Astronomy Problem

Star D has a spectral calss of M2, and star E has a spectral class of B8. How many more times energy per unit surface area does one have over the other?

Originally posted by abejnood
Star D has a spectral calss of M2, and star E has a spectral class of B8. How many more times energy per unit surface area does one have over the other?

Well if the spectral class of m2 is PIx3 then that would make it 4x greater than the spectral class of
b8. So by multiplying the amount of energy per unit would be equal to the amount of the spectral class b8, but still less than the energy given off by, say, M2. The heat factor also comes into consideration which can lower the level of Star D SPECTRAL CLASS M2.

B8 has about 39 times more. This question was in this year's National Science Olympiad.

Originally posted by Balla88
B8 has about 39 times more. This question was in this year's National Science Olympiad.

No, it was in last years at University of Illinois at Urbana Champain. I don't care about the answer, i already knew it, I just want to know HOW they get 39? I can't figure it out. I got the other problems, just not that one.

That is where they held the National Science Olympiad.

EDIT: I didn't mean to say 2006.

Originally posted by abejnood
No, it was in last years at University of Illinois at Urbana Champain. I don't care about the answer, i already knew it, I just want to know HOW they get 39? I can't figure it out....

They probably just ask a brainy physicist like Mr. Trains for the answer. Barring that, they might calculate it based on some assumptions about the surface temperature, similar to....

http://en.wikipedia.org/wiki/Stefan-Boltzmann_law

Originally posted by leisurelysloth
They probably just ask a brainy physicist like Mr. Trains for the answer. Barring that, they might calculate it based on some assumptions about the surface temperature, similar to....

http://en.wikipedia.org/wiki/Stefan-Boltzmann_law

😏 Thank you. 😏

Originally posted by Trains44
😏 Thank you. 😏

So how do they come up with 39X? I did a quick look at spectral
types on goo gal but nothing popped up specifically, except the
old O B A F G K M (Oh Be A Fine Girl, Kiss Me) thing I learned in
8th grade.

Originally posted by Trains44
😏 Thank you. 😏

You're welcome.

Originally posted by sonhouse
So how do they come up with 39X? I did a quick look at spectral
types on goo gal but nothing popped up specifically, except the
old O B A F G K M (Oh Be A Fine Girl, Kiss Me) thing I learned in
8th grade.

Yeah, that's what's the problem with me, too. I can't figure out how they get 39. Why 39? Why not 42? I don't get it 😞

Originally posted by abejnood
Yeah, that's what's the problem with me, too. I can't figure out how they get 39. Why 39? Why not 42? I don't get it 😞

Not 42, thats the answer to EVERYTHING!🙂
Another thing about the 39 times, it seems a bit low since the two
stars are almost opposite sides of the classes, 25,000 deg.
or so on one side V 4000 ish deg C on the other?

Betelgeuse is a B8 star. Rigel is a M2 star.

Betelgeuse has a radius of 1.4 au and a luminosity between 40,000 and 100,000 solars.
Rigel has a radius of 65 au and a luminosity around 50,000 solars.

Therefore Betelgeuse has a surface area of 5.5 × 10^17 km^2 and Rigel has a surface area of 1.2 × 10^21 km^2.

Therefore I get 2.2 * 10^3 as the ratio of energy per unit surface area.

No idea how they got that answer.

Originally posted by XanthosNZ
Betelgeuse is a B8 star. Rigel is a M2 star.

Betelgeuse has a radius of 1.4 au and a luminosity between 40,000 and 100,000 solars.
Rigel has a radius of 65 au and a luminosity around 50,000 solars.

Therefore Betelgeuse has a surface area of 5.5 × 10^17 km^2 and Rigel has a surface area of 1.2 × 10^21 km^2.

Therefore I get 2.2 * 10^3 as the ratio of energy per unit surface area.

No idea how they got that answer.

I looked at the data for Rigel and it shows mass of 17 in units called
M with a square O subscript, could not find the exact meaning but I
suspect it means 17 times of Sol and the size at 60 "R" with the same
subscript. Which I suspect is 60 times the RADIUS if Sol which would
vastly change your numbers. I have heard of big stars, but 65 AU
would make it about 12 BILLION miles across, I think thats patently
impossible. So my guess is its 60 or 65 times 704,000 KM (radius of
sol) or 45 million Km, a more reasonable answer.