Light Years and Parsecs Measures of interstellar distances

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 1016 m 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 brighter than 3 brighter than 4 brighter than 5 brighter than 6

Magnitude difference between stars (m 2 -m 1) Ratio of Intensity of light measured at earth b 1/b 2 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 Taking logs of both sides

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

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

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 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.