E3

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E3 Stellar distances
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Parallax
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(No Transcript)
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Parallax
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Parallax angle
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Parallax angle

• Tan p

P (in rads)
R (1 AU)
d
tan p R/d for small p, tan p p so d R/p
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Parsec

• If the parallax angle is one arcsecond (1 ) the
  distance to the star is called a parsec

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Parsec

• If the parallax angle is one arcsecond (1 ) the
  distance to the star is called a parsec
• d (parsecs) 1
• p (in arcsecs)

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Parsec

• 1 pc 3.26 ly

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Example

• A star has a parallax angle of 0.34 arcsecs. How
  far is the star away from earth in light years?
• d (parsecs) 1 1 2.9pc
• p (in arcsecs) 0.34
• Distance in light years 2.9 x 3.26 9.5 ly

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Converting degrees to arcsecs in radians

• Multiply by 2p to convert to radians
• 360
• Multiply by 1 to convert to arcsecs
• 3600

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Parallax method

• Only useful for close stars (up to 300 ly (100
  pc) as further than that the parallax angle is
  too small (space based telescopes can use this
  method to measure stars up to distances of 500
  pc).

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Apparent and absolute magnitudes
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Hipparchus

• Greek astronomer

• Lived 2000 years ago

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Hipparchus compared the relative brightness of
stars (as seen from earth)

• Brightest star magnitude 1
• Faintest star magnitude 6

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Apparent magnitude and brightness

• Magnitude 1 star is
• 100 times brighter than
• a magnitude 6 star

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• The difference between a magnitude 1 star and a
  magnitude 6 star is 5 steps on the magnitude
  scale and the scale is logarithmic. This means
  that each step equated to a brightness decrease
  of 2.512 since
• (2.512)5100

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Magnitude 1
r5 100
Magnitude 2
Magnitude 3
r 2.512
Magnitude 4
Magnitude 5
Magnitude 6
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Under what conditions?

• Clear sky
• When viewed from earth
• As visible to the naked eye

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Can a star have a magnitude greater than 6?

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Can a star have a magnitude greater than 6?

Yes, but these stars are only visible through a
telescope
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A star of apparent magnitude less than 1
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Negative apparent magnitude?

• They are very bright!!

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Guess the apparent magnitude of Sun

• It is -26.7

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Apparent magnitude

• The apparent magnitude m, of a star of apparent
  brightness b is defined by
• m -(5/2)log (b/b0)
• where b0 is taken as a reference value of 2.52 x
  10-8 W.m-2
• This can also be written as b/b0 2.512-m

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• Question
• Apparent magnitude of Sun is -26.7 and that of
  Betelgeuse is 0.5. How much brighter is Sun than
  Betelgeuse?

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Apparent magnitude of Sun is -26.7 and that of
Betelgeuse is 0.5. How much brighter is Sun than
Betelgeuse?

• Difference in magnitudes is 0.5 - -26.7 27.2
• Each difference in magnitude is a difference of
  2.512 in brightness ((2.512)5100 )
• Therefore the difference in brightness
  2.51227.2
• 7.6 x 1010

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• Sun is 76 billion times brighter than Betelgeuse

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Question 2

• Apparent magnitudes of Andromeda galaxy and Crab
  nebula are 4.8 and 8.4 respectively.
• Which of these is brightest?
• By what factor?

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• Galaxy is brighter
• Difference in apparent magnitudes 8.4 4.8
  3.6
• Difference in brightness therefore 2.5123.6
  27.5 times

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• The Andomeda Galaxy is a
  vast collection
• of stars
• The Crab Nebulae is a debris
• of supernova and is the
• birth place of the new star.

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Apparent magnitude

• Is it a fair way of measuring brightness of a
  star?
• Brightness depends on distance and obeys inverse
  square law

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ABSOLUTE MAGNITUDE

• Let the standard distance be 10 pc
• 1 pc 3.086 x 1016 m
• 3.26 ly
• 206265 AU

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• Absolute magnitude
• is the apparent
• magnitude of a star
• when viewed from
• a distance of 10 pc.

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Absolute magnitude M and apparent magnitude m

• m M 5 log (d/10)
• d is in parsecs!

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Question

• Calculate the absolute magnitude of Sun.
• Apparent magnitude -26.7
• Distance from earth 4.9 x 10-6 pc

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• m M 5 log(d/10)
• -26.7- M 5 log (4.9 x 10-6/10)
• M -26.7 5log(4.9 x 10-7)
• M 4.85

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M 4.85

• This means at a standard distance of 10 parsecs
  the sun would appear to be a dim star.

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Can absolute magnitude be

• Positive ?
• Negative ?
• Any value?

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Spectroscopic parallax
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Spectroscopic parallax

• This refers to the method of finding the distance
  to a star given the stars luminosity and
  apparent brightness. It doesnt use parallax!
  Limited to distances less than 10 Mpc
• We know that b L/(4pd2) so d (L/(4 pb))½

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Spectroscopic parallax - Example

• A main sequence star emits most of its energy at
  ? 2.4 x 10-7 m. Its apparent brightness is
  measured to be 4.3 x 10-9 W.m-2. How far away is
  the star?
• ? 0T 2.9 x 10-3 Km
• T 2.9 x 10-3 / 4.3 x 10-9 12000K

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• T 12000K. From an HR diagram we can see this
  corresponds to a brightness of about 100x that of
  the sun ( 100 x 3.9 x 1026 3.9 x 1028 W)

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Spectroscopic parallax - Example

• Thus d (L/(4 pb))½
• d (3.9 x 1028/(4 x p x 4.3 x 10-9))½
• d 8.5 x 1017 m 90 ly 28 pc

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Using cepheids to measure distance
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Cepheid variables

• At distances greater than Mpc, neither parallax
  nor spectroscopic parallax can be relied upon to
  measure the distance to a star.
• When we observe another galaxy, all of the stars
  in that galaxy are approximately the same
  distance away from the earth. What we really need
  is a light source of known luminosity in the
  galaxy. If we had this then we could make
  comparisons with the other stars and judge their
  luminosities. In other words we need a standard
  candle that is a star of known luminosity.
• The outer layers of Cepheid variable stars
  undergo periodic expansion and contraction,
  producing a periodic variation in its luminosity.

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• Cepheid variable stars are useful to astronomers
  because of the period of their variation in
  luminosity turns out to be related to the average
  absolute magnitude of the Cepheid. Thus the
  luminosity of the Cepheid can be calculated by
  observing the variation in brightness.

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• The process of estimating the distance to a
  galaxy (in which the individual stars can be
  imagined) might be as follows
• Locate a Cepheid variable in the galaxy
• Measure the variation in brightness over a given
  period of time.
• Use the luminosity-period relationship for
  Cepheids to estimate the average luminosity.
• Use the average luminosity, the average
  brightness and the inverse square law to estimate
  the distance to the star.

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Cepheid calculation - Example
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• From the left-hand graph we can see that the
  period of the cepheid is 5.4 days. From the
  second graph we can see that this corresponds to
  a luminosity of about 103 suns (3.9 x 1029 W).

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• From the left hand graph we can see the peak
  apparent magnitude is 3.6 which means we can find
  the apparent brightness from
• b/b0 2.512-m
• b 2.52 x 10-8 x 2.512-3.6 9.15 x 10-10 W.m-2

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• Now using the relationship between apparent
  brightness, luminosity and distance
• d (L/(4pb))½
• d (3.9 x 1029/(4 x p x 9.15 x 10-10))½
• d 5.8 x 1018 m 615 ly 189 pc

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Questions

• Page 512 questions 3, 4, 6, 7, 8, 10, 11, 13,
  14, 15, 16.