anonymous
  • anonymous
could someone please help me get started with this problem. I am lost our galaxy has a diameter of about 100,000 light years. and a thickness of about 2000 light years. our sun is locate about 28000 light years from center of our galaxy, and orbits once every 200 million years. calculate the speed of the sun relative to the center of galaxy in km/s. use C=2 r to determine the orbital distance, and pi=3.14.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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theEric
  • theEric
|dw:1377705239728:dw|It's path is supposed to be a large circle, but I drew it messily, sorry! It's path is the circumference, then. And the problem means to say the the circumference is \(C\) and \(C=2\pi r\). Do you see that? That is the distance, then. And the time the Sun takes to travel that path is 200 million years. So you have the time and the distance, so the speed, \(v\), is found by the \(v=\dfrac{d}{t}\) equation!
theEric
  • theEric
And you want to use \(3.14\) for \(\pi\). And you'll have to convert the distance in light years to the distance in kilometers. In the same way, you'll have to convert the time in years to the time in seconds. Do you know how to convert? You can do it at any time in the problem.
theEric
  • theEric
Google says "1 lightyear =9.4605284 × 10^12 kilometers" and 1 year is "3.15569e7 seconds"

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anonymous
  • anonymous
okay...I'm lost first of all I calculated the speed of light into km like u said. I understand that not sure how to convert time
anonymous
  • anonymous
i calculated the time to be 6.3072 x 10^15...........that's not right
theEric
  • theEric
Well, \(1\ [yr]=3.15569\times 10^7\ [s]\\\ \\\implies 1=\dfrac{3.15569\times 10^7\ [s]}{1\ [yr]}\) So you can multiply your amount of time as \(200\times 10^6\ [yr]\) by the \(1\) above without changing the amount of time!\[200\times 10^6\ [yr]\times\dfrac{3.15569\times 10^7\ [s]}{1\ [yr]}\approx 6.31\times 10^{15}\].
theEric
  • theEric
and that's in seconds.\[200\times 10^6\ \cancel{[yr]}\times\dfrac{3.15569\times 10^7\ [s]}{1\ \cancel{[yr]}}\approx 6.31\times 10^{15}\ [s]\]
anonymous
  • anonymous
so i was right
theEric
  • theEric
Yup! And for distance: \(28.000\times 10^3\ [lyr]\times \dfrac{9.4605284 × 10^{12}\ [km]}{1\ [lyr]}\\\ \\\quad \approx 2.6489\times 10^7 [km]\)
anonymous
  • anonymous
what about seconds
theEric
  • theEric
\(\approx 2.6489\times 10^{17}\ [km]\), sorry.
theEric
  • theEric
The seconds was done above. I wrote in seconds as \([s]\).
theEric
  • theEric
Now, \[v=\frac{2.6489\times 10^{17}\ [km]}{6.31\times 10^{15}\ [s]}\] And now I believe you can finish. Any questions?
anonymous
  • anonymous
that helped a lot. thanks
theEric
  • theEric
You're welcome!
anonymous
  • anonymous
so v= 41.949 km/s is this right
anonymous
  • anonymous
im so confused....where does C come into use
anonymous
  • anonymous
so far this what if done 1 year = (365)(24)(60)(60) = 3.15 x 10^7s 1year = (2.998 x 10^8) (365)(24)(60)(60) =9.454 x 10^12 km the sun takes 200 million years to go around (2 x 10^8)(3.15x 10^7) 6.31 x 10^15 s so the sun takes 6.31 x 10^15 s to travel around the orbit d=(28000 m/s) (9.454 x 10^12km) = (28 x 10^3km)(9.454 x 10^12km) = 2.647 x 10^17 v=d/t = 2.647 x 10^17km / 6.31 x 10^15 = 41.949 km/s is this right so far
theEric
  • theEric
Oh, I'm sorry.. I came back to the problem and forgot the details! That 28,000 lightyears is really not the distance it travels. I'll quote myself from earlier, to maybe regain my dignity! "It's path is the circumference, then. And the problem means to say the the circumference is C and C=2πr. Do you see that? That is the distance, then." \(\qquad\quad \text{circumference}\) \(\qquad\qquad\quad\ \ \downarrow\) So \(v=\dfrac{\text{distance}}{\text{time}}=\dfrac{C}{t}=\dfrac{2\pi r}{t}\) where the radius \(r\) is what you found to be \(d\). That's where the \(C\) comes in. So \(v=\dfrac{2\pi 2.647\times 10^{17}\ [km]}{6.31\times 10^{15}\ [s]}\) with your numbers. That is about \(263.\ [km/s]\) Congrats on finding the distance of a lightyear in kilometers, by the way!
theEric
  • theEric
|dw:1377751467285:dw|
anonymous
  • anonymous
im sorry physics is new to me.... so the distance from center to sun is 2.647 x 10^17....I need half of that for the for the radius C=(2)(pi)( r) =(2)(3.14)(2.647 x 10^17) = 1.66 x 10^18 so distance is my circumference which is 1.66 x 10^18 then.... v=d/t = 1.66x10^18 km / 6.31 x 10^15 s = 263 km/s so the speed of the sun is 263 km/s is this right
theEric
  • theEric
It's alright that you're new! Asking questions will help you, so I'm glad you do that! Also, if you took my answer and used it without looking to find my mistake, you would've been wrong. But you looked, and you thought, and you successfully avoided the bad answer that I accidentally alluded to! That's why I gave you a medal :) And I agree with your answer! It is what I got, too, after I looked at the problem again. And the distance from "center to sun" is 2.647 x 10^17 [km] and is the radius. You said "I need half of that for the for the radius," but you didn't need to change anything about that to get the radius. You didn't change anything, actually. So you got the right answer, I think. Congrats! Take care!

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