# story problem..

• Apr 5th 2010, 04:45 PM
viet
story problem..
Quote:

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.)

Quote:

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.
• Apr 6th 2010, 03:34 AM
HallsofIvy
Quote:

Originally Posted by viet
(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.
• Apr 6th 2010, 07:13 AM
viet
Thanks for your help HallsofIvy. I cuberooted 5x10^11 and got 7937.0