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meri258
Group Title
By looking at distant galaxies, astronomers have concluded that our solar system is circling the center of our galaxy. The hub of this galaxy is located about 2.7 x 10^20 m from our sun, and our sun circles the center about every 200 million years. (convert to seconds). We assume that our sun is attracted by a larger number of stars at the hub of our galaxy, and that the sun is kept in orbit by the gravitational attraction of these stars.
a) calculate the total mass of the stars at the bug of our galaxy.
b) Calculate the approximate number of such stars that are the size of our sun. ~2E30KG
 one year ago
 one year ago
meri258 Group Title
By looking at distant galaxies, astronomers have concluded that our solar system is circling the center of our galaxy. The hub of this galaxy is located about 2.7 x 10^20 m from our sun, and our sun circles the center about every 200 million years. (convert to seconds). We assume that our sun is attracted by a larger number of stars at the hub of our galaxy, and that the sun is kept in orbit by the gravitational attraction of these stars. a) calculate the total mass of the stars at the bug of our galaxy. b) Calculate the approximate number of such stars that are the size of our sun. ~2E30KG
 one year ago
 one year ago

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Shadowys Group TitleBest ResponseYou've already chosen the best response.1
you might want to use this \(P^2= \frac{4 \pi^2 a^3}{G(M1+M2)}\)where M1=mass of the starhub and M2=mass of sun.
 one year ago

meri258 Group TitleBest ResponseYou've already chosen the best response.0
What does "P^2 stand for in that equation?
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
the square of the period. sub it all in and find M1.
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Third_law
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
you're welcome :)
 one year ago

meri258 Group TitleBest ResponseYou've already chosen the best response.0
For part a, I got 2.9E56 Kg, but on the answer key provided by my teacher says 2.9E41Kg. I solved it twice and I am getting the same answer.
 one year ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
nope, i'm getting your answer's key's answer. \(M1+M2=\frac{(2.7 \times 10^{20})^3(4 \pi^2)}{6.673 \times 10^{11}(200000000\times365\times24\times60\times60)} \)
 one year ago
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