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Farr High School

HIGHER PHYSICS

Unit 1

Our Dynamic Universe

Pupil Booklet

Based on booklets by Richard Orr, Armadale Academy

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Section 1

Motion: Equations and Graphs

Suvat

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Vectors and Scalars - Recap It is possible to split up quantities in physics into two distinct groups; those that have a direction assigned [or implied], and those that have no direction. Some are obvious, it makes sense that force has direction; you can push or pull but only in one direction. It would be nonsense to give a direction to, say time; it took 5 seconds East just isn’t right. It is important that you are familiar with which quantity falls into which grouping. Vectors are quantities which have direction and scalars those which don’t. As we go through the course you will build up a bank of scalar and vector quantities. Distance and displacement: Distance is a scalar; it is a measure of how far an object moves. Displacement is a vector; it is a measure of how far away an object is from its starting point in a straight line. It must also indicate the direction between the starting and finishing point. Consider the journey below. A person walks along a path (solid line) from start to end. They will have walked further following the path than if they had been able to walk from start to end in a straight line (dashed line). The solid line denotes the distance = 3km The dashed line denotes the displacement = 2.7 km East Speed and velocity: Speed is a scalar; it is a measure of how far an object moves in unit time (one second). Velocity is a vector; it is a measure of how far away an object has moved from its starting point in a straight line in unit time. It must indicate the direction between the starting and finishing point. The direction of the velocity vector will be the same as that for the displacement vector used in a calculation.

speed =time

distance velocity =time

ntdisplaceme

Hopefully you will have remembered from last year that accelerations happen because of unbalanced forces. Force must have a direction and is always a vector. This means that acceleration must also be a vector with its direction being the same as the unbalanced force.

start end

N 3 km

2.7 km

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Two dimensional Vectors Consider a runner travelling as shown below. The runner takes 2 hours to run from X to Y then on to Z. In this type of situation it is important that you take care to distinguish between what are scalar and vector quantities. a) What distance did the runner travel? Distance is a scalar so direction has no impact on the answer. We simply need to find out how far the runner has actually travelled. 12km from X to Y, then 5km from Y to Z. Total distance then is 12 + 5 = 17 km. b) Calculate the average speed of the runner for their journey. Speed is a scalar so no direction is required.

speed =

time

distance =

2hours

17km = 8.5km/h.

c) Calculate the displacement of the runner from X to Z. Now we are in vector country so direction is required. Remember displacement is the straight line distance between starting and finishing points with direction given. This quantity is called the resultant. This part can be done either by a scale diagram or by using Pythagoras and trigonometry. To find the magnitude of resultant by Pythagoras. R2 = 122 + 52 = 144 + 25 = 169 R = 13km It is now necessary to find the angle that describes the direction of the resultant. Trigonometry can be used here.

TAN =

ADJ

OPP =

12

5 = 0.42 = 22.6°

This angle must now be referenced from a fixed angle. Normally North is used and the angle given as a three figure bearing. The angle between North and the line XZ is 90° – 22.6° = 67.4°, which gives a bearing for the resultant as 067.4°. The displacement is then written as 13km 067.4°. Both magnitude and direction must be given. d) Calculate the average velocity of the runner for the journey.

velocity =

time

ntdisplaceme =

2hours

13km = 6.5km/h

Since velocity is a vector there must be a direction given. The velocity direction is always the same as the displacement direction. So velocity is 6.5km/h 067.4°.

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How do objects move? Not to be confused with why do objects move? We’ll come to that later. You will remember from last year that we can describe the motion of an object in terms of its velocity and acceleration. Because we are physicists we like to write information in a shortened way, the dreaded equations! Just think how much you would have to write down if we didn’t use equations though.

v =

t

s

Instead of: ‘the velocity is equal to the displacement divided by the time taken for that displacement.’ So don’t let me hear you complain about equations again!! Higher level equations of motion: We tend to talk of the three equations of motion 1. v = u + at u = initial velocity 2. s = ut + ½ at2 v = final velocity 3. v2 = u2 + 2as s = displacement a = acceleration t = time The quantities in these equations are all vectors except time. This means that direction is important. Example: If car 1 has a velocity of 4m/s, then we must assign a value of -6m/s for the velocity of car 2 since it is travelling in the opposite direction to car 1. All examples you will encounter can be thought of as moving in a straight line.

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2

4 m/s 6 m/s

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Examples 1. A ball is dropped from rest from a window and it accelerates at 9.8 ms-2 before hitting the ground 1.85 seconds later. a) How high is the window above the ground? b) What is the velocity of the ball as it hits the ground? a) We know u = 0 ms-1 a = 9.8 ms-2 t = 1.85 s

Use s = ut + ½ at2

= 0 + (0.5 x 9.8 x 1.85 x 1.85) = 16.77 m

b) Use v = u + at

= 0 + (9.8 x 1.85) = 18.13 ms-1

2. A car starts from rest and accelerates to a speed of 22ms-1 at a rate of 5 ms-2. How far does it travel in this period? We know u = 0 v = 22 ms-1 a = 5 ms-2

Use v2 = u2 + 2as 222 = 0 + (2 x 5 x s) 484 = 10s

s = 484/10 = 48.4 m Questions 1. A ball is thrown vertically upwards with an initial speed of 12 ms-1. a) Describe what happens to the speed of the ball as it rises to its maximum height. b) How long does the ball take to reach its maximum height? c) What is the maximum height reached by the ball? 2. A motorbike is travelling at 20 ms-1 when he sees a child run into the road 23 m ahead of him. If he takes 2.2 s to stop, will he hit the child?

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Motion-time graphs

In all areas of science, graphs are used to display information. Graphs are an excellent way of giving information, especially to show relationships between quantities. In this section we will be examining three types of motion-time graphs.

Displacement-time graphs

Velocity-time graphs Acceleration-time graphs

If you have an example of one of these types of graph then it is possible to draw a corresponding graph for the other two factors. Displacement – time graphs This graph represents how far an object is from its starting point at some known time. Because displacement is a vector it can have positive and negative values. [+ve and –ve will be opposite directions from the starting point.] Displacement-time graph 1 OA – the object is moving away from the starting point. It is moving a constant displacement each second. This is shown by the constant gradient. What does this mean?

gradient =

x in change

yin change =

time

ntdisplaceme = velocity

We can determine the velocity from the gradient of a displacement time graph. AB – the object has a constant displacement so is not changing its position, therefore it must be at rest. The gradient in this case is zero, which means the object has a velocity of zero [at rest] BC – the object is now moving back towards the starting point, reaching it at time x. It then continues to move away from the start, but in the opposite direction. The gradient of the line is negative, indicating the change in direction of motion.

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Converting displacement time to velocity time graphs The velocity time graph is essentially a graph of the gradient of the displacement time graph. It is important to take care to determine whether the gradient is positive or negative. The gradient gives us the information to determine the direction an object is moving. Velocity-time graphs The area under a speed time graph is equal to the distance travelled by the object that makes the speed time graph. In this course we are dealing with vectors so the statement above has to be changed to: The area under a velocity time graph is equal to the displacement of the object that makes the speed time graph. Any calculated areas that are below the time axis represent negative displacements. It is possible to produce a velocity time graph to describe the motion of an object. All velocity time graphs that you encounter in this course will be of objects that have constant acceleration.

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Scenario: Lydia fires a ball vertically into the air from the ground. The ball reaches its maximum height, falls, bounces and then rises to a new, lower, maximum height. What will the velocity time graph for this motion look like? First decision: will I choose up or down as my positive direction? We will choose up as positive. Now we need to think, what is happening to the velocity? The ball will be slowing down whilst it is moving upwards, having a velocity of zero when it reaches maximum height. The acceleration of the ball will be constant if we ignore air resistance. Part one of graph.

Once the ball reaches its maximum height it will begin to fall downwards. It will accelerate at the same rate as when it was going up. The velocity of the ball just before it hits the ground will be the same magnitude as its initial velocity upwards. Part two of graph The ball has now hit the ground. At this point it will rebound and begin its movement upwards. In reality there will be a finite time of contact with the ground when the ball compresses and regains its shape. In this interpretation we will regard this time of contact as zero. This will result in a disjointed graph. The acceleration of the ball after rebounding will be the same as the initial acceleration. The two lines will be parallel

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Part three of graph

This now is the velocity time graph of the motion described in the original description. Converting velocity-time to acceleration-time graphs We need to consider the gradient of the velocity time graph line. In our example the gradient of the line is constant and has a negative value. This means for the entire time sampled the acceleration will have a single value. All acceleration time graphs you will draw will consist of horizontal lines, either above, below or on the time axis.

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Example The diagram represents the motion of a girl running for a bus (OQ) and then travelling on the bus (QR).

a) How far does the girl run until she gets on the bus?

b) What is the acceleration of the bus between Q and R?

a) Distance = area under graph

= (½ x 2 x 5) + (6 x 5)

= 35m

b) Acceleration = change in velocity

time

= 15 – 5 = 1 ms-2

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Questions

1. The graph shows the journey of a car.

a) What is the acceleration of the car over the first 10 seconds?

b) How far does the car travel in the first 30 seconds?

c) What is the displacement of the car from its starting position at the end of the journey?

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Experiment You are going to investigate the motion of a ‘popper’ toy. The aim is to find the initial velocity of the toy and the time that it is in the air for. You will need to measure the maximum height reached by the popper. This will give us a value for the displacement (s). We know the velocity at the highest point is 0 ms-1 (v) and the acceleration is -9.8 ms-2 (a) as the popper is moving against the force of gravity. We can use the equations of motion to calculate u and t. Repeat the experiment but this time measure the time that the popper is in the air with a stop watch. Calculate the time and compare the two answers. Can you account for any differences? In reality, the initial velocity must be greater than the calculated value. Explain why. We are now going to calculate the initial acceleration of the popper during the ‘pop’. As it restores itself from being turned inside out, the force it exerts against the surface acts over a distance approximately equal to its radius (1.25cm). The acceleration can be found using v2 = u2 + 2as Where u =0

v = the velocity found in the previous calculation and s = 1.25 cm = 0.0125 m We can also find the time of the ‘pop’ using a = v – u t All of this information has come from measuring the maximum height of the popper.

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Freefall When an object is allowed to fall towards the Earth it will accelerate because of the force acting on it due to gravity. This will not be the only force acting on it though. There will be an upwards force due to air resistance. Air resistance increases with speed; you may notice this if you increase your speed when cycling. If an object is allowed to fall through a large enough distance then the air resistance force may increase to become the same magnitude as the force due to gravity acting on the object. When this situation occurs, the forces acting on the object will be balanced. This means the object will fall with constant velocity. This is the terminal velocity of the object, but we will learn more about this later. Questions 1. Draw a diagram to show the forces acting on a 3kg mass at the instant it is released. 2. What will the initial acceleration of the object be? 3. What happens to the magnitude of the air resistance acting on the object as it falls? Explain your answer. 4. Describe what happens to the acceleration of the object as it falls, you must justify your answer. 5. Why can the air resistance not exceed the weight of the object for any length of time? What would happen if this situation did occur?

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Section 2

Forces, Energy and Power

Thrust

Air

resistance

Weight

Unbalanced

Force

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Newton’s Laws

Newton’s first law states that ‘An object will remain at rest, and a moving object will continue with a constant velocity unless an unbalanced force acts on it.’

Newton’s second law allows us to calculate the acceleration of an object which is being acted on by unbalanced forces by using F = ma.

Example The unbalanced force on the car is 6000 – (2000 + 2500) = 1500N to the right. a = F/m = 1500/600 = 2.5 ms-2 to the right

Newton’s third law states that ‘For every action, there is an equal and opposite reaction’.

This law links the action of a force pulling an object with the reaction on the body which is providing the pulling force. Example A car is towing a caravan and the vehicles accelerating at a rate of 1.5 ms-2. The frictional forces on the car and the caravan are 1200N and 1000N respectively. a) What is the thrust produced by the car engine? b) What is the tension in the tow bar? a) Unbalanced force = ma = 1300 x 1.5 = 1950 N So the total thrust = 1950 + 1200 + 1000 = 4150 N b)The unbalanced force required to make the caravan accelerate at 1.5 ms-2

= ma = 500 x 1.5 = 750N The tension in the tow bar is equal to the unbalanced force + the frictional force on the caravan = 750 + 1000 = 1750 N

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Rockets The motion of a rocket once it starts to lift off is complex. We need to consider all the forces acting on the rocket. At the instant of lift off: In order for the rocket to lift off the thrust must be greater than the weight. This produces an unbalanced force which makes the rocket accelerate upwards.

a =

m

FU where FU = Thrust – weight

easy to calculate if we know the thrust and the mass of the rocket. So far so good, but we’ve only moved a couple of centimetres at this point. As the lift off continues: When the rocket begins moving, the air resistance will increase as the speed increases. This acts against motion so the unbalanced force will be reduced. But wait a minute - as the rocket uses up fuel its mass will decrease, reducing the weight and so increasing the unbalanced force. Also, as the rocket gets further from the Earth the gravitational field strength will decrease making the weight smaller again. Also the air will get thinner as altitude increases, reducing the air resistance. All of this is based on the assumption that the thrust remains constant. If it changes, then this becomes an even bigger headache. How do all of these different factors act together to affect the acceleration of the rocket? Essentially, the net effect is that the acceleration increases as the rocket rises.

acceleration

factor giving increase factor giving decrease

decreasing mass, increasing unbalanced force increased velocity, increasing air resistance, reducing unbalanced force

increasing altitude, decreasing weight, increasing unbalanced force

increasing altitude, decreasing air resistance, increasing unbalanced force

Example Shortly after take-off the forces acting on a rocket are as shown in the diagram. Calculate the acceleration of the rocket at this point. m = W/g = 10000/9.8 = 1020.41 kg a = Fu = 12500 – 10000 – 1000 = 1500 = 1.47 ms-1 m 1020.41 1020.41 Tutorial Booklet II Page 12

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Terminal Velocity We have already mentioned how the forces on a falling object change as the object accelerates. Consider a sky diver:

As the sky diver falls, her speed increases.

This increase in speed causes an increase in the frictional forces (air resistance) acting on the sky diver.

Eventually the weight of the sky diver and the air resistance are balanced.

This means that the sky diver will continue to fall at a steady speed, called the terminal velocity.

When the parachute opens, the air resistance increases and the resultant force is upwards, causing her to slow down.

A new, slower terminal velocity is reached. Falling objects are not the only things to reach a terminal velocity. Cars and other vehicles have a top speed due to the increase in the frictional forces which act on them balancing out the thrust from the engine as the vehicle goes faster. Apparent Weight Have you noticed that when you are in a lift you experience a strange feeling when the lift starts to move and as it begins to slow to a stop? However, when the lift is in the middle of its journey you cannot tell if you are moving at all. This is because at the start and end of the journey you will experience an acceleration and consequently an unbalanced force. When you stand on a set of scales the reading on the scales is actually measuring the upwards force. This is the force the scales exert on you. Now this is fine when you are in your bathroom trying to find your weight. Normally, you and your bathroom scales will be stationary and so your weight will be equal to the upwards force [balanced forces]. If you are accelerating upwards in a lift, there is a resultant force upwards so you will appear to weigh more than normal.

If you are decelerating upwards in a lift, there is a resultant force downwards so you will appear to weigh less than normal.

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If you are accelerating downwards in a lift, there is a resultant force downwards so you will appear to weigh less than normal.

If you are decelerating downwards in a lift, there is a resultant force upwards so you will appear to weigh more than normal.

If you are moving at constant speed, the upward and downward forces are balanced and you will weigh the same as if you were weighing yourself in the bathroom at home!

Accelerometers which can use this principle are found in more and more electronic devices. Wii handsets, phones where you can move between functions by shaking the handset, lap tops that know when they’re falling and protect themselves before they hit the ground …… the list goes on and on. Example Imagine a lift accelerating upwards at 2 ms-2. A boy of mass 50kg is in the lift. What is his apparent weight? If the lift is accelerating upwards, there must be an unbalanced force upwards. We know that F = ma so we can work out the unbalanced force upwards. F = 50 x 2 = 100N His normal weight is W = mg = 50 x 9.81 = 490.5N Add the two forces together and his apparent weight will be 590.5N. Questions 1. A person of 60 kg enters a lift. He presses the start button and the lift descends with an acceleration of 1 ms-2. The lift then descends with a steady speed before coming to a rest with a deceleration of 1.2 ms-2 . a) What is the force exerted on the person by the floor when the lift is stationary? b) What is the apparent weight of the person when the lift is accelerating? c) Calculate the apparent weight of the person when the lift is decelerating? 2. A mass of 0.05 kg is suspended inside a lift. The lift starts from rest, accelerates upwards at 0.4 ms-2, moves upwards at a steady speed of 0.6 ms-1 and then decelerates at 0.4 ms-2. Find the readings of the spring balance at each stage of the motion.

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Vector components It is possible to analyse the motion of a projectile by separating its vertical and horizontal motions. This is known as vector resolution and the vertical and horizontal motions are known as components. This technique can be used to simplify the analysis of other types of motion and of forces that act at an angle to the direction of motion.

For projectile motion analysis we need to draw a diagram of the resultant and the components.

Using simple trig.

vH = vcos

vV = vsin

Example

A rugby ball is kicked with a velocity of 25ms-1

at an angle of 65oto the horizontal. What is

(a) the vertical component and

(b) the horizontal component of its velocity?

(a) vV

= vsinƟ

= 25 x sin 65

= 22.66 ms-1

(b) vH

= vcosƟ

= 25 x cos 65

= 10.57ms-1

vH

vV R

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We will consider the example of the mass on a slope and determine the magnitude of the components parallel and perpendicular to the slope.

Enlarging and isolating the forces acting on the mass.

Fllel = mgsinθ

F = mgcosθ

Questions 1. A cannonball is launched from a cannon at a velocity of 100 ms-1 at an angle of 25o to the horizontal. Calculate the initial velocities in the vertical and horizontal directions. 2. A ferry crosses a river that is flowing at 5 ms-1. If the ferry is travelling at 12 ms-1, calculate its resultant velocity. 3. A boy using a force of 250 N pulls a sledge across the snow as shown in the diagram below. Calculate the horizontal and vertical components of this force.

5 ms-1

12 ms-1

40°

250 N

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Potential Energy An object has gravitational potential energy if it is above ground level. The amount of potential energy it has is given by: Potential energy = mass x gravitational field strength x height or Ep

= mgh Example A lift with a mass of 250kg is carrying 3 passengers, each with a mass of 60kg. It travels up to the 2nd floor which is at a height of 20m. How much potential energy is gained by the lift and passengers?

Ep

= mgh = (250 + 60 + 60 + 60) x 9.8 x 20 = 430 x 9.8 x 20 = 84280 J or 84.28 kJ Kinetic Energy All moving objects have kinetic energy. Kinetic energy is a scalar quantity. It is calculated by: Kinetic energy = ½ x mass x velocity2 or E

k

= ½ x m x v2

Example A car with a mass of 600kg is travelling at 12ms-1. How much kinetic energy does it have? E

k

= ½ x m x v2

= ½ x 600 x 122

= 43200 J or 43.2 kJ Work Done and Power Work done is a measure of change in energy. For example, a force does work on an object in order to lift it up and the object gains potential energy. Alternatively, a force does work on an object to change its velocity and hence there is a change in kinetic energy. The unit of work is Newton metres (Nm) or Joules (J).

Power is the rate at which work is done. The unit is Joules per second (Js

-1

) or Watts (W). Power = Energy transferred = work done

time time

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Example A boy of mass 40kg runs up a flight of stairs which are 10m high in 3 seconds. A girl of mass 35kg runs up a flight of stairs which are 13m high in 4 seconds.

(a) Who does the most work? (b) Who has the highest power?

Boy Work done = gain in potential energy = 40 x 9.8 x 10 = 3920J Power = work done = 3920 = 1306.67 W time 3 Girl Work done = gain in potential energy = 35 x 9.8 x 13 = 4459J Power = work done = 4459 = 1114.75 W time 4 So the girl does the most work but the boy has the highest power. Conservation of Energy As an object falls, potential energy is converted into kinetic energy.

potential energy at start = kinetic energy at end Example A diver of mass 55kg climbs up to a diving board 20m high and then dives off. Ignoring the effects of air resistance, what will his speed be just before he hits the water?

Ep at start = Ek at end mgh = ½ mv2

Cancel the mass on each side 9.8 x 20 = 0.5 x v2

v2 = 392 v = 19.80 ms-1

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Section 3

Collisions and Explosions

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Momentum and Impulse Momentum is a measurement relating to the motion of an object. Momentum is defined as the product of the mass and the velocity for an object. Effectively the number obtained gives us an indication of how difficult it would be to stop the object moving. The bigger the momentum the more difficult it would be to stop the object. Consider the following A shopping trolley, mass 5kg travelling at 0.5m/s

An oil tanker, mass 10,000,000kg travelling at 0.01m/s A bullet mass 0.01kg travelling at 500m/s The trolley has a greater velocity than the tanker and a greater mass than the bullet, but would be the easiest (and safest!!) to stop moving. As always we can write an equation to describe the relationship.

p = m x v p = momentum kgms-1 m = mass kg v = velocity ms-1 In any closed system, that is, one which is not acted on by an outside force, momentum is always conserved. That means the total momentum before a collision or explosion is equal to the total momentum after. In reality there will normally be some external force, usually friction. When we conduct experiments we attempt to reduce frictional effects to a minimum. It is important to remember that velocity and momentum are vector quantities and consequently direction is important. Opposite direction, opposite sign.

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Collisions Example 1 Imagine a car of mass 700kg (CAR A) travelling at 30 ms-1 running into the back of a stationary car of mass 500kg (CAR B). The two cars become entangled and move off together. What will be velocity of the cars as they move away? We know that the momentum in a collision is conserved, so the total momentum before the collision is the same as the total momentum after the collision. Before the collision, only car A has momentum. p = mv = 700 x 30 = 21000 kgms-1 After the collision the total momentum is the same as before the collision. p = mv so 21000 = (700 + 500) v v = 21000/1200 v = 17.5 ms-1 Example 2 Car A has a velocity of 20 ms-1 and a mass of 500 kg. It collides with car B which is moving in the opposite direction with a velocity of -30 ms-1 and a mass of 600 kg. If car A rebounds with a velocity of -25 ms-1, what will be the velocity of car B after the collision? Momentum before the collision = momA + momB = (500 x 20) + (600 x -30) = 10000 – 18000 = -8000 kgms-1 Momentum after collision = -8000 = momA + momB

-8000 = (500 x -25) + (600 x vB) -8000 = -12500 + 600vB 600vB = 12500 – 8000 vB = 12500 – 8000

600 vB = 20.83 ms-1 Elastic and Inelastic Collisions If, in a collision, kinetic energy is conserved then that type of collision is known as an elastic collision. When kinetic energy is not conserved the collision is known as inelastic. All real collisions are inelastic, however it is possible to make up a collision where both momentum and kinetic energy are conserved from the data given, so you cannot assume that a collision will always be inelastic in a given question.

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Explosions and Newton’s Laws Far from being some abstract physics law from a text book the law of conservation of momentum has direct consequences in our lives. If you have ever been in a plane or on a jet ski then the motion of these vehicles depends on this law. The propulsion system works by the engine expelling some form of exhaust at high speed in one direction. The conservation of momentum means that something else must move in the opposite direction to conserve momentum. In the cases above that thing is the engine, which is attached to the vehicle causing it to move. This is essentially what is meant by Newton’s third law: every action [force] has an equal and opposite reaction [force]. Questions 1. Trolley A has a mass of 2 kg and is moving with a velocity of 5 ms-1 . It collides head on with a trolley B of mass 4 kg and a velocity of 3 ms-1 in the opposite direction. The trolleys stick together on impact. Calculate their velocity immediately after the collision. 2. Two cars are travelling along a race track. The car in front has a mass of 1200 kg and is moving at

15 m s −1

. The car behind has a mass of 1000 kg and is moving at 25 m s −1

. The cars collide and as a

result of the collision the car in front has a velocity of 20 m s −1

. Determine the velocity of the rear car after the collision.

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Impulse If an object is acted on by an external force its speed may change, resulting in a change in momentum. This change in momentum is also known as impulse. How much an objects speed will change due to an applied force will depend on the size of the force and how long it is applied for. It would be possible to obtain the same impulse with a small force applied over a long time or a large force applied over a short time. Some proposals for interplanetary travel involve very small forces being applied to space vehicles over long periods of time [months] to allow the vehicle to reach a high enough velocity. impulse = change in momentum = force applied x time

p = Ft m(v-u) = Ft F = t

u)-m(v F = ma

Again we have a situation where by considering momentum we arrive at one of Newton’s laws, namely Newton’s second law. It is hardly surprising that moving objects must obey the same basic principles otherwise every situation would need its own set of laws. The analysis of the force acting on an object causing it to change speed can be complex. Often we will examine the force over time in graphical form. The area under a force-time graph represents the change in momentum or impulse. Such a force is not constant. The graph can be used to find the average force.

In the graph above Impulse = Ft = area under graph = ½ x 0.04 x 100 = 2 Ns

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Consider what happens when a ball is kicked. Once the foot makes contact with the ball a force is applied, the ball will compress as the force increases. When the ball leaves the foot it will retain its original shape and the force applied will decrease. This is shown in the graph below. If a ball of the same mass that is softer is kicked and moves of with the same speed as that above, then a graph such as the one below will be produced. The maximum force applied is smaller but the time it is applied has increased. Since the ball has the same mass and moves off with the same speed its impulse will be the same as the original. This would result in a graph of the same area but different configuration.

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This idea is useful in the design of safety features in cars. Essentially the greater the time you can take to decelerate an object, the smaller the force you need to apply. If your face is slowed by the dashboard the time to stop after you make contact with the dashboard will be small, resulting in a large force and a big ‘OUCH’!! Airbags deflate when they are hit; this increases the time for the head to come to rest. The increased time results in a reduced unbalanced force. Less damage will then be done to the head.

Crumple zones work in a similar way to airbags, as the zone is crumpling up, the car continues to move, again increasing the time interval between the collision happening and the car coming to a complete stop.

Example A car with a mass of 500 kg is travelling at 30 ms-1 when a breaking force of 4500 N is applied. Calculate the time taken for the car to stop.

Ft = mv – mu -4500 x t = (500 x 0) – (500 x 30)

t = -15000 4500 t = 3.3 seconds

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Section 4

Gravitation

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Free Fall

When objects fall freely in a gravitational field they accelerate at a rate that corresponds to the gravitational field strength, g. This is free fall. When an object is in free fall it appears to be weightless. For example astronauts in an orbiting space craft seem to be weightless. This is because the sensation of weight is due to the fact that we can feel the reaction force from any surface we are standing on. In free fall there is no sensation of weight. In the orbiting space craft, the space craft and everything in it fall at the same rate i.e. everything is in free fall. Projectiles A projectile can be thought of any object that is moving through the air [or in space] where the only force acting on it is weight. If we consider objects on Earth in the first instance it may be easier to analyse the motion. The diagram above represents a method of analysing projectile motion. At any point in time a projectile will have a velocity, v. This can be split into two rectangular components, vH and vV. By ignoring air resistance there will be no forces acting on the object in the horizontal direction, leading to the object moving at constant speed in that direction. [Newton I] Vertically, the object will accelerate downwards with acceleration 9.8m/s2, again ignoring air resistance. This makes analysis quite simple. The equations to use are: Horizontally: v = s/t since there is a constant speed. Vertically: suvat since there is an acceleration.

vH

vV v

projectile

path

Projectile components

of velocity

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Example

A ball is kicked with an initial velocity of 30ms-1 at an angle of 25o to the horizontal.

(a) What is the maximum height reached by the ball? (b) How long does it take to reach this height? (c) How far away does the ball land?

vv

= vsinƟ vh

= vcosƟ

= 30 x sin 25 = 30 x cos 25

= 12.67 ms-1

= 27.19ms-1

(a) u = 12.67, v = 0, a = -9.8, s = ?

v2 = u2 + 2as

0 = 12.672

+ (2 x -9.8 x s)

19.6s = 160.53

s = 160.53

19.6

s = 8.19m

(c) vh

= 27.19ms-1 , t = 1.29 x 2 = 2.58s, s = ?

s = vh

x t

s = 27.19 x 2.58

s = 70.15m Questions A human cannonball at a circus is fired from the cannon with a muzzle velocity of 20 ms-1 at 30° to the ground, and (hopefully) lands in a safety net that is at the same height as the mouth of the cannon. a) Calculate the horizontal and vertical components of the performer’s velocity. b) How high above the net was he at his highest point? c) How far from the cannon should the net have been placed to safely catch the performer? d) In practice, this distance would have to be slightly shorter. Why?

30°

20 ms-1

32

Newton’s Thought Experiment

Suppose we fire a huge gun horizontally from a high mountain; the projectile will eventually fall to earth, as indicated by trajectory A, because the gravitational force accelerates it directly towards the center of the Earth. As we increase the velocity of our imaginary projectile, it will travel further and further before returning to earth. Finally, Newton reasoned that if the gun fired the projectile with exactly the right velocity, it would travel completely around the Earth, always falling in the gravitational field but never reaching the Earth which curves away at the same rate that the projectile falls. In other words the projectile would now be in orbit around the Earth. Newton concluded that the orbit of the Moon was of exactly the same nature: the Moon continuously "fell" in its path around the Earth because of the acceleration due to gravity, thus producing its orbit.

1. If the speed is low, it will simply fall back to Earth. (A and B)

2. If the speed is the orbital velocity at that altitude it will go on circling around the Earth along

a fixed circular orbit just like the moon. (C)

3. If the speed is higher than the orbital velocity, but not as high as the escape velocity, it will

continue go around Earth in an elliptical orbit. (D)

4. If the speed is very high, it will leave the Earth (E)

On October 4 1957 Newton’s thought experiment became reality with the successful launch of the world’s first artificial satellite, Sputnik. The significance of this launch in terms of its impact on life cannot be underestimated. Low-orbit and geostationary satellites are used for a wide range of applications, including environmental monitoring, communications, military applications and defence, and scientific investigations. Understanding Gravity Sir Isaac Newton (1642–1727) did not discover gravity. People had long been aware that objects when released would fall. Galileo (1564–1642) had carried out experiments on falling objects and put forward a theory on the motion of objects in free fall in his unfinished work De Motu (On Motion). Newton’s significant contribution, ranked ‘among humanity’s greatest achievements in abstract thought’ was to theorise that the force acting locally on an apple could be applied to the universe. Newton developed the universal theory of gravitation. This was published in his best-known work, the Principia. Questions to consider before you progress through this section:

What is gravity?

What is the force of gravity?

What are the effects of gravity?

What do we know about gravity?

How can we make use of gravity?

33

Section 5

Gravity and Mass

34

Gravity and Mass Mass is a measure of how much matter an object contains. This will only change if matter is added to or taken from the object. Gravity is caused by mass, any object that has mass will have its own gravitational field. The magnitude of the field depends on the mass of the object. The larger the mass is, the greater the field strength. The field of an object will then exert a force on any mass in its vicinity. The forces associated with masses smaller than planetary masses are so small that very sensitive equipment needs to be used to measure them.

Gravity is a force that permeates the entire universe; scientists believe that stars were formed by the gravitational attraction between hydrogen molecules in space. The attraction built up, over time, a large enough mass of gas such that the forces at the centre of the mass were big enough to cause the hydrogen molecules to fuse together, generating energy. This is what is happening in the centre of the sun. The energy radiating outwards from the centre of the sun counteracts the gravitational force trying to compress the sun inwards.

In time the hydrogen will be used up, the reaction will stop and the sun will collapse under its own gravity. If you expect to live for 4 or so billion years you could worry about this, otherwise ‘don’t worry, be happy’. When a star is formed there will be debris orbiting around it. This debris may join together to form planets, again due to the gravitational attraction between the particles. This is the explanation that is most commonly used to explain the formation of our own, and other, solar systems. Remember, Newton formulated the laws of gravity, he didn’t invent it. He produced what is known as the universal law of gravitation.

F = 2

21

r

mGm

This relationship shows that the force of attraction between any two masses depends on the magnitude of the masses and the distance between them. G is the universal constant of gravitation and has the value 6.67 x 10-11 m3kg-1s-2. This gives the force of attraction between two pupils of average mass [60kg] sitting 1 metre apart as 2.4 x 10-7N. If there were no resistive forces and one pupil were able to move under the influence of this force towards the other, it would take 22 360s to cover the 1m distance.

1m

35

Another application of the gravitational force is the use made of the ‘slingshot effect’ by space agencies to get some ‘free’ energy to accelerate their spacecraft. Simply put they send the craft close to a planet, where it accelerates due to the gravitational field of the planet. Here’s the clever part, if the trajectory is correct the craft then speeds past the planet with the increased speed. Don’t get it right and you still get a spectacular crash into the planet, could be fun but a bit on the expensive side!

Examples 1. What is the force of attraction between two cars, each of mass 1000

kg, separated by a distance of 1.5 m. F = Gm1m2 = 6.67 x 10-11 x 1000 x 1000 r2 1.52

F = 2.96 x 10-5 N

2. The distance between the Mars and the Sun is 2.28 × 1011

m. The mass of Mars is 3·83 × 1024

kg

and the mass of the Sun is 1·99 × 1030

kg. Calculate the gravitational force between Mars and the Sun.

F = Gm1m2 = 6.67 x 10-11 x 3·83 × 1024 x

1·99 × 1030

r2 (2.28 × 1011)2

F = 9.78 x 1021 N

36

Section 6

Special Relativity

37

Special Relativity This is probably the theory that Einstein is best remembered for. The idea of relativity goes back a long way before Einstein though so as Julie Andrews said in The Sound of Music –‘Let’s start at the very beginning’.

Gallileo was one of the first scientists to consider the idea of relativity. He stated that the laws of physics should be the same in all inertial frames of reference. If you were on a bus moving at constant speed you would experience the laws in the same way as if you were at rest.

Newton followed this up by expanding on Gallileo’s ideas. He also introduced the idea of universal time. He believed that it was the same time at all points in the universe as it was on Earth, not an unreasonable assumption you might think. Then along came the boy Einstein. He had an idea. Would he see his reflection? If he did, then relative to the ground the light would be travelling at 6x108m/s. This caused problems because the speed of light should only depend on the medium it was moving through. He finally came up with a solution: the speed of light is the same for all observers. This means he would measure the speed of light as 3x108m/s and a stationary observer would measure the speed of light as 3x108m/s. This theory led to some other interesting situations. Measured time will change for a moving system depending on who is observing the motion. You leave earth and your twin to go on a space mission. You are in a spaceship travelling at 90% the speed of light and you go on a journey that lasts 20years. When you get back you will find that 46 years will have elapsed on Earth. Your clock will have run slowly compared to one on Earth, however as far as you were concerned the clock would have been working correctly on your spaceship. Mad eh? The good news is that your twin should have 26 more cards and presents to give to you than you have for them!

38

Time dilation explanation Imagine we have a lamp, L, which sends out a pulse of light at the same time as producing a click. The light is reflected from a mirror, a distance D from the lamp, and when it arrives back at the lamp it produces a second click. The time between the clicks is t. The lamp and mirror are moving with velocity v. If an observer moving with the lamp observes the light being reflected they will observe the light travelling a distance of 2D [to the mirror and back] This means the time observed within the system will be

t =

c

2D

where c = speed of light. An observer who is stationary with respect to the system will observe a different scenario. Not only will they see the light move up and down to and from the mirror, they will also observe the light travel to the right as the system moves with velocity v, as demonstrated in the diagram below. The stationary observer will see the light travel a distance 2h between clicks. For them the time between clicks t’ will be

t’ =

c

2h

In this time frame the system will move a distance d = v x t’ where v is the velocity of the system. Using Pythagoras gives a relationship between D, d and h.

h2 = (½d)2 + D2 [equation 1]

t’ =

c

2h h =

2

ct'

d = vt’ ½d = ½vt’

t =

c

2D D =

2

ct

Substituting these values into eqn 1 gives

2

2

ct'

= (½vt’)2 + 2

2

ct

39

If we solve this quadratic for t’ we get: t’ =

where t is the time interval from the moving frame of reference, t’ is the time interval measured in a different frame of reference, c is the speed of light and v is the speed of the moving system.

This is known as the time dilation relationship. This means that the measured time depends on whether or not the observer is moving relative to the timer. Now I hear you ask, “How do we know this? You’re just having us on!” Well there is experimental evidence to support Einstein’s theory. There is a particle known as a muon that is created in the upper atmosphere. It only exists for a short time, a half-life of 1.56 x 10-6s. This means that for every million muons created at a height of 10km only 0.3 should reach the surface of the Earth. In actual fact around 5000 are detected. This is because the ‘muon clock’ runs slowly compared to the observer on Earth and the muon reaches the ground. http://www.scivee.tv/node/2415 A similar effect occurs with the length of a moving object. The term for this is known as length contraction. No surprises then that the length observed from a moving observer (l) will always be less than that measured in the stationary frame of reference (l’). l’ =

The factor is known as the Lorenz factor or the gamma factor and is given the symbol γ.

Why do we not notice these time differences in everyday life?

Looking at the graph on the next page we can see that for small speeds (ie less than 0.1

times the speed of light) the Lorentz factor is approximately 1 and relativistic effects are

negligibly small. Even 0.1 times the speed of light is 300,000 m s –1 or 1,080,000 km h–1 or

about 675,000 mph – a tremendously fast speed compared to everyday life.

However, the speed of satellites is fast

enough that even these small changes

will add up over time and affect the

synchronisation of global positioning

systems (GPS) and television

broadcasters with users on the Earth.

They have to be specially programmed

to adapt for the effects of special.

Very precise measurements of these

small changes in time have been

performed on fast-flying aircraft and

agree with predicted results, within

experimental error.

40

Examples 1. A rocket is travelling at a constant speed of 2.7 x 108 ms-1 compared to an observer on Earth. The pilot measures the journey as taking 240 minutes. How long did the journey take when measured from Earth? t’ = = t’ = 550 minutes 2. A rocket has a length of 25 m when at rest on the Earth. An observer, at rest on the Earth,

watches the rocket as it passes by at a constant speed of 2.0 × 108

ms−1

. Calculate the length of the rocket as measured by the observer. l’ = = l’ = 16.67m Questions 1. A rocket is travelling at a constant speed of 2.5 x 108 ms-1 compared to an observer on Earth. The pilot measures the journey as taking 630 minutes. How long did the journey take when measured from Earth? 2. A spacecraft travels to a distant planet at a constant speed relative to the Earth. A clock on the spacecraft records a time of 1.5 years for the journey while an observer on Earth measures a time of 2.7 years for the journey. Calculate the speed, in ms-1, of the spacecraft relative to the Earth. 3. A rocket has a length of 80 m when at rest on the Earth. An observer, at rest on the Earth,

watches the rocket as it passes by at a constant speed of 2.5 × 108

ms-1. Calculate the length of the rocket as measured by the observer. 4. A spaceship has a length of 100 m when measured at rest on the Earth. The spaceship moves away from the Earth at a constant speed and an observer, on the Earth, now measures its length to be 105 m. Calculate the speed of the spaceship in ms-1.

41

Section 7

The Expanding Universe

42

Doppler and Redshift The Doppler effect is the change in frequency you notice when a source of sound waves is moving relative to you. When the source moves towards you, more waves reach you per second and the frequency heard is increased. If the source moves away from you less waves reach you each second and the frequency heard decreases. You will have noticed this change in pitch as a car comes first towards you then passes and goes away from you. Police radar guns use the Doppler effect to measure the speed of motorists. Doppler is used to measure the speed of blood flow in veins to check for deep vein thrombosis [DVT] in medicine.

Equations

We can consider how this effect occurs. A source produces a sound of frequency fs.

speed of sound = v speed at which source is moving = vs

frequency source = fs observed frequency = fo

The frequency of the source will remain constant. It is the observed frequency that changes. If the movement is towards observer, the waves will 'squash up' Source stationary: In 1 second there will be fs waves produced. The sound will travel a distance of v metres in 1 second. This means there are fs waves in a distance of v metres. Source moving at speed vs towards the observer: In the same way as above, in 1 second fs waves will cover a distance of (v - vs) metres.

43

The observed wavelength of the waves will be

o =

f

v =

s

s

f

)v(v

fo =

o

v

=

s

s

f

)v- (vv

fo = fs

)v - v(

v

s

This gives an observed frequency that is higher than that of the source. If the source is moving away from the observer then the frequency observed will be smaller than that of the source.

fo = fs

)v v(

v

s

A good analogy is shown below:

Imagine a long conveyor belt running at a steady speed. An adult standing about halfway

along deposits sweets onto the belt at a regular rate, say one sweet per second

(frequency).

A child standing at the end of the conveyor belt collects the sweets in a bucket as they fall

off the end. As long as they are both standing still, the child will be collecting the sweets

at the same rate (frequency) as they are being deposited by the adult.

If the adult walks steadily away from the child, still depositing the sweets at the same

rate, the child now receives the sweets at a lower rate (frequency). The sweets wil l be

further spaced out on the conveyor belt (longer ‘wavelength’).

Conversely if the adult walks towards the child, the child will receive the sweets at a

higher rate (frequency) and they will be spaced closer together (shorter ‘wavelength’).

44

Examples 1. A fire engine emits sound waves with a frequency of 1200 Hz from its siren. The fire engine is travelling at 18 ms-1. (a) Calculate the frequency heard by a stationary observer as the fire engine moves towards her. (b) Calculate the frequency heard by the same observer as the fire engine moves away from her.

2. A man standing at the side of the road hears the horn of an approaching car. He hears a frequency of 540 Hz. The horn on the car has a frequency of 510 Hz. Calculate the speed of the car.

The Redshift of a Galaxy

Line Spectra

Each element has its own characteristic line emission spectrum and the corresponding

absorption spectrum. These can be indicated on a diagram as a coloured strip, or as an

intensity versus wavelength graph.

In intensity versus wavelength graphs, spectra outwith the visible region can be

represented.

These characteristic spectra allow elements in the stars and space to be identified.

45

The Discovery of helium

Helium was actually discovered on the Sun through the study of solar spectra, before it was

identified on Earth.

During the total solar eclipse in 1862, Pierre-Jules-Cesar Janssen (1824–1907) noted a

spectral line in the solar prominences that did not match any known element.

The British astronomer Joseph Norman Lockyear (1836–1920), who also worked on solar

spectra, decided that this line must represent a new element.

The new element was named ‘helium’ from the Greek word helios for ‘sun’. This was the

first and only element discovered off the Earth before it was observed on our own planet.

Helium was not identified on Earth until 1895, ie 33 years later.

Redshift

normal

redshifted

When studying distant celestial bodies, their spectra are found to contain the characteristic

absorption lines of elements. However, sometimes the spectral lines are found to have been

moved or shifted towards the red end of the spectrum because of an increase in

wavelength.

This effect is called redshift and, assuming it is due to the Doppler Effect, implies these

bodies are moving away from us. The amount of the shift also allows us to calculate how

fast they are moving away.

Redshift is defined as the change in wavelength divided by the original wavelength, and

given the symbol z.

So, redshift

If there is a decrease in wavelength, ie the line spectrum has moved towards the blue end

of the spectrum, this makes z negative, which means the body is moving towards us. This is

referred to as a blueshift.

46

For bodies travelling at non-relativistic speeds (ie less than 10% of the speed of light) we

can apply the Doppler equation for a stationery observer and a moving source. Using the

Doppler equation for a source moving away from the observer and for light v = c (the speed

of light), we get:

………… equation 1

Since , then and so:

……….. equation 2

From the definition of redshift,

– 1

Substituting from equation 2we get:

– 1

Substituting from equation 1 we get:

Note: This equation only applies to non-relativisticspeeds, say less than 10% of the speed of

light.

47

Section 8

Hubble’s Law

48

Hubble’s Law The astronomer Edwin Hubble noticed in the 1920’s that the light from some distant galaxies was shifted towards the red end of the spectrum.

He examined the spectral lines from various elements and all were shifted by the same amount for each galaxy. This shift was due to the galaxy moving away from the Earth at speed. The bigger the magnitude of the shift the faster the galaxy was moving. Over the course of the few years Hubble examined the red shift of galaxies at varying distances from the Earth. He found that the further away a galaxy was the faster it was travelling. The relationship between distance and speed of galaxy is known as Hubble’s Law.

z =

c

H0 d so v = Hod

where z = redshift =

em

obs

- 1

H0 = the Hubble constant v = the velocity of the galaxy c = speed of light d = distance to galaxy The value of the Hubble constant is not known exactly, however as more accurate measurements are made, especially for the distance to a galaxy, the range of possible values has reduced. It is currently thought to lie between 50 – 80 km/s/Mparsec. 1 Mparsec = 3.2 x 106 light years = 3.1 x 1022m Galaxies are moving away from the Earth and each other in all directions, which suggests that the universe is expanding. This means that in the past the galaxies were closer to each other than they are today. By working back in time it is possible to calculate a time where all the galaxies were in fact at the same point in space. This allows the age of the universe to be calculated. Currently NASA have a value of 13.7 billion years as the age of the universe from this method.

49

Example 1. Light from a distant galaxy is found to contain the spectral lines of hydrogen. The light causing one of these lines has a measured wavelength of 470 nm. When the same line is observed from a hydrogen source on Earth it has a wavelength of 434 nm. (a) Calculate the Doppler shift, z, for this galaxy. (b) Calculate the speed at which the galaxy is moving relative to the Earth. (c) In which direction, towards or away from the Earth, is the galaxy moving?

(a) z =

em

obs

- 1 = – 1 = 0.083

(b) z = vsource so vsource = c x z = 3 x 108 x 0.083 = 2.49 x 107 ms-1

(c) Away, as the light is red-shifted. Measuring Stellar Distances There are a number of methods that astronomers can use to calculate how far away stars are: 1. Parallax In order to calculate how far away a star is, astronomers use a method called parallax. Because of the Earth's revolution about the sun, near stars seem to shift their position against the farther stars. This is called parallax shift. By observing the distance of the shift and knowing the diameter of the Earth's orbit, astronomers are able to calculate the parallax angle across the sky. The smaller the parallax shift, the farther away from earth the star is. This method is only accurate for stars within a few hundred light-years of Earth. When the stars are very far away, the parallax shift is too small to measure. 2. Apparent Brightness The method of measuring distance to stars beyond 100 light-years is to use Cepheid variable stars. These stars change in brightness over time, which allows astronomers to figure out the true brightness. Comparing the apparent brightness of the star to the true brightness allows the astronomer to calculate the distance to the star. This method was discovered by American astronomer Henrietta Leavitt in 1912 and used in the early part of the century to find distances to many globular clusters. Astronomers define star brightness in terms of apparent magnitude (how bright the star appears from Earth) and absolute magnitude (how bright the star appears at a standard distance of 32.6 light years, or 10 parsecs).

50

Section 9

The Expansion of the Universe

51

The Expanding Universe It is generally accepted, based on the evidence given previously, that the universe is expanding. What is not known however is, what is going to happen to the universe in the future? There are essentially two scenarios: 1. Closed universe: the universe will slow its expansion and eventually begin to contract. 2. Open universe: the universe will continue to expand forever. Which of the two scenarios is more likely depends on one factor - what is the mass of the universe? We come back to our old friend gravity. How can we measure the mass of objects in space? You would need a big set of scales. In fact astronomers can relate the orbital speed of galaxies to their masses. The problem is that the masses measured seem to be bigger than the mass that can be accounted for by the number of stars present in a galaxy. This leads to the theory of ‘Dark Matter’. Basically there appears to be stuff there that we can’t see and don’t know what it is, so for the moment give it a name and hope we find out what it actually is later. Now if that was the only thing it might not be too bad, but the universe is expanding at a greater rate than astronomers would expect. It seems that something appears to be opposing the gravitational force.

“Give it a name” I hear you cry!

So they did, they called it Dark Energy. Who said astronomers aren’t creative thinkers?

Dark energy is now believed to make up most of the total content of the universe. Einstein

first proposed the cosmological constant, usually symbolised by ‘lambda’ (Λ), as a

mathematical fix to his theory of general relativity. In its simplest form, general relativity

predicted that the universe must either expand or contract. The accepted wisdom of the

time led Einstein to believe the universe was static, so he added this new term to stop the

expansion. Friedmann, a Russian mathematician, realised that this was an unstable fix,

‘like balancing a pencil on its point’, and proposed an expanding universe model, now

called the Big Bang theory. When Hubble’s study of nearby galaxies showed that the

universe was in fact expanding, Einstein regretted modifying his elegant theory and viewed

the cosmological constant term as his ‘greatest mistake’.

52

Section 10

Big Bang Theory

53

Big Bang Theory The universe started with a sudden appearance of energy which consequently became matter and is now everything around us. There were two theories regarding the universe

The Steady State Universe: where the universe had always been and would always continue to be in existence.

The Created Universe: where at some time in the past the universe was created. Ironically the term ‘Big Bang’ was coined by Fred Hoyle a British astronomer who was the leading supporter of the Steady State theory and who was vehemently opposed to the Big Bang theory. What was the evidence that finally swung the balance towards the Big Bang theory? We first need to consider how it is possible to determine the temperature of distant stars and galaxies.

How hot are the stars?

When you look carefully at the stars, even without binoculars or a telescope, you will see a

range of colours from red through yellow to blue. For example, Betelgeuse (Orion’s armpit)

looks red, Pollux (in Gemini) is similar to the Sun and is yellow, and Rigel has a blue tint.

A star’s colour depends on its surface

temperature. Dark red stars have surface

temperatures of about 2500 K. The surface

temperature of brighter red stars is

approximately 3500 K, yellow stars, like our

Sun, are roughly 5500 K, whilst blue stars range

from about 10,000 to 50,000 K.

Stars appear to the naked eye to be only one

colour but they actually emit a broad spectrum

of colours, easily seen using a prism or

spectrometer.

The thermal radiation spectra have very distinct shapes. The frequency of the light

emitted is determined by the temperature of the star. This idea has been with us for a long time, Jožef Stefan proposed in 1879 that the power irradiated from an object was proportional to its temperature in Kelvin to the fourth power.

P = σT4 where stefans constant, σ = 5.67 x 10-8 Wm-2K-4. What this means is that by examining the spectrum of a distant star, its temperature can effectively be measured.

Visible light is one of seven bands of electromagnetic radiation. These range from the least

energetic, radio waves, to the most energetic, gamma rays. All six bands can be emitted by

stars, but most individual stars do not emit all of them.

54

According to Big Bang theory, the early universe was a very small, hot and dense place, and as it expanded, the gas within it cooled. In 1948 three Physicists theorized that if the Big Bang had actually taken place then there would be a residual background EM radiation, in the microwave region, in every direction in the sky representing a temperature of around 2.7K. Thus the universe should be filled with radiation that is literally the remnant heat left over from the Big Bang, called cosmic microwave background (CMB) radiation. This radiation was finally discovered and in 1978 the discoverers were awarded the Nobel Prize in Physics. In 1989 a satellite was launched to study the background radiation, it was called the Cosmic Background Explorer [COBE]. In 1992 it was announced that COBE had managed to measure fluctuations in the background radiation. This was further evidence to support the Big Bang theory. An image of the fluctuations is shown opposite. Other evidence to support the Big Bang theory includes the relative abundances of hydrogen and helium in the universe.

The most abundant element in the universe is the simplest atom, hydrogen, and there are

many theories to support why there is so much of it. However, just over a quarter of the

ordinary matter in the universe is made up of helium. Some of this comes from hydrogen

fusion in the stars, but this source can only account for about 10% of the observed helium in

the universe. The rest must have already been in existence in the clouds of matter that

formed the galaxies.

The helium has been created by fusion, but not in the stars. The heat for this fusion must

have come from the universe itself. Knowing the current temperature of the Cosmic

Microwave Background lets us know how hot the universe has been in the past, and how

much helium could have been made. The Big Bang theory is again supported by this

evidence. Twenty-five per cent of the universe would have become helium during the

nucleosynthesis era. This brief period of time (less than 5 minutes) after the Big Bang had

temperatures and pressures so high as to allow protons and neutrons to fuse to become

deuterium nuclei and deuterium nuclei to fuse and become helium nuclei.

Another piece of evidence for the Big Bang is the explanation for Olber’s paradox. His paradox was in answer to the question “Why is the sky dark at night?” and is not as obvious as you first imagine.

55

If the universe followed the Steady State model then there should be an even distribution of stars in all directions. All the stars in the universe should be visible. This means the light from the stars should reach Earth and the sky should be bright.

To explain how this comes about imagine you are sitting in the middle of the school dining

hall at lunchtime. If you look in any direction you’re likely to see another student. On a

quiet day, you may be able to see through the gaps and glimpse the walls of the room, but

the busier it gets, the fewer gaps there are. In an infinite dining hall there would be no

gaps at all; in every possible direction a student would be blocking your view.

In an infinite, uniform, unchanging universe the stars are like the students in our infinite

dining hall. This means we would see a star in every direction, and everywhere in the night

sky would be as bright as the surface of the Sun. Any matter blocking our view would be

heated to such a point as to glow with the same radiance or be evaporated away.

Our options to rationalise what we see in reality are either that there are a finite number

of stars or that the universe isn’t unchanging

Big Bang theory takes an alternative view of the universe. The theory suggests that we can

only see a limited number of stars because there was a starting point to the universe.

There may be an infinite number of stars, but we can only see those inside our cosmological

horizon. Not all the tables and chairs are finally arranged yet. The Big Bang theory gives a finite age to the universe, and only stars within the observable universe can be seen. This means that only stars within the distance of 15 000 light years will be observed. Not all stars will be within that range and so the dark sky can be explained.

Summary of the different stages in the early formation of the universe

Up to 10–10 s after the Big Bang – temperature greater than 1015 K

Unknown due to lack of knowledge of the laws of physics at these very high temperatures.

From 10–10 s to 10–4 s (0.1 ms) after the Big Bang – temperature between 1015 K and 1012 K

A dense ‘sea’ of quarks, electrons, neutrinos and photons. Radiation too energetic to allow

formation of protons or neutrons.

From 0.1 ms to 1.0 s – temperature between 1012K and 1010K

Quarks have condensed into protons and neutrons. Radiation too energetic to allow

formation of nuclei.

From 1.0 s to 30,000 years – temperature between 1010 K and 10,000 K

Protons and neutrons have joined to form atomic nuclei. Universe still radiation dominated.

From 30,000 years and 300,000 years – temperature between 10,000 K and 3000 K

Universe is now matter dominated rather than radiation dominated. Otherwise as above.

After 300,000 year to present – temperature between 3000 K and 3 K

Atoms now formed by electrons attaching themselves to nuclei. Photons no longer energetic

enough to knock electrons out of atoms (decoupling).

Photons now cooling to form what we see as microwave background radiation.