## Astronomy C

SciolyHarsh
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### Re: Astronomy C

Did you really need the apparent magnitude and redshift? or am I doing this wrong? yeah i meant to derive it lol but I added the hubble constant for some reason. Anyway, using the Hubble constant you got it right. Your turn
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pb5754
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### Re: Astronomy C

Sorry I completely forgot about this A planet with mass 2.4*10^23 kg orbits a star of mass 2.7 solar masses at a radius of 10 AU with velocity 100 m/s. Find the semimajor axis.
West Windsor-Plainsboro High School South '21

SciolyHarsh
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### Re: Astronomy C

I wasn't sure how exactly to solve this, but I did my best.
Here are my steps: 1. I converted the mass of the planet to solar masses, but it's so small that it's nearly negligible. That said, for the problem's sake, I used it in the calculations.
2. Next, I used the radius of the star to solve for the circumference of the planet. I found it to be 20pi AU.
3. I then converted the velocity from m/s to AU/year, and I solved for the orbital period by finding out how long it would take to orbit the star completely once using the given velocity which I converted to AU/year.
4. I used the orbital period and the masses of both the planet and the star to solve for the semi-major axis using Kepler's Third Law.
My answer for the semi-major axis was 288.394 AU. If I'm wrong, could you post the steps to solve this?
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pb5754
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### Re: Astronomy C

SciolyHarsh wrote:I wasn't sure how exactly to solve this, but I did my best.
Here are my steps: 1. I converted the mass of the planet to solar masses, but it's so small that it's nearly negligible. That said, for the problem's sake, I used it in the calculations.
2. Next, I used the radius of the star to solve for the circumference of the planet. I found it to be 20pi AU.
3. I then converted the velocity from m/s to AU/year, and I solved for the orbital period by finding out how long it would take to orbit the star completely once using the given velocity which I converted to AU/year.
4. I used the orbital period and the masses of both the planet and the star to solve for the semi-major axis using Kepler's Third Law.
My answer for the semi-major axis was 288.394 AU. If I'm wrong, could you post the steps to solve this?
This is how I would do it...
```Use the vis-viva equation (which can be obtained by applying the principle of conservation of mechanical energy): v = sqrt(GM((2/r)-(1/a))), where v = orbital velocity (m/s), M = mass of the star (central body) (kg), r = radius (m), a = semimajor axis (m)
At this point, you can manipulate this equation to obtain the formula a = (GMr)/(2GM-(v^2)r)
Now plugging the values (and remembering to convert to the proper units) in you get a = 5.00 AU.```
West Windsor-Plainsboro High School South '21

SciolyHarsh
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### Re: Astronomy C

Ah! I didn't know about that equation. Thank you for that. I'm not sure what to do now that I've gotten it wrong though
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pb5754
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### Re: Astronomy C

SciolyHarsh wrote:Ah! I didn't know about that equation. Thank you for that. I'm not sure what to do now that I've gotten it wrong though
You can post the next question West Windsor-Plainsboro High School South '21

SciolyHarsh
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### Re: Astronomy C

Alright, so:

The luminosity of Galaxy A is 2.692x10^27 watts. Galaxy A has an apparent magnitude of 6.89.
Galaxy B contains Cepheid variable stars with a period of 0.04252 years. Galaxy B has an apparent magnitude of 4.37.

Calculate the ratio of the distance to Galaxy A to the distance to Galaxy B.
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PM2017
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### Re: Astronomy C

SciolyHarsh wrote:Alright, so:

The luminosity of Galaxy A is 2.692x10^27 watts. Galaxy A has an apparent magnitude of 6.89.
Galaxy B contains Cepheid variable stars with a period of 0.04252 years. Galaxy B has an apparent magnitude of 4.37.

Calculate the ratio of the distance to Galaxy A to the distance to Galaxy B.
Not a solution to this problem you posed, but wow that's a really dim galaxy lol. (like 10-12 orders of magnitude dimmer than most spiral galaxies.)
West High '19
UC Berkeley '23

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SciolyHarsh
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### Re: Astronomy C

I just went with a random number that would be kinda smooth for plugging in. I just realized that its only a couple times brighter than the Sun.
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ET2020
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### Re: Astronomy C

```Galaxy A: 2.692*10^27 W = 7.03 solar luminosities
M = 4.83-2.5log(L) = 2.71
d = 10^([m-M+5]/5) = [b]68.48 pc[/b] (this is an impossibly close galaxy, but you can still do the math with it)

Galaxy B: P (days) = 365.25*0.04252 = 15.53 days
M = -2.81logP - 1.43 = -4.78
Now this is the tricky part . . . you don't state how many stars are in Galaxy B, so I don't know how to solve from here, since the luminosity of the galaxy would be the sum of all the stars in it, and also many of these stars would likely not even be Cepheids. I'll try to solve the problem assuming the galaxy is a single Cepheid with a period of 15 days
d = 10^([m-M+5]/5) = [b]675.2 pc[/b]

68.48/675.2 = ~ [b]0.10[/b]
Did I do it right?```
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