Astronomy

herewegoagain365
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Re: Astronomy

Postby herewegoagain365 » March 8th, 2009, 6:44 pm

At my state test last year, which used a quite weird cartoon theme, said that Cruella de ville found this spot pattern on a dalmation and noted that it looked like a constellation, she also recognized [one of the DSOs]. What constellation was it. They also asked it in reverse, giving us the constellation name and asking which DSO was in it.

They usually only ask you to identify the DSOs from pictures and what constellation the DSOs are in.
Yeah, well, there are always those weird ones. We got a couple of questions at regionals asking what the names of constellations and stars mean. ("What does the word 'Mira' mean?", "What does 'Circinus' mean?", etc.)

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EastStroudsburg13
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Re: Astronomy

Postby EastStroudsburg13 » March 9th, 2009, 11:58 am

Theoretically no, although there3 is some DSO identification, and you may have to ID stars from light curves. Also, sometimes you get an event writer who doesn't follow specs - then, who knows what you will see.
I thought so...That makes me sad, I like identifying constellations and stars! :cry:
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celtics09
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Re: Astronomy

Postby celtics09 » March 9th, 2009, 8:33 pm

I'm still confused about #16 :?:

How do you go from knowing the orbital velocity (in km/s) and wavelengths of a star to the angle of line of sight.

This all seems to complicated. Could someone please explain it in a smooth and comprehensible way.
Thanks
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Re: Astronomy

Postby Flavorflav » March 10th, 2009, 2:39 am

Please explain.
You have actual V from #15, so you just find apparent V from redshift. Apparent V is going to be Vaway or Vx, so Vx/v = cos inclination. I think.
Which part are you having trouble with?

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Re: Astronomy

Postby celtics09 » March 10th, 2009, 4:56 pm

I'm having trouble with the derivation of wavelengths to the calculation of the Apparent magnitude.
I'm not understanding the usage of the formula

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Re: Astronomy

Postby Flavorflav » March 11th, 2009, 2:46 am

What apparent magnitude? You use the redshift formula, z=v/c, where z is the deviation in wavelength. This gives you the velocity away from you, then you do the trig to determine angle.

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Re: Astronomy

Postby celtics09 » March 11th, 2009, 5:11 am

Thanks
I almost got it. What does the 'c' represent?

The speed of light, if not, what other value
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Re: Astronomy

Postby Flavorflav » March 11th, 2009, 6:50 am

It means speed of light.

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Re: Astronomy

Postby celtics09 » March 11th, 2009, 8:58 am

I'm not getting the right answer


what i have so far is,

z = v/c

656.5386 nm - 656.3000 nm = .2386 nm

3E8*.2386 = Aparent Velocity

My original velocity from #15 is 218.2 km/s


Apparent Velocity/ Velocity = .0003


cos inverse of this value yields 89 degrees

not quite the answer


What am I doing wrong
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Re: Astronomy

Postby Flavorflav » March 11th, 2009, 6:36 pm

More than one thing. You have to divide the difference in wavelength by the emitted wavelength before you multiply by C - you want the proportionate change in wavelength, not absolute. Also, I think you should have got .003 with your numbers, not .0003. I think you may have dropped a decimal when you convert m/s to km/s.


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