Transcript E3 – Stellar distances

E3 – Stellar distances
Parallax
Parallax
Parallax angle
Parallax angle
P (in rads)
R (=1 AU)
d
tan p = R/d
» Tan p
for small p, tan p ≈ p
so d = R/p
Parsec
• If the parallax angle is one arcsecond (1 “) the
distance to the star is called a parsec
Parsec
• If the parallax angle is one arcsecond (1 “) the
distance to the star is called a parsec
• d (parsecs) = 1
p (in arcsecs)
Parsec
• 1 pc = 3.26 ly
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
Converting degrees to arcsecs in radians
• Multiply by
• Multiply by
2π to convert to radians
360
1
to convert to arcsecs
3600
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).
Apparent and absolute magnitudes
Hipparchus
• Greek astronomer
• Lived 2000 years ago
Hipparchus compared
the relative
brightness of stars (as
seen from earth)
Brightest star – magnitude 1
Faintest star – magnitude 6
Apparent magnitude and brightness
Magnitude 1 star is
100 times brighter than
a magnitude 6 star
• 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)5=100
Magnitude 1
Magnitude 2
5
r
= 100
Magnitude 3
Magnitude 4
Magnitude 5
Magnitude 6
r = 2.512
* Under what conditions?
• Clear sky
• When viewed from earth
• As visible to the naked eye
Can a star have a magnitude greater than
6?
Can a star have a magnitude greater than
6?
Yes, but these stars are only
visible through a telescope
A star of apparent magnitude less than 1
Negative apparent magnitude?
They are very bright!!
Guess the apparent magnitude of Sun
It is -26.7
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
Question
Apparent magnitude of Sun is -26.7 and that of
Betelgeuse is 0.5. How much brighter is Sun
than Betelgeuse?
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)5=100 )
• Therefore the difference in brightness =
2.51227.2
• = 7.6 x 1010
Sun is 76 billion times brighter than Betelgeuse
Question 2
Apparent magnitudes of Andromeda galaxy and
Crab nebula are 4.8 and 8.4 respectively.
a) Which of these is brightest?
b) By what factor?
Galaxy is brighter
Difference in apparent magnitudes = 8.4 – 4.8 = 3.6
Difference in brightness therefore = 2.5123.6 = 27.5 times
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.
Apparent magnitude
- Is it a fair way of measuring brightness of a
star?
- Brightness depends on distance and obeys
inverse square law
ABSOLUTE MAGNITUDE
Let the standard distance be 10 pc
1 pc = 3.086 x 1016 m
= 3.26 ly
= 206265 AU
Absolute magnitude
is the apparent
magnitude of a star
when viewed from
a distance of 10 pc.
Absolute magnitude M and
apparent magnitude m
m – M = 5 log (d/10)
d is in parsecs!
Question
Calculate the absolute magnitude of Sun.
Apparent magnitude = -26.7
Distance from earth = 4.9 x 10-6 pc
m – M = 5 log(d/10)
-26.7- M = 5 log (4.9 x
-6
10 /10)
M =-26.7 – 5log(4.9 x 10-7)
M = 4.85
M = 4.85
• This means at a standard distance of 10
parsecs the sun would appear to be a dim star.
Can absolute magnitude be
Positive ?
Negative ?
Any value?
Spectroscopic parallax
Spectroscopic parallax
• This refers to the method of finding the
distance to a star given the star’s luminosity
and apparent brightness. It doesn’t use
parallax! Limited to distances less than 10
Mpc
• We know that b = L/(4πd2) so d = (L/(4 πb))½
Spectroscopic parallax - Example
• A main sequence star emits most of its energy
at λ = 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?
• λ 0T = 2.9 x 10-3 Km
• T = 2.9 x 10-3 / 4.3 x 10-9 = 12000K
• 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)
Spectroscopic parallax - Example
Thus d = (L/(4 πb))½
d = (3.9 x 1028/(4 x π x 4.3 x 10-9))½
d = 8.5 x 1017 m = 90 ly = 28 pc
Using cepheids to measure distance
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.
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.
•
•
•
•
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.
Cepheid calculation - Example
• 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).
• 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
• Now using the relationship between apparent
brightness, luminosity and distance
• d = (L/(4πb))½
• d = (3.9 x 1029/(4 x π x 9.15 x 10-10))½
• d = 5.8 x 1018 m = 615 ly = 189 pc
Questions
Page 512 questions 3, 4, 6, 7, 8, 10, 11, 13, 14,
15, 16.