Question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

Answer 1

Here is the full question

Suppose there are 10,000 civilizations in the Milky Way Galaxy. If the civilizations were randomly distributed throughout the disk of the galaxy, about how far (on average) would it be to the nearest civilization?

(Hint: Start by finding the area of the Milky Way's disk, assuming that it is circular and 100,000 light-years in diameter. Then find the average area per civilization, and use the distance across this area to estimate the distance between civilizations.)

Answer:

1000 light-years (ly)

Explanation:

If we go by the hint; The area of the disk can be expressed as:

$A = \pi (\frac{D}{2})^2$

where D = 100, 000 ly

Let's divide the Area by the number of civilization; if we do that ; we will be able to get 'n' disk that is randomly distributed; so ;

$d= \frac{A}{N} =\frac{\pi (\frac{D}{2})^2 }{10, 000}$

The distance between each disk is further calculated by finding the radius of the density which is shown as follows:

$d = \pi r^2 e$

$r^2_e= \frac{d}{\pi}$

$r_e = \sqrt{\frac{d}{\pi} }$

replacing d = $\frac{\pi (\frac{D}{2})^2 }{10, 000}$ in the equation above; we have:

$r_e = \sqrt{\frac{\frac{\pi (\frac{D}{2})^2 }{10, 000}}{\pi} }$

$r_e = \sqrt{\frac{(\frac{D}{2})^2 }{10, 000}}$

$r_e = \sqrt{\frac{(\frac{100,000}{2})^2 }{10, 000}}$

$r_e = 500 ly$

The distance (s) between each civilization = $2(r_e)$

= 2 (500 ly)

= 1000 light-years (ly)