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Re: Astronomy
Posted: March 8th, 2009, 6:44 pm
by herewegoagain365
Glacierguy1 wrote: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.)
Re: Astronomy
Posted: March 9th, 2009, 11:58 am
by EastStroudsburg13
Flavorflav wrote: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!
Re: Astronomy
Posted: March 9th, 2009, 8:33 pm
by celtics09
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
Re: Astronomy
Posted: March 10th, 2009, 2:39 am
by Flavorflav
Flavorflav wrote:Glacierguy1 wrote: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?
Re: Astronomy
Posted: March 10th, 2009, 4:56 pm
by celtics09
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
Thanks
Re: Astronomy
Posted: March 11th, 2009, 2:46 am
by Flavorflav
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.
Re: Astronomy
Posted: March 11th, 2009, 5:11 am
by celtics09
Thanks
I almost got it. What does the 'c' represent?
The speed of light, if not, what other value
Re: Astronomy
Posted: March 11th, 2009, 6:50 am
by Flavorflav
It means speed of light.
Re: Astronomy
Posted: March 11th, 2009, 8:58 am
by celtics09
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
Re: Astronomy
Posted: March 11th, 2009, 6:36 pm
by Flavorflav
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.