## Astronomy C

PM2017
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### Re: Astronomy C

I just realized that editing a pre-existing post will not give notifications, so here is this post booster...
West High '19
UC Berkeley '23

Go Bears!

Knyte_Xjn
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### Re: Astronomy C

PM2017 wrote:A student at West High is angry he lost the tryouts for astronomy to PM2017. For this reason, he decides to throw PM2017's laptop into a black hole and seeing as he is actually better than PM2017 (and was unfairly disfavoured!) somehow observes that the laptop is, for some reason a perfect blackbody, and is normally blue (assume 470.00 nm). As the laptop passes through the event horizon, the student observes that the laptop seems to have a wavelength of 933.89 nm.

The student recalls from his studying that the mass of a black hole and the radius of its event horizon are related linearly. He dejectedly realizes that he can now use the wavelength data to find this relationship, but is too depressed to do so. Can you find the relationship between the mass of the black hole, in solar masses, and radius of a black hole, in kilometers?

Assume that G = 6.67e-11, c = 2.99e8, and a solar mass is 2.00e30 kg.

`V = sqrt((2*G*M)/R)`
```V = sqrt((2*G*M)/R)
Since you are assuming that a linear relationship exists between mass and radius, substitute M*k for R.
V = sqrt((2*G*M)/(M*k)) = sqrt((2*G)/k)
vrec = λapp/λtrue - 1
vrec = c * (933.89nm / 470nm - 1) ~ 2.95*10^8 m/s
2.95*10^8 m/s= sqrt((2*G)/k)
vrec^2 = 8.71*10^16 m^2/s^2 = (2*G)/k
8.71*10^16 m^2/s^2 = (1.33*10-10 kg m^3 ^-2)/k
8.71*10^16  = 1.33*10^-10 m/kg / k
k = 1.33*10^-10m/kg/8.71*10^16 ~ 1.53*10^-27 m/kg
1.53*10^-27 m/kg * 1km/1000m * 2.00*10^30kg/1 Msol
~ 3.06 km/Msol
R = 3.06 km/Msol * M```
R. I. P. 01/20/2019

PM2017
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### Re: Astronomy C

Knyte_Xjn wrote:
PM2017 wrote:A student at West High is angry he lost the tryouts for astronomy to PM2017. For this reason, he decides to throw PM2017's laptop into a black hole and seeing as he is actually better than PM2017 (and was unfairly disfavoured!) somehow observes that the laptop is, for some reason a perfect blackbody, and is normally blue (assume 470.00 nm). As the laptop passes through the event horizon, the student observes that the laptop seems to have a wavelength of 933.89 nm.

The student recalls from his studying that the mass of a black hole and the radius of its event horizon are related linearly. He dejectedly realizes that he can now use the wavelength data to find this relationship, but is too depressed to do so. Can you find the relationship between the mass of the black hole, in solar masses, and radius of a black hole, in kilometers?

Assume that G = 6.67e-11, c = 2.99e8, and a solar mass is 2.00e30 kg.

`V = sqrt((2*G*M)/R)`
```V = sqrt((2*G*M)/R)
Since you are assuming that a linear relationship exists between mass and radius, substitute M*k for R.
V = sqrt((2*G*M)/(M*k)) = sqrt((2*G)/k)
vrec = λapp/λtrue - 1
vrec = c * (933.89nm / 470nm - 1) ~ 2.95*10^8 m/s
2.95*10^8 m/s= sqrt((2*G)/k)
vrec^2 = 8.71*10^16 m^2/s^2 = (2*G)/k
8.71*10^16 m^2/s^2 = (1.33*10-10 kg m^3 ^-2)/k
8.71*10^16  = 1.33*10^-10 m/kg / k
k = 1.33*10^-10m/kg/8.71*10^16 ~ 1.53*10^-27 m/kg
1.53*10^-27 m/kg * 1km/1000m * 2.00*10^30kg/1 Msol
~ 3.06 km/Msol
R = 3.06 km/Msol * M```
West High '19
UC Berkeley '23

Go Bears!

Name
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### Re: Astronomy C

PM2017 wrote:A student at West High is angry he lost the tryouts for astronomy to PM2017. For this reason, he decides to throw PM2017's laptop into a black hole and seeing as he is actually better than PM2017 (and was unfairly disfavoured!) somehow observes that the laptop is, for some reason a perfect blackbody, and is normally blue (assume 470.00 nm). As the laptop passes through the event horizon, the student observes that the laptop seems to have a wavelength of 933.89 nm.

The student recalls from his studying that the mass of a black hole and the radius of its event horizon are related linearly. He dejectedly realizes that he can now use the wavelength data to find this relationship, but is too depressed to do so. Can you find the relationship between the mass of the black hole, in solar masses, and radius of a black hole, in kilometers?

Assume that G = 6.67e-11, c = 2.99e8, and a solar mass is 2.00e30 kg.

`V = sqrt((2*G*M)/R)`
Assuming you don't have to use the given values/equations
```You can simplify r=2GM/c^2
2G/c^2= 1.49e-27
1.49e-27*M= r in terms of kg and meters
Convert that in terms of solar masses and km
1.49e-27/1000*2e30 = 2.98
Radius (in km) = 2.98*Mass (in solar masses)```
Syosset HS '21
Favorite Past Events: Microbe, Invasive, Matsci, Fermi
Events!: Astro, Code, Fossils
```Cornell/LISO/Duke/MIT/Regionals/Brown/SOUP
Astro: 2/4/2/10/2/2/3
Code: 9/1/2/1/1/1/1
Fossils: 16/13/6/9/2/3/8
1 fermi cornell, 10 orni LISO, 4 green gen regionals```

PM2017
Member Posts: 520
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State: CA
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### Re: Astronomy C

Name wrote:
PM2017 wrote:A student at West High is angry he lost the tryouts for astronomy to PM2017. For this reason, he decides to throw PM2017's laptop into a black hole and seeing as he is actually better than PM2017 (and was unfairly disfavoured!) somehow observes that the laptop is, for some reason a perfect blackbody, and is normally blue (assume 470.00 nm). As the laptop passes through the event horizon, the student observes that the laptop seems to have a wavelength of 933.89 nm.

The student recalls from his studying that the mass of a black hole and the radius of its event horizon are related linearly. He dejectedly realizes that he can now use the wavelength data to find this relationship, but is too depressed to do so. Can you find the relationship between the mass of the black hole, in solar masses, and radius of a black hole, in kilometers?

Assume that G = 6.67e-11, c = 2.99e8, and a solar mass is 2.00e30 kg.

`V = sqrt((2*G*M)/R)`
Assuming you don't have to use the given values/equations
```You can simplify r=2GM/c^2
2G/c^2= 1.49e-27
1.49e-27*M= r in terms of kg and meters
Convert that in terms of solar masses and km
1.49e-27/1000*2e30 = 2.98
Radius (in km) = 2.98*Mass (in solar masses)```
I feel rather silly now. Sorry Knyte and all others who tried fro leading you in the wrong way. The worst part is, I've used this derivation before, and completely forgot...
West High '19
UC Berkeley '23

Go Bears!

Knyte_Xjn
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### Re: Astronomy C

PM2017 wrote:
Knyte_Xjn wrote:
PM2017 wrote:A student at West High is angry he lost the tryouts for astronomy to PM2017. For this reason, he decides to throw PM2017's laptop into a black hole and seeing as he is actually better than PM2017 (and was unfairly disfavoured!) somehow observes that the laptop is, for some reason a perfect blackbody, and is normally blue (assume 470.00 nm). As the laptop passes through the event horizon, the student observes that the laptop seems to have a wavelength of 933.89 nm.

The student recalls from his studying that the mass of a black hole and the radius of its event horizon are related linearly. He dejectedly realizes that he can now use the wavelength data to find this relationship, but is too depressed to do so. Can you find the relationship between the mass of the black hole, in solar masses, and radius of a black hole, in kilometers?

Assume that G = 6.67e-11, c = 2.99e8, and a solar mass is 2.00e30 kg.

`V = sqrt((2*G*M)/R)`
```V = sqrt((2*G*M)/R)
Since you are assuming that a linear relationship exists between mass and radius, substitute M*k for R.
V = sqrt((2*G*M)/(M*k)) = sqrt((2*G)/k)
vrec = λapp/λtrue - 1
vrec = c * (933.89nm / 470nm - 1) ~ 2.95*10^8 m/s
2.95*10^8 m/s= sqrt((2*G)/k)
vrec^2 = 8.71*10^16 m^2/s^2 = (2*G)/k
8.71*10^16 m^2/s^2 = (1.33*10-10 kg m^3 ^-2)/k
8.71*10^16  = 1.33*10^-10 m/kg / k
k = 1.33*10^-10m/kg/8.71*10^16 ~ 1.53*10^-27 m/kg
1.53*10^-27 m/kg * 1km/1000m * 2.00*10^30kg/1 Msol
~ 3.06 km/Msol
R = 3.06 km/Msol * M```
Extremely sorry for the extremely late post!
1. (a) What is the minimum speed required for a celestial object to free itself from the gravitational attraction of a massive object?
(b) You observe a planet orbiting a massive star with a mass of 74 Msol. Calculate the value mentioned in 1a for this planet to derail from its orbit and move away from the star's gravitational influence given that the radius of the orbit is 3.4 AU.
(c) An object given ________(answer to 1a) will continually move away from the other object, its speed __________ approaching _________. (fill in the blanks)
R. I. P. 01/20/2019

Name
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### Re: Astronomy C

Knyte_Xjn wrote:
Extremely sorry for the extremely late post!
1. (a) What is the minimum speed required for a celestial object to free itself from the gravitational attraction of a massive object?
(b) You observe a planet orbiting a massive star with a mass of 74 Msol. Calculate the value mentioned in 1a for this planet to derail from its orbit and move away from the star's gravitational influence given that the radius of the orbit is 3.4 AU.
(c) An object given ________(answer to 1a) will continually move away from the other object, its speed __________ approaching _________. (fill in the blanks)
```1. Escape velocity
2. sqrt(2GM/r) is sqrt(2*6.67e-11*1.48e32kg/5.1e11m) = sqrt(3.87e10)=196754 m/s
3. a object given [u]escape velocity[/u] will continually move away from the other object, its speed [u]gradually[/u] approaching [u]0[/u]```
Syosset HS '21
Favorite Past Events: Microbe, Invasive, Matsci, Fermi
Events!: Astro, Code, Fossils
```Cornell/LISO/Duke/MIT/Regionals/Brown/SOUP
Astro: 2/4/2/10/2/2/3
Code: 9/1/2/1/1/1/1
Fossils: 16/13/6/9/2/3/8
1 fermi cornell, 10 orni LISO, 4 green gen regionals```

Knyte_Xjn
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### Re: Astronomy C

Name wrote:
Knyte_Xjn wrote:
Extremely sorry for the extremely late post!
1. (a) What is the minimum speed required for a celestial object to free itself from the gravitational attraction of a massive object?
(b) You observe a planet orbiting a massive star with a mass of 74 Msol. Calculate the value mentioned in 1a for this planet to derail from its orbit and move away from the star's gravitational influence given that the radius of the orbit is 3.4 AU.
(c) An object given ________(answer to 1a) will continually move away from the other object, its speed __________ approaching _________. (fill in the blanks)
```1. Escape velocity
2. sqrt(2GM/r) is sqrt(2*6.67e-11*1.48e32kg/5.1e11m) = sqrt(3.87e10)=196754 m/s
3. a object given [u]escape velocity[/u] will continually move away from the other object, its speed [u]gradually[/u] approaching [u]0[/u]```
Parts a and b are correct, but for part c, the second blank should be asymptotically approaching 0 to emphasize that the object's speed will never actually be 0. Your turn! R. I. P. 01/20/2019

Name
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### Re: Astronomy C

1. A star with the same radius and twice the temperature of our sun is found. What is its luminosity (in solar luminosity)?
2. The parallax of a type 1a supernova was found to be .05 milliarcsec. What its distance (in parsecs)?
3. What is the supernovas apparent magnitude?
4. The apparent magnitude of the star in part 1 and the supernova in part 2/3 is the same. What is the distance of the star?
Syosset HS '21
Favorite Past Events: Microbe, Invasive, Matsci, Fermi
Events!: Astro, Code, Fossils
```Cornell/LISO/Duke/MIT/Regionals/Brown/SOUP
Astro: 2/4/2/10/2/2/3
Code: 9/1/2/1/1/1/1
Fossils: 16/13/6/9/2/3/8
1 fermi cornell, 10 orni LISO, 4 green gen regionals```

PM2017
Member Posts: 520
Joined: January 20th, 2017, 5:02 pm
State: CA
Location: Enjoying College! :D
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### Re: Astronomy C

Name wrote:1. A star with the same radius and twice the temperature of our sun is found. What is its luminosity (in solar luminosity)?
2. The parallax of a type 1a supernova was found to be .05 milliarcsec. What its distance (in parsecs)?
3. What is the supernovas apparent magnitude?
4. The apparent magnitude of the star in part 1 and the supernova in part 2/3 is the same. What is the distance of the star?
```[strikethrough]1. 8 (stefan boltzmann)
2. 20,000 (1000/parallax)
3. -2.8 (distance modulus; abs mag is -19.3)
4. really close, or 0.859 parsecs (Abs mag = 2.53, because the 2.5th root of 8 is ~2.30 4.83 - that is 2.53; apparent magnitude is -2.83)[/strikethrough]```
Sorry, I was being lazy and did all this on a computer and no paper or pencil, and I just felt lazy, so lazy in fact that I decided to make this sentence a run on sentence.

EDIT: OK, so I came back later and wow I messed everything up. The hide functionality has now been fixed. Apparently, I thought that 2^4 = 8 (???). Thirdly, my equation seems to have been wrong. I have no idea what I was doing...
West High '19
UC Berkeley '23

Go Bears!

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