Light Years and Parsecs
Measures of interstellar distances
The Light Year
• The distance that light travels in 1 year is a
light year
1 light year = 9.46 x 1015 metres.
Radius of Earth
orbit (1 a.u.)
The parsec
This angle is
equal to 1 second
of arc (1/36000)
x
The distance x is one parsec
i.e. the parsec is the distance from the sun at which the radius of the
Earth orbit subtends an angle of one second of arc.
The name is an abbreviation of the term “parallax second”
1 parsec = 3.09 x 1016m or 3.26 light years
Absolute Magnitude
The apparent brightness of stars conveys
no information about their distance from
us. Some of the brightest stars here are
more distant than the faintest
Apparent and Absolute Magnitude
• The apparent magnitude gives us
information about how bright a star
appears to be from Earth.
• It gives us no information about the how
bright the star actually is!
• We need another idea to compare the
actual brightness of the stars.
• This is what the idea of absolute
magnitude does.
Absolute Magnitude
• If the stars were equally distant then their
relative brightness would give us a true
comparison of their brightness.
• Absolute magnitude gives us the value of
a star’s brightness at a standard distance
of 10 parsecs
Absolute Magnitude Formula
• We know that the magnitude scale is a
logarithmic scale
1
2
3
5
4
6
x 2.512
x 2.512
x 2.512
x 2.512
x 2.512
brighter
than 2
brighter
than 3
brighter
than 4
brighter
than 5
brighter
than 6
Magnitude difference
between stars
(m2-m1)
Ratio of Intensity of light
measured at earth b1/b2
1
2.512
2
(2.512)2 = 6.31
3
(2.512)3 = 15.85
4
(2.512)4 = 39 .82
5
(2.512)5 = 100
10
(2.512)10 = 104
15
(2.512)15 = 106
20
(2.512)20 = 108
From This table we can see it
can be determined that the
relationship between the two
quantities is
b1
( m2  m1 ) / 5
 100
b2
Taking logs of both sides
b1
m2  m1  2.5 log( )
b2
The Absolute Magnitude Formula
• Now where M is the
apparent magnitude
of the star brought to
a distance of 10
parsecs and B the
intensity of light
received from the star
at that distance and m
and b are the original
values
m  M  2.5 log(
B
b
)
From the inverse square law:
P
b
,
2
4D
P
B
4 10 2
B d 
 
b  10 
Combining this equation with
Finally
Where D is the standard
distance of 10 parsecs
2
m  M  2.5 log(
B
)
b
2
d 
m  M  2.5 log  
 10 
d 
m  M  5 log  
 10 
Example
• Capella is a bright nearby star. Its apparent
magnitude is +0.05 and its distance is 14
parsecs. What is its absolute magnitude.
• Compare this value to the absolute magnitude of
the Sun(+4.8). How many magnitudes is Capella
brighter than the Sun and therefore calculate the
how many times more power is emitted by
Capella than our Sun.
Answer
M  0.05  5 log( 1.4)  0.7
Capella is 5.5 magnitudes brighter than the Sun
How much more powerful than the Sun?
b1
m2  m1  5.5  2.5 log( )
b2
b1
 158
b2
So Capella is about 160 times more powerful than the Sun.