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I have 3 questions on my homework where I can't find any information in the notes or book. Can anyone help?

1. a star's brightness decreases with the square of the distance from the star. the power/ unit area reaching the Earth from the Sun equals 1370 watts/m2. What is the solar power / unit area reaching Saturn, which is at a distance of 9.5 AU

2. Some stellar remnants (essentially dead stars) can have surface temperatures as high as 30,000 K and yet have a luminocity of only 0.05 x the solar luminocity. What can explain this?
-they are very very small
-they are very very distant from earth
-they radiate in an undectable part of the electromagnetic spectrum

3. Consider a star with a surface temp of 21,500 K. In what part of the electromagnetic spectrum would the dominant spectral wavelength be observed?
-microwave
-infrared
-visible
-ultraviolet

2007-10-14 11:30:07 · 2 answers · asked by Sarah 4 in Science & Mathematics Astronomy & Space

2 answers

1) The question gives you all the information you need to calculate the answer, unless you need to be reminded that Earth is 1AU from the Sun.

2) Luminosity refers to absolute, not apparent magnitude, so distance is not the answer. Read up on white dwarfs.

3) You need two things:
1) Wien displacement law - λ (max) = 2.898 x 10^6 nm/T
2) A chart of the electromagnetic spectrum: http://en.wikipedia.org/wiki/Image:Electromagnetic-Spectrum.png

2007-10-14 11:49:16 · answer #1 · answered by injanier 7 · 0 0

Ok I'll help you out a little

1. Do the math, that's the only way you'll learn.

2 and 3, sunlight has a Kelvin temperature of 5,500 - 6,500 and blue white in character, household lights are about 2,800 kelvin and orange or yellow in character.

Now, extrapolate. That's the only way you'll learn.

Orange yellow to white blue.

Use your brain

That's why God gave it to you!

Look at the Electro Magnetic Spectrum

Orange-Yellow to White Blue

Extrapolate what 30,000 and 21,500 Kelvin would look like (or not look like, as the case may be and that's a hint) based upon the two Kelvins I have provided.

2007-10-14 12:33:45 · answer #2 · answered by Anonymous · 0 0

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