<p>Without checking the numbers in detail, according to Wikipedia, the volume of the <a href="http://en.wikipedia.org/wiki/Observable_universe">observable universe</a> is about $3.5\cdot 10^{80} \mbox{ m}^3$, and the volume of <a href="http://en.wikipedia.org/wiki/Earth">Earth</a> is about
$1.08321\cdot 10^{21} \mbox{ m}^3$.</p>

<p>By dividing the two volumes we get a factor of $3.2\cdot 10^{59}$, or written as decimal number: The observable comoving volume of the universe is about
320,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000-times the volume of Earth.</p>

CC BY-SA 4.0
2019-05-13T11:01:03.897Z
That's about the same as the number of atoms in the largest stars.
Steve Linton
CC BY-SA 4.0
2019-05-14T05:46:53.853Z
@SteveLinton So the Earth is to the universe as an atom is to a very large star.
Florin Andrei
CC BY-SA 3.0
2014-05-13T17:14:57.137Z
dats alot, woah!
Math chiller
CC BY-SA 3.0
2014-05-14T16:44:50.697Z
Your answer assumes we are pulverizing the earth to *completely* fill the volume of a universe-sized container. Without getting into the complicated math behind forming optimal latices of congruent spheres, you should multiply your answer by a factor of pi/(3*sqrt(2)) or about 0.74048. The Kepler Conjecture says that is the highest density that can be achieved by any arrangement of spheres. Oh, and since the observable universe is also expanding at an accelerated rate, you should also update your answer every few hundred millions years just to be safe. Just saying.
Robert Cartaino
CC BY-SA 3.0
2014-05-15T09:37:45.853Z
@RobertCartaino That's why I just provided two valid digits; so the numbers should be valid more or less next week, too. ;) Btw. sorry for pressing Earth into a cube, next time I'll be more careful.
Gerald
CC BY-SA 3.0
2014-05-13T16:09:38.560Z
2014-05-13T16:09:38.560Z
Gerald
<p>I was pondering our insignificance, when I wondered - how much smaller is our planet then the (observable) universe? And being as I don't know how to do the math, I'm asking it here.</p>

<p>So how many of our planet (in space it occupies - i.e. ignoring space between the space between the spheres) can fit inside the known/observable universe?</p>

CC BY-SA 3.0
2014-05-13T23:40:04.103Z
Really to me this question is pointless. I mean, the answer is obviously going to be a ridiculously big number, so what does it changes to you if it is 10^50 or 10^100??
harogaston
CC BY-SA 3.0
2014-05-14T01:49:46.950Z
i was trying to find something to compare it to, in order to better understand it and relay it.
Math chiller
<earth><universe><size>
How many earths fit in the observable universe?
Math chiller
# How many earths fit in the observable universe?

(Asked by: username_0 on 2014-05-13)

I was pondering our insignificance, when I wondered - how much smaller is our planet then the (observable) universe? And being as I don't know how to do the math, I'm asking it here.

So how many of our planet (in space it occupies - i.e. ignoring space between the space between the spheres) can fit inside the known/observable universe?

**Tags:** earth, universe, size
> **Comments:**
> **username_1**: Really to me this question is pointless. I mean, the answer is obviously going to be a ridiculously big number, so what does it changes to you if it is 10^50 or 10^100??  
> **username_0**: i was trying to find something to compare it to, in order to better understand it and relay it.  

---

## 1 answers

### Answer by username_2 on 2014-05-13. Score: 16 | Accepted answer

Without checking the numbers in detail, according to Wikipedia, the volume of the [observable universe](http://en.wikipedia.org/wiki/Observable_universe) is about $3.5\cdot 10^{80} \mbox{ m}^3$, and the volume of [Earth](http://en.wikipedia.org/wiki/Earth) is about
$1.08321\cdot 10^{21} \mbox{ m}^3$.

By dividing the two volumes we get a factor of $3.2\cdot 10^{59}$, or written as decimal number: The observable comoving volume of the universe is about
320,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000-times the volume of Earth.

> **Comments:**
> **username_0**: dats alot, woah!  
> **username_5**: Your answer assumes we are pulverizing the earth to *completely* fill the volume of a universe-sized container. Without getting into the complicated math behind forming optimal latices of congruent spheres, you should multiply your answer by a factor of pi/(3*sqrt(2)) or about 0.74048. The Kepler Conjecture says that is the highest density that can be achieved by any arrangement of spheres. Oh, and since the observable universe is also expanding at an accelerated rate, you should also update your answer every few hundred millions years just to be safe. Just saying.  
> **username_2**: @RobertCartaino That's why I just provided two valid digits; so the numbers should be valid more or less next week, too. ;) Btw. sorry for pressing Earth into a cube, next time I'll be more careful.  
> **username_3**: That's about the same as the number of atoms in the largest stars.  
> **username_4**: @SteveLinton So the Earth is to the universe as an atom is to a very large star.  

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