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.

The parsecRadius 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

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 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

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

5/)(

2

1 12100 mm

bb

Taking logs of both sides

)log(5.22

112 b

bmm

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

)log(5.2bB

Mm

From the inverse square law:

22 104,

4 PB

DPb

Where D is the standard distance of 10 parsecs

2

10

d

bB

Combining this equation with )log(5.2bB

Mm 2

10log5.2

dMm

Finally

10

log5 dMm

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.

Answer7.0)4.1log(505.0 M

)log(5.25.52

112 b

bmm

Capella is 5.5 magnitudes brighter than the Sun

How much more powerful than the Sun?

1582

1 bb

So Capella is about 160 times more powerful than the Sun.