What would be the average distance between stars if the universe was a sphere 12000 light years in diameter?

Question

For the purposes of this question, you can make the following assumptions:

The visible universe contains 70 sextillion stars (aka "70 Billion Trillion" or, if you prefer, a 7 followed by 22 zeros ). (That's the total number of visible stars in the known universe, according to a recent study by Australian astronomers.)

All of the stars in the visible universe are packed into a sphere of diameter 12000 light years.

Assume all stars are the same size as our sun.

All stars are evenly distributed throughout the sphere.

So the question is:

What would be the average distance between stars? (Or, if you prefer, how many stars per cubic light year?)

Answer 1

Stars per cubic light year??

70 x 10^22 / 12000 = 5.8333 x 10^19 stars per cubic light year.

Answer 2

I'm going to first convert and do everything in meters, just to keep it neat.  Your new universe has a diameter of 1.14e20 meters, and a radius of 5.68e19meters.  It has a total volume of 4/3 * pi * r^3 = 7.66e59 meters cubed.

If 70e21 stars are in there, that leaves about 1.09e37 cubic meters for each star (just division there).  If the stars are on a regular grid, then each star has a cube with sides of 2.2e12  meters (I find that by taking the cube root of 1.09e37.)  

The distance from the Sun to Uranus (for comparison) is 3e12 meters.  So the stars would be closer together than the size of our solar system.   I think this would lead to everything merging into one giant black hole, and then to another "big bang" type explosion.

Alternatively, we could just do the stars per cubic lightyear thing, which is easier.

4/3*pi*r^3 = 9.05e11 cubic light years

7e22 stars / 9.05e11 cubic light years = 7.74e10 stars per cubic light year.  That's 77.4 billion stars per cubic light year!

Answer 3

speed of light:

186000 mps or 299790 km/s

converted to LY:

186000 X 31536000 (seconds in 1 year) = 5.86e+12 LY (miles)

for km: 1.6 X 5.86e+12 = 9.3851136e+12 LY (km)

12000 LY in KM = 9.3851136e+12 X 12000 = 1.126213632e+17 km

               in MILES =  5.86e+12 X 12000 = 7.0388352e+16 miles

Sun's diameter: 1.4 million kilometers (870,000 miles)

Sun's diameter in LY: 870000/5.86e+12 = 1.484e-7 LY

  into volume in LY: 4/3 X PI X (0.742e-7)^3 =  1.711e-21 LY

for the 12000 LY, it would have a volume of 4/3XPIX6000^3

                       = 9.047e+11 LY

I can calculate the number of volumes of Suns which can fit into the 12000LY expanse...but I understand the tricky part of this question. I can't give you the perfect answer, I don't have the equation.

12000 LY cubed 'volume' expanse divided by Sun's Volume (in LY):

9.047e+11 LY / 1.711e-21 LY = 5.286e+32

You would be able to fit 5.286e+32 Sun's in that expanse of 12000LY (diameter).

You want 7e+22 stars fitting.

7e+22 X 1.711e-21 = 119.77 LY for all stars in volume

9.047e+11 LY -  119.77 LY = 9.046e+11 LY of space cubic left over

Divide 'space' by number of stars:   9.046e+11 LY / 7e+22 = 1.292e-11 of volume per star of empty space

convert to diameter: V = 4/3 x PI x r^3 -----> D = 2 ((0.75V/PI)^0.333) = 2 ((0.75 x  1.292e-11/PI)^0.333) = 2.911e-4 LY

Space diameter + Sun diameter = 2.911e-4 LY + 1.484e-7 LY = 2.913e-4 LY

Solar system diameter = 79 AU...convert to 92e+6 X 79 = 7.268e+9 miles

Convert to LY = 1.491e-7 LY

Star and space =  2.911e-4 LY in diameter

Solar system = 1.240e-3 LY  in diameter

You would be able to fit 4.259 stars (single star and space volume) in the region of the solar system.  That's too packed man!

If this were the case, we would have more solar energy than needed. The planet orbits may not even exist...since collision is quite likely. If there was a habital world it would probably never see the night. Lots of energy going around. Civilization would either never come to be or the be very evolved to handle the pressures of having more stars than needed.

Answer 4

Let's calculate the volume of the Sun (an average star).

7.0 * 10^5 km radius  So, volume is V = 4 / 3 * pi * R^3 = 1.44 * 10^18 km^3

For 7.0 * 10^22 stars the total volume of all stars is:

V = 4 / 3 * pi * R^3 = 1.0 * 10^41 km^3

Now let's calculate the volume of a space 12,000 light years in diameter.

R = 1.2 * 10^4 * 9.46 * 10^12 = 1.14 * 10^17 km

V = 4 / 3 * pi * R^3 = 6.2 * 10^51 km^3

Dividing you determine that the 12,000 Light year volume is  6.2 * 10^10  times as large as the total of all of the stars.

Now let's assume that all of the stars have a planetary system like our solar system.

Radius of our solar system is about 4.5 * 10^9 km so the volume is-

V = 4 / 3 * pi * R^3 = 3.8 * 10^29 km^3

Times all of the stars is: 3.8 * 10^29 * 7.0 * 10^22 = 2.57 * 10^52

OOPS...  What this says is that with the stars evenly packed there would be 4.3 stars within our solar system.  Kinda close.

Answer 5

Keep in mind that the average distance from each star to its closest neighbor is far, far less than the average distance between every possible pair of stars in the Universe. The average distance between every pair of stars would presumably be around half the distance to the farthest stars; if the Universe is about 156 billion light years in diameter and is shaped like a four-dimensional sphere, then this distance would be about 122 billion light years. On the other hand, the average distance from each star to its closest neighbor is more like two or three light years, so if you compressed the Universe to a diameter of 12000 light years and scaled everything down appropriately, the average distance from each star to its nearest neighbor would be about 1.5 million to 2 million kilometers. Note however that this is NOT the same as the answer would be if you distributed the stars evenly throughout the sphere, because in reality stars are packed into galaxies with relatively empty space between them. If you distributed 7*10^22 stars in a sphere with a radius of 12000 light years, you'd be putting them into about 904 billion cubic light years. Divide 7*10^22 by 904 billion cubic light years and you get about 77 billion stars per cubic light year. If you arranged them in a three-dimensional grid, the distance from each star to its six nearest neighbors would be around 2.2 billion kilometers.

Hopefully that answers your question. :)

Answer 6

Volume of sphere = (4/3)(π R3).= 4π R3/ 3

R = 12000

R3 = 1,728,000,000,000

π = 3.1415926535897932384626433832795

So, volume of space in example is:

V= 4 x 3.1415926535897932384626433832795 x 1,728,000,000,000/                                          3      

= 7,238,229,473,870.883621417930355076

N = number of stars

    = 70000000000000000000000      

Stars per cubic light year = N/V

= 70000000000000000000000/

   7,238,229,473,870.883621417930355076

= 9,670,873,278.1533624165120342327049

~ 9,670,873,278 stars per ly3

If stars were squashed into this small a universe, there would be a lot of light in the sky, but I'm not sure what the average separation of stars would be - perhaps too small for the formation of planets?