Astronomy: Speed at the edge of the galaxy

AI Thread Summary
The discussion revolves around calculating the velocity of a star at the edge of the galaxy, given the number of stars and the galaxy's radius. The user outlines their approach, assuming a uniform distribution of stars and needing to find the force acting on the star to derive its velocity. They express difficulty in solving a specific integral related to the force calculation and seek assistance. Another participant suggests looking into the Shell Theorem, noting it applies to spherically symmetric distributions, while also questioning the implications of using an exponential distribution. The conversation highlights the complexities of gravitational calculations in astrophysics and the challenges of computational tools like Mathematica.
razidan
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Homework Statement


The number of stars in the galaxy is N=10^12 and the radius of the galaxy is Rgalaxy = 20 kPc
Let m be the average mass of a star in the galaxy.
what is the velocity of a star at the edge of the galaxy (relative to the center of the galaxy)

Homework Equations



The Attempt at a Solution


I had to make 2 assumptions here:
1) uniform distribution of stars, ## σ=\frac{Nm}{\pi R^2}## .
2) the star in question is at a distance of R+##\epsilon## from the center, and then take the limit of ##\epsilon ## goes to 1 light year (or any other distance that is way smaller then R).

In order to solve the question I need to find the force, F, on the star and solve ##F=\frac{mv^2}{R}## for v.

In order to find the force, one has to solve the integral ##F=Gmσ \int\limits_0^R\int_0^{2\pi}\frac{rdrd\theta}{(R+\epsilon)^2+r^2-2rRCos(\Theta)}##.
And that is where I'm stuck.
Ideas?

Thanks,
Raz.
 
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phyzguy said:
Try looking up the Shell Theorem.
Hmm... Haven't thought of that. But what if the distribution is exponential (##e##)? will the shell theorem still hold?
My intuition says that different distributions will result in different speeds.
 
The shell theorem always holds as long as the distribution is spherically symmetric.
 
phyzguy said:
The shell theorem always holds as long as the distribution is spherically symmetric.
Thanks. that simplifies things...
any ideas as to why Mathematica won't solve that integral, though?
 
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