If all the matter in our solar system were distributed evenly throughout a cylindrical section of space, the thickness of the sun and the diameter of, say, the orbit of Neptune, how dense would it be?
Same question for the Milky Way galaxy - how dense would it be if the matter were evenly dispersed throughout the a cylinder the height of the thickest region, and the diameter of the widest span?
Just to help others:
And: “the Sun […] contains 99.86% of the system’s known mass”
Just to clarify - define the solar system as the objects within the described cylinder - no need to deal with the Kuiper belt and Oort cloud (although feel free to do this as a separate exercise, if you like).
Sun: Mean diameter 1.392×109 m
Neptune: Average distance: 4.55 x 1012 m
Mass of sun: 1.9891 × 1030 kg
Volume = pi r^2 h = pi * ( 4.55 x 1012 m / 2 )^2 * 1.39210^9 = 7.20410^33
Density = 1.989110^30 kg / 7.20410^33 =** 276*10^-6 kg/m^3
**
Wow. Solar system is much denser than I thought. It’s more than gram per four cubic meters.
Thanks. So one kilogram of matter is going to occupy about 3,600 cubic metres?
Milky Way Diameter = 100,000 light years
Thickness: 1,000 light years
Volume = 2.510^12 light years = 6.6510^60 m^3
Mass = 5.810^11 Suns = 5.810^11 * 1.989110^30 kg/Sun = 1.1510^42
Density = 1.1510^42 kg / 6.6510^60 m^3 = **173 * 10^-21 kg/m^3
**
Divide that by 12 if you want to use a thickness of 12,000 for the Milky Way to include gas instead of just stars.
If I hadn’t forgotten to multiply by pi. :smack:
It’s closer to 11.4
Updated numbers:
Volume = pi r^2 h = pi * ( 4.55 x 1012 m / 2 )^2 * 1.39210^9 = 22.6310^33
Density = 1.989110^30 kg / 22.6310^33 = 87.9*10^-6 kg/m^3
So, it’s actually closer to one gram every 11,400 m^3, once you take into account the whole forgetting to multiply by pi thing.
Something like that. I used 1.99 instead of 1.9891 for the mass. I figure with the precision we’re using, it’s as good of an estimation as any.
Heck, you want amusement? How dense is the Sun?
Mass of Sol=1.9891x10^30kg
Radius of Sol=6.96x10^8m
So rho(Ithink that’s the one)={1.9891x10^30kg}/{4/3π[(6.96x10^8m)^3]}={1.9891x10^33g}/{1.412x10^33(cm^3)}=1.408 g/cm^3
That is, the Sun is not that much denser than water. And since it’s REALLY dense toward the core, the part we see is more like brightly glowing fog. It gets even more fun when you consider the Sun as a red giant in ~5 billion years, with a radius of about 100 times what is now, and considerably less mass.
Should any arithmetic be in error, this office will deny all knowledge.
The sun is 50% more dense than my brain. Surprising 
Santo, you were also dividing the orbital radius by 2 for some unaccountable reason. OK, maybe you thought it was the orbital diameter.
Using the semimajor axis of Neptune (Santo actually used the aphelion distance, which is somewhat larger) and adding in the masses of the gas giants (they change the answer in the 4th significant digit), I get
Volume: 3.14159 * 1.392e9 * (4.503e12)^2 = 8.867e34 m^3
Mass: 1.9891e30 + 1.8986e27 + 5.6846e26 + 8.6810e25 + 1.0243e26 = 1.9918e30 Kg
Density: 2.246e-5 Kg/m^3
So your Kg of matter will occupy about 44,500 cubic meters.
Compared to the solar system though, you are incredibly dense. Thick, one might even say.
This MUST be from my previous self-subjection to chemicals that are not supposed to be used for human consumption…
CAN YOU, given the information contained ONLY within the movie itself, work out a decent guesstimate of the specific gravity of the material used to construct the mothership in Independence Day?
Don’t worry, the result is silly. We all have our slow days…
Boulder, eh? I remember that place…a hotbed of subversion, perversion, and HORROR!