Astronomy C

[Name]

Let's look at 3C273 and AGNs
1. What type of quasar is 3C273 (radio-loud or quiet)? What does being a radio-loud quasar mean? Which type of quasar is more common?
2. Some quasars create the illusion of multiple quasars in close proximity, yet is only one quasar. What causes this illusion?
3. Classify the following AGNs on the type of galaxy they are most likely to appear in- Seyfert 1, Seyfert 2, Quasars, Radio galaxies (Ex. Spiral Galaxy)
4. Seyfert 1 and Quasar AGNs are observed to have broad emission lines. What causes these emission lines?

Click to Show Answer

1. 3C 273 is radio-loud. Radio-loud quasars have high amounts of radio emission, but radio-quiet quasars are more common.
2. Gravitational lensing 
3. Seyfert 1 & 2 - spiral, Radio galaxies - elliptical, Quasars - elliptical?
4. The broad line region - fast-moving gas clouds which absorb & re-emit light from the accretion disk as broad emission lines because of Doppler broadening (due to their high speeds).

Yup, your turn

[AstroClarinet]

Let's see if I'm doing this cosmology stuff right.
1. What's the difference between the cosmological principle & the perfect cosmological principle? What evidence is there against the perfect cosmological principle?
2. If a galaxy has a comoving distance of 5.32 Mpc and had a proper distance of 3.02 Mpc when the observed light was emitted from it, find the redshift of the light.
3. Find the temperature of the CMB at the time the light was emitted from the galaxy in #3.

[nobodynobody]

Let's see if I'm doing this cosmology stuff right.
1. What's the difference between the cosmological principle & the perfect cosmological principle? What evidence is there against the perfect cosmological principle?
2. If a galaxy has a comoving distance of 5.32 Mpc and had a proper distance of 3.02 Mpc when the observed light was emitted from it, find the redshift of the light.
3. Find the temperature of the CMB at the time the light was emitted from the galaxy in #3.

1. THe cosmological principle is that the universe is homogenous and isotropic throughout space, while the perfect cosmological principle states that the universe is homogenous and isotropic through space as well as time. The evidence against that is anything related to the evolution of the universe.
2. 5.32/3.02 - 1 -> z = .76
3. a = 3.02/5.32 = 0.57
.57 * temperature = 2.7k(present day)
temperature = 4.73k
I hope I did that correctly

[AstroClarinet]

Let's see if I'm doing this cosmology stuff right.
1. What's the difference between the cosmological principle & the perfect cosmological principle? What evidence is there against the perfect cosmological principle?
2. If a galaxy has a comoving distance of 5.32 Mpc and had a proper distance of 3.02 Mpc when the observed light was emitted from it, find the redshift of the light.
3. Find the temperature of the CMB at the time the light was emitted from the galaxy in #3.

1. THe cosmological principle is that the universe is homogenous and isotropic throughout space, while the perfect cosmological principle states that the universe is homogenous and isotropic through space as well as time. The evidence against that is anything related to the evolution of the universe.
2. 5.32/3.02 - 1 -> z = .76
3. a = 3.02/5.32 = 0.57
.57 * temperature = 2.7k(present day)
temperature = 4.73k
I hope I did that correctly

Looks right to me!

[nobodynobody]

Hello!
In a distant system, there is a central star (Mass of 2.3 solar masses) with 2 exoplanets, each of negligible mass. Exoplanet 1 is orbiting from .5 AU away, while the other (exoplanet 2) is orbiting at 9.6 AU from the central star. Assume the orbits to be clockwise (when the right-hand rule direction is in the same direction as south) , circular, and coplanar. An intelligent lifeform (that dwells on exoplanet 2) decides they want to explore the other exoplanet. They have decided to take the most efficient transfer that would use the least amount of energy to get there.

1. What is this transfer called?
2. Assuming the spacecraft is free from the gravitational pull of the exoplanets (the only gravity is from the star, lol), what velocity would they need to travel at when they are at exoplanet 2?
3. How long would it take to reach exoplanet 1?
4. Turns out, this planet is almost identical to earth, the only difference being its distance from the star and its direction of orbit. It rotates the same way, has the same axial tilt, has the same poles, defines east/west the same way, and they use the exact same altitude-azimuth coordinate system that earth does! To the surprise of the aliens, launching at the exact moment of the equinox will give them a perfect amount of time to reach exoplanet1. There are observers at the precise location of the equinox, eager to watch the spaceship depart. When the central star is directly overhead, where in the sky should they look to find exoplanet 1 (altitude-azimuth coordinates)?
5. What cardinal direction will the spaceship launch in (from the perspective of the same observers)?

[astronomybuff]

Disclaimer: I SUCK with coordinates

1. Hohmann transfer

2. I believe this would just be the vis-viva equation, which can be derived by using energy principles. v^2=GM(2/r-1/a), where r is the distance from the orbiting body and a is the semimajor axis. Now, I have to be a bit careful. I think a is the average of the distances and r is 9.6. Just curious, is there a shortcut for AU or do you just have to convert? Anyway, plugging in I got 14, 938.7 m/s. Yeah, I probably did that wrong...

3. We can use Kepler's third law. T^2=a^3, since we are in AU. A is the average of the distances or 5.05 AU. Cubing this and then taking the square root, I got 11.3 years, BUT we want the time it takes to get there, not the time of orbit, which is half of the ellipse, so 11.3/2=5.65 years.

4. 45 degrees altitude 90 degrees azimuth?

5. East definitely not just a guess

[nobodynobody]

Disclaimer: I SUCK with coordinates

1. Hohmann transfer

2. I believe this would just be the vis-viva equation, which can be derived by using energy principles. v^2=GM(2/r-1/a), where r is the distance from the orbiting body and a is the semimajor axis. Now, I have to be a bit careful. I think a is the average of the distances and r is 9.6. Just curious, is there a shortcut for AU or do you just have to convert? Anyway, plugging in I got 14, 938.7 m/s. Yeah, I probably did that wrong...

3. We can use Kepler's third law. T^2=a^3, since we are in AU. A is the average of the distances or 5.05 AU. Cubing this and then taking the square root, I got 11.3 years, BUT we want the time it takes to get there, not the time of orbit, which is half of the ellipse, so 11.3/2=5.65 years.

4. 45 degrees altitude 90 degrees azimuth?

5. East definitely not just a guess

1. Correct!

2. You set up the equation correctly, but I got a different number. One of us probably a calculation error. I'm not aware of any shortcut either, I just converted.

3. Remember that the central star has a mass of 2.3 solar masses (use m*p^2=a^3)

4. Azimuth is correct (it will be either 90 degrees west or east of north, depending on where the exoplanet is in its orbit). I'm not sure about the attitude. I haven't done the calculation for myself yet, but I have a feeling that it will be a lot less (because the exoplanet is relatively very close to the central star).

5. Nope, but it's a very good guess (unless I'm wrong lol)

[astronomybuff]

Disclaimer: I SUCK with coordinates

1. Hohmann transfer

2. I believe this would just be the vis-viva equation, which can be derived by using energy principles. v^2=GM(2/r-1/a), where r is the distance from the orbiting body and a is the semimajor axis. Now, I have to be a bit careful. I think a is the average of the distances and r is 9.6. Just curious, is there a shortcut for AU or do you just have to convert? Anyway, plugging in I got 14, 938.7 m/s. Yeah, I probably did that wrong...

3. We can use Kepler's third law. T^2=a^3, since we are in AU. A is the average of the distances or 5.05 AU. Cubing this and then taking the square root, I got 11.3 years, BUT we want the time it takes to get there, not the time of orbit, which is half of the ellipse, so 11.3/2=5.65 years.

4. 45 degrees altitude 90 degrees azimuth?

5. East definitely not just a guess

1. Correct!

2. You set up the equation correctly, but I got a different number. One of us probably a calculation error. I'm not aware of any shortcut either, I just converted.

3. Remember that the central star has a mass of 2.3 solar masses (use m*p^2=a^3)

4. Azimuth is correct (it will be either 90 degrees west or east of north, depending on where the exoplanet is in its orbit). I'm not sure about the attitude. I haven't done the calculation for myself yet, but I have a feeling that it will be a lot less (because the exoplanet is relatively very close to the central star).

5. Nope, but it's a very good guess (unless I'm wrong lol)

Ohh for number 3 yeah, thought it was the Sun lol. I probably converted incorrectly for 2. Yeah, I haven't learned much about coordinates. I'll post a problem soon. .

[astronomybuff]

Two satellites are launched at a distance R from a planet of negligible radius. The first satellite launches at speed v and enters a circular orbit. The second satellite is launched at a speed of v/2. What is the minimum distance between the second satellite and the planet over the course of its orbit?

[0sm0sis]

Yay orbits!

Click to Show Answer

Because the orbit is circular, we can equate the centripetal acceleration to the gravitational force to find that .

From there, we can use the Vis-Viva equation to find the semimajor axis of the new orbit. The equation is , but we must use  in place of . This gives us



Which simplifies to:



From there we should use reasoning with how orbits work. Because the velocity is perfectly tangential from the beginning, the initial distance from the star is either the perigee or apogee. Because a larger velocity is required to have circular motion for this position, this distance is probably the apogee, the furthest distance from the star.

We also know that the perigee distance + apogee distance = 2a (this can be seen easily by drawing a diagram of an elliptical orbit). Therefore, , meaning that our perigee  is: