby mountains experience greater gravitational force compared to places over less dense rock or water.

Thirdly, as gravitational force depends on altitude, places with greater elevation such as mountain ranges experience less gravitational force, compared to areas at or below sea level.

Remember- Pendulum experiment, errors in the experiment, factors affecting the strength of Earth’s gravity.

## Gather secondary information to predict the value of acceleration due to gravity on other planets

Just pick and choose a few values to memorise. If they give you a question in the exam regarding the different accelerations they’ll most likely give you a table of values and ask you to do calculations with it. Don’t spend long on this point. Also, Pluto is no longer officially a planet.

 Planet Gravitational Acceleration ($m/s^2%$) Mercury 4.07 Venus 8.90 Earth 9.80 Mars 3.84 Jupiter 24.83 Saturn 10.50 Uranus 8.45 Neptune 11.20

## Define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field

Again, you need to understand this section. A question may focus on why potential energy takes a negative value, and you need to be able to comprehensively explain and justify why. The reason the dotpoint is defined as a very large distance away is because this is equivalent to a point outside the field. Gravitational fields, like many fields, have no theoretical maximum range and theoretically exist at an infinite distance away from an object. In practice, because gravitational fields obey inverse square law and decrease in strength rapidly as distance increases, at large distances the field is for all intents and purposes nonexistent. Regardless, there is technically no point in the universe outside a gravitational field, hence a very large distance away is used.

Gravitational potential energy is defined as the work done to move an object from a point a very large distance to a specified point in the gravitational field. The work done is the energy input provided by the gravitational field to the object as it falls to that particular distance. $E_p = -\frac{Gm_1m_2}{r}$ is a more accurate definition because it takes into account the weakening of gravitational fields at a distance, and also results in objects far away out of the field having no energy, rather than the simpler definition $E_p = mgh$ where at an infinite distance, there is infinite potential energy.

Remember- Potential energy is negative, and is the work done in moving an object from an infinite distance to a point within the field.