Astronomy C
 Unome
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Re: Astronomy C
1. What DSO does this data come from?
2. What spectral regions are each of these graphs showing?
3. This graph shows two phases. Based on the graph, over how many phases was data actually collected?
4. The bottom graph is phaseshifted from the other two. Considering the properties of the system that produced this data, why is this so?

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Re: Astronomy C
PM2017 wrote:Yeah, I'm sorry, those were some really bad questions... I don't know what exactly I was thinking.Unome wrote:1. I ended up with 1.7E29 W, no idea if it works though. I used P = 0.5(pdot)(mass)(initial velocity squared), where I computed the initial velocity from v = 2(pi)(r). 2. Calculating via L = (m)(r^2)(1/p) since angular velocity = 1/p, I got p_initial = 28935 days. 3. Rotational period =/= pulsation period so... idk
Just ignore myvquestions and just continue the thread (although for number 1. and therefore 2. I got vastly different answers, which I will post when I get home.)
Again, apologizing for those terrible questions.
[math]E = \frac{1}{2}I\omega^2[/math] [math]E = \frac{1}{5}mr^2 * \frac{1}{p^2}[/math] [math]\frac{dE}{dt} = \frac{1}{5} mr^2 (\frac{2}{p^3} \frac{dp}{dt})[/math] [math]dE = \frac{2mr^2}{5p^3}dp[/math] [math]dE = \frac{2mr^2}{5p^3} \dot{P}[/math] Then plug in values and I got 6.2676E30 Watts
Last edited by jonboyage on April 18th, 2018, 8:20 am, edited 2 times in total.
I was in a bin
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 PM2017
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Re: Astronomy C
That's how I did it.jonboyage wrote:PM2017 wrote:Yeah, I'm sorry, those were some really bad questions... I don't know what exactly I was thinking.Unome wrote:1. I ended up with 1.7E29 W, no idea if it works though. I used P = 0.5(pdot)(mass)(initial velocity squared), where I computed the initial velocity from v = 2(pi)(r). 2. Calculating via L = (m)(r^2)(1/p) since angular velocity = 1/p, I got p_initial = 28935 days. 3. Rotational period =/= pulsation period so... idk
Just ignore myvquestions and just continue the thread (although for number 1. and therefore 2. I got vastly different answers, which I will post when I get home.)
Again, apologizing for those terrible questions.[math]E = \frac{1}{2}Iw^2[/math] [math]E = \frac{1}{2}mr^2 * \frac{1}{p^2}[/math] [math]\frac{dE}{dt} = \frac{1}{2} mr^2 (\frac{2}{p^3} \frac{dp}{dt})[/math] [math]dE = \frac{mr^2}{p^3}dp[/math] [math]dE = \frac{mr^2}{p^3} \dot{P}[/math] Then plug in values and I got 1.5669E31 Watts
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Re: Astronomy C
[hide]
1) PM2017: looks good
2) yep: d2 = d1 * sqrt(2.512^m2m1) => d2 = 34.16 pc.
3) This question was kind of obscure and probably won't come up this year (maybe not ever on a scio astro test)
I should have mentioned that the GMC was in virial equilibrium.
The way I did it was: using linewidthsize correlation you can get dispersion velocity:
(dispersion velocity) = (0.72pm0.03)(R/pc)^(0.5pm0.05) (km/s)
= 9.44 km/s
Then use virial theorem:
GM/R = (dispersion velocity)^2
M = (R * (dispersion velocity)^2) / G
= ((9.46e17 m)*(9440 m s^1)^2) / (6.67e11 m^3 kg^1 s^2)
= ((9.46e17)*(8.9e7) m^3 s^2) / (6.67e11 m^3 kg^1 s^2)
= ((9.46e17)*(8.9e7) kg) / (6.67e11)
= 1.26e36 kg
= 1.26e36 / 1.9891 * 10^30 Msun
= 6.346e5 Msun
[/hide]
1) PM2017: looks good
2) yep: d2 = d1 * sqrt(2.512^m2m1) => d2 = 34.16 pc.
3) This question was kind of obscure and probably won't come up this year (maybe not ever on a scio astro test)
I should have mentioned that the GMC was in virial equilibrium.
The way I did it was: using linewidthsize correlation you can get dispersion velocity:
(dispersion velocity) = (0.72pm0.03)(R/pc)^(0.5pm0.05) (km/s)
= 9.44 km/s
Then use virial theorem:
GM/R = (dispersion velocity)^2
M = (R * (dispersion velocity)^2) / G
= ((9.46e17 m)*(9440 m s^1)^2) / (6.67e11 m^3 kg^1 s^2)
= ((9.46e17)*(8.9e7) m^3 s^2) / (6.67e11 m^3 kg^1 s^2)
= ((9.46e17)*(8.9e7) kg) / (6.67e11)
= 1.26e36 kg
= 1.26e36 / 1.9891 * 10^30 Msun
= 6.346e5 Msun
[/hide]
 c0c05w311y
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Re: Astronomy C
1. DEM L241 2. From top to bottom, radio, xray, gamma 3. 2? I'm basing this on how many dots appear to be on top of each other. However I'm really not sure if this is how the question was supposed to be answered. 4. The bottom graph is gamma, and the system is an xray/gamma binary. So the gamma emissions come from the acceleration of particles around the neutron star, and the periodicity is due to its orbit (more particles in the area when its closer to the companion in its elliptical orbit). However I'm blanking on how there would be a phase shifted emission in radio and xray. Can someone explain?
 Unome
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Re: Astronomy C
Mostly correct. For 3, you could tell by reading the source, but the easiest way is to notice that both phases have exactly identical data, MEs, etc.  it's one phase of data, doubled to make it easier to interpret. Your turn.
 Adi1008
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Re: Astronomy C
Starting this up again:
An expanding ring of material from a supernova appears to be shaped as an ellipse due to its inclination with Earth (assume that the ring, in reality, expanded uniformly so it is actually circular). Its major axis is twice as long as its minor axis. What is the inclination of the ring?
An expanding ring of material from a supernova appears to be shaped as an ellipse due to its inclination with Earth (assume that the ring, in reality, expanded uniformly so it is actually circular). Its major axis is twice as long as its minor axis. What is the inclination of the ring?
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 PM2017
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Re: Astronomy C
Aren't there two potential answers (30 degrees or 60 degrees)Adi1008 wrote:Starting this up again:
An expanding ring of material from a supernova appears to be shaped as an ellipse due to its inclination with Earth (assume that the ring, in reality, expanded uniformly so it is actually circular). Its major axis is twice as long as its minor axis. What is the inclination of the ring?
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 Adi1008
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Re: Astronomy C
PM2017 wrote:Aren't there two potential answers (30 degrees or 60 degrees)Adi1008 wrote:Starting this up again:
An expanding ring of material from a supernova appears to be shaped as an ellipse due to its inclination with Earth (assume that the ring, in reality, expanded uniformly so it is actually circular). Its major axis is twice as long as its minor axis. What is the inclination of the ring?
It depends on where the angle is being measured from, which I forgot to specify T_T
If the angle between the plane in which the ring lies is 30 degrees with the horizontal or 60 degrees with the vertical it'd be correct
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Re: Astronomy C
Starting this up for next year's topic:
Suppose an astronomer discovers an elliptical galaxy whose major axis is 2 times longer than its minor axis.
a. What would the classification of this galaxy be?
b. What assumption do you make when classifying the galaxy about its orientation when viewed from Earth?
Suppose an astronomer discovers an elliptical galaxy whose major axis is 2 times longer than its minor axis.
a. What would the classification of this galaxy be?
b. What assumption do you make when classifying the galaxy about its orientation when viewed from Earth?
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