Download Section 6 The Expanding Universe The Doppler Effect

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.
2.
3.
4.
State what is meant by the Doppler effect and give examples.
Copy the Doppler equations for moving sources.
Describe briefly how Cepheid variables are used to measure the
distances to galaxies.
Copy Hubble's formula (make sure that you understand the units).
Evidence for the Expanding Universe
Rotational velocity
1.
2.
Describe how the mass of galaxies can be calculated.
What problem did this produce?
Seeing the light
1.
2.
3.
Describe how light can be used to calculate the mass of a galaxy.
What problem did this produce?
How much of the universe is thought to be dark matter?
Dark matter
1.
Explain what is meant by MACHOs and WIMPs.
MACHOs
1.
2.
3.
4.
Describe 2 examples of MACHOs.
Briefly describe how gravitational lensing can be used to indicate the
presence of a MACHO.
Briefly describe how black holes can be detected.
State the evidence for and against MACHOs.
WIMPs
1.
2.
State that WIMPs are non-baryonic and their prime candidates.
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 stars .............................................................................................. 16
WIMPs ....................................................................................................... 17
Debate the arguments ............................................................................... 20
The Expanding Universe Problems .............................................................. 21
Solutions .................................................................................................... 31
Content Statements
Contents
Notes
Contexts
a) The Doppler
Effect and
redshift of
galaxies.
The Doppler Effect is
observed in sound and
light. For sound, the
apparent change in
frequency as a source
moves towards or
away from a stationary
observer should be
investigated.
Doppler Effect in terms
of terrestrial sources
e.g. passing
ambulances.
Investigating the
apparent shift in
frequency using a
moving sound source
and datalogger.
Applications include
measurement of speed
(radar),
echocardiogram and
flow measurement.
The Doppler Effect
causes similar shifts in
wavelengths of light.
The light from objects
moving away from us is
shifted to longer
wavelengths – redshift.
The redshift of a galaxy
is the change in
wavelength divided by
the emitted
wavelength. For slowly
moving galaxies,
redshift is the ratio of
the velocity of the
galaxy to the velocity
of light.
(Note that the Doppler
Effect equations used
for sound cannot be
used with light from
Measuring distances to
distant objects.
Parallax measurements
and data analysis of
apparent brightness of
standard candles.
The unit ‘Particles and
Waves’ includes an
investigation of the
inverse square law for
light. Centres may
wish to include this
activity in this topic.
In practice, the units
used by astronomers
include light-years and
parsecs rather than SI
units.
Data analysis of
measurements of
1
fast moving galaxies
because relativistic
effects need to be
taken into account.)
b) Hubble’s Law.
c)
galactic velocity and
distance.
The revival of
Einstein’s cosmological
constant in the context
of the accelerating
universe.
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 age
of the Universe.
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 Sun 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
2
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.
3
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 sour ce 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
ultrasound’, 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 wav es
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 v s and its frequency is f s . 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, λ obs .
λ obs
=
v
fs
vs
fs
=
1 (v - v s )
fs
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The observed frequency,
f obs
=
Thus f obs =
and f obs =
v
=
λ obs
fs
fs
v
v
1 (v - v )
s
fs
=
fs
v
(v - v s )
(v - v s )
for a source moving towards a
stationary observer,
v
(v + v s )
for a source moving away from a
stationary observer.
The two formulae above can also be written as
v
f
= fs
(v ± v s )
Note: the fractional change in wavelength z may be calculated using
z = (λ obs - λ rest ) / λ rest
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 u niverse 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 Sl ipher 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 Edwin Hubble.
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 astronomy. (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 Milky Way. 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 galax ies 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 c andles’,
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 star,
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 H o 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 accept ed 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 cosmo logy. 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 tired light 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 Big Bang 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 radio 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 radiation 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 w ith 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 gala xy 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 t hat,
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.
10
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 theorie s 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 matte r. 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 appea r 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’.
11
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
electron 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 st ars 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 black hole 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 dis tance,
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 MA CHO.
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.
15
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 a nd
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.
16
In January 1995, a team of American and Japanese scientists announced
‘compelling evidence’ for the existence of a massive black hole at th e
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 t o
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, astronomer s
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
17
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.
19
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
20
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 f s . When the source
is moving towards a stationary observer, the observer hears a
____A_____ frequency f o . When the source is moving away from a
stationary observer, the observer hears a ____B_____ frequency f o .
This is known as the _____C____ ____D_____.
Match each letter with the correct word from the list below:
2.
Doppler
effect
quieter
softer
higher
louder
lower
Write down the expression for the observed frequency f o , detected
when a source of sound waves in air of frequency f s moves:
(a)
towards a stationary observer at a constant speed, v s
(b)
away from a stationary observer at a constant speed, v s .
21
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
by stationary
observer / Hz
Frequency of
source / Hz
Speed of source
moving towards
observer / m s 1
(a)
400
10
(b)
400
850
(c)
1020
(d)
2125
2000
170
200
4.
Speed of source
moving away from
observer / m s 1
10
20
5
(e)
(f)
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.
22
6.
7.
A student is standing on a station platform. A train approaching the
station sounds 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.
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.
23
11.
12.
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.
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.
14.
A bat 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.
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.
24
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.
17.
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.
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 .
25
When a star is moving away from the Earth
rest . This is known as the ____B_____ shift.
observed
is ____A_____ than
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
20.
blue
longer
red
shorter
towards.
In the table shown, calculate the value of each missing quantity.
Fractional change in
wavelength, z
Wavelength of light
on Earth rest / nm
Wavelength of light
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
26
Hubble’s law
In the following questions, when required, use the approximation for
H o = 2·4 × 10
1.
18
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.
3.
Convert the following distances in metres into distances in light years.
(a)
Approximate distance from the Earth to our Sun = 1·44 × 10 11 m.
(b)
Approximate distance from the Earth to next nearest star Alpha
Centauri = 3.97 × 10 16 m.
(c)
Approximate distance from the Earth to a galaxy in the
constellation of Virgo = 4·91 × 10 23 m.
In the table shown, calculate the value of each missing quantity.
Speed of galaxy relative
to Earth / m s 1
Approximate
distance from Earth
to galaxy / m
Fractional change
in wavelength, z
(a)
7.10 × 10 22
(b)
(c)
1.89 × 10 24
(d)
1·70 × 10 6
(e)
(f)
2·21 × 10 6
(g)
(h)
27
4.
5.
6.
7.
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?
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.
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.
A distant star is travelling directly away from the Earth at a speed
of 2·4 × 10 7 m s 1 .
(a)
Calculate the value of z for this star.
28
(b)
8.
A hydrogen line 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.
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 × 10 7 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 × 10 14 Hz.
11.
(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.
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.
29
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.
30
Solutions
The Doppler effect and redshift of galaxies
1.
A
B
C
D
higher
lower
Doppler
effect
2a.
f o = f s [v / (v - v s )]
2b.
f o = f s [v / (v + v s )]
3a.
f s = 400 Hz
v = 340 ms -1
v s = 10 ms -1
f o = f s [v / (v - v s )]
f o = 400 [340 / (340 - 10)]
f o = 412 Hz
3b.
f s = 400 Hz
v = 340 ms -1
v s = 10 ms -1
f o = f s [v / (v + v s )]
f o = 400 [340 / (340 + 10)]
f o = 389 Hz
3c.
f o = 850 Hz
v = 340 ms -1
v s = 20 ms -1
f o = f s [v / (v - v s )]
850 = f s [340 / (340 - 20)]
f s = 850 / [340 / (340 - 20)]
f s = 800 Hz
3d.
f o = 1020 Hz
v = 340 ms -1
v s = 5 ms -1
f o = f s [v / (v + v s )]
1020 = f s [340 / (340 + 5)]
f s = 1020 / [340 / (340 + 5)]
f s = 1035 Hz
3e.
f o = 2125 Hz
f s = 2000 Hz
v = 340 ms -1
f o = f s [v / (v - v s )]
2125 = 2000 [340 / (340 - v s )]
2125 / 2000 = 340 / (340 - v s )
340 / (2125 / 2000) = 340 - v s
v s = 20 ms -1
3f.
f o = 170 Hz
f s = 200 Hz
v = 340 ms -1
f o = f s [v / (v + v s )]
170 = 200 [340 / (340 + v s )]
170 / 200 = 340 / (340 + v s )
340 / (170 / 200) = 340 + v s
v s = 60 ms -1
31
4.
The frequency of the sound would increase and decrease.
5a.
f s = 1000 Hz
v = 340 ms -1
v s = 20 ms -1
f o = f s [v / (v - v s )]
f o = 1000 [340 / (340 - 20)]
f o = 1063 Hz
5b.
f s = 1000 Hz
v = 340 ms -1
v s = 20 ms -1
f o = f s [v / (v + v s )]
f o = 1000 [340 / (340 + 20)]
f o = 944 Hz
6a.
f s = 200 Hz
v = 340 ms -1
v s = 25 ms -1
f o = f s [v / (v - v s )]
f o = 200 [340 / (340 - 25)]
f o = 216 Hz
6b.
f s = 200 Hz
v = 340 ms -1
v s = 25 ms -1
f o = f s [v / (v + v s )]
f o = 200 [340 / (340 + 25)]
f o = 186 Hz
7.
f o = 470 Hz
f s = 450 Hz
v = 340 ms -1
f o = f s [v / (v - v s )]
470 = 450 [340 / (340 - v s )]
470 / 450 = 340 / (340 - v s )
340 / (470 / 450) = 340 - v s
v s = 14 ms -1
8.
f o = 540 Hz
f s = 500 Hz
v = 340 ms -1
f o = f s [v / (v - v s )]
540 = 500 [340 / (340 - v s )]
540 / 500 = 340 / (340 - v s )
340 / (540 / 500) = 340 - v s
v s = 25.2 ms -1
f o = f s [v / (v + v s )]
f o = 500 [340 / (340 + 25.2)]
f o = 466 Hz
9.
f s = 540 Hz
r = 0.75m
v = 340 ms -1
v s = 10 ms -1
Highest
f o = f s [v / (v - v s )]
f o = 540 [340 / (340 - 10)]
f o = 556 Hz
32
Lowest
f o = f s [v / (v + v s )]
f o = 540 [340 / (340 + 10)]
f o = 525 Hz
10a.
v s = 20 ms -1
f s = 300 Hz
v = 340 ms -1
f o = f s [v / (v - v s )]
f o = 300 [340 / (340 - 20)]
f o = 318.75 Hz
v = fλ
340 = 318.75 x λ
λ = 1.07 m
v s = 20 ms -1
f s = 300 Hz
v = 340 ms -1
f o = f s [v / (v + v s )]
f o = 300 [340 / (340 + 20)]
f o = 283.33 Hz
11a.
v s = 10 ms -1
f s = 1000 Hz
v = 340 ms -1
v = fλ
340 = 283.33 x λ
λ = 1.2 m
f o = f s [v / (v + v s )]
f o = 1000 [340 / (340 + 10)]
f o = 971 Hz
11b.
v s = 10 ms -1
f s = 1000 Hz
v = 340 ms -1
f o = f s [v / (v - v s )]
f o = 1000 [340 / (340 - 10)]
f o = 1030 Hz
12.
f o / f s = 0.9
v = 340 ms -1
f o = f s [v / (v - v s )]
0.9 = 340 / (340 - v s )
340 / (0.9) = 340 - v s
v s = 37.8 ms -1
10b.
33
13a.
v s = 3.60 ms -1
v = 340.00 ms -1
f s = 350 kHz = 350000
Hz
13b.
As the bat's speed decreases the frequency observed decreases.
v s decreases = v - v s increases = v / (v - v s ) decreases = f o decreases
13c.
v s = 3.60 ms -1
fo =
f s = 350 kHz = 350000 f o =
Hz
fo =
-1
v = 340 ms
fo =
14a.
v s = 54 kmh -1
v s = 54 kmh -1
v s = 54x10 3 mh -1
v s = 54x10 3 / (60 x 60) ms -1
v s = 15 ms -1
14b.
v s = 15ms -1
f s = 1500 Hz
v = 340 ms -1
f o = f s [v / (v - v s )]
f o = 1500 [340 / (340 - 15)]
f o = 1569 Hz
14c.
v s = 15ms -1
f s = 1500 Hz
v = 340 ms -1
f o = f s [v / (v + v s )]
f o = 1500 [340 / (340 + 15)]
f o = 1437 Hz
15.
f o = 640 Hz
f s = 600 Hz
v = 340 ms -1
f o = f s [v / (v - v s )]
640 = 600 [340 / (340 - v s )]
640 / 600 = 340 / (340 - v s )
340 / (640 / 600) = 340 - v s
v s = 21.3 ms -1
16a.
f s = 2200 Hz
r = 0.8 m
16b.
f s = 2200 Hz
v s = 15.1 ms -1
v = 340 ms -1
s = 2πr
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
Minimum
f o = f s [v / (v + v s )]
f o = 2200 [340 / (340 + 15.1)]
The bat hears the echo, the sound wave is
moving towards it.
f o = f s [v / (v - v s )]
f o = 350000 [340 / (340 - 3.60)]
f o = 353745.54 Hz
f o = 354 kHz
f s [v / (v + v s )]
350000 [340 / (340 + 3.6)]
346332.95 Hz
346 kHz
34
f o = 2106 Hz
16.c
f s = 2200 Hz
v s = 15.1 ms -1
v = 340 ms -1
Maximum
f o = f s [v / (v - v s )]
f o = 2200 [340 / (340 - 15.1)]
f o = 2302 Hz
17.
f o towards = 460 Hz
f o away = 410 Hz
v = 340 ms -1
s = 3x10 3 m
Moving towards
f o = f s [v / (v - v s )]
460 = f s [340 / (340 - vs )]
460 / [340 / (340 - v s )] = f s
Moving away
f o = f s [v / (v + v s )]
410 = f s [340 / (340 + vs )]
410 / [340 / (340 + v s )] = f s
460 / [340 / (340 - v s )] = 410 / [340 / (340 + vs )]
460 / 410 = [340 / (340 - v s )] / [340 / (340 + v s )]
460 / 410 = (340 + v s ) / (340 - v s )
460 x (340 - v s ) = 410 x (340 + v s )
(460 x 340) - 460v s = (410 x 340) + 410v s
(460 x 340) - (410 x 340) = 410v s + 460v s
[(460 x 340) - (410 x 340)] / (410+ 460) = v s
v s = 19.5 ms -1
s = vst
3x10 3 = 19.5 x t
t = 154s
18.
v s = 10 ms -1
f s = 1100 Hz
f o = 1200 Hz
19.
A
B
C
D
E
f o = f s [v / (v - v s )]
1200 = 1100 [v / (v - 10)]
1200 x (v - 10) = 1100v
1200v - 12000 = 1100v
1200v - 1100v = 12000
v = 12000 / (1200 - 1100)
v = 120 ms -1
longer
red
shorter
blue
away
35
20a.
λ rest = 365x10 -9 m
λ obs = 402x10 -9 m
z
z
z
z
=
=
=
=
(λ obs - λ rest ) / λ rest
(402x10 -9 - 365x10 -9 ) / 365x10 -9
0.101
1.01x10 1
20b.
λ rest = 434x10 -9 m
λ obs = 456x10 -9 m
z
z
z
z
=
=
=
=
(λ obs - λ rest ) / λ rest
(456x10 -9 - 434x10 -9 ) / 434x10 -9
0.0507
5.07 x10 2
20c.
z = 8.00x10 -2
λ rest = 486x10 -9 m
z = (λ obs - λ rest ) / λ rest
8.00x10 -2 = (λ obs - 486x10 -9 ) / 486x10 -9
(8.00x10 -2 x 486x10 -9 ) + 486x10 -9 = λ obs
λ obs = 5.249 x10 -7 m
λ obs = 525 nm
20d.
z = 4.00x10 -2
λ rest = 656x10 -9 m
z = (λ obs - λ rest ) / λ rest
4.00x10 -2 = (λ obs - 656x10 -9 ) / 656x10 -9
(4.00x10 -2 x 656x10 -9 ) + 656x10 -9 = λ obs
λ obs = 6.82 x10 -7 m
λ obs = 682 nm
20e.
z = 5.00x10 -2
λ obs = 456x10 -9 m
z = (λ obs - λ rest ) / λ rest
5.00x10 -2 = (456x10 -9 - λ rest ) / λ rest
5.00x10 -2 x λ rest = (456x10 -9 - λ rest )
456x10 -9 = 5.00x10 -2 λ rest + λ rest
456x10 -9 = λ rest (5.00x10 -2 + 1)
λ rest = 456x10 -9 / (5.00x10 -2 + 1)
λ rest = 4.34 x10 -7 m
λ rest = 434 nm
20f.
z = 1.00x10 -1
λ obs = 402x10 -9 m
z = (λ obs - λ rest ) / λ rest
1.00x10 -1 = (402x10 -9 - λ rest ) / λ rest
1.00x10 -1 x λ rest = (402x10 -9 - λ rest )
402x10 -9 = 1.00x10 -1 λ rest + λ rest
402x10 -9 = λ rest (1.00x10 -1 + 1)
λ rest = 402x10 -9 / (1.00x10 -1 + 1)
λ rest = 3.65 x10 -7 m
λ rest = 365 nm
36
Hubble’s law
1a.
1 light year
s = vt
s = 3.0x10 8 x (1 x 365 x 24 x 60 x 60)
s = 9.46x10 15 m
1b.
50 light years
s = vt
s = 3.0x10 8 x (50 x 365 x 24 x 60 x 60)
s = 4.75x10 17 m
1c.
100000 light years
s = vt
s = 3.0x10 8 x (100000 x 365 x 24 x 60 x 60)
s = 9.46x10 20 m
1d.
16000000000 light
years
s = vt
s = 3.0x10 8 x (16000000000 x 365 x 24 x 60 x 60)
s = 1.51x10 26 m
2a.
d = 1.44x10 11 m
number of light years = distance / distance travelled in 1
light year
number = 1.44x10 11 / 3.0x10 8 x (1 x 365 x 24 x
60 x 60)
= 1.52x10 -5 light years
2b.
d = 3.97x10 16 m
number of light years = distance / distance travelled in 1
light year
number = 3.97x10 16 / 3.0x10 8 x (1 x 365 x 24 x
60 x 60)
= 4.2 light years
2c.
d = 4.91x10 23 m
number of light years = distance / distance travelled in 1
light year
number = 4.91x10 23 / 3.0x10 8 x (1 x 365 x 24 x
60 x 60)
= 5.19x10 7 light years
3a.
d = 7.10x10 22 m
H o = 2.4x10 -18 s -1
v = Hod
-18
v = 2.4x10
x 7.10x10 22
5
v = 1.7x10 ms
3b.
c = 3.0x10 8 ms -1
5
v = 1.7x10 ms
-1
-1
z = v/c
z = 1.7x10 5 / 3.0x10 8
z = 5.67x10 -4
37
3c.
d = 1.89x10 24 m
H o = 2.4x10 -18 s -1
v = Hod
24
22
v = 1.89x10 x 7.10x10
6
v = 4.54x10 ms
3d.
c = 3.0x10 8 ms -1
6
-1
6
-1
v = 4.54x10 ms
3e.
3f.
v = 1.70x10 ms
H o = 2.4x10 -18 s -1
c = 3.0x10 8 ms -1
6
v = 1.70x10 ms
3g.
6
v = 2.21x10 ms
H o = 2.4x10
3h.
-1
-1
-18
s -1
c = 3.0x10 8 ms -1
6
v = 2.21x10 ms
-1
-1
z = v/c
z = 4.54x10 6 / 3.0x10 8
z = 1.51x10 -2
v = Hod
6
1.70x10 = 2.4x10
24
d = 1.89x10 m
-18
xd
z = v/c
z = 1.70x10 6 / 3.0x10 8
z = 5.67x10 -3
v = Hod
6
2.21x10 = 2.4x10
23
d = 9.21x10 m
-18
xd
z = v/c
z = 2.21x10 6 / 3.0x10 8
z =7.37x10 -3
4a.
λ obs = 466x10 -9 m
λ rest = 434x10 -9 m
z = (λ obs - λ rest ) / λ rest
z = (466x10 -9 - 434x10 -9 ) / 434x10 -9
z = 7.37x10 -2
4b.
c = 3.0x10 8 ms -1
z = 7.37x10 -2
z=v/c
7.37x10 -2 = v / 3.0x10 8
v = 2.21x10 7 ms -1
4c.
Away, as the observed wavelength is longer than the rest wavelength.
5a.
λ rest = 505x10 -9 m
λ obs = 530x10 -9 m
c = 3.0x10 8 ms -1
z = (λ obs - λ rest ) / λ rest
z = (530x10 -9 - 505x10 -9 ) / 505x10 -9
z = 4.95x10 -2
z=v/c
4.95x10 -2 = v / 3.0x10 8
v = 1.49x10 7 ms -1
38
5b.
v = 1.49x10 7 ms -1
H o = 2.4x10 -18 s -1
v = Hod
1.49x10 7 = 2.4x10 -18 d
d = 6.21x10 24 m
6a.
v = 0.074c ms -1
v = 0.074 x 3x10 8
v = 2.22x10 7 ms -1
6b.
v = 2.22x10 7 ms -1
H o = 2.4x10 -18 s -1
v = Hod
2.22x10 7 = 2.4x10 -18 d
d = 9.25x10 24 m
7a.
v = 2.4x10 7 ms -1
c = 3.0x10 8 ms -1
z=v/c
2.4x10 7 = v / 3.0x10 8
z = 8.0x10 -2
7b.
λ obs = 530x10 -9 m
z = 8.0x10 -2
z = (λ obs - λ rest ) / λ rest
8.0x10 -2 = (530x10 -9 - λrest ) / λ rest
8.0x10 -2 x λ rest = (530x10 -9 - λ rest )
530x10 -9 = 8.0x10 -2 λ rest + λ rest
530x10 -9 = λ rest (8.0x10 -2 + 1)
λ rest = 530x10 -9 / (8.0x10 -2 + 1)
λ rest = 4.10 x10 -7 m
λ rest = 410 nm
8a.
λ rest = 489x10 -9 m
λ obs = 538x10 -9 m
c = 3.0x10 8 ms -1
z = (λ obs - λ rest ) / λ rest
z = (538x10 -9 - 489x10 -9 ) / 489x10 -9
z = 1.00x10 -2
z=v/c
1.00x10 -2 = v / 3.0x10 8
v = 3.0x10 7 ms -1
8b.
v = 3.0x10 7 ms -1
H o = 2.4x10 -18 s -1
v = Hod
3.0x10 7 = 2.4x10 -18 d
d = 1.25x10 25 m
d = 1.25x10 25 /3.0x10 8 x 365 x 24 x 60 x 60
d = 1.32x10 9 light years
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9.
d = 1x10 10 light years
H o = 2.4x10 -18 s -1
d = 1x10 10 x 3.0x10 8 x 365 x 24 x 60 x 60
d = 9.4608x10 25 m
v = Hod
v = H o = 2.4x10 -18 x 9.4608x10 25
v = 2.27x10 7 ms -1
10a.
c = 3.0x10 8 ms -1
f rest = 5.00x10 14 Hz
v = fλ
3.0x10 8 = 5.00x10 14 x λ
λ = 6.00x10 -7 m
10b.
c = 3.0x10 8 ms -1
λ rest = 6.00x10 -7 m
v = 3.0x10 7 ms -1
z=v/c
z = 3.0x10 7 / 3.0x10 8
z = 0.1
z = (λ obs - λ rest ) / λ rest
0.1 = (λ obs - 6.00x10 -7 ) / 6.00x10 -7 ]
(0.1 x 6.00x10 -7 ) + 6.00x10 -7 = λ obs
λ obs = 6.6x10 -7 m
v = fλ
3.0x10 8 = f x 6.6x10 -7
f = 4.55x10 14 Hz
11a.
λ rest = 486x10 -9 m
λ obs = (486x10 -9
+20x10 -9 )m
c = 3.0x10 8 ms -1
z = (λ obs - λ rest ) / λ rest
z = (506x10 -9 - 486x10 -9 ) / 486x10 -9
z = 4.11x10 -2
z=v/c
4.11x10 -2 = v / 3.0x10 8
v = 1.23x10 7 ms -1
11b.
H o = 2.4x10 -18 s -1
v = 1.23x10 7 ms -1
v = Hod
1.23x10 7 = 2.4x10 -18 x d
d = 5.125x10 24 m
d = 5.125x10 24 / (3.0x10 8 x 365 x 24 x 60 x 60)
d = 5.42x10 8 m
d = 542 million light years
40
12.
λ rest = 656x10 -9 m
λ obs = 660x10 -9 m
c = 3.0x10 8 ms -1
z = (λ obs - λ rest ) / λ rest
z = (660x10 -9 - 656x10 -9 ) / 656x10 -9
z = 6.098x10 -3
z=v/c
6.098x10 -3 = v / 3.0x10 8
v = 1.83x10 6 ms -1
13.
v = 2 kms -1 = 2000
ms -1
λ rest = 486.1x10 -9 m
c = 3.0x10 8 ms -1
z=v/c
z = 2000 / 3.0x10 8
z = 6.67x10 -6
6.67x10 -6 = (λ obs - λ rest ) / 486.1x10 -9
wavelength shift = (λ obs - λ rest ) = 6.67x10 -6 /
486.1x10 -9
wavelength shift = 3.24x10 -12
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