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North Berwick High School

Department of Physics

Higher Physics

Unit 1 Our Dynamic Universe Section 6 The Expanding Universe Section 6 The Expanding Universe Note Making

Make a dictionary with the meanings of any new words.

The Doppler Effect and Hubble 1. State what is meant by the Doppler effect and give examples. 2. Copy the Doppler equations for moving sources. 3. Describe briefly how Cepheid variables are used to measure the distances to . 4. Copy Hubble's formula (make sure that you understand the units).

Evidence for the Expanding Universe Rotational velocity 1. Describe how the mass of galaxies can be calculated. 2. What problem did this produce?

Seeing the light 1. Describe how light can be used to calculate the mass of a . 2. What problem did this produce? 3. How much of the universe is thought to be dark matter?

Dark matter

1. Explain what is meant by MACHOs and WIMPs.

MACHOs 1. Describe 2 examples of MACHOs. 2. Briefly describe how gravitational lensing can be used to indicate the presence of a MACHO. 3. Briefly describe how holes can be detected. 4. State the evidence for and against MACHOs. WIMPs 1. State that WIMPs are non-baryonic and their prime candidates. 2. State the evidence for and against Wimps.

Section 6 The Expanding Universe

Contents

Content Statements ...... 1 The Doppler Effect ...... 4 Evidence for the Expanding Universe...... 10 Measuring the mass of galaxies ...... 10 Rotational velocity ...... 10 Seeing the light ...... 11 Dark matter ...... 13 MACHOs vs. WIMPs ...... 13 MACHOs ...... 14 Detecting MACHOs ...... 15 Searching with Hubble ...... 15 Gravitational lensing ...... 15 Circling ...... 16 WIMPs ...... 17 Debate the arguments ...... 20 The Expanding Universe Problems ...... 21 Solutions ...... 31

Content Statements

Contents Notes Contexts

a) The Doppler The Doppler Effect is Doppler Effect in terms Effect and observed in and of terrestrial sources of light. For sound, the e.g. passing galaxies. apparent change in ambulances. as a source Investigating the moves towards or apparent shift in away from a stationary frequency using a observer should be moving sound source investigated. and datalogger. Applications include measurement of speed (), echocardiogram and .

The Doppler Effect Measuring distances to causes similar shifts in distant objects. of light. measurements The light from objects and data analysis of moving away from us is apparent brightness of shifted to longer standard candles. wavelengths – redshift. The unit ‘Particles and The redshift of a galaxy ’ includes an is the change in investigation of the divided by inverse square law for the emitted light. Centres may wavelength. For slowly wish to include this moving galaxies, activity in this topic. redshift is the ratio of the velocity of the In practice, the units galaxy to the velocity used by astronomers of light. include light-years and (Note that the Doppler parsecs rather than SI Effect equations used units. for sound cannot be Data analysis of used with light from measurements of

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fast moving galaxies galactic velocity and because relativistic distance. effects need to be taken into account.) The revival of Einstein’s cosmological constant in the context of the accelerating universe. b) Hubble’s Law. Hubble’s Law shows the relationship between the recession velocity of a galaxy and its distance from us. Hubble’s Law leads to an estimate of the . c) Evidence for the Measurements of the expanding velocities of galaxies Universe. and their distance from us lead to the theory of the expanding Universe. Gravity is the force which slows down the expansion. The eventual fate of the Universe depends on its mass. The orbital speed of the and other stars gives a way of determining the mass of our galaxy. The Sun’s orbital speed is determined almost entirely by the gravitational pull of matter inside its orbit. Measurements of the mass of our galaxy and others lead to the conclusion that there is significant mass which

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cannot be detected – dark matter.

Measurements of the expansion rate of the Universe lead to the conclusion that it is increasing, suggesting that there is something that overcomes the force of gravity – dark energy.

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Section 6 The Expanding Universe

The Doppler Effect

The Doppler effect is the change in frequency observed when a source of sound waves is moving relative to an observer. When the source of sound waves moves towards the observer, more waves are received per second and the frequency heard is increased. Similarly as the source of sound waves moves away from the observer less waves are received each second and the frequency heard decreases.

Examples of the Doppler effect are: a car horn sounding as it passes a stationary observer; a train whistling as it passes under a bridge. In ‘Doppler ’, blood flowing in the body is measured using ultrasound reflections from the arteries.

There are two situations to be considered when deriving the apparent frequency associated with the relative movement of a source of sound waves and an observer.

Either the source moves relative to the observer, or the source is stationary and the observer moves.

Source moves relative to a stationary observer

The source is moving at speed vs and its frequency is fs . At positions in front of the moving source the waves 'pile up'.

Let v = speed of sound and the source produces waves of wavelength λ;

λ = v fs

An observer in front of the moving source will receive waves of a shorter wavelength, λobs.

v vs 1 λobs = - = (v - vs) fs fs fs

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The observed frequency,

v v v fobs = = = fs λobs 1 (v - vs) (v - vs) f s

v Thus fobs = f s for a source moving towards a (v - vs) stationary observer,

v and fobs = fs for a source moving away from a (v + vs) stationary observer.

The two formulae above can also be written as v f = fs (v ± vs)

Note: the fractional change in wavelength z may be calculated using z = (λobs - λrest) / λrest

Complete the exercise to find the velocity of the galaxy M31 at the http://imagine.gsfc.nasa.gov/YBA/yba-intro.html website.

Einstein’s famous general theory of relativity implied that the universe should be either expanding or contracting. However, the accepted scientific wisdom at that time was that the universe was of fixed dimensions and had been so for all time. This would lead Einstein to include in his theory what he has described as ‘the biggest blunder of my life’, by including a cosmological constant that brought a balance to the universe and kept it stable.

Working entirely independently from this, American Vesto Slipher had discovered that light from stars showed similar characteristics to the Doppler shift of sound waves. When light was travelling to an observer the wavelength

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would appear to contract and create a blue shift in the wavelength. However, Slipher was to also discover that the stars were all moving away from the Earth and created a cosmic red shift. Slipher’s observations did not receive much notice, although they were essential to the pioneering work of .

The name Hubble is most famous now for the telescope in orbit around the Earth. But where does the name come from? Edwin Hubble was one of the greats of 20th century . (See the science timeline section to find out more of his story.)

At the end of World War I the number of known galaxies in the universe totalled one: our own . Everything that was observable was believed to be part of the Milky Way or unimportant puffs of gas at the periphery of the universe. Hubble was to demonstrate that there are many more galaxies. Currently astronomers believe that there are possibly 140 billion galaxies in the universe. In other words, if a galaxy was a chocolate raisin, there are enough to fill the SECC.

Hubble’s two driving pursuits in his work were to discover the age and the size of the universe. To do this he had to use stars known as ‘standard candles’, stars whose luminosity can be reliably calculated and used as a reference point to measure the luminosity and relative distance of other stars. However, it was not Hubble who found these reference stars. The term ‘standard candles’ was coined by Henrietta Swan Leavitt, who noticed that a particular type of , known as a Cephid variable, had a constant frequency pulse. Leavitt discovered that by comparing the relative magnitudes of Cephids at different points it was possible to calculate where they were in relation to each other. By knowing the relative distances of the standard candles there was a practical way to measure the large-scale universe. Today, Cepheid variables remain one of the best methods for measuring distances to galaxies and are vital to determining the expansion rate (the Hubble constant) and age of the universe.

So, by knowing the luminosity of a source it is possible to measure the distance to that source by measuring how bright it appears to us: the dimmer it appears the farther away it is. Thus, by measuring the period of these stars (and hence their luminosity) and their apparent brightness, Hubble was able to show that some nebulae were not clouds within our own galaxy, but were external galaxies far beyond the edge of the Milky Way. In 1923 he demonstrated that one of these clouds was in fact a huge collection of stars, a galaxy. This galaxy, for which Hubble was to use the term ‘nebula’, was a hundred thousand light years across and over nine hundred thousand light years away. This meant that the universe must be much, much larger than previously thought.

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Hubble’s second revolutionary discovery was based on comparing his measurements of the Cepheid-based galaxy distance determinations with measurements of the relative velocities of these galaxies using Slipher’s red shift work. What he was able to show was that the further away a galaxy was the faster it was moving away from us.

Using data from this graph he was able to produce the following relationship: v = Hod

Where v is the speed at which a galaxy moves away from us and d is its distance from us. The constant of proportionality Ho is now called the Hubble constant. The common unit of velocity used to measure the speed of a galaxy is km s–1, while the most common unit for measuring the distance to nearby galaxies is called the megaparsec (Mpc), which is equal to 3.26 million light years or 30,800,000,000,000,000,000 km! Thus the units of the Hubble constant are (km s–1)/Mpc.

As you can imagine from the data in the graph, the value of Hubble’s constant has been a subject of some debate. Initial attempts at calculating the age of the universe found it to be younger than the Earth. Values have ranged from as low as 50 (km s–1)/Mpc to as high as 100 (km s–1)/Mpc.

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Improved measurements narrowed the range and the currently accepted value comes from WMAP: 73.5 (km s–1)/Mpc (give or take 3.2 (km s–1)/Mpc). This measurement is completely independent of traditional measurements using Cepheid variables and other techniques.

This discovery marked the beginning of the modern age of cosmology. Belgian George Lemaitre had used Einstein’s general theory and had predicted the outcome of Hubble’s work to produce what he called his ‘fireworks theory’, which suggested that the universe started at a fixed point, underwent a massive expansion and is still expanding now. He did not predict the linear relationship, and the combination of his and Hubble’s work was to become known as Hubble’s law.

To view a timeline of this work use the following link: http://www.timelineindex.com/content/view/1207.

During the 1930s other ideas were proposed as non-standard cosmologies to explain Hubble’s observations, including the Milne model, the oscillatory universe (originally suggested by Friedmann, but advocated by Einstein and Richard Tolman) and Fritz Zwicky’s hypothesis.

After World War II, two distinct possibilities emerged. One was Fred Hoyle’s steady-state model, whereby new matter would be created as the universe seemed to expand. In this model, the universe is roughly the same at any point in time. The other possibility was Lemaître’s theory, advocated and developed by George Gamow, who introduced big bang nucleosynthesis and whose associates, Ralph Alpher and Robert Herman, predicted the cosmic microwave background (CMB). It is an irony that it was Hoyle who coined the name that would come to be applied to Lemaître’s theory, referring to it as ‘this big bang idea’ in derision during a 1950 BBC broadcast.

For a while support was split between these two theories. Eventually, the observational evidence, most notably from radio source counts, began to favour the latter. The discovery of cosmic microwave background in 1964 secured the Big Bang as the best theory of the origin and evolution of the cosmos. Much of the current work in cosmology includes understanding how galaxies form in the context of the Big Bang, understanding the physics of the universe at earlier and earlier times, and reconciling observations with the basic theory.

The way that Hubble’s expansion law predicts the movement of the galaxies is important: the speed of recession is proportional to distance. Imagine an onion loaf being baked. If every part of the loaf expands by the same amount in a given interval of time, then the onion pieces would move away (recede)

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from each other with exactly a Hubble-type expansion law. In a given time interval, an onion piece close to another would move relatively little, but a distant onion piece would move relatively farther – and the same behaviour would be seen from any onion piece in the loaf. In other words, the Hubble law is just what one would expect for a homogeneous expanding universe, as predicted by the Big Bang theory. Moreover no onion piece or galaxy occupies a special place in this universe – unless you get too close to the edge of the loaf, where the analogy breaks down.

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Evidence for the Expanding Universe

Measuring the mass of galaxies

Rotational velocity

We have seen how Hubble used Doppler shift to determine the recessional velocity of galaxies. However, by using Doppler shift in a slightly different way, scientists can learn much about how galaxies move. They know that galaxies rotate because, when viewed edge-on, the light from one side of the galaxy is blue-shifted and the light from the other side is red-shifted. One side is moving towards the Earth, the other is moving away. The speed at which the galaxy is rotating can also be calculated from how far the light is shifted. Knowing how fast the galaxy is rotating, the mass of the galaxy can be found mathematically.

When scientists looked closer at the speeds of galactic rotation, they found something strange. Classical physics would determine that the individual stars in a galaxy should act similarly to the planets in our solar system – the greater the distance from the centre, the slower they should move. But the results from Doppler shift measurements reveal that the stars in many galaxies do not slow down at farther distances. In fact, the stars move at speeds that should see them escape the galaxy’s gravitational field; there is not enough measured mass to supply the gravity needed to hold the galaxy together. (See: http://imagine.gsfc.nasa.gov/docs/teachers/galaxies/imagine/act_modeling.ht ml.)

This would suggest that a galaxy with such high rotational speeds in its stars contains more mass than is predicted by calculations. Scientists theorise that, if the galaxy was surrounded by a halo of unseen matter, the galaxy could remain stable at such high rotational speeds.

You can have a go at weighing the Milky Way at the following website: http://imagine.gsfc.nasa.gov/docs/teachers/galaxies/imagine/student_weighin g.html.

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Seeing the light Astronomers can also use measurements of how much light there is to determine the mass of a galaxy (or a cluster of galaxies). By measuring the amount of light reaching the Earth, scientists can estimate the number of stars in the galaxy. Knowing the number of stars in the galaxy, scientists can mathematically determine the mass of the galaxy.

Fritz Zwicky used both methods described here to determine the mass of the Coma cluster of galaxies over half a century ago. Using our second technique, Zwicky estimated the total mass of a group of galaxies by measuring their brightness. But when the other method was used to compute the mass of the same cluster of galaxies, his calculations came up with a number that was 400 times greater than his original estimate. The discrepancy in the observed and computed masses is now known as ‘the missing mass problem’. The high rotational speeds that suggest a halo reinforce Zwicky’s findings. Zwicky’s findings were little used until the 1970s, when scientists began to realise that only large amounts of hidden mass could explain many of their observations. Scientists also realised that the existence of some unseen mass would also support theories regarding the structure of the universe. Today, scientists are searching for the mysterious dark matter, not only to explain the gravitational motions of galaxies but also to validate current theories about the origin and the fate of the universe.

Repeatedly using different methods to establish the masses of galaxies has found discrepancies that suggest that approximately 90% of the universe is matter in a form that cannot be seen, known as dark matter. Some scientists think dark matter is in the form of massive objects, such as black holes, that are situated around unseen galaxies. Other scientists believe dark matter to be subatomic particles that rarely interact with ordinary matter.

Dark matter is the term given to matter that does not appear to be emitting electromagnetic radiation, i.e. matter that cannot be seen. Scientists can infer that the dark matter is there from observations of its effects, but they cannot directly view it. Bruce H. Margon, chairman of the astronomy department at the University of Washington, told the New York Times, ‘It’s a fairly embarrassing situation to admit that we can’t find 90 per cent of the universe’. This problem has scientists scrambling to try and find where and what this dark matter is. ‘What it is is any body’s guess,’ adds Dr Margon. ‘Mother Nature is having a double laugh. She’s hidden most of the matter in the universe, and hidden it in a form that can’t be seen’.

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Examine the evidence for the missing matter here: http://imagine.gsfc.nasa.gov/docs/teachers/galaxies/imagine/act_evidence.ht m

Look at an explanation of how we can search for dark matter here: http://www.ted.com/talks/patricia_burchat_leads_a_search_for_dark_energy. html.

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Dark matter

MACHOs vs. WIMPs

So how do we look for dark matter? It can’t be seen or touched: we know of its existence by implication. It has been speculated that dark matter could be anything from tiny subatomic particles having 100,000 times less mass than an to black holes with masses millions of times that of the Sun. This has divided scientists into two schools of thought as they consider possible candidates for dark matter. These have been dubbed MACHOs (massive astrophysical compact halo objects) and WIMPs (weakly interacting massive particles). Although these acronyms are amusing, they can help you remember which is which. MACHOs are the big, strong dark matter objects ranging in size from small stars to super massive black holes. MACHOs are made of ‘ordinary’ matter, which is called baryonic matter. WIMPs, however, are small weak subatomic dark matter candidates, and are thought to be made of material other than ordinary matter, called non-baryonic matter. Astronomers search for MACHOs and particle physicists look for WIMPs. (Baryonic matter is made up of hydrogen and helium atoms, non-baryonic is made up of subatomic particles.)

Astronomers and particle physicists disagree about what they think dark matter is. Walter Stockwell, of the dark matter team at the Center for Particle Astrophysics at University of California at Berkeley, describes this difference:

The nature of what we find to be the dark matter will have a great effect on particle physics and astronomy. The controversy starts when people made theories of what this matter could be and the first split is between ordinary baryonic matter and non-baryonic matter.

Since MACHOs are too far away and WIMPs are too small to be seen, astronomers and particle physicists have devised ways of trying to infer their existence.

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MACHOs Massive compact halo objects are non-luminous objects that make up the halos around galaxies as suggested by Zwicky. In the first instance MACHOs are thought to be brown dwarf stars or black holes. Their existence was predicted by theory long before there was any proof. The existence of brown dwarfs was predicted by theories that describe star formation. Albert Einstein’s general theory of relativity famously predicted black holes, but the idea was first suggested by John Michell based on Newton’s corpuscular theory of light.

Brown dwarfs, like the Sun, are made from hydrogen, but they are usually much smaller. Stars like our Sun form when a mass of hydrogen collapses under its own gravity and the intense pressure initiates a nuclear reaction, emitting light and energy. Brown dwarfs differ from normal stars insofar as because of their relatively low mass they do not have enough gravity to ignite when they form. A brown dwarf therefore does not become a ‘real’ star; it is merely an accumulation of hydrogen gas held together by gravity. Brown dwarfs do give off some heat and a small amount of light.

Black holes, unlike brown dwarfs, are created by an overabundance of matter. A star made of hydrogen forms helium when the hydrogen atoms collide in the star. Eventually, the hydrogen fuel is used up and the gas starts to cool. All that matter ‘collapses’ under its own enormous gravity into a relatively small area. The is so dense that anything that comes too close to it, even light, cannot escape the pull of its gravitational field. Stars at a safe distance, beyond the event horizon, will circle around the black hole, much like the motion of the planets around the Sun. It was first believed that black holes emitted no light – that they were truly black. However, it is now believed that high-energy particles are ejected in jets along the axis of rotation of a black hole.

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Detecting MACHOs

Astronomers are faced with quite a challenge in detecting MACHOs. They must detect, over astronomical distances, things that give off little or no light. However, the task is becoming easier as astronomers create more refined telescopes and techniques for detecting MACHOs.

Searching with Hubble Using the Hubble Space Telescope, astronomers can detect brown dwarfs in the halos of our own and nearby galaxies. However, the images produced do not reveal the large numbers of brown dwarfs that astronomers hoped to find. ‘We expected [the Hubble images] to be covered wall to wall by faint, red stars,’ reported Francesco Paresce of the Johns Hopkins University Space Telescope Science Institute in the Chronicle of Higher Education. Research results disappointed: calculations based on the Hubble research estimate that brown dwarfs constitute only 6% of galactic halo matter.

Gravitational lensing Astronomers use a technique called gravitational lensing in the search for dark matter halo objects. Gravitational lensing occurs when a massive dark object passes between a light source, such as a star or a galaxy, and an observer on the Earth. The gravitational field is so large that it causes the light to bend. The object focuses the light rays, causing the intensity of the light source to apparently increase. Astronomers diligently search photographs of the night sky for the telltale brightening that indicates the presence of a MACHO.

So why doesn’t a MACHO block the light? How can dark matter act like a lens? The answer is gravity. Albert Einstein proved in 1919 that gravity bends light rays. He predicted that a star that was positioned behind the Sun would be visible during a total eclipse. Einstein was correct: the Sun’s gravitational field bent the light rays coming from the star and made it appear next to the sun.

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Not only can astronomers detect MACHOs with the gravitational lens technique, but they can also calculate the mass of the MACHO by determining distances and the duration of the lens effect. Although gravitational lensing has been known since Einstein’s demonstration, astronomers have only begun to use the technique to look for MACHOs in the past 20 years.

Gravitational lensing projects include the MACHO project (America and Australia), the EROS project (France) and the OGLE project (America and Poland).

Circling stars Another way to detect a black hole is to notice the gravitational effect that it has on objects around it. When astronomers see stars circling around a gravitational mass, but cannot see what that mass is, they suspect a black hole. Then by carefully observing the circling objects, the astronomers can conclude that a black hole does exist.

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In January 1995, a team of American and Japanese scientists announced ‘compelling evidence’ for the existence of a massive black hole at the American Astronomical Society meeting. Led by Dr Makoto Miyosi of the Mizusawa Astrogeodynamics Observatory and Dr James Moran of the Harvard- Smithsonian Center for Astrophysics, this group calculated the rotational velocity from the Doppler shifts of circling stars to determine the mass of a black hole. This black hole has a mass equivalent to 36 million times that of our sun. While this finding and others like it are encouraging, MACHO researchers have not turned up enough brown dwarfs and black holes to account for the missing mass. Thus most scientists concede that dark matter is a combination of baryonic MACHOs and non-baryonic WIMPs.

Evidence for MACHOs: Astronomers have observed objects that are either brown dwarfs or large planets around other stars using the properties of gravitational lenses.

Evidence against MACHOs: While they have been observed, astronomers have found no evidence of a large enough population of brown dwarfs that would account for all the dark matter in our Galaxy.

WIMPs In their efforts to find the missing 90% of the universe, particle physicists theorise about the existence of tiny non-baryonic particles that are different from what we call ‘ordinary’ matter. The prime candidates include neutrinos, axions and neutralinos. Subatomic WIMPs are thought to have mass (whether they are light or heavy depends on the particle), but usually only interact with baryonic matter gravitationally; they pass right through ordinary matter. Since each WIMP has only a small amount of mass, there needs to be a large number of them to account for the missing matter. That means that millions of WIMPs are passing through ordinary matter, the Earth and you and me, every few seconds. Although some people claim that WIMPs were proposed only because they provide a ‘quick fix’ to the missing matter problem, neutrinos were first ‘invented’ by physicists in the early 20th century to help make particle physics interactions work properly. They were later observed, and physicists and astronomers now have a good idea how many neutrinos there are in the universe. However, they are thought to be without mass – in 1998 one type of

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neutrino was discovered to have a mass, but it was insufficient for the neutrinos to contribute significantly to dark matter.

According to Walter Stockwell, astronomers also concede that at least some of the missing matter must be WIMPs. ‘I think the MACHO groups themselves would tell you that they can’t say MACHOs make up the dark matter’. The problem with searching for WIMPs is that they rarely interact with ordinary matter, which makes them difficult to detect.

Axions are particles which have been proposed to explain the absence of an electrical dipole moment for the neutron. They thus serve a purpose for both particle physics and for astronomy. Although axions may not have much mass, they would have been produced abundantly in the Big Bang. Current searches for axions include laboratory experiments and searches in the halo of our galaxy and in the Sun.

Neutralinos are members of another set of particles that has been proposed as part of a physics theory known as supersymmetry. This theory is one that attempts to unify all the known forces in physics. Neutralinos are massive particles (they may be 30–5000 times the mass of the proton), but they are the lightest of the electrically neutral supersymmetric particles. Astronomers and physicists are developing ways of detecting the neutralino, either underground or searching the universe for signs of their interactions.

Evidence for WIMPS: Theoretically, there is the possibility that very massive subatomic particles, created in the right numbers and with the right properties in the first moments of time after the Big Bang, are the dark matter of the universe. These particles are also important to physicists who seek to understand the nature of subatomic physics.

Evidence against WIMPS: The neutrino does not have enough mass to be a major component of dark matter. Observations have so far not detected axions or neutralinos.

There are other factors that help scientists determine the mix between MACHOs and WIMPs as components of the dark matter. Recent results by the WMAP satellite show that our universe is made up of only 4% ordinary matter. This seems to exclude a large component of MACHOs. About 23% of our

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universe is dark matter. This favours the dark matter being made up mostly of some type of WIMP. However, the evolution of structure in the universe indicates that the dark matter must not be fast moving, since fast-moving particles prevent the clumping of matter in the universe. So while neutrinos may make up part of the dark matter, they are not a major component. Particles such as the axion and neutralino appear to have the appropriate properties to be dark matter, but they have yet to be detected.

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Research and prepare to debate the arguments

http://imagine.gsfc.nasa.gov/docs/teachers/galaxies/imagine/act_dark_matter .html

Related links

http://www.bbc.co.uk/science/space/deepspace/darkmatter/

http://www.darkmatterphysics.com/

http://astro.berkeley.edu/~mwhite/darkmatter/dm.html

http://science.nasa.gov/astrophysics/focus-areas/what-is-dark-energy/

http://map.gsfc.nasa.gov/universe/uni_matter.html

http://www.astronomytoday.com/cosmology/darkmatter.html

Evidence against the expanding universe

http://www.etheric.com/Cosmology/redshift.html

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The Expanding Universe Problems

The Doppler effect and redshift of galaxies

In the following questions, when required, use the approximation for speed of sound in air = 340 m s 1.

1. In the following sentences the words represented by the letters A, B, C and D are missing:

A moving source emits a sound with frequency fs. When the source is moving towards a stationary observer, the observer hears a ____A_____ frequency fo. When the source is moving away from a stationary observer, the observer hears a ____B_____ frequency fo. This is known as the _____C______D_____.

Match each letter with the correct word from the list below:

Doppler effect higher louder lower

quieter softer

2. Write down the expression for the observed frequency fo, detected when a source of sound waves in air of frequency fs moves:

(a) towards a stationary observer at a constant speed, vs

(b) away from a stationary observer at a constant speed, vs.

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3. In the table shown, calculate the value of each missing quantity (a) to (f), for a source of sound moving in air relative to a stationary observer.

Frequency heard Frequency of Speed of source Speed of source by stationary source / Hz moving towards moving away from observer / Hz observer / m s 1 observer / m s 1

(a) 400 10

(b) 400 10

850 (c) 20

1020 (d) 5

2125 2000 (e)

170 200 (f)

4. A girl tries out an experiment to illustrate the Doppler effect by spinning a battery-operated siren around her head. The siren emits sound waves with a frequency of 1200 Hz.

Describe what would be heard by a stationary observer standing a few metres away.

5. A police car emits sound waves with a frequency of 1000 Hz from its siren. The car is travelling at 20 m s 1.

(a) Calculate the frequency heard by a stationary observer as the police car moves towards her.

(b) Calculate the frequency heard by the same observer as the police car moves away from her.

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6. A student is standing on a station platform. A train approaching the station its horn as it passes through the station. The train is travelling at a speed of 25 m s 1. The horn has a frequency of 200 Hz.

(a) Calculate the frequency heard as the train is approaching the student.

(b) Calculate the frequency heard as the train is moving away from the student.

7. A man standing at the side of the road hears the horn of an approaching car. He hears a frequency of 470 Hz. The horn on the car has a frequency of 450 Hz.

Calculate the speed of the car.

8. A source of sound emits waves of frequency 500 Hz. This is detected as 540 Hz by a stationary observer as the source of sound approaches.

Calculate the frequency of the sound detected as the source moves away from the stationary observer.

9. A whistle of frequency 540 vibrations per second rotates in a circle of radius 0·75 m with a speed of 10 m s 1. Calculate the lowest and highest frequency heard by a listener some distance away at rest with respect to the centre of the circle.

10. A woman is standing at the side of a road. A lorry, moving at 20 m s 1, sounds its horn as it is passing her. The horn has a frequency of 300 Hz.

(a) Calculate the wavelength heard by the woman when the lorry is approaching her.

(b) Calculate the wavelength heard by the woman when the lorry is moving away from her.

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11. A siren emitting a sound of frequency 1000 vibrations per second moves away from you towards the base of a vertical cliff at a speed of 10 m s 1.

(a) Calculate the frequency of the sound you hear coming directly from the siren.

(b) Calculate the frequency of the sound you hear reflected from the cliff.

12. A sound source moves away from a stationary listener. The listener hears a frequency that is 10% lower than the source frequency.

Calculate the speed of the source.

13. A flies towards a tree at a speed of 3·60 m s 1 while emitting sound of frequency 350 kHz. A moth is resting on the tree directly in front of the bat.

(a) Calculate the frequency of sound heard by the bat.

(b) The bat decreases its speed towards the tree. Does the frequency of sound heard by the moth increase, decrease or stays the same? Justify your answer.

(c) The bat now flies directly away from the tree with a speed of 4·50 m s 1 while emitting the same frequency of sound. Calculate the new frequency of sound heard by the moth.

14. The siren on a police car has a frequency of 1500 Hz. The police car is moving at a constant speed of 54 km h 1.

(a) Show that the police car is moving at 15 m s 1.

(b) Calculate the frequency heard when the car is moving towards a stationary observer.

(c) Calculate the frequency heard when the car is moving away from a stationary observer.

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15. A source of sound emits a signal at 600 Hz. This is observed as 640 Hz by a stationary observer as the source approaches.

Calculate the speed of the moving source.

16. A battery-operated siren emits a constant note of 2200 Hz. It is rotated in a circle of radius 0·8 m at 3·0 revolutions per second. A stationary observer, standing some distance away, listens to the note made by the siren.

(a) Show that the siren has a constant speed of 15·1 m s 1.

(b) Calculate the minimum frequency heard by the observer.

(c) Calculate the maximum frequency heard by the observer.

17. You are standing at the side of the road. An ambulance approaches you with its siren on. As the ambulance approaches, you hear a frequency of 460 Hz and as the ambulance moves away from you, a frequency of 410 Hz. The nearest hospital is 3 km from where you are standing.

Estimate the time for the ambulance to reach the hospital. Assume that the ambulance maintains a constant speed during its journey to the hospital.

18. On the planet Lts, a nattra moves towards a stationary ndo at 10 m s 1. The nattra emits sound waves of frequency 1100 Hz. The stationary ndo hears a frequency of 1200 Hz.

Calculate the speed of sound on the planet Lts.

19. In the following sentences the words represented by the letters A, B, C, D and E are missing: Copy and complete the paragraphs in your notes.

A hydrogen source of light gives out a number of emission lines. The wavelength of one of these lines is measured. When the light source is on the Earth, and at rest, the value of this wavelength is rest. When the same hydrogen emission line is observed, on the Earth, in light coming from a distant star the value of the wavelength is observed.

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When a star is moving away from the Earth observed is ____A_____ than rest. This is known as the ____B_____ shift.

When the distant star is moving towards the Earth observed is ____C_____ than rest. This is known as the ____D_____ shift.

Measurements on many stars indicate that most stars are moving ____E_____ from the Earth.

Match each letter with the correct word from the list below:

away blue longer red shorter towards.

20. In the table shown, calculate the value of each missing quantity.

Fractional change in Wavelength of light Wavelength of light wavelength, z on Earth rest / nm observed from star, observed / nm

(a) 365 402

(b) 434 456

8·00 × 10 2 486 (c)

4·00 × 10 2 656 (d)

5·00 × 10 2 (e) 456

1·00 × 10 1 (f) 402

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Hubble’s law

In the following questions, when required, use the approximation for

18 1 Ho = 2·4 × 10 s

1. Convert the following distances in light years into distances in metres.

1 light year

50 light years

100, 000 light years

16, 000, 000, 000 light years

2. Convert the following distances in metres into distances in light years.

(a) Approximate distance from the Earth to our Sun = 1·44 × 1011 m.

(b) Approximate distance from the Earth to next nearest star Alpha Centauri = 3.97 × 1016 m.

(c) Approximate distance from the Earth to a galaxy in the constellation of Virgo = 4·91 × 1023 m.

3. In the table shown, calculate the value of each missing quantity.

Speed of galaxy relative Approximate Fractional change to Earth / m s 1 distance from Earth in wavelength, z to galaxy / m

(a) 7.10 × 1022 (b)

(c) 1.89 × 1024 (d)

1·70 × 106 (e) (f)

2·21 × 106 (g) (h)

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4. Light from a distant galaxy is found to contain the spectral lines of hydrogen. The light causing one of these lines has a measured wavelength of 466 nm. When the same line is observed from a hydrogen source on Earth it has a wavelength of 434 nm.

(a) Calculate the Doppler shift, z, for this galaxy.

(b) Calculate the speed at which the galaxy is moving relative to the Earth.

(c) In which direction, towards or away from the Earth, is the galaxy moving?

5. Light of wavelength 505 nm forms a line in the spectrum of an element on Earth. The same spectrum from light from a galaxy in Ursa Major shows this line shifted to correspond to light of wavelength 530 nm.

(a) Calculate the speed that the galaxy is moving relative to the Earth.

(b) Calculate the approximate distance, in metres, the galaxy is from the Earth.

6. A galaxy is moving away from the Earth at a speed of 0·074 c.

(a) Convert 0·074 c into a speed in m s 1.

(b) Calculate the approximate distance, in metres, of the galaxy from the Earth.

7. A distant star is travelling directly away from the Earth at a speed of 2·4 × 107 m s 1.

(a) Calculate the value of z for this star.

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(b) A in the spectrum of light from this star is measured to be 443 nm. Calculate the wavelength of this line when it observed from a hydrogen source on the Earth.

8. A line in the spectrum from a hydrogen atom has a wavelength of 489 nm on the Earth. The same line is observed in the spectrum of a distant star but with a longer wavelength of 538 nm.

(a) Calculate the speed, in m s 1, at which the star is moving away from the Earth.

(b) Calculate the approximate distance, in metres and in light years, of the star from the Earth.

9. The galaxy Corona Borealis is approximately 1 000 million light years away from the Earth. Calculate the speed at which Corona Borealis is moving away from the Earth.

10. A galaxy is moving away from the Earth at a speed of 3·0 × 107 m s 1. The frequency of an emission line coming from the galaxy is measured. The light forming the same emission line, from a source on Earth, is observed to have a frequency of 5·00 × 1014 Hz.

(a) Show that the wavelength of the light corresponding to the emission line from the source on the Earth is 6·00 × 10 7 m.

(b) Calculate the frequency of the light forming the emission line coming from the galaxy.

11. A distant quasar is moving away from the Earth. Hydrogen lines are observed coming from this quasar. One of these lines is measured to be 20 nm longer than the same line, of wavelength 486 nm from a source on Earth.

(a) Calculate the speed at which the quasar is moving away from the Earth.

(b) Calculate the approximate distance, in millions of light years, that the quasar is from the Earth.

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12. A hydrogen source, when viewed on the Earth, emits a red emission line of wavelength 656 nm. Observations, for the same line in the spectrum of light from a distant star, give a wavelength of 660 nm. Calculate the speed of the star relative to the Earth.

13. Due to the rotation of the Sun, light waves received from opposite ends of a diameter on the Sun show equal but opposite Doppler shifts. The relative speed of rotation of a point on the end of a diameter of the Sun relative to the Earth is 2 km s 1. Calculate the wavelength shift for a hydrogen line of wavelength 486·1 nm on the Earth.

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Solutions

The Doppler effect and redshift of galaxies

1. A higher B lower C Doppler D effect

2a. fo = fs [v / (v - vs)]

2b. fo = fs [v / (v + vs)]

3a. fs = 400 Hz fo = fs [v / (v - vs)] -1 v = 340 ms fo = 400 [340 / (340 - 10)] -1 vs = 10 ms fo = 412 Hz

3b. fs = 400 Hz fo = fs [v / (v + vs)] -1 v = 340 ms fo = 400 [340 / (340 + 10)] -1 vs = 10 ms fo = 389 Hz

3c. fo = 850 Hz fo = fs [v / (v - vs)] -1 v = 340 ms 850 = fs [340 / (340 - 20)] -1 vs = 20 ms fs = 850 / [340 / (340 - 20)] fs = 800 Hz

3d. fo = 1020 Hz fo = fs [v / (v + vs)] -1 v = 340 ms 1020 = fs [340 / (340 + 5)] -1 vs = 5 ms fs = 1020 / [340 / (340 + 5)] fs = 1035 Hz

3e. fo = 2125 Hz fo = fs [v / (v - vs)] fs = 2000 Hz 2125 = 2000 [340 / (340 - vs)] -1 v = 340 ms 2125 / 2000 = 340 / (340 - vs ) 340 / (2125 / 2000) = 340 - vs -1 vs = 20 ms

3f. fo = 170 Hz fo = fs [v / (v + vs)] fs = 200 Hz 170 = 200 [340 / (340 + vs)] -1 v = 340 ms 170 / 200 = 340 / (340 + vs ) 340 / (170 / 200) = 340 + vs -1 vs = 60 ms

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4. The frequency of the sound would increase and decrease.

5a. fs = 1000 Hz fo = fs [v / (v - vs)] -1 v = 340 ms fo = 1000 [340 / (340 - 20)] -1 vs = 20 ms fo = 1063 Hz

5b. fs = 1000 Hz fo = fs [v / (v + vs)] -1 v = 340 ms fo = 1000 [340 / (340 + 20)] -1 vs = 20 ms fo = 944 Hz

6a. fs = 200 Hz fo = fs [v / (v - vs)] -1 v = 340 ms fo = 200 [340 / (340 - 25)] -1 vs = 25 ms fo = 216 Hz

6b. fs = 200 Hz fo = fs [v / (v + vs)] -1 v = 340 ms fo = 200 [340 / (340 + 25)] -1 vs = 25 ms fo = 186 Hz

7. fo = 470 Hz fo = fs [v / (v - vs)] fs = 450 Hz 470 = 450 [340 / (340 - vs)] -1 v = 340 ms 470 / 450 = 340 / (340 - vs ) 340 / (470 / 450) = 340 - vs -1 vs = 14 ms

8. fo = 540 Hz fo = fs [v / (v - vs)] fs = 500 Hz 540 = 500 [340 / (340 - vs)] -1 v = 340 ms 540 / 500 = 340 / (340 - vs ) 340 / (540 / 500) = 340 - vs -1 vs = 25.2 ms

fo = fs [v / (v + vs)] fo = 500 [340 / (340 + 25.2)] fo = 466 Hz

9. fs = 540 Hz Highest r = 0.75m fo = fs [v / (v - vs)] -1 v = 340 ms fo = 540 [340 / (340 - 10)] -1 vs = 10 ms fo = 556 Hz

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Lowest fo = fs [v / (v + vs)] fo = 540 [340 / (340 + 10)] fo = 525 Hz

-1 10a. vs = 20 ms fo = fs [v / (v - vs)] fs = 300 Hz fo = 300 [340 / (340 - 20)] -1 v = 340 ms fo = 318.75 Hz

v = fλ 340 = 318.75 x λ λ = 1.07 m

-1 10b. vs = 20 ms fo = fs [v / (v + vs)] fs = 300 Hz fo = 300 [340 / (340 + 20)] -1 v = 340 ms fo = 283.33 Hz

v = fλ 340 = 283.33 x λ λ = 1.2 m -1 11a. vs = 10 ms fo = fs [v / (v + vs)] fs = 1000 Hz fo = 1000 [340 / (340 + 10)] -1 v = 340 ms fo = 971 Hz

-1 11b. vs = 10 ms fo = fs [v / (v - vs)] fs = 1000 Hz fo = 1000 [340 / (340 - 10)] -1 v = 340 ms fo = 1030 Hz

12. fo / fs = 0.9 fo = fs [v / (v - vs)] -1 v = 340 ms 0.9 = 340 / (340 - vs ) 340 / (0.9) = 340 - vs -1 vs = 37.8 ms

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-1 13a. vs = 3.60 ms The bat hears the echo, the sound is v = 340.00 ms-1 moving towards it. fs = 350 kHz = 350000 fo = fs [v / (v - vs)] Hz fo = 350000 [340 / (340 - 3.60)] fo = 353745.54 Hz fo = 354 kHz

13b. As the bat's speed decreases the frequency observed decreases. vs decreases = v - vs increases = v / (v - vs ) decreases = fo decreases

-1 13c. vs = 3.60 ms fo = fs [v / (v + vs)] fs = 350 kHz = 350000 fo = 350000 [340 / (340 + 3.6)] Hz fo = 346332.95 Hz -1 v = 340 ms fo = 346 kHz

-1 -1 14a. vs = 54 kmh vs = 54 kmh 3 -1 vs = 54x10 mh 3 -1 vs = 54x10 / (60 x 60) ms -1 vs = 15 ms

-1 14b. vs = 15ms fo = fs [v / (v - vs)] fs = 1500 Hz fo = 1500 [340 / (340 - 15)] -1 v = 340 ms fo = 1569 Hz

-1 14c. vs = 15ms fo = fs [v / (v + vs)] fs = 1500 Hz fo = 1500 [340 / (340 + 15)] -1 v = 340 ms fo = 1437 Hz

15. fo = 640 Hz fo = fs [v / (v - vs)] fs = 600 Hz 640 = 600 [340 / (340 - vs)] -1 v = 340 ms 640 / 600 = 340 / (340 - vs ) 340 / (640 / 600) = 340 - vs -1 vs = 21.3 ms

16a. fs = 2200 Hz s = 2πr r = 0.8 m s = 2 x π x 0.8 s = 5.027 m In one second it does three revolutions, so the total distance = 3 x 5.027 = 15.07m v = s / t v = 15.07 / 1 v = 15.1 ms-1 16b. fs = 2200 Hz Minimum -1 vs = 15.1 ms fo = fs [v / (v + vs)] -1 v = 340 ms fo = 2200 [340 / (340 + 15.1)]

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fo = 2106 Hz

16.c fs = 2200 Hz Maximum -1 vs = 15.1 ms fo = fs [v / (v - vs)] -1 v = 340 ms fo = 2200 [340 / (340 - 15.1)] fo = 2302 Hz

17. fo towards = 460 Hz Moving towards fo away = 410 Hz fo = fs [v / (v - vs)] -1 v = 340 ms 460 = fs [340 / (340 - vs)] 3 s = 3x10 m 460 / [340 / (340 - vs)] = fs

Moving away fo = fs [v / (v + vs)] 410 = fs [340 / (340 + vs)] 410 / [340 / (340 + vs)] = fs

460 / [340 / (340 - vs)] = 410 / [340 / (340 + vs)] 460 / 410 = [340 / (340 - vs)] / [340 / (340 + vs)] 460 / 410 = (340 + vs) / (340 - vs) 460 x (340 - vs) = 410 x (340 + vs) (460 x 340) - 460vs = (410 x 340) + 410vs (460 x 340) - (410 x 340) = 410vs + 460vs [(460 x 340) - (410 x 340)] / (410+ 460) = vs -1 vs = 19.5 ms

s = vst 3x103 = 19.5 x t t = 154s

-1 18. vs = 10 ms fo = fs [v / (v - vs)] fs = 1100 Hz 1200 = 1100 [v / (v - 10)] fo = 1200 Hz 1200 x (v - 10) = 1100v 1200v - 12000 = 1100v 1200v - 1100v = 12000 v = 12000 / (1200 - 1100) v = 120 ms-1

19. A longer B red C shorter D blue E away

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-9 20a. λrest = 365x10 m z = (λobs - λrest) / λrest -9 -9 -9 -9 λobs = 402x10 m z = (402x10 - 365x10 ) / 365x10 z = 0.101 z = 1.01x101

-9 20b. λrest = 434x10 m z = (λobs - λrest) / λrest -9 -9 -9 -9 λobs = 456x10 m z = (456x10 - 434x10 ) / 434x10 z = 0.0507 z = 5.07 x102

-2 20c. z = 8.00x10 z = (λobs - λrest) / λrest -9 -2 -9 -9 λrest = 486x10 m 8.00x10 = (λobs - 486x10 ) / 486x10 -2 -9 -9 (8.00x10 x 486x10 ) + 486x10 = λobs -7 λobs = 5.249 x10 m λobs = 525 nm

-2 20d. z = 4.00x10 z = (λobs - λrest) / λrest -9 -2 -9 -9 λrest = 656x10 m 4.00x10 = (λobs - 656x10 ) / 656x10 -2 -9 -9 (4.00x10 x 656x10 ) + 656x10 = λobs -7 λobs = 6.82 x10 m λobs = 682 nm

-2 20e. z = 5.00x10 z = (λobs - λrest) / λrest -9 -2 -9 λobs = 456x10 m 5.00x10 = (456x10 - λrest) / λrest -2 -9 5.00x10 x λrest = (456x10 - λrest) -9 -2 456x10 = 5.00x10 λrest + λrest -9 -2 456x10 = λrest (5.00x10 + 1) -9 -2 λrest = 456x10 / (5.00x10 + 1) -7 λrest = 4.34 x10 m λrest = 434 nm

-1 20f. z = 1.00x10 z = (λobs - λrest) / λrest -9 -1 -9 λobs = 402x10 m 1.00x10 = (402x10 - λrest) / λrest -1 -9 1.00x10 x λrest = (402x10 - λrest) -9 -1 402x10 = 1.00x10 λrest + λrest -9 -1 402x10 = λrest (1.00x10 + 1) -9 -1 λrest = 402x10 / (1.00x10 + 1) -7 λrest = 3.65 x10 m λrest = 365 nm

36

Hubble’s law

1a. 1 light year s = vt s = 3.0x108 x (1 x 365 x 24 x 60 x 60) s = 9.46x1015 m

1b. 50 light years s = vt s = 3.0x108 x (50 x 365 x 24 x 60 x 60) s = 4.75x1017 m

1c. 100000 light years s = vt s = 3.0x108 x (100000 x 365 x 24 x 60 x 60) s = 9.46x1020 m

1d. 16000000000 light s = vt years s = 3.0x108 x (16000000000 x 365 x 24 x 60 x 60) s = 1.51x1026 m

2a. d = 1.44x1011 m number of light years = distance / distance travelled in 1 light year number = 1.44x1011 / 3.0x108 x (1 x 365 x 24 x 60 x 60) = 1.52x10-5 light years

2b. d = 3.97x1016 m number of light years = distance / distance travelled in 1 light year number = 3.97x1016 / 3.0x108 x (1 x 365 x 24 x 60 x 60) = 4.2 light years

2c. d = 4.91x1023 m number of light years = distance / distance travelled in 1 light year number = 4.91x1023 / 3.0x108 x (1 x 365 x 24 x 60 x 60) = 5.19x107 light years

22 3a. d = 7.10x10 m v = Hod -18 -1 -18 22 Ho = 2.4x10 s v = 2.4x10 x 7.10x10 v = 1.7x105 ms-1

8 3b. c = 3.0x10 ms-1 z = v/c 5 -1 8 v = 1.7x10 ms z = 1.7x105 / 3.0x10 z = 5.67x10-4

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24 3c. d = 1.89x10 m v = Hod -18 -1 24 22 Ho = 2.4x10 s v = 1.89x10 x 7.10x10 v = 4.54x106 ms-1

8 3d. c = 3.0x10 ms-1 z = v/c 6 -1 8 v = 4.54x10 ms z = 4.54x106 / 3.0x10 z = 1.51x10-2

6 -1 3e. v = 1.70x10 ms v = Hod -18 -1 -18 Ho = 2.4x10 s 1.70x106 = 2.4x10 x d d = 1.89x1024 m

8 3f. c = 3.0x10 ms-1 z = v/c 6 -1 8 v = 1.70x10 ms z = 1.70x106 / 3.0x10 z = 5.67x10-3

6 -1 3g. v = 2.21x10 ms v = Hod -18 -1 -18 Ho = 2.4x10 s 2.21x106 = 2.4x10 x d d = 9.21x1023 m

8 3h. c = 3.0x10 ms-1 z = v/c 6 -1 8 v = 2.21x10 ms z = 2.21x106 / 3.0x10 z =7.37x10-3

-9 4a. λobs = 466x10 m z = (λobs - λrest) / λrest -9 -9 -9 -9 λrest = 434x10 m z = (466x10 - 434x10 ) / 434x10 z = 7.37x10-2

8 4b. c = 3.0x10 ms-1 z = v / c z = 7.37x10-2 7.37x10-2 = v / 3.0x108 v = 2.21x107 ms-1

4c. Away, as the observed wavelength is longer than the rest wavelength.

-9 5a. λrest = 505x10 m z = (λobs - λrest) / λrest -9 -9 -9 -9 λobs = 530x10 m z = (530x10 - 505x10 ) / 505x10 8 -2 c = 3.0x10 ms-1 z = 4.95x10

z = v / c 4.95x10-2 = v / 3.0x108 v = 1.49x107 ms-1

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7 -1 5b. v = 1.49x10 ms v = Hod -18 -1 7 -18 Ho = 2.4x10 s 1.49x10 = 2.4x10 d d = 6.21x1024 m

6a. v = 0.074c ms-1 v = 0.074 x 3x108 v = 2.22x107 ms-1

7 -1 6b. v = 2.22x10 ms v = Hod -18 -1 7 -18 Ho = 2.4x10 s 2.22x10 = 2.4x10 d d = 9.25x1024 m

7a. v = 2.4x107 ms-1 z = v / c 8 7 8 c = 3.0x10 ms-1 2.4x10 = v / 3.0x10 z = 8.0x10-2

-9 7b. λobs = 530x10 m z = (λobs - λrest) / λrest -2 -2 -9 z = 8.0x10 8.0x10 = (530x10 - λrest) / λrest -2 -9 8.0x10 x λrest = (530x10 - λrest) -9 -2 530x10 = 8.0x10 λrest + λrest -9 -2 530x10 = λrest (8.0x10 + 1) -9 -2 λrest = 530x10 / (8.0x10 + 1) -7 λrest = 4.10 x10 m λrest = 410 nm

-9 8a. λrest = 489x10 m z = (λobs - λrest) / λrest -9 -9 -9 -9 λobs = 538x10 m z = (538x10 - 489x10 ) / 489x10 8 -2 c = 3.0x10 ms-1 z = 1.00x10

z = v / c 1.00x10-2 = v / 3.0x108 v = 3.0x107 ms-1

7 -1 8b. v = 3.0x10 ms v = Hod -18 -1 7 -18 Ho = 2.4x10 s 3.0x10 = 2.4x10 d d = 1.25x1025 m

d = 1.25x1025 /3.0x108 x 365 x 24 x 60 x 60 d = 1.32x109 light years

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9. d = 1x1010 light years d = 1x1010 x 3.0x108 x 365 x 24 x 60 x 60 -18 -1 25 Ho = 2.4x10 s d = 9.4608x10 m

v = Hod -18 25 v = Ho = 2.4x10 x 9.4608x10 v = 2.27x107 ms-1

10a. c = 3.0x108 ms-1 v = fλ 14 8 14 frest = 5.00x10 Hz 3.0x10 = 5.00x10 x λ λ = 6.00x10-7 m

10b. c = 3.0x108 ms-1 z = v / c -7 7 8 λrest = 6.00x10 m z = 3.0x10 / 3.0x10 v = 3.0x107 ms-1 z = 0.1

z = (λobs - λrest) / λrest -7 -7 ] 0.1 = (λobs - 6.00x10 ) / 6.00x10 -7 -7 (0.1 x 6.00x10 ) + 6.00x10 = λobs -7 λobs = 6.6x10 m

v = fλ 3.0x108 = f x 6.6x10-7 f = 4.55x1014 Hz

-9 11a. λrest = 486x10 m z = (λobs - λrest) / λrest -9 -9 -9 -9 λobs = (486x10 z = (506x10 - 486x10 ) / 486x10 +20x10-9 )m z = 4.11x10-2 8 c = 3.0x10 ms-1 z = v / c 4.11x10-2 = v / 3.0x108 v = 1.23x107 ms-1

-18 -1 11b. Ho = 2.4x10 s v = Hod v = 1.23x107 ms-1 1.23x107 = 2.4x10-18 x d d = 5.125x1024 m

d = 5.125x1024 / (3.0x108 x 365 x 24 x 60 x 60) d = 5.42x108 m d = 542 million light years

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-9 12. λrest = 656x10 m z = (λobs - λrest) / λrest -9 -9 -9 -9 λobs = 660x10 m z = (660x10 - 656x10 ) / 656x10 8 -3 c = 3.0x10 ms-1 z = 6.098x10

z = v / c 6.098x10-3= v / 3.0x108 v = 1.83x106 ms-1

13. v = 2 kms-1 = 2000 z = v / c ms-1 z = 2000 / 3.0x108 -9 -6 λrest = 486.1x10 m z = 6.67x10 8 c = 3.0x10 ms-1 -6 -9 6.67x10 = (λobs - λrest) / 486.1x10 -6 wavelength shift = (λobs - λrest) = 6.67x10 / 486.1x10-9 wavelength shift = 3.24x10-12

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