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Rise and fall of the clockwork universe
Models and rules
Candidates should be able to: confirm or to determine mathematical relationships, eg systems which may or may not produce proportional, exponential, simple harmonic, inverse or inverse square relationships; these should incorporate the skills and techniques developed throughout course quality of data obtained is taken into account;
• identify, given appropriate data, the consequences of systematic error and uncertainty in measurements and the need to reduce these, eg in measurement of astronomical distances related to calculations of the size or age of the universe, and the effect of increased resolution and greater range of observations on those calculations;
•use computers to create and manipulate simple models of physical systems and to evaluate the strengths and weaknesses of the use of computer models in analysis of physical systems, eg the approximations and simplifications necessary in computer models, the ability of powerful computers to model very complex systems;
Candidates should be able to identify and describe:
the nature and use of mathematical models, eg models of random variation producing mathematical relationships, iterative and iconic models to predict behaviour in systems too complex for simple analysis; systematic analysis, using physical principles, to produce mathematical relationships;
changes in established scientific views with time, eg the explanatory power of Newton’s contributions to mechanics and gravitation, the implications of special relativity, the cosmological view of the universe (which is being continually refined), the statistical nature of the kinetic theory of gases and of the Boltzmann factor;
an issue arising from scientific research and development, stating more than one viewpoint that people might have about it, eg the understanding gained from space exploration and its incidental benefits, such as prestige, international co-operation and development in related disciplines, and drawbacks, such as cost and diversion of finds from other research.
•This module covers the core physics of radioactive decay and decay of charge on a capacitor, energy and momentum, the harmonic oscillator and circular orbits. The field model is developed through consideration of gravitational fields. The idea of differential equations and their solution by numerical or graphical methods is built up gradually using finite difference methods.
Recommended prior knowledge •The work is a continuation of the AS course as well as picking up on some ideas from GCSE. Candidates are expected to:
•• show knowledge and understanding of digital images as arrays of binary digits encoding pixel values, and of their manipulation and storage (AS 1.1.1 Imaging and signalling);
•• show knowledge and understanding of digital communication as a series of binary digits encoding sampled signal values, and of the frequency spectrum of a signal (AS 1.1.1 Imaging and signalling);
Recommended prior knowledge •• show knowledge and understanding of current as a flow of charged particles and potential difference as energy per unit charge, and be able to perform circuit calculations involving current, potential difference, resistance and energy (AS 1.1.2 Sensing);
•• know that radioactivity arises from the breakdown of an unstable nucleus and that there are three main types of radioactive emission with different penetrating powers (Sc7c);
•• know the relationship between force and work (AS 2.2 Space time and motion);
•• know the quantitative links between kinetic energy, potential energy and work (AS 2.2 Space time and motion);
•• know about the bodies in the solar system and that gravitational forces determine the movement of planets, moons, comets and satellites (Sc8c).
•
Charge
Charge is a property of a particle-measured in coloumbs
*if a particle has charge it will experience a force in an electric field
*Moving charge causes a magnetic field force*charge can be either positive or negative-and repel/attract opposites
Static electricity-caused by the transfer of electrons (and electrons only; not protons!) from one object to another
*The charged atoms are ions
-easiest way to charge an object with static e is with friction
*If we allow the charge that builds up in the static electricity to flow current will flow
The conservation of charge - it is not possible to destroy or create charge-it can only be cancelled out
Coulomb is the SI unit of electrical charge. A charge of 1C passes a point when a current of 1A flows for 1s.
Charge
*Charge is carried by tiny electron particles –negatively charged
-electrons actually flow from negative to positive
*One coulomb of charge is made from (6x 1018
) alot of electrons
* charge on electrons is very very small and is –ve -1.6 x 10-19
Therefore current is the movement of charged particles.
Current
*For current to flow:
-Must be complete circuit
-Energy must be supplied
*if these are met current flows from the positive end around the circuit to the negative end
Eg light bulb lights up if current flows
where Q=difference in charge ( C)
t=per second and I= Current (Amperes)
Sum of current in = sum of current out
In a series circuit-Current is the same everywhere- so the rate of flow of the particles is constant throughout
In a parallel circuit- Current is split between the branches of the circuit as the number of charged particles flowing cannot change
*Current is the rate of flow of charge and is measured in coloumbs per second also known as amperes
-the amount of charge which flows past a point every second is current at that point
We can measure current via an ammeter but in a seriesAmpere is the SI unit of electric current.
Charge carrier- any charged particles, such as electrons, responsible for a current
In a parallel circuit, the potential difference across each branch of the circuit is equal to the EMF, as the same 'force' is pushing along each path of the circuit. The number of charge carriers (current) differs, but the 'force' pushing them (voltage) does not.
Potential divider-a circuit in which two or more components are connected in series to a supply
Potential Difference As charge travels around a circuit, each coulomb of charge has less potential energy, so the voltage (relative to the power source) decreases
The difference between the voltage at two points in a circuit is known as potential difference, and can be measured with a voltmeter.
In a series circuit, the total voltage (EMF) is divided across the components, as each component causes the voltage to decrease, so each one has a potential difference. The sum of the potential differences across all the components is equal to the potential difference (but batteries have their own 'internal resistances', which complicates things slightly-internal resistance)
Series-a term used when components are connected end-to-end in a circuit.
A potential divider may work by combining a variable resistor such as an LDR or thermistor with a constant resistor, as in the diagram below.
As the resistance of the variable resistor changes, the ratio between the resistances changes, so the potential difference across any given resistor changes
Alternatively, a potential divider may be made of many resistors. A 'wiper' may move across them, varying the number of resistors on either side of the wiper as it moves, as in the following diagram:
In many cases, you will be told to assume that the internal resistance of the power source is negligible, meaning that you can take the total potential difference as the EMF of the power source.
Resistance (Ohms)-how many volts to drive a current per amp
The current through a component depends on what voltage and resistance which is a property of an artefact
The greater resistance the more voltage needed to make the current flow
higher resistance the more energy needed to push the same number of electrons through part of the circuit
Voltage over current
Resistance can vary for example with temperature
Conductance ( Siemens)A measure of how well an artefact (electrical component, not a material) carries electric current – the oposite of resistance
Current over volts/p.d.
As charged particles try to make their way around a circuit they encounter resistance to their flow- eg electrons colliding with atoms in a metal
ResistorsResistors are components which are designed to have a fixed resistance
• For a fixed resistance current and voltage are proportional to each other
Ohms lawA potential difference across an artefact constructed from ohmic conductors is equal to theproduct of the current running through the component and resistance of component V=IR
Any resistor that obeys Ohm's Law is called an ogmic resistor,( dont- non-ohmic)
Ohms law- the current in a metallic conductor is directly proportional to the potential difference across its ends.
Resistor-an electrical component whose resistance in a circuit remains constant, independent of current or potential difference
if we apply Ohm's law, remembering that the current is constant throughout a series circuit:
Multiply by current (I):
So, just as the resistances in series add up to the total resistance, the potential
differences add up to the total potential difference. The ratios between the resistances are equal to the ratios between the potential differences. In other words, we can calculate the potential difference across a resistor using the formula:
Electrical resistance- the ratio of potential difference to current
Power is the rate of transfer of energy (the rate of work done)
Measured in Watts – joules per second
As Power= ( currrent) x (voltage)Power= coulombs per sec x energy per charge
It measures how much potential energy is converted into other forms of energy- heat, light etc by a component or circuit in one secondIt is is due to the drop in the potential energy (/voltage) of charge
Where P= power, E= energy transferred (J) and t= seconds
I= Current( A), V= potential difference or EMF
As the power is the amount of energy changing per second the amount of energy being given out per second will equal the component giving out energyKilowatt-hour - the energy transferred by a
1 kW device in a time of 1 hour, 3.6MJ.
Power-the rate at which energy is transferred or the rate at which work is done.
P= V2 R
Voltage is the amount of energy transferred per coloumb
Or p.d across the component
Volts are Joules per Coulomb- energy converted per unit of charge removed
1 volt= 1 joule of energy moving 1 coloumb of charge through the component
Where V= Voltage ( V )E= different in potential energy ( j )Q= Charge (coulombs)
If there are several components in the circuit then the voltage from the power supply is shared
This includes in parallel circuits they will have the same voltage across them
-Higher voltage = more energy transferred by these
Potential difference is the difference between voltage at two points on a circuitAs charge travels around a circuit each coulomb of charge has less potential energy and so the voltage relative to the power source decreases
Potential difference is the difference between voltage at two points on a circuitAs charge travels around a circuit each coulomb of charge has less potential energy and so the voltage relative to the power source decreases
Potential difference-the energy lost per unit charge by charges passing through a component
Ammeters are assumed to have zero resistance. When we put them in series with other components we assume they do not add to the total resistance and therefore change the current they are supposed to be measuring.
Voltmeters are assumed to have infinite resistance. When we put them in parallel with other components we assume they do not change the resistance of the circuit and therefore the voltage they are supposed to be measuring.
Resistor NetworksMost circuits you come across will contain a number of resistive components. Sometimes it is necessary to be able to calculate the effective resistance or conductance of a combination of resistors. Here are the formulae for simple combinations.
Internal Resistance
You have a bulb at home which apparently needs 9.0V to work. You connect your cell across it. The bulb lights up but you suspect that it is not quite as bright as you thought. A voltmeter across the bulb reads 8.7V. What's going on?
We appear to be losing 0.3V but only when a current is being drawn from the cell
E = 9VV = 8.7Vv = 9 - 8.7 = 0.3 VI = 0.4 AR = V / I = 8.7 / 0.4 = 21.8r = v / I = 0.3 / 0.4 = 0.75
Imagine you go to a shop to buy a 9V cell. In the shop you check the terminal p.d. with a voltmeter and, sure enough, it reads 9.0V
The more current is drawn from the cell the more voltage we lose
The cell is giving 9 J to every coulomb that flows through it but some of this is being transferred getting through the cell itself,
after all it has resistance too.
We can think of the cell as being another resistive component in the circuit, the 9V is being shared between it and the external circuit.
Batteries just like other components in an electric circuit, have a resistance. This resistance is known as internal resistance.
This means that applying Ohm's law (V = IR) to circuits is more complex than simply feeding the correct values for V, I or R into the formula.Meaning we use the following formula:
PD across battery = EMF of battery - voltage to be accounted forLet us replace these values with letters to give the simpler formula:Vexternal = E - VinternalSince V = IR:Vexternal = E - IRinternalYou may also need to use the following formula to work out the external potential difference, if you are not given it:Vexternal = IΣRexternal
The existence of internal resistance is indicated by measuring the potential difference across a battery. This is always less than the EMF of the battery. This is because of the internal resistance of the battery
also remember the effects of using resistors in both series and parallel circuits.
Internal resistance- the resistance of an emf source. The internal resistance of a battery is due to its chemicals.
how the p.d. across a cell varies with the current drawn from it
In most questions you have to do internal resistance will not be a factor and you can assume that the voltage of the power supply is fixed. It should be obvious when you do need to consider it.
Imagine I need a voltage of 6V but I only have a 9V cell. What can I do
I could set up a circuit like this. The 9V is shared between the two resistors so this circuit is called a potential divider.The larger resistor gets more voltage. In fact the ratio of the voltages is equal to the ratio of the resistances.
i.e. 6 / 3 = 200 / 100
Potential DividersA potential divider, or potentiometer, consists of a number of resistors, and a voltmeter. The voltage read by the voltmeter is determined by the ratio of the resistances on either side of the point at which one end of the voltmeter is connected.To understand how a potential divider works, let us consider resistors in series. The resistances add up, so, in a circuit with two resistors:
Radio Microwaves Infra RedVisible Light Ultra Violet X Rays Gamma Rays
These waves interact with matter in different ways, hey have different properties and so have different uses.
c = 3 x 108 m/sAll travel at the same speed in a vacuum - speed of light-They can also travel through empty space and dont need a medium
They too have frequency and wavelengthThey are also all transverse waves- vibrations at right angles
EM WAVES MOVE ENERGY-They are oscillating E field and B field both of which are perpendicular to each other to the direction of propagation of the wave-a apace containing E electric and B magnetic fields usually at zero but can be disturbed by movement of charged objects- eg a spark- this creates a disturbance in these fields which travels away from the disturbance at the speed of light
Progressive waves
distribute energy from a point source to a surrounding area. They move energy in the form of vibrating particles or fields.
Mechanical waves are any waves that move through a medium. For example, water waves
Two types of progressive waves – longitudunal and transverse
WAVE basicsLONGITUDINAL TRANSVERSE
Oscillations are parallel to the direction the wave is traveling in
Oscillations are perpendicular to the direction the wave is traveling in
e.g. P waves, sounde.g. water waves, S waves, all
electromagnetic waves
There are two different types of progressive waves
WAVE basicsDisplacement, X, metres- How far a point on the wave has moved from its undisturbed position
Amplitude, a, metres- maximum displacement
Wavelength, λ, metres- the length of one whole wave, from crest to crest, or trough to trough
Period,T, seconds- the time taken for a whole vibration
Frequency , f, hertz- the number of vibrations per second passing a given point
Phase difference- the amount by which one wave lags behind another wave in degrees/ radians
The greater the amplitude of a wave then the more energy it is carrying
one hertz is equal to one wave per second
These are different!!!!!!!!!!!!!!!!!!!!
This is a displacement-time graph
-Note time period from trough to trough
-Amplitude is maximum displacement from central equilibrium position
-One complete wave will still be 360o
-Peak to trough = 180o
This is a displacement-distance graph
-Note wavelength from trough to trough
-Amplitude is maximum displacement from central equilibrium position
-One complete wave will still be 360o
-Peak to trough = 180o
You should be able to use this equation LEARN
Amplitude = maximum displacement Wavelength = distance between adjacent points which move in phase ( or simply distance from peak to peak ) Frequency = number of waves which pass a point every second
In addition…
radio and TV waves have such a high frequency that the kilohertz (kHz) or even the megahertz (MHz) are better units.
1 kHz = 1,000 Hz1 MHz = 1,000,000 Hz
WAVE basics
Speed = distance = wavelength = wavelength / (1/frequency) = frequency x wavelength
time time period
WAVE basics
Wavelength of blue light 470 nm
Wavelength of red light 700nm
The wave equation
Velocity= frequency x wavelengthMetres & hz
FREQUENCY= 1hZ ( OR 1/S ) 1 PERIOD
A wave is a disturbance by which energy is transferred
This plots the distance through the medium on the x-axis, and the amount of disturbance on the y-axis. The amount of disturbance is known as the amplitude. Wave amplitudes tend to oscillate between two limits, as shown. The distance in the medium between two 'peaks' or 'troughs' (maxima and minima on the waveform) is known as the wavelength of the wave.
Can be shown as displacement distance graphs or displacement time
Distribute energy from a point source to surrounding area
•RF 1.1 Creating models
•This section first considers models where the rate of change of a quantity is proportional to that quantity. It then goes on to consider the model of simple harmonic motion.
•The central importance of the exponential function develops candidates’ skills in Application of Number.
•Radioactive decay and capacitor discharge provide a context, but the point of view is broader, looking at these as examples of any kind of change where the change is proportional to the amount. Candidates have the opportunity to study examples of their own choice, for example, the action of heart defibrillators or camera flash units.
10.1 What if?
I know that models are simplified descriptions of reality, obeying definite rules
I know that exponential changes are ones in which the rate of change of a quantity is proportional to that quantity
Models
* We model things, they are simple and are created to show how real systems work
*give predictions
* model variations
* examples show radioactive decay, decay of charge on a capacitator, oscillating systems and how they resonate
*we build computational models to answer what if questions
* made with objects and rules but a simplified artificial version of reality
* They can show unexpected 0 and predict
* one form of a model can be adapted for different problems 00
A model is a set of assumptions that simplifies and idealises a particular problem
Designed specifically to give insight representation of a physical system
Models of a structure of a system- rutherfords atom Models of a process- rate of decay
When you have a model you can study its behaviour: try it under various conditions. In this way you discover its properties!
Although these properties may be necessary consequences of the type of model you are using (same kinds of models behave the same way), you're making assumptions
and using maths to study the actual nature of the model
By using assumptions you can write equations to describe calculations and predictions. Without the assumptions there would be too many factors to
consider
- By altering the conditions we can show outcomes and predictions for real life
- amending and extending models for one process mean that the model can be used to describe another unrelated process.
- Assumptions mean that equations are used to describe the process.
- The assumptions make it easier as there are less factors to consider
-Some aspects of the real world but artificial
-chaotic behaviour not allowing long term predictions
-computational models show visual displays,tackle larger complex problems that cannot be dealt with mathematically and showing how a wide range of changes turn out
Topics in this section are all based on models using differentiation for rate of changeSo easily modified and extended for different processesRadioactive decay and capacitor discharge
Models use the basic assumptions- the bare essential features for example
A population of rabbits, becoming more unstable when the population is higher.
Using two rules:- a rabbit breeds a new rabbit with a certain probability- a rabbit dies, with a certain probability
Observe that the rules are simple, stripped down. No males or females or other
Some models include actual scale models (wind tunnels).Mathematical models to determine for example a bridge designer models to determine possible stresses in each part of the bridge.
Simplified and idealisedAlso predictions may be compared to real life experimental results to test the validity of the model, which will be then modified due to this
Kinetic theoryKinetic theory of gases model assumptions as hard elastic spheres, ignoring any structure molecules have , to obtain a relationship for pressure and volume (of ideal gas)
Examples are models of radioactive decay, decay of stored electric charge and models of oscillating systems and how they resonate
10.2 Stocks and flows
I know that the charge on a capacitor of capacitance C at potential difference V is Q = CV so that C = Q/V and V = Q/C. The unit of capacitance is the Farad
I know that the differential equation for exponential change of a quantity Q is dQ = kQ dt with exponential growth if k is positive and exponential decay if k is negative.
I know that the differential equation for discharge of charge Q on a capacitance C through resistance R is dQ = -Q
dt RC
I know that the solution of the differential equation for discharge of a capacitor is Q = e -t/RC Qo
I know that the corresponding solution for radioactive decay is N = e -λt No
I know that the time constant RC is the time for the charge to reduce to 1/e ≈ 0.37 of its former value.
I know that the half-life of radioactive decay t1/2 is equal to loge 2 ≈ 0.693 of the time constant 1/λ.
I know that the energy stored on a capacitor = (1/2)QV= (1/2)CV2= (1/2)Q2/C
CapacitorsCharging and discharging
Capacitors store electrical charge
Capacitance is a measure of charge stored per volt
C = Q Charge, coulombs , V is p.d in volts
V and Capacitance in farads (F) I F=CV-1
Usually 1F is too big, pico,nano or micro farads are used
- useful as store up electric charge for use, amount and rate its released can be controlled
Flash photographyCharge stored in a capacitor flows through tube of xenon gas which emits a bright flash , the capacitor has to discharge really quickly and give a short pule of high current
DefibrillatorsElectric shock for heart, circuit can vary how much charge, released in short controlled burst
Back-up power supplies-computersIn case of power cut, large capacitors store charge while power is on then release charge slowly. Capacitors designed to discharge over hours, maintaining a steady flow of charge.
P.d across capacitor is proportional to charge stored
A capacitor acts as a store of opposite charges that are kept separated.
Storing charge / storing waterPotential difference depends on the quantity stored in both cases.
Graph for current
Current is the rate of flow of charge,
-Q is calculated from Q = It-measured in μC-V in Volts-Gradient of this graph is Capacitance=Q/V
-Electrons flow onto plate connected to negative battery terminal, negative charge builds up
-This build up of negative charge repels electrons off the plate connected to the positive terminal, making that plate positive. As these e- are attracted to this plate
The net charge stored is (+Q) + (-Q) = 0.
Because these charges have been pulled apart, the capacitor stores energy.
Charging Capacitors- When switched flicked to the left charge builds up on the plates if capacitor.
- Electrical energy provided by the battery is stored by the capacitor
Charging: As charge accumulates on the surfaces of capacitor being charged,Not a constant flow but one which gradually decreases as charge accumulates.
Charges by connecting to battery so current flows until capacitor is fully charged then stops
An equal but opposite charge builds up on each plate causing a potential difference between the plates. No charge can flow between the plates because they are separated by an insulator (dialectric)
Charging Capacitors
When fully charged
Initially the current is high but as charge builds up on the plates electrostatic repulsion makes it harder for electrons to be deposited
When the p.d across capacitor equals across battery current becomes zero. Capacitor is fully charged
Discharging Capacitors
Discharging: When a capacitor discharges charge flows in the opposite direction, and charge falls as the p.d. across the capacitor falls. Charge, current and p.d. all follow this exponential pattern
To discharge a capacitor take out the battery and reconnect the circuit to resistor.
- When a charged capacitor is connected across a resistor, the p.d drives the current through the circuit-The current flows in the opposite direction from the charging current-The capacitor is fully discharged when the p.d across the plates and the current in the circuit are both zero
- If switch is flicked the energy stored on plates will discharge through the bulb (converting electrical energy to light and heat, work is done)
- work is done removing charge from one plate and depositing charge to the other. The energy for this comes from the electrical energy of the battery, charge x p.d
Charge, current and p.d. all follow this exponential pattern
Discharging Capacitors
Discharge rate proportional to charge remainingThrough the origin
-Graphs show the less charge remaining the lower rate of discharge
The time taken to charge/discharge depends on:- The capacitance of capacitor ( C) Affects amount of charge that can be transferred at a given voltage
-The resistance of the circuit ( R) This affects the current in the circuit
Discharging
Q = Qoe-(t/RC)
Where Q is charge left, and Q0 the
charge when fully charged
Negative as discharging
dQ = - Q dt RC
Where Q is charge remaining, R is resistance, C capacitanceExact solution from equation
Discharging/Charging graphs are same for Charge, and Potential difference
Graphs are exponential.
The current time graph for charging and discharging is same as voltage-time graph for discharging
exponential change because the rate of loss of charge is proportional to the amount of charge left.
Equations for the discharging of a capacitor are same for charge current and Potential difference
The discharge of a capacitor:
Charge Q = Q0 e-t/RC
Voltage V = Q/C and C is constant, so V = V0 e-t/RC
Also the current I through resistor R is I = V/R which gives I = I0 e-t/RC
Current
From adjusting the variable resistor you can keep the charging current constant, (although impossible when nearly fully charged), recording the p.d regularly until it equals the battery p.d.
Energy stored by capacitor is area under p.d against charge
Not Q-V graph-straight line through origin-directly proportional
E= 1 QV 2
-Using equation Q =CV substituting
E= 1 CV2
2
E is proportional to V2
- work is done removing charge from one plate and depositing charge to the other. The energy for this comes from the electrical energy of the battery, charge x p.d
Exponential decay of charge
What if.. ..current flowing through resistance is proportional to potentialdifference and potential difference is proportional to chargeon capacitor?
capacitance C
current Iresistance R
potent ial difference, V
Potential difference Vproportional to charge Q
V = Q/C
Rate of f low of chargeproportional to potentialdif ference
I = dQ /dt = V/R
flow of charge decreases chargerate of change of charge proportional to charge
dQ /dt = –Q /RC
t im e for half chargeto decay is large ifresistance is largeand capacitance islarge
Charge decays exponentially if current is proportional to potentialdifference, and capacitance C is constant
Q
t
current I = dQ/dt
charge Q
Radioactive decay times
Time constant 1/λat time t = 1/λN/N0 = 1/e = 0.37 approx.t = 1/λ is the time constant of the decay
Half-life t1/2
0
N0
N0/2
N0/e
dN/d t = –λ N N/N0 = e–λt
t = 0 t = t1/2
t = time constant 1/λ
In 2 = loge 2
The half-life t1/2
is related to the decay constant λ
N/N0 = = – exp(–λ t1/2)
In = –λt1/2
t1/2
= ln 2 = 0.693λ λ
12
at time t1/2 number N becomes N0/212
Although the motion of electrons in the circuit will lead to random fluctuations in current, the number of charge carriers is so large that the variation appears continuous and smooth. The flow of charge is in fact a statistical average, subject to random variation, but these effects are too small to be noticed.
CARE IS NEEDED whilst charging capacitors to high voltages. It is important to make sure that capacitors are fully discharged after use.
Charge on a capacitor decreases exponentially-When a capacitor is discharging the amount of charge left on the plates falls exponentially with time- Same length of time for charge to halve, no matter how much charge you start with...
Time constant
When = (RC)
then Q = 1 Q
0 e
Where 1/e = 1/ 2.718 =0.37
So time constant is the time taken for the charge on a discharging capacitor to fall to 37% or on charging to rise to 63%The larger the resistance in series to a capacitor the longer it takes to charge or discharge
-t/RC
Q
Qe
0
= t-
N
N λe0
= Similar equations for radioactive decay and charge on capacitor
Equations for energy stored
E = Q0V0
E = CV02
E = Q02/C
12
12
12
Energy stored on capacitor = QV12
Energy delivered at p.d. V when a small charge δQ flows δE = V δQ
Energy δE delivered by same charge δQ falls as V falls
Energy delivered = charge × average p.d.
Energy delivered = Q0 V0
capacitor discharges add up strips to get triangle
V1
V2
V0
V0 /2
δQ δQ Q0charge Q charge Q
Capacitance, charge and p.d.
Q0 = CV0
V0 = Q0 /CC = Q/V
Q0V012
energyδEdelivered= V2 δQ
energyδEdelivered= V1 δQ
energy = area =
12
discharge is exponential and so these calculations are very approximate
Capacitors in parallel
Two small capacitors in parallel can be thought as being the same as one big capacitor
Adding capacitors in parallel increases the space available to store charge and will therefore increase the capacitance
The pd across each capacitor is the same as the total pd
As the capacitors or in parallel they each have the same voltage across them, so Vs cancel
Capacitors in series
In series each capacitor will have the same amount of charge stored because the charge from the first one travels to the second one and so on.
The total charge stored is the charge that was moved from the cell, which equals the charge that arrived at the first capacitor, which equals the charge that arrived at the second and the next
So QT = Q1 = Q2= Q3
The voltage of the circuit is spread out amongst the capacitors so each one has a fraction of the total
This is similar to resistors in parallel
So combinations of capasitors is opposite to those of resistors
Radioactivity and
exponential decay
I know that exponential changes are ones in which the rate of change of a quantity is proportional to that quantity
I know that radioactive decay is exponential, if the random nature of the decay is smoothed out, such that dN = -λN
dt
I know that the half-life of the radioactive decay is the (constant) time taken for the number of decaying nuclei to be halved
I know that the corresponding solution for radioactive decay is N = e -λt No
I know that the time constant RC is the time for the charge to reduce to 1/e ≈ 0.37 of its former value.
I know that the half-life of radioactive decay t1/2 is equal to loge 2 ≈ 0.693 of the time constant 1/λ.
Radioactive decayUnstable atoms are radioactive-too many neutrons, not enough or too much energy in the nucleus compared to protons in nucleus
-Unstable atoms break down by releasing energy and or particles until they reach a stable form this process is called radioactive decay
-Radioactive decay is a random process- cant tell when or which atom will decay
In some types of atom, the nucleus is unstable, so decays into a more stable atom. This radioactive decay is completely spontaneous. Forms new elementsHeating, subjecting to high pressure or strong magnetic fields won't affect the rate of decay.
Particles that ionise other atoms strongly have a low penetrating power, because they lose energy each time they ionise an atom
Alpha-can be expressed as-Alpha particles have a large charge(+2), so ionise other atoms easily (that they pass). -alpha particles lose energy rapidly as they travel, through ionising. So small range in air.
Beta-beta particle has Mass of one elctron= 0 (a charge of -1)-Protons & neutrons made of smaller particles "quarks"-Under certain conditions, a neutron can decay to produce a proton plus an electron. -The proton stays in the nucleus, whilst - the electron leaves the atom with high energy/speed, → beta particle.-the atomic mass is unchangedthe atomic number increases by 1.
Gamma rays – EM waves-no mass or charge change-Do not ionise other atoms- don’t lose much energy as travel as don’t interact with much matter-Therefore high penetrating power, and very long range-no such thing as pure gamma source, only by using aluminum with beta source-Used as tracer in medicine
- in very heavy elements (Uranium and Radium)-as have too many protons to be stable. They can become more stable by emitting an alpha particle.
-occurs in very "neutron-rich" elements, ,elements are typically created in nuclear reactors.Having too few protons and too many neutrons to be stable. They can thus become more stable by emitting a beta particle.-interact less readily with other atoms than alpha particles cause less ionisation and have a longer range in air. -less charge also less ionising
-Occurs after alpha/beta still has energy- excited state-loses this energy by emitting a pulse of very high frequency em radiation- gamma ray
Slow , high ionising ability , low penetrating power
Fast, medium ionising ability, medium penetrating power, stopped by aluminium
Very fast (c ) , cant ionise high penetrating power, stopped by lead High energy photon
Modelled by exponential decay Based on a very large number of atoms as a large sample shows a pattern
on overall behaviour
Cant predict the decay of an individual atom but can predict how many atoms will decay in a given time
If you plot number of atoms that decay each second against time you always get an exponential decay curve-models radioactive decay
The only reason we can do any calculations on radioisotopes is because there are huge numbers of atoms in most samples so we can use statistics to accurately predict what's most likely to happen.
There is background radiation everywhere! From cosmic rays (sun) to health related procedures such as x-rays. A geiger counter registers a count of fragments of radioactive elements.
This is taken into account in some occasions.
Over large time interval and sample to smooth decay
Smoothed-out radioactive decay
Consider only the smooth form of the average behaviour.In an interval dt as small as you please:
probability of decay p = λ dtnumber of decays in time dt is pNchange in N = dN = –number of decays
dN = –pNdN = –λN dt
time t
∆t
∆N
∆N
∆t
Actual, random decay
Simplified, smooth decay
time t
rate of change= slope
probability p of decay in short time ∆t is proportional to ∆t:
p = λ ∆ taverage number of decays in time ∆t is pN∆t short so that ∆N much less than Nchange in N = ∆N = –number of decays
∆N = –pN∆N = –λN ∆t ∆N = –λN∆t
dNdt=
dN = –λNdt
Actual, random decay fluctuates. The simplified model smoothsout the fluctuations
Real experimental graph shows the effects of randomness, remembenr as exponential decay is just a model
Background radiation is also taken into account.
Ionising radiation is dangerous. Therefore we can find absorbed dose ( energy absorbed per kg of tissue) measured in grays or j kg-1
The number of nuclei decaying in a time t = pN where p is the probability of decay and N is the number of undecayed nuclei at that time.
p = λ ∆t (probability of decay in a fixed time interval)
Rate of decay Measured by decay constant (lamda) in s-1
Decay constant measures how quickly isotope will decay
(probability of a nucleus decaying per second)
Bigger means more likely to decay (faster rate of decay- number of atoms)
Probability that an atom will decay per second
Activity of sample-Number of unstable atoms that decay each second
-Activity ∝ size of sample N
Therefore activity-time graph an exponential curve due to as time goes on and number unstable decreases activity decreases
Activity=λN in Bq Becquerels 1Bq = 1 decay per second
As atoms decay the sample size gets smaller so the activity falls, shallowerdN = - λ N
dt
that this is a differential equation in which the rate of change of N is directly proportional to N.
Graph of unstable atoms remaining in sample N against time
Gradient is negative
Gradient is the change in the number of radioactive nuclei remaining in a given time (rate of decay) which must also be negative
Rate of decay = dN = - λN dt
As atoms decay sample gets smaller so activity falls and graph gets shallower and shallower
Differential equations- describe rate of change-number of undecayed atoms in sample or decay of charge of a capacitor has similar equation
Half life T1/2 of an isotope -Its the average time it takes for the number of undecayed atoms to halve-Not found by counting atoms but measuring time it takes for activity to halve
Longer half life means stays radioactive for longer
Same graph for count rate, activity or number of atoms
Same shape for capacitor decay-When actually measuring remember background needs to be subtracted from activity to give source activity
can be found from graph:
-Where t=0 , go to half of original value, half life is the time at this-t3-t2 is same difference as t1 is as half life is constant
Or equation:T
1/2= ln 2
λ
Decay constant and half life are constant for a sampleN= N
0 e-ʎt
Number of atoms remaining from original , decay constant and time, remember negativeIs an exact solution of differential equation number of undecayed atoms
The half-life of radioactive decay t1/2 is equal to loge 2 ≈ 0.693 of the
time constant 1 / λ.
Clocking radioactive decay
Half-life
time t
N0
N0
/2
N0
/4
N0 /8
t1/2
t1/2
t1/2
t1/2
Activity
time tt1/2
halves every half-life
slope = activity =dNdt
Measure activity. Activity proportional to number N left
Find factor F by which activity has been reduced
Calculate L so that 2L = F
L = log2F
age = t1/2
L
Radioactive clock
In any time t the number N is reduced by a constant factor
In one half-life t1/2
the number N is reduced by a factor 2
In L half-lives the number N is reduced by a factor 2L
(e.g. in 3 half-lives N is reduced by the factor 23 = 8)
number N ofnuclei halvesevery time tincreases byhalf-life t
1/2
dN/dt = - λ N has a solutionN = N0 e-λt. When N = N0/2, N0/2 = N0 e-
λt 1/2 and so e-λt 1/2 = ½. By taking logs of both sides we have λ t 1/2 = ln 2 where t 1/2 represents the half-life.
◦ In one half-life, N falls by a factor of 2. ◦ The number of nuclei is reduced by a factor F which equals 2L in L half-lives. ◦ Since F = 2L we can use L = log2F to find L.◦ The time elapsed would then be t1/2 L.◦ If time elapsed and L are both known, t1/2 can be calculated. Note that L does NOT have to be a whole number of half-lives.
F = 2L L = log2F age = t1/2 L
Modelling decay
Discharging capacitors Radioactive isotopes
Decay equation is Q= Q0e-(t/RC) Decay equation is N= N
0e- tʎ
The quantity that decays is Q, the amount of charge left on the plates of the capacitor
The quantity that decays is N, the number of unstable nuclei remaining
Initially , the charge on plates is Q0
Initially , the number of nuclei is N0
It takes RC seconds for amount of charge remaining to fall to 37% of its initial value
It takes 1/ʎ seconds for the number of nuclei remaining to fall to 37% of initial value
The time taken for the amount of charge left to decay by half (the half life) is t
1/2 = ln2 x RC
The time taken for the number of nuclei left to decay by half (the half life) is t
1/2 = ln2/ ʎ
Exponential changes are ones in which the rate of change of a quantity is proportional to that quantity.
Radioactive decay is exponential, if the random nature of the decay is smoothed out
08/12/14
Plotting the log of an exponential function will produce a straight line graph
As the graphs contain e you can use ln graphs to show linearity
N= N0 e-ʎt
ln (N)= -ʎt
Iteration methods to solveN= N
0 e-ʎt using dN = - λN
dt
A similar method can be applied to different equations of the same type
Time seconds N dN/dt ∆N New N
Equal intervals Always initial before time
Using the equation and the original value
Found from the value of dN/dt multiplied by dt (time interval)
New value from previous calculations N +∆N, used as next original value
Through the origin as proportional
dQ/dt = kQ as the general form of the differential equation for exponential change
where Q can represent any quantity. If k is positive we have exponential growth and if k is negative we have exponential decay
Simple harmonic
motion and oscillators
10.3 Clockwork models
I know that the period of a harmonic oscillator is independent of its amplitude
I know that the variation of displacement of a harmonic oscillator with time is sinusoidal, having the general form s = A sin (2πft + φ) where φ is a phase angle. The expressions s = A sin (2πft) and s = A cos (2πft) are often convenient.
I know that the motion of a harmonic oscillator is governed by the differential equationd2s = a = -(2πf)2s with (2πf)2s = k dt2 m in the case of a mass m restrained by springs of spring constant k.
I know that the acceleration of a simple harmonic oscillator is proportional to its displacement and always acts towards the equilibrium (zero displacement) position.
I know that there are fixed phase relationships between the variations of displacement, force, acceleration and velocity. In particular there is a phase difference of π/2 between displacement and velocity, and between velocity and acceleration.
SIMPLE HARMONIC MOTION
When a bodies acceleration is directly proportional to its displacement but in the opposite direction
Acceleration towards the midpoint and opposite direction to displacement
-Its an idealised time of oscillating motion for objects vibrating or oscillating
-May or may not be real simple harmonic motion but analysis of motion and predictions as Simple harmonic motion is a reasonable approximation.
Radian mode used for calculations
Pendulums show simple harmonic motion, as does this mass
time period stays the same even when amplitude changes
Displacement, velocity and acceleration are all vectors so direction is important.
As you can see when a is max V is 0, when V is max a is 0. At the bottom the mass stops momentarily, velocity is 0
During a full oscillation.Displacement is a sin curve, velocity a cosine curve and acceleration is a negative sin curve (or phase difference of 180o)- The differential or gradient of the previous When it passes through the middle ,
equilibrium position, it has max velocity, zero displacement and min acceleration.It has the most acceleration when changing direction
Showing that acceleration is proportional to displacement but opposite direction (negative)
Amplitude A The maximum displacement from the equilibrium position of vibrating object
m
Period T The time for one complete oscillation- there and back s
Frequency F The number of oscillations in one second Hz s-1
Displacement x or y How far the object is from its equilibrium position m
The first time a mass returns to equilibrium position half a cycle has been reached
When an oscillating object reaches max displacement it changes direction immediately speed=0 then travels fast through equilibrium position
x = A sin (2πft))
ω = 2πf = 2π its a constant related
T to frequency, rad s-1
Properties are functions of time, in radians
Displacement varies sinusoidally with time- if it starts at the equilibrium position (displacement starts at zero)If starts at extremes (highest point or lowest point) its a cos wave as its a translated sin waveIf it starts at the minima then it is a negative cos wave (upside down cos wave) Basically as soon as it starts the wave which represents is displacement from go
Constant time period even though amplitude changes
Or X = (- ) A cos (2πft)
Or v = Aω cos (ωt)
VelocityEquation from derivative or gradient of displacementDepending on what displacement is
When sin(wt) =1 when wt= π (or multiple of)
Given in booklet, so max when x =0
ω = 2πf = T its a constant related to frequency, rad s-1
Acceleration acts in opposite direction to displacementWhich is why should be negative in equation
Acceleration of the mass will be proportional to displacement but in opposite direction
F=-kx (k spring constant) f=ma -kx = maa = - (k/m)x
a = - w2x w2 = k/m w=(k/m)0.5 = 2πf
Or
Or –Aw2Sin(wt)
Always trueUnless starts at min point, check
Differentiate the displacement twice
a α –s the constant is (2πf)2:a = - (2 πf)2 s
08/12/14
Simple harmonic motion (SHM)
The restoring force is directly proportional to the displacement from the equilibrium position and is always directed towards it.
We can summarise this with the equation F = - ks.
With mass m and spring constant k the acceleration is given by
a = F = - k s m m
We can write this as the second-order equation
d2s = – k s. dt2 m
Period of a harmonic oscillator is independent of amplitude and vecoity
T = 1/f = 2π (m/k)0.5 therefore
Obtained from a proportional to F
From f=ma
Therefor when object suspended from string and displaed Restoring force proportional to displacement
T2 = 4π 2(m/k)
Plotting mass against T2 Gives straight line, gradient (4π 2)/k
Showing that T2 proportional to mT is not affected by Amplitude
between displacement and velocity and between velocity and acceleration
If x = A cos 2πftv = dx = -2πfA sin 2πft dta = dv = - (2πf)2A cos 2πft = - (2πf)2x. dt
we know that a = - k x. m
Hence, (2πf)2 = k mand f = k 2π m
It follows that T = 2π (m/k).
The equations confirm the phase difference of π/2 between s, v and a.
T ∝ √m and T ∝ 1/√k
and f ∝ √k and f ∝ 1/√m
or more succinctly, T ∝ √(m /k)
and f ∝ √(k/m).
Mass
T
The stiffer K
T is unaffected by A amplitude
T
Stiffness
The heavier m
T
Amplitude
From origin Tending at both axis, never touching
08/12/14A person uses a rope swing to get across a stream. They run and grab the rope at A and swing to the other side at B, 6m from the start.They hang on to the rope until it stops at the end of the swing at C, 6.5 m from the start.The centre of the swing (and of the stream), D, is now 3.5 m away, so the amplitude of the swing is 3.5 m.
a) If the frequency of the swing is 0.2 Hz, what was the person's speed as they passed the opposite bank of the stream at B, 3 m from the centre of the swing?b) What was the maximum speed during the swing and where exactly was the person then?
Force, acceleration, velocity and displacement
If this is how the displacement varieswith time...
... the velocity is the rate of changeof displacement...
... the acceleration is the rate ofchange of velocity...
...and the acceleration tracks the forceexactly...
... the force is exactly opposite tothe displacement...
Phase differences Time traces varies with time like:
π/2 = 90°
π/2 = 90°
π = 180°
zero
displacement s
force F = –ks
displacement s
cos 2πft
same thing
–sin 2πft
–cos 2πft
–cos 2πft
cos 2πft
acceleration = F/m
ve locity v
Graphs of displacement, velocity, acceleration and force against time have similar shapes butdiffer in phase
There are fixed phase relationships between the variations of displacement, force, acceleration and velocity. In particular, there is a phase difference of π / 2 between displacement and velocity, and between velocity and acceleration.
When vmax : displacement=0 Zero acceleration
when displacement max:Velcocity =0 Acceleration is maximum negative or positive(towards equilibrium)
If a=0 no resultant force,At equilibrium position x=0-speed at maximumObserved by differential
Language to describe oscillations
+A
0
–Aperiodic time Tphase changes by 2π
Aangle ωt
Sinusoidal oscillation
time t
amplitude A
f turns persecond
ω = 2πf radian per second
2π radianper turn
Phasor picture
s = A sin ωt
Periodic time T, frequency f, angular frequency ω:
f = 1/T unit of frequency Hz ω = 2πf
Equation of sinusoidal oscillation:
s = A sin 2πft s = A sin ωt
P h a s e d i f f e r e n c e π / 2
s = A s i n 2 π f ts = 0 w h e n t = 0
s = A c o s 2 π f ts = A w h e n t = 0
t = 0
sand falling from a swinging pendulum leavesa trace of its motion on a moving track
A sinusoidal oscillation has an amplitude A, periodic time T, frequency f = and a definate phase1T
s = A sin (2πft + ф) Where ф is the phase angle
There are fixed phase variations between displacement, force, acceleration and velocity
Dynamics of a harmonic oscillator
How the graph starts
force changesvelocity
0
time t
δt
zero initialvelocity
velocity would stayzero if no force
How the graph continues
force of springs accelerates mass towards centre,but less and less as the mass nears the centre
time t
0
trace straighthere because nochange ofvelocity
no force at centre:no change of velocity
trace curvesinwards herebecause ofinwardschange ofvelocity
change of velocitydecreases asforce decreasesnew velocity
= initial velocity+ change ofvelocity
Constructing the graph
if no force, same velocityand same change indisplacementplusextra change indisplacement fromchange of velocity dueto force
= –(k/m) s (δt)2
change in displacement = v δt
δt
δt
extra displacement= δv δt
change of velocity δv= acceleration × δtδv = –(k/m) s δt
because of springs:force F = –ks
acceleration = F/macceleration = –(k/m) s
Health warning! This simple (Euler) method has a flaw. It always changes thedisplacement by too much at each step. This means that the oscillator seemsto gain energy!
extra displacement
This is a graphical representation of rates of change, and how they change with time. You now have a formula for drawing a graph of displacement against time for a simple harmonic oscillator so you may be asked to try out this technique at this stage. Remember that with all step-by-step methods there are errors introduced at each stage.
Changing rates of change
change in ds = d(ds) = dv dt
= a dt2
The first derivative ds/dt says how steeply the graph slopesThe second derivative d2s/dt2 says how rapidly the slope changes
dtd s = v d t
v = dsdt
s
t
dtds = v dt
dt
ds = (v + dv) dt
a = dvdt
t
s
dtv dt
dtv dt
dv dt
change in ds = d(ds) = dv dt = a dt2
s
t
new slope = new rate of change of displacement
= new velocity (v + dv)
new ds = (v + dv) dt
dv = a dt
slope = rate of change of displacement
= velocity v
= acceleration a
d ds d2sdt dt dt 2 = a( )=
rate of changeof velocity
rate of changeof slope
=
Simple harmonic motionModel of simple harmonic motion of a mass and a spring system
Where x is displacement, k spring constant and m mass. Used to predict the frequency of oscillation (example in a car suspension)
From second derivative of displacement, first derivative of velocity
◦ Many, but not all oscillators are isochronous.
◦ If the oscillations are isochronous the period does not depend on the amplitude.
◦ Oscillatory systems have an equilibrium position.
◦ The restoring force is always directed towards the equilibrium position.
◦ The time trace (displacement against time graph) is cyclic in nature but the amplitude
and period may change with time.
Hookes lawHookes law
Where F is the tension acting on the spring.e =extension = (l-lo); l is the stretched length and lo is original length, and.k is the gradient of the graph above. It is known as the spring constant.
-Force (or the tension) is equal to the spring constant multiplied by the extension
-Hookes law is obeyed during the elastic region of a property (or sometimes called the limit of proportionality)
-The greater the value of Spring constant the stiffer the spring
-during this elastic region the value of this spring constant doesn’t change unless you change the shape of the spring or the material that the spring or wire is made of.
a force extension graph
•The gradient represents the spring constant and so the gradient is constant showing hookes law obeyed
•The material will demonstrate elastic behaviour before it gets to P (the limit of proportionality)
•Point P shows it has exceeded the limit of proportionality by showing a curve (non- constant gradient) meaning hookes law is not obeyed here so the material will not show elastic behaviour past point p
•the material will experience deformation and so not return to its original shape at point P
The extension of an elastic object is directly proportional to the force applied to it
The force constant-the ratio of force to extension for a spring or a wire.
Motion of a harmonic oscillator
large force to left
large displacement to right
zero velocity
mass m
displacementagainst time
velocityagainst time
forceagainst time
right
left
small displacement to right
small velocityto left
mass msmall force to left
right
left
large velocityto left
mass mzero net force
right
left
small displacement to left
smallvelocity
mass m
small force to right
right
left
large displacement to left
mass m
large force to right
zero velocityright
left
Everything about harmonic motion follows from the restoring forcebeing proportional to minus the displacement
10.4 Resonating
I know that the energy stored in a stretched spring, at extension x if the stretching force F = kx is ½kx2
I know that the energy stored in a mechanical oscillator is the sum of its potential energy ½kx2 and its kinetic energy ½mv2. At maximum extension and zero velocity, all the energy is stored in potential energy of the spring; at zero extension and maximum velocity all the energy is stored as kinetic energy of the moving mass.
I know that an oscillator driven by a sinusoidally varying force responds strongly and has a large amplitude if the force varies at or close to its natural frequency. This is resonance.
I know that the range of frequencies over which a resonator responds with large amplitude varies with the amount of damping. The width of the resonant response curve increases as the damping increases.
Free and forced
vibrations
Learning outcomes• The energy stored in a stretched spring, at extension x if the stretching force F = kx is ½kx2 • The energy stored in a mechanical oscillator is the sum of its potential energy ½kx2 and its kinetic energy ½mv2. At maximum extension and zero velocity, all the energy is stored in potential energy of the spring; at zero extension and maximum velocity all the energy is stored as kinetic energy of the moving mass.• An oscillator driven by a sinusoidally varying force responds strongly and has a large amplitude if the force varies at or close to its natural frequency. This is resonance.• The range of frequencies over which a resonator responds with large amplitude varies with the amount of damping. The width of the resonant response curve increases as the damping increases.
The relationship between force to extend a spring, and extension, determines the energy stored.
Energy stored in a spring
no forcework F1 δxforce F1
δ x
larger force
extension x
area below graph= sum of (force ×change in displacement)
extra areaF1 δx
total area Fx
21
unstretched
F1
Energy supplied
energy stored in stretchedspring = kx2
21
small change δxenergy supplied = F δx
stretched to extension x by force F:energy supplied = Fx2
1
spring obeysHooke’s law: F = kx
F = 0x = 0
F = kx
Energy stored in a stretched spring is kx22
1
00
x
Energy flow in an oscillator
displacement
time
time
s = A sin 2πft
PE = kA2 sin22πft
0
0
potential energy= ks21
2
potential energy
energy in stretched spring
energy carried by moving mass
time
time
0
0
kinetic energy= mv21
2
velocity
mass andspringoscillate
vmax
A
Avmax vmax
The energy stored in an oscillator goes back and forth between stretched spring and movingmass, between potential and kinetic energy
from spring tomoving mass
from movingmass to spring
from movingmass to spring
from spring tomoving massenergy in
stretched spring
energy inmoving mass
kinetic energy
vmax = 2πfA
v = vmax cos 2πft
KE = mvmax cos22πft212
12
Constant energy as long as none is lost to surroundings. Total energy is constant, equal to the sum of KE and PE.
Every half cycle the energy changes from PE to KE to PE again.
Forced vibrations happen when there is an external driving force
The frequency of this force is called the driving frequency
Resonance happens when driving frequency = natural frequency
When the driving frequency approaches the natural frequency the system gains more and more energy from the driving force and so vibrates with rapidly increasing amplitude, When this happens system is resonating
PE KE
Total Constant
-A A -A A
-A A
Sharp peaks
UnderdampedSystem remains oscillatingCritical dampingSmallest amount of damping requiredOverdampedMore than critical, takes longer to reach equilibrum
Damping displacement over time graphAMPLITUDE decays EXPONENTIALLY WITH TIME
PERIOD stays the SAME
Amplitude decreases every cycle due to damping
At start certain amount, lost every cycle due to friction or air res.
Damping uses up energy to stop oscillaions
There is no longer a periodic force acting on (free oscillator) then energy will be quickly dissipated and reach rest quicker
Resistances are added, shock absorbers can be used
Damping doesn’t reduce the frequency of the driven oscillation
Objects which oscillate freely (pendulums) have a frequency called the NATURAL FREQUENCY
Objects may also have FORCED OSCILLATION –> Made to oscillate so their frequency is called the DRIVING FREQUENCY
As the frequency approaches the natural frequency the amplitude increases
Amplitude-frequency graphFor damping will never cross axis
The more damping the slightly lower amplitude even for natural frequency
If energy is being removed from the system the amplitude of the oscillations must become smaller amplitude decreases with time, higher damping means faster reduction in this
Resonance is when a free vibration is allowed to oscillatr at natural frequency => if the driving force oscillates at natural frequency
Determined by natural frequency, the amplification
A longer pendulum has a larger time period, and a smaller natural frequency
Amplitude is at maximum when frequency is fo – this is resonance
A periodic force needs to be on it to carry on vibrations due to lost energy
This is a forced frequency oscillation, if it equals the naural frequency or a whole multiple of it amplitude will grow
Resonance at f0 is the biggest amplitude
During vibrations with resoncace can be dangerous
Pendulums are only linear if small angle, otherwise oscillations decrease amplitudes, at longer amplitudes/angles motion is not sinosoidual
The frequency of un-damped oscillations in a systm allowed to oscillate on its own is fo
The frequency of undamped oscillations in a system, which has been allowed to oscillate on its own, called the natural frequency fo
Need to apply a periodic force, giving a periodic force gives a forced frequency. If the forced frequency equals the natural frequency of a system (or a whole number multiple of it) then the amplitude of the oscillations will grow, called resonance.
The energy sloshes back and forth between being stored in a spring, and carried by the mass.
An oscillating mass between springs can be described in terms of the energy exchanges taking place – between elastic energy stored in the springs and kinetic energy carried by the moving mass. The exchange happens because of the action of the force; energy transferred is force × distance.
The amplitude of the oscillator is maximum when the frequency of the varying force matches the natural frequency of the oscillator.
You should be aware of the effect of damping. The width of the resonance curve increases as the damping increases.
In a SHO energy can be POTENTIAL O KINETIC
This potential could be elastic, gravitational or electrical
Total energy at any time remains constant in an undamped system and is the sum of others
When the driving frequency matches the natural frequency of an oscillator the amplitude of oscillation can rise dramatically. This is resonance. This experiment gets you to measure how the amplitude of an oscillator changes with the frequency of the driver. If time is short, the class may be divided so that half use a damped oscillator and half an undamped one. It is important to measure the natural frequency so that the frequency at maximum amplitude can be compared with it.
Having seen a few examples of resonance and produced sketch graphs of amplitude and frequency it will be worthwhile to model what has been observed using the same techniques met earlier in the chapter. The concepts of damping, driving force and energy changes are crucial here and all brought out clearly in the activity. A simple harmonic oscillator is subject to two other forces – a periodic driving force and a drag force, depending on velocity. The model allows you to simulate experiments done in the laboratory, but perhaps more importantly, to see how such a situation can be modelled, building the model out of many well understood fragments. The work on resonance can be concluded by considering the readings and discussing the part resonance plays in the modern world.
Resonant response
10
5
0
1
Example: ions in oscillat ing electric field
ions in a crystalresonate andabsorb energy
Oscillator driven by oscillating driver
electricf ield
+ – + –
low damping:large maximum responsesharp resonance peak
frequency/natural frequency
0 0.5 1 1.5 2.0
10
5
0
1
frequency/natural frequency0 0.5 1 1.5 2.0
more damping:smaller maximum responsebroader resonance peak
R e s o n a n t r e s p o n s e i s a t m a x i m u m w h e n t h e f r e q u e n c y o f a d r i v e r i s e q u a l t o t h e n a t u r a l f r e q u e n c yo f a n o s c i l l a t o r
narrow rangeat peakresponse
12
wider rangeat peakresponse
12
The energy stored in a spring when stretched at extension PE= 0.5 k x2
When stretching with F =kx
The amplitude of a wave shows how much energy the oscillator has.
PE= 0.5 kx2 energy is proportional to amplitude2
Therefore if Amplitude is halved energy is quartered.
Different amounts of damping have different effects
DampingWhen overdamping occurs the more damping on a system the slower it will lose energy to its surroundings
Due to friction of the oscillating body and particles in the air etc, the amplitude of vibrations becomes progressively smaller as energy is lost.
If the energy is being removed from the system, the amplitude of oscillations becomes smaller, they re being dampedAmplitude decreases with timeThe higher the damping the faster oscillations reduce in size
Critical damping is the damping required to make the oscillations stop in the quickest possible time without going to the equilibrium position
Car suspensions are damped to stop them bouncing for a long time.However, if over damped then the car may jolt uncomfortably every time the car goes over a bump in the road. Over damping also means that there is a long delay before the suspension can react to any more bumps.
Plastic deformation of ductile materials reduces the amplitude of oscillations in the same ways as damping. As the material changes shape, it absorbs energy, so the oscillation will become smaller.
•RF 1.2 Out into space
This section develops ideas about gravitational field and potential. Free fall as ‘zero gravity’, space
flight and planetary orbits are considered. Momentum, kinetic and potential energies and
conservation laws are covered.
There are extended opportunities to use modelling software.
Space flight and astronomical data provide a context and there are opportunities to discuss how
ideas developed by Galileo, Kepler and Newton make it possible to do such things as weigh the
Earth, explain tides and broadcast TV globally by satellites..
MomentumMomentumP = mvUnits: kgms-1 or Ns
-Momentum is a vector so direction of v
Conserved -before and after collision *(whether coalesce or not)
-take into account directions
Total before = total after (provided that no external forces - for example, friction - act on the system).
Questions may ask for the change in momentum or total momentum rtfq!
Rockets use momentum conservation, by emitting gas in a direction the system will balance the change in momentum by velocity in the opposite direction. The change in momentum will be the velocity of the gas and mass of it.
The thrust of a jet is the momentum carried per second (the same as forceThe thrust of a jet is the momentum carried per second (the same as force)
Explosions conserve momentum!the momentum before an explosion is zero so after the explosion must therefore be zero as well
Momentum is always conserved in a closed system where no external forces act. Momentum is always conserved, even in inelastic
Jets and rockets
rocket velocity Vincreases by ∆Vin time ∆t
rocket mass M
mass ∆m ejectedin time ∆t
momentum carried by gas plusmomentum change of rocket = 0
for jet:v ∆m = –∆p
for rocket:∆p = M ∆V
momentum carriedaway by jet:∆p = v ∆m in time ∆t
∆p
gas velocity v
change of momentum ofrocket:∆p = M∆V in time ∆t
–∆p
Rocket thrust = –v ∆m∆t
M∆V = –v ∆m
∆V = –v ∆m
M
thrust = ∆p
= M ∆V
= –v ∆m
∆t ∆t ∆t
equal andopposite
The equation F = m∆v = mv - mu
∆t ∆t
rewritten as F∆t = m∆v. = mv - mu
Impulse = change in momentum= F∆t Ns or Kgm/s
impulse of the force is the change of momentum. You should see why momentum is sometimes given the units of Ns.
the size of force and time it acts affects the change in momentum
The equation F = m∆v = mv - mu
∆t ∆t
rewritten as F∆t = m∆v. = mv - mu
Impulse = change in momentum= F∆t Ns or Kgm/s
impulse of the force is the change of momentum. You should see why momentum is sometimes given the units of Ns.
the size of force and time it acts affects the change in momentum
change of momentum is determined by the impulse In golf more momentum is transferred to the ball if the club is in contact for longer, or crumple zones
Area under is impulse
consider collisions from several inertial frames of reference (observers moving at constant velocity relative to each other), together with an argument from symmetry.
+v
Progress Mir
–v
Two crafts approach one another and dock together
View 1 Observation craft hovers where the craft willmeet
The same event looks differentfrom two different points of view
video of collision seen from observation craft
observation craft
+v
Progress Mir
–v
View 2 Observation craft travels alongside M ir
observation craft
video of collision seen from observation craft
+2v +v
Progress Mir
+v –v
Progress Mir
–v
One event seen from two points of view
Before collision
After collision
momentum before= +mv – mv = 0
Momentum is different in the two views of the same event, but ineach case: momentum after = momentum before
momentum after= 0
+v
Progress
velocity of this frame relative to frame above
+2v
Before collision
Progress
momentum before= +m(2v) = +2mv
momentum after= (2m)v = 2mv
After collision
Mir
–v
Mir
velocity = 0
Progress Mir
velocity = 0
MirProgress
2m
+v
Newton’s third law arises:Momentum is always conserved
Inertial massis the mass of a body determined by its momentum. Determined by this second law, from the acceleration when the force is not due to gravity. f = ma f= ∆p ∆t m = f = ∆p a a∆t
Use the conservation of momentum to find the fraction of KE lost
equal masses, elastic collision
before
velocity v velocity zero
after
velocity zero velocity v
after
before
totalmomentum
during
equal masses, elastic collision
before
velocity v velocity –v
after
velocity –v velocity v
after
before
totalmomentum
during
0
0
equal masses, inelastic collision
before
after
both velocitieszero
after
before
totalmomentum
during
0
0velocity v velocity –v
unequal masses, inelastic collision
before
velocity v velocity zero
after
velocity a littleless than v
after
before
totalmomentum
during
unequal masses, elastic collision
before
velocity v velocity zero
after
velocity a littleless than v
velocity muchless than v
after
before
totalmomentum
during
unequal masses, elastic collision
before
velocity v velocity zero
after
velocity muchless than v
velocity up to2v
after
before
totalmomentum
during
Perfectly elastic collisions-momentum is conserved-Kinetic energy is conserved (in the same form)-So: Relative speed of approach is relative speed of separation (same speed)
Inelastic collisions-momentum is conserved-Kinetic energy is not conserved (in the same form)-So Speeds can change after
work and energyWork is done when energy is transferredWork done = Force x Distance IN THE DIRECTION OF DISTANCE MOVED
Therefore if force right angles to motion no work is done
The Force is the one that causes displacement
Work done= Fcos θ x dArea under force-distance graph is work done
You must have energy to do work,The conservation of energy applies.
Work done is therefore the change in energy due to a force moving a distance
Used in kinetic energy, the work done is the change (energy transferred)
Energy is the capacity of a body to do work
Total Work done = work done overcoming friction + work done against gravity.
If a force hinders the displacement of an object it does work against it or it does negative work.
The area under the graph between two values of r, however, does tell us how much work would be needed.
with gravitational fields is that the force is not constant so we cannot use W = F x dFrom Newton's law of gravitation the weight of an object varies with 1/r2
so, to move an object away from the Earth the force needed would also vary with 1/r2
W = ∑ F dr
e work required W = m ∆V
On a graph of g against r it is the area under the graph between any point and infinity ( g is after all the force that would act on 1kg)
Power is the rate at which work is done. (Watts) work Time
Joules = Nm , kgm2s-2 GPE= mgh dependent on mass/heightEnergy stored due to lifting against gravitational field
PE stored = work done to lift
EPE = 0.5 kx2
Springs follow hookes law (f=kx )
if there is no EPE it is in equilibrium position (when not stretched/compressed )
KE = 0.5 mv2
KE stored of a moving object = work done to accelerate from stationary to that speed
Ke equation can be derived from f =ma and work done
Work done to lift a loaded cart up inclined plane at constant speed is equal to the PE change of cart,
As its energy transferred, from 0 PE
Mechanical energy shows total energy in a system conserved
If external forces (friction, air res) do work energy is not conserved
Only forces can be internal forces
The pendulum is an undamped system and mechanical energy can be used
Newtons 1st law Velocity is constant if resultant Force or a =0
- STAYS CONSTANT, STATIONARY OR MOVING
Newtons laws
For example objects moving in a circle arent going in a straight line and so forces arent balance, no work is done as the force acts perpendicular to motion.
"An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force
If an object continues in a straight line at constant velocity, all forces acting on the object are balanced."
Newtons laws
2nd law that rate of change of momentum is proportional to force
F = ma = ∆p ∆t
Force is the rate of change of momentum
The time of contact therefore affects impact force.Force is determined by how quickly momentum changes.
You should see that this is the physics behind the use of egg boxes to protect eggs, seat belts and crumple zones in cars to protect the occupants, parachute landing
hammering in a nail, a steel hammer head so that the hammer head loses its momentum over a short time interval and the force exerted is very large. However, rubber headed hammers used for laying concrete paving slabs!
Newtons laws
We use f= ma for constant mass and average force (v/t = a )
3rd law Equal and opposite forces (same type), due to the conservation of momentumprinciple of conservation of momentum:
∆p1 + ∆p2 = 0 ∆p1 = - ∆p2 F1∆t = - F2∆t. Notice that the force acts for the same time on each of the objects! From this we can say that F1 = - F2
come in equal and opposite pairsSame type, for example weight (due to gravity) isn't the same as air resistance
Terminal velocity is where constant speed is reached as no more acceleration possible as resultant =0
Pressure= force/ area Pa or N/m2
Pressure is useful because it can be used to transmit forces from one place to another using liquids and gases.
friction (or drag) increases as speed increases.
Circular motionTo make an object move in a circular path, a force must act perpendicular to velocity
No work is done as the force acts perpendicular to motion
There is a resultant force, (centripetal) as CHANGES SPEED AND DIRECTION
a = v2 = w2r r F = mv2 from f = ma r
V = 2πr = 2 πrf frequency = (1/t)
Tw =2π = 2 πf = ∆ϴ constant measured in rad sec-1
T ∆t
Equations are not given
Derived from newton's laws, Circular motion shows: ● The body at constant speed doesn’t slow down / speed up. ● The effect of a constant force perpendicular to velocity makes direction change at constant rate so circular motion.
● centripetal force does no work on that body. It's at right angles to motion
All equations in radians
v = ωr
s = rθ
One full rotation is 2π radians
Angular speedMeasures the angle of a complete circle covered per second
Equations can be derived from the uses of general equations
Acceleration towards centre of circular orbit
velocity turns throughangle ∆θ as planetgoes along circularpath in short time ∆t
circular pathradius rspeed v
A
B
v2
v ∆tr
v1
r
r
radius turns through ∆θ velocity turns through ∆θ
arc AB = distance in time∆t at speed varc AB = v ∆t
change of velocity ∆vtowards centre of circle
= ∆θ
A
B
r
speed v∆θ
v2
v1
∆v
∆θ
multiply by v: v v ∆t
r= ∆v
divide by ∆t:v2
r= acceleration
∆vv
∆v∆t
=
≈
∆θ = arc ABr
v ∆tr
∆vv
∆θ ≈
∆θ
Acceleration towards centre = v2
r
If there is no centripetal force object moves at tangent
Acceleration acts towards the centre
Centripetal acceleration proportional to v squared
Speed vv
v
v
v ∆v
∆θ∆θ
speed v carries objectround angle ∆θ in time ∆t
∆v = v ∆θ
v
v∆θ
Speed v2 2
22
v2
v2
∆v4
∆v4 = ( ) ( )v
2∆θ2
∆θ2
speed v / 2 carries objectround angle ∆θ / 2 in time ∆t
Centripetal acceleration proportional to v squared
Speed vv
v
v
v ∆v
∆θ∆θ
speed v carries objectround angle ∆θ in time ∆t
∆v = v ∆θ
v
v∆θ
Speed v2 2
22
v2
v2
∆v4
∆v4 = ( ) ( )v
2∆θ2
∆θ2
speed v / 2 carries objectround angle ∆θ / 2 in time ∆t
The centripetal force is the resultant force . In planets it will be due to gravity (the gravitational force)
Orbital time 2 α orbital radius 3
Assume sun at the centre of solar system, mean orbit
For simplicity we can assume that the Sun is at the centre of the solar system. Taking from a spreadsheet data on mean orbit
Display Material 30O OHT 'Kepler’s third law'
Kepler discovered mars had an elliptical path so introduced that .Kepler: Geometry rules the Universe
Sun
Mars
Astronomy
Orbit of Mars an ellipse with Sun at a focus
Geometry
ab
focus focus
planet
Ellipse: curve such that sum of a and b is constant
Law 1: a planet moves in an ellipse with the Sun at one focus
the sun physically causes the motion of the planetsSeen from earth the planets move differently
Showing on linear scales the orbit period rises faster than mean orbit radius, showing the third law
0Mercury Venus Earth Mars Jupiter Saturn Uranus
20
40
60
80
time of orbit
Size of orbit - time taken to orbit
0Mercury Venus Earth Mars Jupiter Saturn Uranus
5
10
15
20
size of orbit
Orbital time increases more rapidly than orbital radius.
Speed of orbit depends on radius and mass of larger bodyFrom a forces point if the sun attracts a planet there will be a component of force in the direction(/or opposite to) of travel.Therefore planets KE increases as PE falls. (as gets closer)-a comet moves faster when closer to the sun
Orbital time 2 α orbital radius 3
Assume sun at the centre of solar system, mean orbit
Gravity-decreases by inverse square law
-exerts a force descibed by the law of gravity (called weight )
-provides centripetal force for orbits
--has a ptential energy whch determines escape velocity
VG = - Gm/r
The change in gravitational potential (ΔVG) is the change in energy per unit mass when it is moved from one point to another in the field.
Therefore: ΔVG = ΔE/m
The gravitational potential gradient is defined as the rate of change of gravitational potential with distance in the field. This is equal to the gravitational field intensity (g = EG) at that point.
Therefore: g = EG = -ΔVG/Δr
gravitational field strength is the same as the gravitational potential gradient.
gravitational field strength is the rate at which the field changes as the distance from the mass producing the field changes.
using g = - ∆V/∆r and g = - G M / r2 integrate between points r1 and r2 and then use boundary condition that the potential is zero when r2 is infinity.
NEWTON'S LAW OF UNIVERSAL GRAVITATION
Every particle of matter in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of their separation. (the inverse square law)
F α M1m2 F α 1 r2
Fgrav = - GMm r 2
Fgrav α Mm r 2
For point masses
Gravitational Force The force of gravity provides the unbalanced force causing free-fall. (object near surface experiences free-fall acceleration of 9.8 ms-2)
The force keeping moons planets in orbit is also due to gravity.
F = - G m1m2
r2 G = 6.67 × 10–11 N m2 kg–2
The very small value of G means that the force involved is only significant for large masses (astronomical scale)
Negative as force is always attractive
Distances from the centre of mass- point mass (centre point)
We assume point masses, that the centre of mass in the middle
Gravitational force F = mgUsed when gravitational field g is uniform (constant g) - Strength of F decreases as you get further, as g decreases as you get further
force is not a property of the field. It depends on what is inside the field.
Gravitational field strength is g, it is non uniform
g = F mg = F = - GMm m r2
g = -Gm r2.
acceleration due to gravity and field strength at a point are the same, though different units (ms-2 and Nkg-1).
Line density shows field strength, similarly line length can also show field strength
vector quantity (magnitude & direction at every point in space)negative,towards centre of large mass
At that point*
Radial field
The field strength at a point depends solely on position and masses;
increase mass =increase gravitational field strength .
increase the distance = strength reduces
( For example if you go twice as far from a mass, its gravitational field is ¼ as strong).
g ∝ 1 . r2.
g mass∝ .
gravitational field strength is the force per unit mass at a point in the field
Put simply it is the rate at which the field changes as the distance from the mass producing the field changes.
g is a property of the field. Independent of object in the field
Gravitational potential Gravitational potential around a point mass
r
GMV −=
In a radial field
From infinity it starts at zero, and moving back towards earth it becomes more negative
even though it has a negative sign. It doesn't have a direction, only a magnitude. – so scalar
Vg is measured in J kg-1
and is a scalar quantity
Gravitational potential is gravitational potential energy per unit mass at a point in a field.The negative of the work done per kg by an object when it moves from infinity (where potential is zero) to a point in a field
gravitational potential gradient = Gravitational field strength at that point
Note that a change in gravitational potential energy ∆Ep = mg∆h
so a change in gravitational potential difference ∆Vgrav = ∆Ep / m = g∆h
The unit of gravitational potential energy is the joule (J) whereas the unit of gravitational potential difference is the joule per kilogram (Jkg-1)
Gravitational potential in a radial fieldDifference from uniform field to the radial field. The way that field
strength varies with distance in a radial field g ∝ 1/r2. Potential in a radial field varies as 1/r.
-finding the area under the g against r graph, by integration
-establish the idea of field strength as a potential gradient g = - ∆V/∆r. Consider the analogy with hills and contour lines: at each point the size and direction of the force depends on the steepness of the slope.
Gravitational potential energy in a radial field
Work would have to be done to separate the two masses to infinity and that work would represent the magnitude of the potential energy
work done in moving a mass m from the Earth’s surface to infinity :
Gravitational potential is:“the work done per kg by an object when it moves from infinity to a point in a field".
Gravitational field and gravitational potential
Field = – dV/drIf V = GM /r, thenfield = – dV/dr = – GM/r2
If the potential varies as 1/r then the field variesas 1/r 2
Assume 1/r variation of potential and calculate differencein potential
Difference in potential ∆V between r and r + ∆ris:
Vgrav at r + ∆r
Vgrav
at r
0
rradius r
0 r + ∆r
at radius r + ∆r :
field g = – slope ∆V/∆r
∆r
∆V
at radius r :
field g = – ∆V∆r
V = – GMr
V = – GMr + ∆r
– GMr + ∆r
∆V = – GMr – ( )
∆V = – GM (r + ∆r) + GMr
r (r + ∆r)
i f ∆r is small: ∆V = GM∆r
r2
Thus: field g = – GMr 2
Gravitational potential in a uniform field-instead of analysis of forces of a body through gravitational field, consider energy changes.
-gravitational potential energy leads to gravitational potential, a scalar field description. - They are independent of mass as because when equating they cancel out for example: mgh = ½ m v2 so h = v2/2g which is independent of mass.
Area under the g against h graph shows the work done per unit mass in moving vertically through a uniform gravitational field, gravitational potential difference between the two points in the field.
Area under graph = gravitational potential V
satellites can remain in space without always using energy to stay there. They follow equipotentials.
G r a p h o f g r a v i t a t i o n a l f i e l d a g a i n s t 1 / r 2
0.0
–0.1
–0.2
–0.3
–0.4
–0.50 50 100 150 200
distance r from centre of Earth/106 m
pair A
pair B
pair Cpair D
Velocities are onlyapproximately directed awayfrom Earth. Acceleration is thusonly approximately equal to g
Graph of gravitational field against r
0.0
–0.1
–0.2
–0.3
–0.4
–0.50 0.0010
pair A
pair Bpair C
pair D
0.0005 0.0015
Graph approximates to astraight line through theorigin.Gravitational fieldproportional to 1/r2
r–2/(106 m)–2
shape of equipotential around the earth, same potential
-Field lines are at right angles to equipotential as no work is done unless there is a component of displacement in the direction of the force (of gravity).
-Equal differences get further and further away from each other
radial gravitational field magnitude of field is same at any given distance from centre
The acceleration of the Moon and the inverse square law
The acceleration of the Moon is simply diluted Earth gravity. The acceleration measures the gravitational field
Diluting Earth’s gravity
Moon’s orbitradius = 384000 km
Acceleration of the Moon
Time1 Moon month = 27.3 days
= 27.3 × 24 × 3600 s= 2.35 × 106 s
v = 24.1 × 108 m
Speed
2.35 × 106 s
v = 1020 m s–1
Travelling aroundthe Earth once amonth
orbit radius384000 km
Moon
Earth
Distancecircumference = 2π 3.84 × 108 m
= 24.1 × 108 m
r a t i o = 6 4 0 0 k m
R a t i o o f E a r t h ’ s r a d i u st o M o o n ’ s o r b i t
= 1
6 0
384000 km
g = 9.8 m s–2 at surface
orbit radius384000 km
Moon
Earthdilutedgravitationalpull
a = v2m
a = (1020 m s–1)2
Acceleration
3.84 × 108 m
a = 0.0027 m s–2
Gravity diluted byinverse square law
Earth’s surface:g = 9.8 m s–2
At Moon’s orbit:
acceleration = 9.8 m s–2
602
a = 0.0027 m s–2
Earth’s radiusradius = 6400 km
acceleration foundfrom motion
same accelerationfound from inversesquare law
r
Geostationary satellite
M = mass of Earth
m = mass of satelli teR = radius of satellite orbitv = speed in orbit
G = gravitational constant= 6.67 × 10–11 N kg–2 m2
= 5.98 × 1024 kg
T = time of orbit= 24 hours = 86400 s
Calculating the radius of orbit
Gravitational force onsatellite
GMm
R2
insert values of G, M and T : R = 4.2 × 104 km
R = 6.6 × radius of Earth (6400 km)
Force producingacceleration to centre
mv2
R
Forces are equal:
multiply by R:
mv2
R
GMm
R2=
divide by m: v2
R
GM
R2=
v2 GMR
=
speed in orbit dependson time of orbit andradius
v 2πRT
=
v2 4π2R2
T2=
equal
equal
satellite orbit turns at samerate as Earth turns
orbit radiusR = 42000 km
gravitationalforce
N
S
Kepler’s thirdlaw deduced
equate express ions for v2: GM
R
4π2R2
T2=
rearrange to calculate R: GMT 2
4π2 R3=
Geostationary satellite
M = mass of Earth
m = mass of satelli teR = radius of satellite orbitv = speed in orbit
G = gravitational constant= 6.67 × 10–11 N kg–2 m2
= 5.98 × 1024 kg
T = time of orbit= 24 hours = 86400 s
Calculating the radius of orbit
Gravitational force onsatellite
GMm
R2
insert values of G, M and T : R = 4.2 × 104 km
R = 6.6 × radius of Earth (6400 km)
Force producingacceleration to centre
mv2
R
Forces are equal:
multiply by R:
mv2
R
GMm
R2=
divide by m: v2
R
GM
R2=
v2 GMR
=
speed in orbit dependson time of orbit andradius
v 2πRT
=
v2 4π2R2
T2=
equal
equal
satellite orbit turns at samerate as Earth turns
orbit radiusR = 42000 km
gravitationalforce
N
S
Kepler’s thirdlaw deduced
equate express ions for v2: GM
R
4π2R2
T2=
rearrange to calculate R: GMT 2
4π2 R3=
Geostationary satellites are important for communications purposes.
r / m
Descriptions using forces
Gravitational field strength
perkilogram
Gravitational equations and graphs
g = – GMr2
r / m
Gravitational forces
for amass
r / m
Descriptions using energies
Gravitational potential
V = – GMr
F = – GMmr2
r / m
Gravitational potential energy
E = – GMmr
gradient of curve
area under curve
Gravitational potential energy Egrav = mgh = m (- GM ) h (Using equation for g)
r 2 Epotential grav = - GMm always negative
r
As r ∞ Epotential grav 0
It’s the energy of an object due to its position in a gravitational fieldNear earth is mgh, simple energy changes as constant g
This formula is from the law of gravity and work done against gravity to bring a mass to a given point in space (due to the inverse square of gravity)
The zero of grav. energy is at an infinite distance, therefore its negative after this as gravity does positive work as the mass approaching.
Basically the mass is trapped by the force of gravity until it can provide enough energy to escape, (escape velocity)
Therefore negative until gains positive energy to escape
Objects in gravitational field have potential to do work if they have a force
acting on them.
Gradient at a tangent is the gravitational force at that point
If larger force curve steeper this occurs at further distances
Curve tends to zero as r increases
The force becomes zero at infinity , therefore potential energy becomes zero at infinity
Moving back towards earth from infinity decreases the PE becoming more negative.
Further away means more potential energy
Work is done to bring an object from infinity to radius r
When Apollo 11 made the homeward journey from the Moon, its speed increased as its kinetic energy increased as a result of a decrease in potential energy. The total energy remained constant. We have Etotal = Epotential + Ekinetic so … Ep = Et - Ek Since Vgrav = Ep / m Vgrav = Et / m - Ek / m Et / m = constant - ½ v2
we can use – ½ v2 to represent gravitational potential and it will be accurate except for a constant that is linked to the total energy. A graph of – ½ v2 against 1 / r should produce a straight line, but that line would only go through the origin if the total energy were zero.
Gravitational field and gravitational potential energy
Uniform gravitational field
Field p icture
force on mass in grav itationa l field = mg
field = force/m ass = g
40 J kg–1
30 J kg–1
20 J kg–1
10 J kg–1 1 m
field strength = g
potential energy change= force × dis tance = mg∆h
gravitational potential difference= potential change/mass= mg∆h/m= g∆h
Potential energy p icture
mass mpotential energychange ∆E
mass m
lift by height ∆h
∆h
field g = slope
gravitationalpotentialdifference
∆h
displacement upwards
area g ∆h =gravitationalpotentialdifference
∆h
displacement upwards
Field is slope of potential hill Potential difference is area under graph
g
force mg
The field is the rate of change of potential with displacement
This is how escape velocity is determined
Gravitational potential wellf ield down potential slope
level in well ispotential energyper kg
∆Vgrav
∆r
radius r
00
Gravitational potential and radius
∆r
radius r
00
g
Gravitational field and radius
The field is the slope of the graph of potential against radiusThe difference in potential is the area under the graph of field against radius
∆r
field g= –slope ∆Vgrav/∆r
∆Vgrav
∆r
g∆Vgrav = area g∆r
Total energy and escapeThe total energy of a space craft Etotal = Epotential grav + Ekinetic
To escape from the potential well, Ep + Ek ≥ 0To stay within the potential well, Ep + Ek < 0To reach infinity with a speed of Ep + Ek = 0.
(Ep is negative)
So to escape.. Ep + Ek ≥ 0 -GMm + ½ mv2 ≥ 0 r - GM + ½ v2 ≥ 0 r This leads to v2 ≥ 2 GM / r
(on launch:)
masses cancel
To escape to infinity from the Earth’s surface, the body must projected with this amount of initial energy as kinetic energy.
Not affected by mass of the object!
A body in powered flight does not have to rely upon its initial KE to escape from the Earth, so does not need reach the escape velocity
area under a force-distance graph, which represents the work done for a particular mass moved against a force. If this process was reversible, it would be possible to store energy by doing work against a force, increasing ‘potential energy’.
Here you see the vital starting point that ∆Vg = area g∆r which follows from the definition of ∆Vg given earlier. The relationship g = - ∆V/∆r follows since a positive value of ∆r causes an increase in potential making ∆V positive and making - ∆V/∆r negative. This is right as the field strength g acts in the opposite direction to r.
The integration to find the area under the g-r graph is shown below. Gravitational potential due to a spherical mass We can find the change in gravitational potential by using calculus.
The diagram shows a spherical mass M. To move a unit mass from A to B, a small distance ∆ x, a force must be applied to the right. Force applied to the unit mass = G M / x2 to the right.
The work done to move the unit mass from A to B will be
.2
xx
GMW ∆×=
The total work done in moving the unit mass from some point P to infinity will be:
.d2 r
GM
r
GMGM
x
GMx
x
GM
rr=
−−
∞−=
−=
∞∞
∫ Gravitational potential at infinity is therefore higher than at P. Since the definition implies potential at infinity is zero, the potential at any point P will be given by:
.r
GMVg −=
If the sphere has a radius R, the gravitational potential at its surface will be:
.R
GM−
•RF 1.3 our place in the universe
This section covers a descriptive and mainly qualitative outline of the main features of the
observable universe consistent with the hot big bang model of its origin. The ideas of the
universality of the speed of light and the relativistic consequence of time dilation are introduced.
Radar
-used to determine distances within our solar system -pulse transmitted and signal reflected-Using the speed of light
Signal received is weaker as when hits objects small part reflected and part scattered.
Could be affected by noise, and obstacles (rare)
Assume c both ways, and time taken to arrive is the same as time back
Assume linear path (true for vacuum), as curved in atmosphere due to refractive index
C x ( ∆t ) =d 2
Vrel = d2 – d1
∆t
Strength α 1 . Inverse square law although
d4 reflected so power of 22
Size of observable universe= age of universe x c
When looking at the universe we can only see as far back to the beginning
First pulse
2nd pulse
Find the difference in distance between a time interval using two pulses
Angle will never be less than 45o (speed of light)
Angles of pulses are at 45o as at c
1 lightyear = distance travelled by light in a vacuum in one year
1 parsec = distance of an object which has a parallax of 1” of arc(subtends at angle of 1 sec of earth = (1/3600)o
1o = 60’ (minutes) = 36000” (seconds)
Au is mean distance from earth to sun
Distance is measured by reflecting a radar pulse then measuring the time sent and the time receivedback . The clock used travels w ith you and the radar equipment (at rest).
Assumptions: speed c is constant so reflection occurs halfway through the out-and-back time speed c is not affected by the motion of the distant objec t
time t/s 10
8
6
4
2
0
2 4 6
radar pulseout and back
x /c = 4 s
your worldlinetravelling in time
distance x/c
radar pulseout and back
pulse out
pulseback
4sback
4sout
sloping worldlineof object movingrelativeto you
time t/s 10
8
6
4
2
0
2 4 6
radar pulseout and back
x /c = 3 s
dis tance x /c
pulse out
pulseback
3 sback
3 sout
worldlineof objec tnotmovingrelative toyou goesvertically
Space time diagrams
Radar measures distances using light-travel timesThe constant speed of light is assumed
the object moves past the observer at t = 0 and the first radar pulse is sent out at that time.
time t/s10
8
6
4
2
03 ls 6 ls
distance x/c
worldline ofmoving object
worldline ofobserver
5
pulse 2 back t2
(t2 – t1)12
pulse is reflected
pulse 2 out t1
pulse 1 out and back instantly
distance at moment of reflection
from pulse travel time
= c (t2 – t1)12
from object travel time
= v (t2 + t1)12
t2 – t1
t2 + t1=
v
c
t2=
1 + v/c
t1 1 – v/c
pulse takes 6 s to go and returndistance = 3 light-s
pulse sent at t = 2 s must havebeen reflected at t = 5 s
t2 = 8 s
t1 = 2 s
t2= 4t1
c
vc
8 s – 2 s 3=
8 s + 2 s=
5
(t2 + t1)12
Two-way radar speed measurment (simplified)
Speed is measured by comparing the interval between returning pulses with the interval at which they weresent
distance is c(t2 – t1)/2 and as v(t2 + t1)/2.
If the first pulse went out and returned at t=0
Thereforeinterval between pulses coming back = ∆t back = 1 + v/c interval between pulses sent out ∆t out 1- v/c
When calculated it will show that ∆t back is larger than ∆t out by a factor of K2 (the Doppler effect occuring twice due to reflection) (doppler stretching factor2)
∆t back = k2 x ∆t out
And that..∆t back = k2 = 1 + v/c ∆t out 1- v/c
When reflected k2
For there and back, Doppler factor k∆tback= k x k∆tout = k2∆tout
time t 10
8
6
4
2
02 6
distance x/c
Let the one-way Doppler shift = k
Pulses sent from A at intervals ∆twill arrive at B at intervals k ∆tout
k2=
1 + v/c
1 – v/c
4
pulses return toobserver at intervals
∆tback = k × k∆tout
pulses sent outat intervals ∆tout
The same shift k must apply to pulsessent from B to A
Therefore:pulses arrive back at A at intervals∆tback = k × k∆tout
Two-way Doppler shift
One-way Doppler shift
k =1 + v/c1 – v/c
The two-way Doppler shift = k2pulses arrive atmoving object atlarger intervalsk∆tout
A B
Doppler shift – two-way and one-way
The Doppler shift k is the observed quantity that measures the speed of a remote object
We then use these equations differently in special relativity. As time dilation affects the Doppler effect.
Reflected
Parallax smaller angle means further
-looking at things from different angles
-suitable for close stars (closer than 100 light years) as affected by assumptions of geometry of earths orbit(angles become too small)
-the parallax is the angle between position and length (distant stars, angle is smaller)-1 parsec subtends at angle of 1 sec of earth
Cepheid variables
-stars of the same type with same period, peridocially dim and brighten uniformly
-from nearby galaxies otherwise too dim to be observed
-not all stars same brightness
Intensity = power r2
Plamp = Psun
rlamp2 rsun
2
In light years brightness α 1 . d2
Inverse square law, EM wave light (not reflected so power of 2)
The Doppler effectObserved when a source of waves moves relative to observer
Apparent shift in frequency (relative to observer)
As distance between observer and source decreasing/increasing waves reach quicker/slower
Explains shift in EM waves from moving stars
∆λ = v . When v<<c
λ cWavelength ratio = Velocity ratio
Derived from v = fλ = λ . T
∆λ = vT λoriginal = cT λobserved =λoriginal + ∆λ = cT + vT
When speed of source waves is c
Depends on1The laws of physics are the same for all observers moving at a constant velocity. 2The speed of light is the same for all observers. It does not depend on the speed of the source or the speed of the observer.
wavelength
wavelengthlarger by ∆λ
The wavelength increases when the source travels away from the receiver
λ =cf
=cT
λ+c
cT∆λ= + vT
=v
λ cT1 +
source and receiver both at rest
c = speed of lightf = frequencyT = t ime for 1 cycle
receiver
wavefronts
source
∆λ=
cv
λ
λ + ∆λ = cT + vT
sourcemovingand receiverat rest
velocity v
sourcetravelsdistance vTin eachcycle
receiver
Doppler shift
Time dilation
sitting beside the clock clock travelling past you at speed v
mirrors d = cτ apart
time out and back(1 tick) = 2τclock recordswristwatch time τ
d = cτ cτ cτcτ
x = vt x = vt
You see the light take a longer pathbut the speed is still cSo the time t is longer.The moving clock ticks more slowly.
time dilation
t = γtwith
Pythagoras’ theorem:
(cτ)2 = (cτ)2 – x2
τ2 = t2 – (x/c)2
substitute distancex = vt
τ2 = t2 (1 – v2/c2)
gives
Time dilation is a consequence of the constant speed of light
The light clock
1 – v2/c 2√1
γ =1 – v2/c 2√
τ t =
Time dilation due to ɣ
t = t0 ɣ where to is original
This equation can be applied similarly to others using the original, proper value and
ɣ to give observedLength, time, mass and thereforeenergy
Relativistic factor affects..length, time dilation, relativistic mass
When v<<c (v2/c2) = 0 so =1 ɣ
Relative to the observer only..Different observers see different values
mo is rest mass
so E0 is rest energy
Impossible to say that two distinct events happen at the same time in different points in space because observers see differently. It is relative.Equivalence of inertial frames, where the laws of physics apply equally within. And that speed of light is constant . Are used to combat this.
But it DOES NOT APPLY TO WAVELENGTH OR VELOCITYAS THIS IS AFFECTED BY THE DOPPLER EFFECT WHICH IS THEN AFFECTED BY TIME DILATIONFor example relative velocity
Vobserved = vemitted x
(The doppler effect is the same)
Relativistic energyAs an object approaches speed of light relativistic mass increases (as v increases affecting KE and therefore energy and therefore mass)
Using E= mc2 (the idea that energy is affected by mass and mass by energy)
and p = mv
If rest energy Eo = moc2 E= ɣ mc2
p =mv ɣ
The equation using the relativistic factor can be applied to mass
Where m0 is the rest mass and m is relativistic mass (affected by special relativity)m = m0 ɣ
Of course when =1 rest mass is ɣrelativistic mass
Relativistic doppler effect
k = λrecieved = λemitted
For a large value of z, (v/c) 1
Ratio of frequencies is also
f observed = kfo
a universal speed limitusing E= mcɣ 2 (when v=0 gives rise to E=mc2
And therefore γ factor approaches infinity as v approaches c, it would take an infinite amount of energy to accelerate an object with mass to the speed of light. The speed of light is the upper limit for the speeds of objects with positive rest mass]
Relativistic doppler effectThe change in frequency and wavelength of light when taking into account the effects of special relativity. Including time dilation effects of SR.
λrecieved = rrecieved = k λoriginal roriginal λ +∆ λ = 1 + v . = 1+z
λ c
When v<<c
From z = ∆λ λ∆λ = v . When v<<c
λ c
Obviously the equation for relativistic doppler effect can be used as if v << c the factor will be 1 and therefore simplify to this equation for non relativistic. they give very similar values when v « c.
Non relativistic Doppler effect ..
λrecieved = k λoriginal
Λ +∆ λ = (1 + ɣ v ) = λ c
= 1+z
From z = ∆λ λ
Relativistic doppler effect ..
Doppler effect for sound (non relativistic) when v<<c
Vobserved = Vemitted
Doppler effect for light or EM waves (relativistic) when v closer to c
Vobserved = Vemitted
x
Vobserved =
Vemitted
Relativistic effects
7
6
5
4
3
2
1
4
3
2
1
00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v/c
relativ ist ic Do ppler shift γ (1 + v /c )
0
non- relativ ist ic Doppler shif t 1 + v /c
identical for v << c
γ ~ 1 for all v << c
γ is always > 1
γ increases towardsinfin ity as v approaches c
(1 – v2 /c 2)
1 γ =
Relativistic and non-relativistic Doppler shifts
Relativistic factor γ
~
0 0.1 0.2 0 .3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v/c
Relativistic Doppler shift is:
λ+∆λλ
=1+v/c1–v/c
= 11+v/c1+v/c
multiply by
giving
λ+∆λλ
=(1–v/c)(1+v/c)
1+v/c
which is
λ+∆λλ
=(1–v2/c2)
1+v/c
it is neater to write this as
λ+∆λλ
= γ(1+v/c)
with =1–v2/c2
1λ
At low speeds relativistic and non-relativistic results are the same.At speeds near c the relativistic factor γ increases rapidly
Relativistic effects
7
6
5
4
3
2
1
4
3
2
1
00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v/c
relativ ist ic Do ppler shift γ (1 + v /c )
0
non- relativ ist ic Doppler shif t 1 + v /c
identical for v < < c
γ ~ 1 for all v << c
γ is always > 1
γ increases towardsinfin ity as v approaches c
(1 – v2 /c 2)
1 γ =
Relativistic and non-relativistic Doppler shifts
Relativistic factor γ
~
0 0.1 0.2 0 .3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v/c
Relativistic Doppler shift is:
λ+∆λλ
=1+v/c1–v/c
= 11+v/c1+v/c
multiply by
giving
λ+∆λλ
=(1–v/c)(1+v/c)
1+v/c
which is
λ+∆λλ
=(1–v2/c2)
1+v/c
it is neater to write this as
λ+∆λλ
= γ(1+v/c)
with =1–v2/c2
1λ
At low speeds relativistic and non-relativistic results are the same.At speeds near c the relativistic factor γ increases rapidly
Relativistic effects
7
6
5
4
3
2
1
4
3
2
1
00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v/c
relativ ist ic Do ppler shift γ (1 + v /c)
0
non- relativ ist ic Doppler shif t 1 + v /c
identical for v < < c
γ ~ 1 for all v << c
γ is always > 1
γ increases towardsinfin ity as v approaches c
(1 – v2 /c 2)
1 γ =
Relativistic and non-relativistic Doppler shifts
Relativistic factor γ
~
0 0.1 0.2 0 .3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v/c
Relativistic Doppler shift is:
λ+∆λλ
=1+v/c1–v/c
= 11+v/c1+v/c
multiply by
giving
λ+∆λλ
=(1–v/c)(1+v/c)
1+v/c
which is
λ+∆λλ
=(1–v2/c2)
1+v/c
it is neater to write this as
λ+∆λλ
= γ(1+v/c)
with =1–v2/c2
1λ
At low speeds relativistic and non-relativistic results are the same.At speeds near c the relativistic factor γ increases rapidly
Relativistic effects
7
6
5
4
3
2
1
4
3
2
1
00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v/c
relativ ist ic Do ppler shift γ (1 + v /c )
0
non- relativ ist ic Doppler shift 1 + v /c
identical for v < < c
γ ~ 1 for all v << c
γ is always > 1
γ increases towardsinfin ity as v approaches c
(1 – v2 /c 2)
1 γ =
Relativistic and non-relativistic Doppler shifts
Relativistic factor γ
~
0 0.1 0.2 0 .3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v/c
Relativistic Doppler shift is:
λ+∆λλ
=1+v/c1–v/c
= 11+v/c1+v/c
multiply by
giving
λ+∆λλ
=(1–v/c)(1+v/c)
1+v/c
which is
λ+∆λλ
=(1–v2/c2)
1+v/c
it is neater to write this as
λ+∆λλ
= γ(1+v/c)
with =1–v2/c2
1λ
At low speeds relativistic and non-relativistic results are the same.At speeds near c the relativistic factor γ increases rapidly
Relativistic effects
7
6
5
4
3
2
1
4
3
2
1
00 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v/c
relativ ist ic Do ppler shift γ (1 + v /c )
0
non- relativ ist ic Doppler shift 1 + v /c
identical for v << c
γ ~ 1 for all v << c
γ is always > 1
γ increases towardsinfin ity as v approaches c
(1 – v2 /c 2)
1 γ =
Relativistic and non-relativistic Doppler shifts
Relativistic factor γ
~
0 0.1 0.2 0 .3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
v/c
Relativistic Doppler shift is:
λ+∆λλ
=1+v/c1–v/c
= 11+v/c1+v/c
multiply by
giving
λ+∆λλ
=(1–v/c)(1+v/c)
1+v/c
which is
λ+∆λλ
=(1–v2/c2)
1+v/c
it is neater to write this as
λ+∆λλ
= γ(1+v/c)
with =1–v2/c2
1λ
At low speeds relativistic and non-relativistic results are the same.At speeds near c the relativistic factor γ increases rapidly
evidence for the expansion and origin of the Universe. cosmological red-shift and the cosmic microwave background, as evidence of the expansion of the Universe from a ‘hot big bang’ origin. (No reference is made to the relative proportions of the elements in the early Universe.) -at large red shifts, we are not looking at ‘recession velocities’ The space of the whole Universe is expanding, stretching wavelengths with it. Thus Doppler shifts, which do indicate relative velocities for nearer objects, now indicate an expansion of space, of scale factor k. -Hubble established the relation between red-shift and distance but the value for the Hubble constant. A larger value of H0 would mean a more rapidly expanding Universe.
(v = H0r )
Red shiftA measure of expansion due to λ increase, scale factor k is (1 + z) ALSO AFFECTED BY RELATIVITYThe theory of relativity gives the relativistic Doppler shift k as
Where k= 1+zA red shift of z gives an expansion scale of 1+z (or k)
( )( )cv
cvk
/1
/1
−+=
λλ∆=z
Z = -1 (as s.f. is z+1)
From z = ∆λ and k = 1+ z λ
Cosmological redshift is caused solely by the expansion of the universe. The value of the red shift indicates the recession velocity.
For small distances /velocities v<<c z= (v/c)
At larger distances and velocities : special relativity is taken into account and therefore
In doppler shift the wavelength of the emission depends on the motion of the object, directions (if towards wavelength shiften blue end, if away red end). But in (cosmological) red shift the wavelength that was originally emitted is lengthened as it travels through expanding space. Basically it is due to the expansion of space itself not from the motion of an individual body
The further away a galaxy the larger its redshift
Evidence that galaxies are moving away
V= Hod where v is the recessional velocity and H is a constant
Going backwards in time distant galaxies are further but moving faster therefore in the past must have been closer together
all matter and space started at a single point
Red shiftLight from distant stars and galaxies show spectra to be shifted towards the red end. This is a doppler shift *(of light)
Indicating that galaxies are moving away. Using results and distance showing that distant galaxies are moving away faster; the universe is expanding?
Red shift is recessional velocity.
Recessional velocity is proportional to distance.
Hubbles law*** -Measured red shifts are stated in terms of a Z parameter and scale factor k
v = H0r
A galaxy at d from us takes time t=d/v from us backwards, using hubbles law we
find that 1 = estimate for age of universe Ho
Hubble’s law and the age of the Universe
Hubble found that the further away a galaxy is, thelarger its redshift.
He interpreted this to mean that distant galaxies arereceding from us.
For a galaxy a distance d from us, Hubble wrote
v = H d
where v is the speed of a galaxy away from us andH is a constant called Hubble’s constant.
Run the Universe backwards in time......distant galaxies are further away but are movingfaster...
...in the past galaxies must have been closertogether...
...even further back, all the matter and space in theUniverse was concentrated at a single point.
t = dv
dHd
1H
= =
This time is independent of d and v and tells us how long ago the Universe wasa single point - this is the age of the Universe.
Strictly, in a reversed Universe, the galaxies accelerate as they fall together sothat the ‘Hubble time’, 1/H, gives an upper limit for the age of the Universe.
A galaxy distance d from us takes a time t = d/v to reach us in a reversedUniverse. From Hubble’s law:
this is an upper limit for estimate as is in reverse working backwords (as they get further they increase, so we have just used their maximum velocity to find the max estimate)
If we assume that the speed v was constant, the Hubble time would represent the time since galaxies were close together. This time represents the age of the Universe. Although universe must be younger than this time.
The ‘age of the Universe’
distance to galaxy
speed of recession is directly proportional to distance v = H0r
H0 is the Hubble constant
units of H0 are speed
= 1
distance time
H0
1= Hubble time
= timeunits ofH0
1
If speed v were constant, thenH0
1vr = time since galaxies were close together=
+
+
+
+
++
+
+
Wavelengths are red shifted because spacetime stretches as the light travels through it.The expansion of space is related to the cosmological red shift
The cosmological red-shift...
Think of an electromagnetic wave drawn on the balloon, travelling from one galaxy to another.A light wave travels from galaxy 1 to galaxy 2. The galaxies are a distance de mitte d apartwhen the wave is emitted.
The Universe expands.
λ emitted
d emitted
Space is stretched and the wave with it.
λreceived
d rece ived
When the light is received, the galaxies are a distance d received apart.
At such large values of z, the redshift is mainly the cosmological redshift, and not a valid measure of the actual recessional velocity of the object with respect to us.
The expanding Universe according to general relativity
According to generalrelativity, the Big Bangwas not an explosion ofmatter into empty space.
Both space and matter came into existence together about 14 billion years ago.
Imagine a balloon with dots drawn onit to represent galaxies, and slowlyblow it up.
As the balloon grows, thespace between thegalaxies grows. Thegalaxies do not movewithin the surface.
In the real Universe - unlike in the balloon model - thegalaxies themselves do not grow: their gravity holdsthem together.
two key pieces of evidence for the expansion and origin of the Universe: cosmological red-shift and the cosmic microwave background.
Cosmological red-shift
Light travels from one galaxy to another, as the Universeexpands
lightem itted
λemitted
The Universe expands...the photons travel
Remitted
Robserved
Space stretching Wavelength stretching
Remitted
Robserved
λemitted
λobserved
λobserved
λobserved λemitted + ∆λ=light observed
∆λλemitted
= Z
Robserved
Remitted
λobserved
λemitted=
Robserved
Remitted
λemitted + ∆λλemitted
=
Robserved
Remitted λemitted= 1 +
∆λ
Robserved
Remitted= 1 + z
Cosmic microwave background radiation
Interstellar space isfilled with a photon‘gas’ (and someatoms). Thetemperature of thisgas is proportional tothe energy of thephotons.
The energy of aphoton is proportionalto its frequency.Therefore thetemperature of thephoton gas isproportional to thefrequency of theradiation.
The cosmic microwave background radiation
...13 billion yearsafter the BigBang.Temperature: 2.7 KTypical wavelengthof radiation: 1 mm
The Universeexpands, stretchingthe wavelength ofthe photons. Thegreater thewavelength, thelower thefrequency. Thetemperature of thephoton gas falls.
+
+
+
+
+ +
+ +
In the beginning......there was the Big Bang...... the Universe is filled witha plasma of elementaryparticles, all exchangingenergy with photons ofelectromagnetic radiation.
The cosmic microwave background radiation
...300 000 years after the Big Bang...
Temperature: 3000 KTypical wavelength of radiation: 1 µm
As the temperaturefalls, atoms form aselectrons are held inorbit around nuclei ofprotons andneutrons.
The Universebecomes transparentto photons which nolonger interact soeasily with atoms andso travel unaffectedthrough the Universe.
+
+
+
Evidence that the Universe has evolved from an initial uniform, hot dense state comes from the existence of the cosmic microwave background.
The decoupling of the radiation – the Universe becomes transparent to electromagnetic radiation.
Evolution from hot to cool
This was interpreted as the expanded (red shifted) radiation from the time around 300 000 years after the ‘big bang’ when ionised atoms recombined into atoms, and photons no longer interacted strongly with them.
This gives evidence of an evolving Universe; one that was not the same in the past as it is now it had a beginning.
The expansion of the Universe stretches the wavelength of the radiation, decreasing its frequency and therefore reducing the energy density and lowering the temperature.
Temp α energy
energy α frequency
Frequency α (1/ λ)
Temp α (1/ λ)
zR
R +=1THEN
NOW
The expansion of the Universe
now:
cluster ofgalaxies
typical distance that increases withthe expansion of the Universe, R0
The cosmic microwav e background radiation
R0
typical wavelength, λ0
typical wavelength, λt
λ0
λ t
then:
typical distance between clusters, R t
Rt
R0 > R t
λ0 > λt
so put λ0 = λt + ∆λ
R0
R t
λ0
λ t λ t
= λ t + ∆λ
= 1 + ∆λλ t
R0
R t1 + z
=
=
This is the same effect as is seen in the red-shift of distant galaxies.
Stars generally seem to have the same ratio of helium to hydrogen (10:90). But in old stars this is all they contain, while in younger there are traces of heavy elements.
These heavy elements are thought to have been made in the stellar explosions, called supernovae, at the end of a stars life. Universe mustve been much smaller and hotter, nuclear fusion of hydrogen led to the ratio of He:H
Led to BMR
But there are disagreements in the expansion of the universeDark matter-unseen but has a lot of massCosmic background- flunctuations are small, and due to gravity accentuating any flunctuations the universe mustve been even more isotropic, but there has to have been some variation in density for matter to cluster in galaxiesAge- Critical density for the universe W
