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PETE GREENWOOD
Module 5 - Astrophysics Option
Resolving power
“A measure of how good your eyes (or any optical instrument)
are at distinguishing as separate two distant objects with a small angular separation.”
When light passes through an aperture it diffracts and produces a pattern on the opposite
side to the source. Generally, the smaller the aperture the greater the diffraction. For a
circular aperture the pattern is like this:
snake.phys.lsu.edu/ ~gclayton/astrostuff.html
The resolving power (Θ) is given by: x/y where x is the distance between the two objects
and y is the distance between the objects and the eye (or optical instrument).
Θ = x/y
For Example: If two car headlights are 1.5m apart and we can distinguish the two lamps at
a distance of 5km the resolving power is given by:
1.5 / 5000 = 3x10-4 rad
Rayleigh Criterion for resolving two sources
If the diffraction patterns of two light sources overlap we can say that they are “close” to
each other (although this may only be in terms of angular separation).
When the maximum of one diffraction
pattern overlaps with the minimum of
another we can say the two sources are
just resolvable, this is when the angle of
separation; Θ = λ/d.
repairfaq.ece.drexel.edu/.../ module5.html
Provided the angular separation of two
sources is greater than λ/d the two sources can be resolved (we can see them as two
separate sources). Where d is the diameter of the aperture and λ is the wavelength of the
electro-magnetic radiation.
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PETE GREENWOOD
Telescopes
For best resolution the diameter of the entrance aperture must be as large as possible, this
is achieved by using optically smooth (defects are smaller that the wavelength of light)
concave reflectors.
Radio-Telescopes
These “look” at and receive radio waves instead of light-waves. Since radio-waves have a
much much much longer wavelength than light as large a collector as possible is needed
to obtain the maximum resolving power (Θ=λ/d) for the telescope.
For example:
Radio telescopes at Joderal Bank have a diameter of 76m, if we estimate the wavelength
of the radio waves to be 1m the resolving power is:
1/76 = 1.3x10-2
The above telescope could resolve a 10p coin (2.4cm diameter) at 16km since the angle
subtended by the coin is:
2.4x10-2 / 16x104 = 1.5x10-6 rad
This gives a resolving power of:
550x10-4 / 5 = 1.1x10-7
Radio astronomy is often favoured over traditional forms of astronomy since:
 Observations can be made during the day
 Objects can be viewed that are optically invisible
 Quasars can be observed
 The gathering power is huge (proportional to the diameter2)
 The process is not impeded by the dust and gas of our atmosphere
 Wire mesh can be used for the reflector (provided the holes are no bigger than
1/20th of the wavelength of the radio waves to be studied) this make the reflector
lighter and less affected by wind and rain
Quantum efficiency of detectors
Quantum efficiency is the ratio of the number of responses in the detecting medium to the
number of arriving photons (or quanta of radiation).
Photographic film is 4% efficient
The eye is 1% efficient
Using photographic film to capture images of space means that faint sources must be
tracked for a long time or large aperture instruments must be used. Tracking must end a
sunrise!
Charge Coupled Devices (CCD’s) as used in digital cameras have a quantum efficiency
of up to 70%! A CCD is a silicon chip that is divided into picture elements (pixels). The
following process summarises the mode of action for a CCD:
1. Photons from the image hit the CCD
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PETE GREENWOOD
2. Electrons are released (the number released is proportional the to brightness/
intensity of the light)
3. The camera produces an image identical to the electron pattern
4. When the exposure is complete the charge is processed to produce an image
Cosmology
The evolution of the universe has been deduced by observing the light from stars. Stars
consist of the same elements as exist on Earth, this provides evidence that all the stars
and planets originated in the same place. In a laboratory, the emission lines in the
spectrum of hydrogen have characteristic wavelengths resulting in characteristic colours.
These same lines can be observed in the light from stars. Often the spectra from stars
appear to have slightly different wavelengths; this is due to the Doppler Effect.
The Doppler Effect
In sound, relative motion of a source and observer causes an apparent change in the
frequency of the sound detected by the observer. This change is due to The Doppler Effect
and is a function of the relative velocity of the source and observer:
Formula:
Δf / f = v/c
v
velocity of approach
Δf
change in frequency
f
frequency of emission
(calculated in lab)
c
speed of e/m waves
Δf = (v/c) x f
The equivalent for wavelength is:
Δλ / λ = -v/c
Δλ = (-v / c) x λ
or
From this we can see that the wavelength is apparently shortened when object and
observer are getting closer together.
Binary Systems
In these systems two stars orbit about their common centre of mass. If it is possible to
observe them in their plane of orbit the Doppler shift from each star will change during their
orbit. These changes may be superimposed on a gross effect caused by the recession of
the system as a whole.
Hubble’s Law
Hubble found a link between the velocity (v) of recession of distant galaxies and their
distance (d) from us. By plotting a graph showing the velocity of recession against their
distance from us he was able to show a dramatic relationship; The further away a galaxy
is, the higher its velocity. Hubble was also able to show that the motion of the galaxies
was not random, but in general the galaxies are all moving apart from each other.
v is proportional to d
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PETE GREENWOOD
This simple proportional relationship leads to:
v = H0 x d
Where H0 is Hubble’s constant (65km.s-1.Mpc-1)
v is usually in km.s-1 and d is in megaparsec (Mpc)
There are two explanations for Hubble’s law where the light from a star shows red-shift:
1.
The star may be moving away from us, like many stars in our galaxy and
subsequently showing a red-shift (an example of the Doppler effect).
Or
2.
Space itself may be expanding. If this were the case, all the stars and galaxies
would be moving with it and therefore they would all appear to be receding from one
another.
It is this second explanation that we use as it also supports the BIG BANG theory which
we use to explain the existence of our universe.
The Parsec (pc)
"Parallax of one arc second."
The astronomical unit (a.u.) is the distance between the Earth and the Sun (150 000 000
km)
www.definity-systems.net/ ~apw/astro/units.html
One parsec (pc) is the perpendicular distance from the Earth and Sun you must be for the
angular separation of the Earth and Sun to be 1arc.second.
1 arc.second = 1 / 3600 of a degree
(1° = 60min = 60sec therefore 1° = 1 x (60minx60sec) = 3600 arc.seconds)
1pc = 3.26ly
1 light year (ly) = the distance light travels in a year
Parallax
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PETE GREENWOOD
Parallax is defined as ½ the angle through which a stars direction changes as the earth
moves from 1 extremity of its orbit to another (measured in arc seconds).
The age of the universe
Using Hubble’s constant (H0) as 65km.s-1.Mpc-1 we can estimate the age of the universe
since:
H0 = speed of recession / distance
H0 = v / d
We can estimate the age of the universe to be 1 / H0.
Since 1 Mpc = 3.1x1022 m, the age of the universe is approximately:
3.1x1022 m / 65000 m.s-1
= 4.8x1017 s
To calculate the age of the universe in years we need to divide by:
(60s x 60min x 24hrs x 365.25 days)
= 31557600 Number of seconds in a year
Which gives us:
4.8x1017 / 31557600
= 1.5x1010 Years = 15 Billion Years
Optics - lens ray diagrams
When drawing a lens ray diagram for a Bi-convex lens the following points must be
remembered:
 A ray of light passing through the optical centre of the lens does not deviate from
its path
 All rays of light parallel to the principle axis pass through the principle focus
Focal Length
Lens Power
=f
= metres
= 1/f = Dioptres
(m)
(D)
Object outside 2f
A real image is created between f and 2f, it is inverted
and smaller than the object.
Used in cameras and in the eye whilst reading
Object at 2f
A real image is
created at 2f, the
image is the same size as the
object and inverted.
Used in photocopiers
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PETE GREENWOOD
Object between f and 2f
The real image is outside 2f, inverted and magnified.
Used in projectors, and as the objective lens in
microscopes.
Object at f
The image is at infinity as parallel rays of light are
formed.
Used in spotlights and headlamps.
Object between f and the lens
A virtual image is formed on the same side of the lens
as the object. The image is the same way up as the
object and magnified.
Used in magnifying glass’, eye lenses’ of optical
instruments for distant viewing e.g. telescopes.
http://www.glenbrook.k12.il.us/gbssci/phys/Class/refrn/u14l5da.html
Real images can be focused onto a screen whereas
virtual images cannot; virtual images are the only images
that are the right way up.
The thicker the lens is in the middle compared with the edges, the shorter is its the
focal length.
Objects far away – telescopes
When looking at objects at huge distances any two rays of light coming from that object
will appear parallel. An image is subsequently produced in the focal plane of the objective
lens; it is this image that is focused by the second “eye-piece” lens.
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PETE GREENWOOD
The eyepiece lens is positioned so that the image from the objective lens is at its focal
length. Therefore the image from the eyepiece lens is at infinity, and the telescope is said
to be in “normal adjustment” (this will not strain the eye through prolonged usage).
webs.mn.catholic.edu.au/.../ hsc_astrophysics.htm
The
final
image produced by the eyepiece lens is inverted, but this is not an issue as space has no
up / down since the orientation of an object is only relative to gravity.
Lens (and mirror) Formula
http://www.physics.carleton.ca/~watson/1000_level/Waves_and_Optics/Gifs/Convex_Lens_Formula.gif
The thin lens formula is: 1/u + 1/v = 1/f
Any units can be used for this equation provided the same units are used throughout the
equation.
Rules for the lens equation:
 If the object is real the u is positive (if the object is virtual then u is negative)
 If the image is real than v is positive (if the image is virtual then v is negative)
 If the lens is convex f is positive (if the lens is concave then f is negative)
Manipulation
1/u + 1/v = 1/f
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PETE GREENWOOD
Multiply both sides by v:
Therefore:
v/u + v/v = v/f
v/u = v/f -1
Linear Magnification
Linear magnification equals
Or
M = v/f -1
M = v/u
Linear magnification is not very useful for application with telescopes since the distance
from the object cannot easily be calculated. For these situations we use angular
magnification.
Angular Magnification
Angular magnification is calculated using two angles, β and α and the equation:
angular magnification = β / α
α
The angle subtended at the unaided eye by two rays of light form the top and
bottom of an object
β
The angle subtended at the unaided eye by the image from the telescope
© Peter Greenwood 2004
By creating triangles from; the angles α and β, the respective focal lengths and the image
height we can say that:
tan α = Image height / Focal length Objective
tan β = Image height / Focal length Eye-piece
Because α and β are very small we can say that:
(When α and β are in radians)
Therefore:
tan α/ β is equal to α/ β
α = Image height / Focal length Objective
α = I / Fo
β = Image height / Focal length Eye-piece
β = I / Fe
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PETE GREENWOOD
Angular Magnification manipulation
We can re-arrange the equation for angular magnification as follows:
M=β/α
M = (I / Fe) / (I / Fo)
M = (I / Fe) x (Fo / I)
M = Fo / Fe
If:
And we substitute in the values from above:
Which means that:
By cancelling the I’s we get:
To get the largest angular magnification for telescopes it is best to use an objective lens
with a large focal length and an Eyepiece lens with as small a focal length as possible.
There are however limits to this as large lens suffer from both chromatic (distortions of the
colours) and spherical aberration (distortion of the shape) due to the diffraction that goes
on within the lens.
Lens (and mirror) Equation
(1 / u) + (1 / v) = (1 / f)
u = distance from object to lens
v = distance from image to lens
f = focal length of the lens
If the image is real then v is positive
If the image is virtual then v is negative
For a converging lens (or mirror) the focal length is positive
For a diverging lens (or mirror) the focal length is negative
Limitations of Refracting telescopes
Spherical Aberration may occur, this is where imperfections in the glass due to impurities
and the irregular forming of the lens due to forces acting on the glass during cooling (for
example gravity) cause distortions of the image as the lens is not a perfect curve (not all
light is transmitted through the lens).
Due to the different wavelengths and frequencies of the visible spectrum the different
colours of light are refracted by different amounts (blue more than red) this is known as
chromatic aberration and can results with coloured fringes appearing around objects.
The advantage of a refracting telescope is that it is a sealed unit and therefore there is no
need for cleaning. Refracting telescopes can be made smaller and are cheaper than
reflecting telescopes.
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PETE GREENWOOD
Reflecting telescopes
Cassegrain telescope
This design uses two mirrors and a convex eyepiece
lens. The primary mirror is a (parabolic) concave shape
that focus’ incoming rays of light onto the secondary
mirror, a (parabolic) convex shape.
The magnification of a reflecting telescope is given by:
M = Fo / Fe where Fo is the equivalent focal
length of the two mirrors combined and Fe is the focal
length of the eye-piece lens.
As shown in the diagram below the secondary mirror is
placed within the focal length of the primary mirror. This secondary mirror reflects the
image to the eyepiece where a convex lens focuses the image into the eye of the
observer.
http://zebu.uore
gon.edu/~js/glo
ssary/reflecting
_telescope.html
The
Advantage
s of a
reflecting
telescope
are:
 No
chr
omatic aberration as no refraction occurs (except in the eye piece)
 No spherical aberration (if a parabolic mirror is used)
 Greater resolving power as larger apertures can be used
Disadvantages of a reflecting telescope:
 Maintenance is needed due to the open nature of the telescope (dusting the mirror)
Spectral Classes of Stars
Stars have been classified into groups according to the dominant lines in their absorption
and emission spectrum. The classes had the following letters ascribed to them by the
ancient Greeks:
O
B
A
F
G
K
M
(Oh Be A Fine Girl Kiss ME!)
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PETE GREENWOOD
Hertzsprung-Russell Diagram
imagine.gsfc.nasa.gov/. ../LC_main_p8.html
Absolute magnitude is the magnitude of the star at 10pc (large negative numbers
indicate the greatest magnitude)
Most modern stars are plotted on the graph based upon their luminosity and surface
temperature.
Main Sequence stars are the most common; these are shown in the central curve of the
graph which extends from highly luminous blue (top left) to the to very dim red (bottom
right) stars. Stars leave this main sequence when the dominant fusion progress changes;
when this occurs they will become one of the following stars:
Red Giants and Supergiants are relatively cool and therefore emit a small amount of light
per unit surface area. This means that be as bright as they are they must have a very large
surface area.
White Dwarfs are very hot dim stars; they are very dense due to a small diameter
(suggested by their low luminosity). These are quite common but difficult to observe.
Large stars can suffer catastrophic collapse that causes a sudden temperature rise which
may result in a supernova explosion which can result in the expulsion of a large fraction of
the stars mass. If a supernova explosion occurs the remaining mass of the star will
collapse under its own gravity getting smaller and smaller until it is no more than a point of
infinite density.
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PETE GREENWOOD
The origin of Absorption Spectra
In general absorption spectra consist of a continuous emission spectra crossed by dark
lines at certain points that correspond to emission spectral lines of certain elements that
exist between the source of the continuous spectra and the observer.
In a simple case, an atom of an element may undergo excitation by the absorption of a
photon. On return to its ground state the photon is re-emitted in a random direction. The
observed spectrum is therefore darkened at the point corresponding to the wavelength of
the photon:
instruct1.cit.cornell.edu/.../ astro101/lec13.htm
If the surface of a star is very hot, high energy photons will leave its surface. This will
cause absorption spectra to be observed at the short wave end of the continuum.
In a class O star, the temperature is high enough to ionise significant numbers of helium
atoms. Helium absorption spectra are seen from these stars. Hydrogen in the first excited
state (n=2) can give rise to visible absorption lines (Balmer series) because n=2 becomes
the “normal” ground state at high temperatures.
In cooler stars, absorption spectra are more obvious at the long wavelength parts of the
spectra. These are caused by absorption from molecules and transitions in molecular
vibrational energy states.
Finding Spectral Class by Dominant Absorption Lines
Firstly use the H-R Diagram to find the absolute magnitude (M) and examine the
absorption spectra to measure the apparent magnitude.
Using:
We can rearrange to give:
m-M = 5log(d/10)
(m-M)/5 = log(d/10)
Therefore:
10[ (m-M)/5 ] = d/10
10 x 10[ (m-M)/5 ] = d
10[ {(m-M)/5}+1 ]
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PETE GREENWOOD
Magnitude and Intensity
Magnitude is a measure of how bright a star is. There are, however, two different ways of
indicating a stars magnitude; apparent magnitude and absolute magnitude.
The scale we use to measure magnitude is based on that created by the ancient Greeks
which ran from 1 to 6. On the ancient Greek scale 1 was the brightest star they could
view and 6 was the dimmest. The problem with this scale is that with the creation of
optical aids (such as the telescope) we can now see stars that are much further away than
those the ancient Greeks could see. These stars now visible to us fall on both sides of this
scale; some are dimmer than the dimmest stars the ancient Greeks could view, whilst
others have absolute magnitudes much greater than those seen by the ancient Greeks.
The Brightest stars visible to us are those with the largest negative magnitudes and the
dimmest stars have the highest positive numbers. We can now see stars with a range of
values from -25 through to +25.
Apparent magnitude is how bright a star is when viewed from any location, this means
that apparent magnitude is relative to the location of the star from the viewpoint. A star of
apparent magnitude 2 is two and a half times brighter than a star of apparent
magnitude 1, just as a star of apparent magnitude 3 is two and a half times brighter than a
star of apparent magnitude 2. This logarithmic scale is used as the eye responds to light in
a logarithmic fashion.
Apparent magnitude and light intensity (I) are linked by the equation:
m = -2.5 x log I + const.
The above equation is used to compare the apparent magnitudes of 2 stars with their
intensities (in J.S-1.m-2). The use of an unknown constant does not matter since the
constant will cancel out when comparing two stars of known intensities or magnitudes:
m1 = -2.5 x log I1 + const.
m2 = -2.5 x log I2 + const.
m2 - m1 = -2.5 x log I2+ const. - -2.5 x log I1+ const.
m2 - m1 = -2.5 (log I2 - log I1)
m2 - m1 = -2.5 log (I2 / I1)
NB: this is the most useful form of the equation!!
We can re-arrange the above equation as follows:
Since:
m2 - m1 = -2.5 log (I2 / I1)
We can say:
(m2 - m1) / 2.5 = log (I2 / I1)
Therefore:
10x x [(m2 - m1) / 2.5] = (I1 / I2)
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PETE GREENWOOD
For Example:
“Compare the apparent magnitudes of the Andromeda Galaxy (m=4.8) and the Crab
Nebula (m=8.4)”
(m2 - m1) / 2.5 = log (I2 / I1)
(8.4 - 4.8) / 2.5 = log (I2 / I1)
10x (8.4 - 4.8) / 2.5 = 10x 1.44
therefore
=
27.54
This means that the Andromeda Galaxy is 27.54 times as intense as the Crab Nebula.
Absolute Magnitude is used as a method of comparing the magnitudes of distant stars
using a common scale; it is the brightness of a star when it is at a distance of 10 parsecs
from the sun. It is important to remember that although this 10 pc is measured from the
sun and not the earth, the difference in the two measurements is negligible due to the
huge distance the stars are from the earth.
Absolute magnitude can be calculated by using the spectral class of the star and crossreferencing this with the main sequence band on the H-R diagram, this method is open to
huge variation due to the fact that the main sequence is a band on the H-R diagram and
not a line.
Another method is to use the equation that relates the absolute and apparent magnitude
(although this is only reliable for stars closer than 10Mpc (10 000 000pc):
m
- M = 5 x log (d/10)
By re-arranging the equation on the previous page we can calculate the distance a star is
from us (if we know the absolute and apparent magnitudes):
d = 10 x 10(m - M) x 1/5
Alpha Centuri has apparent magnitude 0.1 and is at a distance of 1.32pc from the sun, find
the absolute magnitude:
m - M = 5 x log (d/10)
0.1 - M = 5 x log (1.32/10)
M = 0.1 - 5 x log (1.32/10)
M = 4.497 or 4.5 (2 d.p.)
The above result is very similar to our sun.
Compare the absolute magnitudes of the two stars Sirius (apparent magnitude -1.46 at a
distance of 2.7pc from the sun) and Betelgeuse (apparent magnitude +0.50 at a distance
of 94pc from the sun)
M = m - 5 log (d/10)
Sirius
Betelgeuse
M = -1.46 - 5 log (2.7/10)
M = +0.50 - 5 log (94/10)
= 1.38
= -4.37
This shows that Betelgeuse is the brightest star in terms of absolute magnitude (although
not, in this case, in terms of apparent magnitude)
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PETE GREENWOOD
Deriving: m - M = 5 x log (d/10)
If i is the intensity of light reaching the earth from a star from a star when it is d pc from
Sun, and I is the intensity when the same star is 10pc from the Earth we can say that
since:
Intensity is proportional to 1/distance2
By using:
i is proportional to 1 / dpc2
I is proportional to 1 / 10pc2
and
We can see that:
(I / i) = (d / 10)2
Now if we take logs of both sides:
log (I / i) = log (d / 10)2
Which is the same as:
(log I - log i) = 2 log d/10
Since:
m = -2.5 x log i + const.
and
M = -2.5 x log I = const.
M - m = -2.5 (log I - log i)
From previous page:
M - m = -2.5 (log I - log i)
M - m = -2.5 x 2 x log (d/10)
M - m = -5 x log (d/10)
Therefore:
m - M = 5 x log (d/10)
Escape velocity and Black holes
Escape velocity is the instantaneous launch velocity that would just allow an object to
escape the gravitational influence of another object.
It can be shown that:
Ve = √ (2GM/R)
(For an object of mass M, radius R. where G is the gravitational constant 6.67x10-11 N
m2 kg-2)
For massive objects of a small radius (high density) the velocity of escape (V e) becomes
very large. For a black hole Ve ≥ c (velocity of light).
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PETE GREENWOOD
For a more massive black hole Ve ≥ c at a distance above the physical surface.
The radius of the sphere formed by the points where the
velocity of escape is equal to the velocity of light is
known as the event horizon.
The radius of this sphere is called the Swarzschild
Radius. Within this, the velocity of escape is greater
than the speed of light i.e. light
cannot escape.
At the Schwarzschild radius Ve = c therefore c = √
(2GM/R) this can be rearranged to give the equation in
terms of the Swarzschild Radius: R = 2GM / C2
© Peter Greenwood 2004
Black Body Radiation and Wien’s Displacement Law
A black body is a perfect absorber; it will absorb all wavelengths of radiation falling on it. At
room temperature it will appear black since it does not reflect lights.
A black body is capable of radiating or emitting all wavelengths of radiation provided its
temperature is high enough.
The graph here shows the energy distribution
from a black body
λ indicates the wavelength of most intense
radiation emitted from a black body, it is not
the radiation of longest wavelength. This is
often referred to as λmax.
As the temperature of different stars increases,
the wavelength of the most intense radiation
increases (shifts to the right of the scale).
www.fofweb.com/Subscription/ Science/Helicon.a...
At 4000K the whole visible spectra is emitted
by black bodies. The emission is however weighted towards the red end of the spectra, at
much higher temperatures (around 20000K) the spectra is shifted towards the blue end of
the spectra.
λmax x Tk = const.
constant = 0.00289m.k.
m.k. = metre-kelvin
“Show that a temperature of around 4000K is needed for λmax to be in the visible
spectrum.”
T= 4000K
λmax needs to be approx:
550x10-9m for the MIDDLE of the spectrum
640-740x10-9 for the RED end of the spectrum
580x10-9 for the YELLOW area of the spectrum
420-480x10-9 for the BLUE end of the spectrum
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PETE GREENWOOD
λmax x Tk = 0.00289m.k.
λmax = 0.00289m.k / Tk.
λmax = 0.00289m.k / 4000K
λmax = 4128.6K
Stefan’s Law
Luminosity
= Power from all the surface of a star
= Energy per second per m2 x total surface area of a star
= Joule seconds (J.s-1)
or
Watts (W)
Stephan’s law states:
Power (P) is proportional to Surface Area (A)
Power (P) is proportional to Surface Temp4 (T4)
P = σ x A x T4
The total power radiated by a star black body is proportional to the fourth power of is
absolute temperature.
Where σ is Stephan’s constant σ = 5.67x10-8 W.m-2.K-4
Quasars
Quasars (quasi-stellar objects) are extremely luminous objects at immense distances
(upward of 3 billion light years) from earth. They are extraordinary because they appear to
be very small but they outshine whole galaxies.
Quasars were discovered by analysing the location of a strong radio source. The
spectrum examined exhibited a series of lines that no-body could identify. These spectral
lines were found to be the result of incredible red-shift which indicated the quasar was
moving away from Earth at 15% of the speed of light (45000km.s -1). Upon further
examination quasars were confirmed to look like one singular star instead of a whole
galaxy, indeed the size of the quasars was so small that the angular diameter of the
objects could not be measured.
Quasars are:
Billions of light years away
As bright as a thousand galaxies
Much smaller than a galaxy
This presents the following uncertainties or controversies:
For quasars to be as bright as they are they must be consuming far more
energy than could be contained in object of their size unless it is far denser
than anything we have come across (except black-holes).
There is some scepticism about the use of Red-shifts to determine the
distance of quasars from the Earth.
Some quasars seem to be receding at incredible speeds of up to 90% of the
speed of light.
Notes produced by Peter Greenwood: reproduced by his kind permission.
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