Astronomy C

User avatar
SciolyHarsh
Member
Member
Posts: 37
Joined: May 20th, 2018, 5:44 pm
Has thanked: 0
Been thanked: 0

Re: Astronomy C

Post by SciolyHarsh » January 15th, 2019, 3:55 pm

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
2017-2018 Events: Chemistry Lab, Dynamic Planet, Microbe Mission, Experimental Design, Rocks and Minerals

2018-2019 Events: Dynamic Planet, Astronomy, Sounds of Music, Circuit Lab, Geologic Mapping

pb5754
Member
Member
Posts: 464
Joined: March 5th, 2017, 7:49 pm
Division: C
State: NJ
Pronouns: He/Him/His
Has thanked: 17 times
Been thanked: 30 times

Re: Astronomy C

Post by pb5754 » January 24th, 2019, 3:23 pm

Sorry I completely forgot about this :oops:
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

User avatar
SciolyHarsh
Member
Member
Posts: 37
Joined: May 20th, 2018, 5:44 pm
Has thanked: 0
Been thanked: 0

Re: Astronomy C

Post by SciolyHarsh » February 5th, 2019, 4:13 pm

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?
2017-2018 Events: Chemistry Lab, Dynamic Planet, Microbe Mission, Experimental Design, Rocks and Minerals

2018-2019 Events: Dynamic Planet, Astronomy, Sounds of Music, Circuit Lab, Geologic Mapping

pb5754
Member
Member
Posts: 464
Joined: March 5th, 2017, 7:49 pm
Division: C
State: NJ
Pronouns: He/Him/His
Has thanked: 17 times
Been thanked: 30 times

Re: Astronomy C

Post by pb5754 » February 6th, 2019, 2:22 pm

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

User avatar
SciolyHarsh
Member
Member
Posts: 37
Joined: May 20th, 2018, 5:44 pm
Has thanked: 0
Been thanked: 0

Re: Astronomy C

Post by SciolyHarsh » February 6th, 2019, 3:04 pm

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
2017-2018 Events: Chemistry Lab, Dynamic Planet, Microbe Mission, Experimental Design, Rocks and Minerals

2018-2019 Events: Dynamic Planet, Astronomy, Sounds of Music, Circuit Lab, Geologic Mapping

pb5754
Member
Member
Posts: 464
Joined: March 5th, 2017, 7:49 pm
Division: C
State: NJ
Pronouns: He/Him/His
Has thanked: 17 times
Been thanked: 30 times

Re: Astronomy C

Post by pb5754 » February 6th, 2019, 3:27 pm

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

User avatar
SciolyHarsh
Member
Member
Posts: 37
Joined: May 20th, 2018, 5:44 pm
Has thanked: 0
Been thanked: 0

Re: Astronomy C

Post by SciolyHarsh » February 6th, 2019, 6:54 pm

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.
2017-2018 Events: Chemistry Lab, Dynamic Planet, Microbe Mission, Experimental Design, Rocks and Minerals

2018-2019 Events: Dynamic Planet, Astronomy, Sounds of Music, Circuit Lab, Geologic Mapping

User avatar
PM2017
Member
Member
Posts: 522
Joined: January 20th, 2017, 5:02 pm
Division: Grad
State: CA
Location: Enjoying College! :D
Has thanked: 19 times
Been thanked: 2 times

Re: Astronomy C

Post by PM2017 » February 6th, 2019, 8:23 pm

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

Go Bears!

User avatar
SciolyHarsh
Member
Member
Posts: 37
Joined: May 20th, 2018, 5:44 pm
Has thanked: 0
Been thanked: 0

Re: Astronomy C

Post by SciolyHarsh » February 6th, 2019, 8:32 pm

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.
2017-2018 Events: Chemistry Lab, Dynamic Planet, Microbe Mission, Experimental Design, Rocks and Minerals

2018-2019 Events: Dynamic Planet, Astronomy, Sounds of Music, Circuit Lab, Geologic Mapping

User avatar
ET2020
Member
Member
Posts: 67
Joined: April 16th, 2018, 11:35 am
Division: C
State: NY
Location: FM
Has thanked: 0
Been thanked: 0

Re: Astronomy C

Post by ET2020 » February 26th, 2019, 8:39 pm

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?
Fayetteville Manlius High School
2020 Events: Astronomy, Code, Dynamic, PPP

Post Reply

Return to “2019 Question Marathons”

Who is online

Users browsing this forum: No registered users and 1 guest