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Math Help - story problem..

  1. #1
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    story problem..

    assume that the galaxy is a group of 400 billion stars distributed along a cubic lattice, each star separated from it's nearest by about 5LightYear. If 100 currently signaling civilization are randomly distributed somewhere in this galaxy
    (hint: let N be the number of stars in cubic galaxy; N = 400x10^9. If each star is separated by a distance D = 5LY, then the total volume of the cubic galaxy is Vgal= ND^3. Now the number of civilizations in the galaxy is n_{civ} = 100. Let each one of them be surrounded by an "empty" cube of volume V_{civ} = x^3 in which there are no other civilizations--on the avg. So x is the avg distance between these civilization.)

    How far away in LY is the nearest civilization likely to be?
    from the hint I got Vgal = (400x10^9)(5^3) = 2^{13}.
    I'm not sure how to solve the rest.
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  2. #2
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    Quote Originally Posted by viet View Post
    (hint: let N be the number of stars in cubic galaxy; N = 400x10^9. If each star is separated by a distance D = 5LY, then the total volume of the cubic galaxy is Vgal= ND^3. Now the number of civilizations in the galaxy is n_{civ} = 100. Let each one of them be surrounded by an "empty" cube of volume V_{civ} = x^3 in which there are no other civilizations--on the avg. So x is the avg distance between these civilization.)



    from the hint I got Vgal = (400x10^9)(5^3) = 2^{13}.
    I'm not sure how to solve the rest.
    Vgal= (4*10^9)(125)= 500*10^9= 5*10^11 which is NOT 2^{13} so even that is wrong.

    Divide that by 100 to get 5*10^9 as the average volume of a "civilization". x is the cube root of that number.
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  3. #3
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    Thanks for your help HallsofIvy. I cuberooted 5x10^11 and got 7937.0
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