## Astronomy C

PM2017
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State: CA

### Re: Astronomy C

1. W49B
a. Iron is only in half the remnant, and other things including silicon and sulfur were evenly spread. Also,  It shows x-ray emission from chromium and manganese.
b. The SN is type 1b/1c probably, which means the progenitor was probably an LBV/WR with a lot of mass loss. Wikipedia says the progenitor was probably around 25 solar masses.
c. Given a period of 25 days, we can calculate the absolute magnitude using the period luminosity relationship. M = -2.8log(25 days) - 1.43 gives an absolute magnitude of M = -5.34. Using m - (-5.34) = -5+5log(7972 pc) gives m = 9.17.

2. HR 5171A
a. B, C, F, i think
b.  its currently a yellow hypergiant, which is a pretty unstable state that is probably between red supergiant and blue supergiant stages, which explains why we see so few of them. I don't know if HR5171A is moving towards a bluer or redder stage, but it could become a blue supergiant, a red supergiant, or (very unlikely, it might be able to go supernova directly). It could shed it's outer envelope and become a WR binary. Eventually it will explode in a type 2 (probably P but it depends on how much mass loss in later stages) or type 1b/1c (especially if there is more mass loss such as if it enters a WR stage) supernova, and probably leave behind a black hole.
c. 1315 to 1575 solar radii

3. Geminga
a. gamma
b. 273 ms
c. 13.51

By the way, we should do something where we all log on and rapid fire this stuff, one after another. How does 1/21/18, at 3 PM PST (6 PM eastern) sound?
West High '19
UC Berkeley '23

Go Bears!

c0c05w311y
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Location: Carnegie Mellon University

### Re: Astronomy C

Question: What does this graph describe? Explain why the Gamov peak exists for nuclear fusion.

(Sorry that this question isn't the greatest but I wanted to post something before going to sleep. Btw, I should be available at that time for the rapid-fire session! I think its a great idea)

EDIT: i just noticed that this is usually spelled Gamow. I don't know if Gamov was a typo or an alternate spelling.
Last edited by c0c05w311y on January 23rd, 2018, 3:01 am, edited 1 time in total.

PM2017
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Posts: 491
Joined: January 20th, 2017, 5:02 pm
State: CA

### Re: Astronomy C

Question: What does this graph describe? Explain why the Gamov peak exists for nuclear fusion.

(Sorry that this question isn't the greatest but I wanted to post something before going to sleep. Btw, I should be available at that time for the rapid-fire session! I think its a great idea)

EDIT: i just noticed that this is usually spelled Gamow. I don't know if Gamov was a typo or an alternate spelling.
I can't access this link from my school because its imgur. Can we find a different image sharing website?
West High '19
UC Berkeley '23

Go Bears!

c0c05w311y
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Joined: February 20th, 2017, 7:56 pm
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Location: Carnegie Mellon University

### Re: Astronomy C

I edited my post using an imgbb link, does that work for you?

PM2017
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Posts: 491
Joined: January 20th, 2017, 5:02 pm
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### Re: Astronomy C

I edited my post using an imgbb link, does that work for you?
It definitely does, thank you!
West High '19
UC Berkeley '23

Go Bears!

PM2017
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Posts: 491
Joined: January 20th, 2017, 5:02 pm
State: CA

### Re: Astronomy C

Question: What does this graph describe? Explain why the Gamov peak exists for nuclear fusion.

(Sorry that this question isn't the greatest but I wanted to post something before going to sleep. Btw, I should be available at that time for the rapid-fire session! I think its a great idea)

EDIT: i just noticed that this is usually spelled Gamow. I don't know if Gamov was a typo or an alternate spelling.
Lol, here we go again. Time for another sloppy answer in an attempt to revive this thread. (I'll get back to this point in a moment)

This graph represents the Gamow window and the Gamow peak.

With classical physics, fusion would occur practically never, due to the Coulomb barrier. However, quantum tunneling allows atoms to fuse. I won't pretend to fully understand Gamow peak here, but there is essentially a drop-off in Maxwellian distribution, and tunneling, but in opposite directions. The Gamow peak exists at a point where both are very low.
I honestly don't know exactly why it exists, so this may be the first incorrect answer on this forum, but I needed to revive the thread.

Speaking of which, I say we implement a point system, where people get points for correctly answered questions.
I propose 15 points per question, if answered within 24 hrs, 10 points if in 48 hrs, 7 if in 72, 5 if in 96, 3 if in a week (168 hrs), and 1 anytime after that.
I'll make a google sheet, and keep it updated (I might ask for people to PM accounts so I can share with you guys, so updating continues happening)
West High '19
UC Berkeley '23

Go Bears!

c0c05w311y
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Posts: 19
Joined: February 20th, 2017, 7:56 pm
State: PA
Location: Carnegie Mellon University

### Re: Astronomy C

Thats a good explanation for quantum tunneling, but im not quite sure what you mean by "in opposite directions" and where "both are very low".

For quantum tunneling you got it: it allows particles to get close enough to fuse while their thermal energy doesn't under classical physics.
Probability to fuse is dependent on energy of the particles, their mass (easier to fuse lighter elements), and their charge (cause its about energy). The gamov peak exists because, although atoms are much more likely to fuse at higher energies, there are also far fewer particles at those higher energies. So the gamow peak is simply the narrow range of energies where a significant amount of particles fuse due to overlap between the two lines on the graph.

Sorry that this isn't really the kind of question you would expect to see on a test, but I like explanation questions because they are good at helping people learn things. This is a semi-obscure topic, but I thought it was pretty cool.

PM2017
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State: CA

### Re: Astronomy C

I really hope this isn't just a thing between c0c0 and me...

Anyways...

A traveler 29 km above the singularity of a black hole is falling into the black hole, In desperation, he sends out an SOS signal in radio waves. Although the signal reaches his planet, he does not escape the black hole. What is the greatest possible mass of the black hole?

Given this mass, how much longer will this black hole exist before succumbing to Hawking radiation?
West High '19
UC Berkeley '23

Go Bears!

jonboyage
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### Re: Astronomy C

I really hope this isn't just a thing between c0c0 and me...

Anyways...

A traveler 29 km above the singularity of a black hole is falling into the black hole, In desperation, he sends out an SOS signal in radio waves. Although the signal reaches his planet, he does not escape the black hole. What is the greatest possible mass of the black hole?

Given this mass, how much longer will this black hole exist before succumbing to Hawking radiation?
I'll take a stab at it
Part 1:

$c=\sqrt{\frac{2GM}{R}} \Rightarrow M=\frac{c^2R}{2G}=\frac{(3*10^8)^2*29000}{2*6.67*10^{-11}}=1.96*10^{31}kg=9.84M_{\odot}$

Part 2:

$T_{bh}=\frac{\hbar c^3}{8\pi kGM}=6.13*10^{-8}(\frac{M_\odot}{M})$

$\frac{dE}{dt}=L_{bh}=4\pi R_{s}^2\sigma T_{bh}^4=4\pi R_{s}^2\sigma (\frac{\hbar c^3}{8\pi kGM})^4=4\pi R_{s}^2\sigma (6.13*10^{-8}(\frac{M_\odot}{M}))^4$

$E=Mc^2 \Rightarrow M=\frac{E}{c^2} \Rightarrow \frac{dM}{dt}=\frac{1}{c^2} \frac{dE}{dt}=\frac{L_{bh}}{c^2}=\frac{4\pi R_{s}^2\sigma (6.13*10^{-8}(\frac{M_\odot}{M}))^4}{c^2}=1.037*10^{-47}\frac{kg}{s}$
(very very very slow)

$\frac{1.96*10^{31}kg}{1.037*10^{-47}\frac{kg}{s}}=t=1.89*10^{78}s=5.99*10^{70}yr$
I was in a bin

Rustin '19
UPenn '23

PM2017
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Posts: 491
Joined: January 20th, 2017, 5:02 pm
State: CA

### Re: Astronomy C

I really hope this isn't just a thing between c0c0 and me...

Anyways...

A traveler 29 km above the singularity of a black hole is falling into the black hole, In desperation, he sends out an SOS signal in radio waves. Although the signal reaches his planet, he does not escape the black hole. What is the greatest possible mass of the black hole?

Given this mass, how much longer will this black hole exist before succumbing to Hawking radiation?
I'll take a stab at it
Part 1:

$c=\sqrt{\frac{2GM}{R}} \Rightarrow M=\frac{c^2R}{2G}=\frac{(3*10^8)^2*29000}{2*6.67*10^{-11}}=1.96*10^{31}kg=9.84M_{\odot}$

Part 2:

$T_{bh}=\frac{\hbar c^3}{8\pi kGM}=6.13*10^{-8}(\frac{M_\odot}{M})$

$\frac{dE}{dt}=L_{bh}=4\pi R_{s}^2\sigma T_{bh}^4=4\pi R_{s}^2\sigma (\frac{\hbar c^3}{8\pi kGM})^4=4\pi R_{s}^2\sigma (6.13*10^{-8}(\frac{M_\odot}{M}))^4$

$E=Mc^2 \Rightarrow M=\frac{E}{c^2} \Rightarrow \frac{dM}{dt}=\frac{1}{c^2} \frac{dE}{dt}=\frac{L_{bh}}{c^2}=\frac{4\pi R_{s}^2\sigma (6.13*10^{-8}(\frac{M_\odot}{M}))^4}{c^2}=1.037*10^{-47}\frac{kg}{s}$
(very very very slow)

$\frac{1.96*10^{31}kg}{1.037*10^{-47}\frac{kg}{s}}=t=1.89*10^{78}s=5.99*10^{70}yr$
I don't have my binder on me, and I don't have much time to check either, but it looks correct. I'll tell you the definitive answer later, but feel free to continue the marathon.
West High '19
UC Berkeley '23

Go Bears!