1. Light Years and Parsecs
Measures of interstellar distances

2. The Light Year
The distance that light travels in 1 year is a light year
1 light year = x 1015 metres.

3. 1 parsec = 3.09 x 1016m or 3.26 light years
The parsec
Radius of Earth orbit (1 a.u.)
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

4. Absolute Magnitude

5. 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

6. 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.

7. 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

8. Absolute Magnitude Formula
We know that the magnitude scale is a logarithmic scale 1 2 3 4 5 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

9. Taking logs of both sides
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 = 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
Taking logs of both sides

10. 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

11. From the inverse square law:
Where D is the standard distance of 10 parsecs
Combining this equation with
Finally

12. Example
Capella is a bright nearby star. Its apparent magnitude is 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.

13. Answer Capella is 5.5 magnitudes brighter than the Sun
How much more powerful than the Sun?
So Capella is about 160 times more powerful than the Sun.