Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Astrophysics distribution question

  1. May 11, 2006 #1
    "A star, with mass 5.40*10^30 kg, revolves about the center of the Milky Way galaxy, which is 3.93*10^20 m away, once every 3.41*10^8 years. Assuming that each of the stars in the galaxy has a mass equal to that of our star, that the stars are distributed uniformly in a sphere about the galactic center, and that our star is essentially at the edge of that sphere, estimate roughly the number of stars in the galaxy."

    I don't really understand how I am supposed to answer this question. Using simple mathematics I can calculate the distance travelled of this star, and the speed with which it travels, but I do not know what else I need to do. I don't see where the mass comes into it, is volume and/or density relevant? I suspect this might be something to do with the Maxwell-Boltzmann distribution, am I in the right direction? Thanks.
  2. jcsd
  3. May 11, 2006 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, it has nothing to do with the MB distribution.

    The idea is this: the centripetal force is provided by the force of gravity, right? So [itex] {m v^2 \over r } = {G m M \over r^2}[/itex], right? Now, what is the meaning of m and M? m is the mass of the star and it cancels out. M is the mass of *all the stars within a radius r form the center of the galaxy*. Therefore, if you divide this mass by the mass of a single star, you get the total number of stars (assuming they all have the same mass) *within the radius r* (that is all the stars that are closer to the center of the galaxy than the star you are considering). This is why they had to add that the star is at the edge so that you may say that the number you get is the total number of stars in that galaxy. Now that you know the toatl number of star, it`s easy to calculate their density (assuming it's uniform), right?

  4. May 11, 2006 #3
    Right, that does makes sense, but I just tried, and got an answer which is clearly wrong. Here's what I did:

    (m*v^2)/r = (G*m*M)/r^2
    m = 5.40*10^30 kg
    G = 9.81 ms^-2
    r = 3.93*10^20 m
    v = d/t
    v = (2*pi*r)/t
    v = (2*pi*3.93*10^20)/(3.41*10^8*365.25*24*60*60)
    v = (2.469*10^21)/(1.076*10^16)
    v = 229,460 ms^-1
    (v^2)/r = (G*M)/r^2
    M = (v^2*r^2)/(G*r)
    M = (v^2*r)/G
    M = (229,460^2*3.93*10^20)/9.81
    M = 2.109*10^30 kg
    Number of stars = M/m = 2.109*10^30/5.40*10^30
    Number of stars = 0.39

    This clearly can't be right? The areas I may have possibly gone wrong in are a) my value for G; should it be 9.81? or b) my manipulation of the formula to have M as the subject. HELP!
  5. May 11, 2006 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No, G is the constant appearing in the universal of gravitation, 6.67x10^-11 Nm^2/kg^2. The rest seems ok to me (notice that you don't even need the mass of the star to find M, it's only when you find the number of stars that m is needed)
  6. May 11, 2006 #5
    Ah that yielded the correct answer. Many thanks for your help buddy.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook