# A math question for those who like BIG numbers **URL:** \ **Category:** Factual Questions\ **Created:** [August 22, 2004, 5:14pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983 "2004-08-22T17:14:37Z")\ **Posts on this page:** 19\ **Page:** 1
**Author:** ![Aceospades](https://avatars.discourse-cdn.com/v4/letter/a/848f3c/32.png) [@Aceospades](https://boards.straightdope.com/u/Aceospades)\ **Post date:** [August 22, 2004, 5:14pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/1 "2004-08-22T17:14:37Z")
If you started off with one higrigen atom and doubled it every second (1,2,4,8,16,ect…), how long would it take to fill up a space as big as the known universe. The hydrogen would be in gas form, but at a density of right before it turned liquid. Bonus question. Start off with a neutron and start doubling it (in essence creating a neutron star). How long would that take? ---
**Author:** ![spingears](https://avatars.discourse-cdn.com/v4/letter/s/ebca7d/32.png) [@spingears](https://boards.straightdope.com/u/spingears)\ **Post date:** [August 22, 2004, 5:27pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/2 "2004-08-22T17:27:51Z")
Higriger is not listed in the periodic table. You may need a “Quibic” quantum computer to solve. More _homework_ for dopers? :dubious: Didn’t the instructor furnish the data? ---
**Author:** ![ZebraShaSha](https://avatars.discourse-cdn.com/v4/letter/z/eada6e/32.png) [@ZebraShaSha](https://boards.straightdope.com/u/ZebraShaSha)\ **Post date:** [August 22, 2004, 5:52pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/3 "2004-08-22T17:52:44Z")
First, it is hydrogen. Second, atomic mass does not equal volume. Third, there are many different radii I could use. I am going to use atomic radius for this problem. Fourth, hydrogen is diatomic. In nature you will not find one hydrogen atom by itself but rather two bonded together. With all that being said, here is my take on it. Volume of known universe (assuming it is spherical): 1.9 × 10^33 cubic light years 1 light year is roughly equal to 9.46 trillion km Atomic radius of one hydrogen atom: 25 picometers or 2.5 × 10^-14 km (this will help us later) Equation of a circle: 4/3_pi_r^3 The volume of a hydrogen atom would there for be 6.54498 x 10^-41 cubic km We now want an exponential equation ln(Y/X)=t\*k When t=2 we have ln(1.309 x 10^-40/6.54498 x 10^-41)=2\*k Solving for k we now have a constant of 0.34657359 ln(9460000000000/2.5 × 10^-14)=0.34657359_t 61.1979941=0.34657359_t 176.580085=t Or, about 2.94300142 minutes. Why does that not feel right? ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [August 22, 2004, 5:56pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/4 "2004-08-22T17:56:00Z")
You’d never be able to fill the universe to that density of _anything_, as it would collapse under its own gravitation long before the universe was anywhere near full. However, we can make a few assumptions and ignore physics completely. If we assume by “known universe” you mean the sphere enclosing the visible universe, that we’re dealing with a sphere with a radius of about 15 billion LY. Your requirement about the particle density is a bit vague. For clarity, we’ll use STP, at which (according to [this page](http://whyfiles.org/017planet/gas_cloud2.html)) that there are about 700 billion billion hydrogen atoms per cm[sup]3[/sup] of pure hydrogen gas. To fill up a sphere 15 billion LY in radius to this density would take 1.17 x 10[sup]78[/sup] atoms. Doubling once per second starting at one would take a mere 260 seconds! (2^260 = 1.85 x 10[sup]78[/sup]. **Spingears** : :rolleyes: ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [August 22, 2004, 5:58pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/5 "2004-08-22T17:58:18Z")
> [@ZebraShaSha](#): > > Or, about 2.94300142 minutes. Why does that not feel right? Seems ok to me. Even though we used slightly different assumptions, our answers agree to well within an order of magnitude. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [August 22, 2004, 6:01pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/6 "2004-08-22T18:01:39Z")
> [@ZebraShaSha](#): > > Fourth, hydrogen is diatomic. Is hydrogen diatomic in all situations? I seem to remember reading that nitrogen and oxygen are not diatomic inside stars. ---
**Author:** ![ZebraShaSha](https://avatars.discourse-cdn.com/v4/letter/z/eada6e/32.png) [@ZebraShaSha](https://boards.straightdope.com/u/ZebraShaSha)\ **Post date:** [August 22, 2004, 6:05pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/7 "2004-08-22T18:05:57Z")
CURSE ME! It will take that long to fill up ONE cubic light year. I knew I made a mistake. Here it is only better: ln((9460000000000_1.9 × 10^33 cubic light years)/2.5 × 10^-14)=0.34657359_t 137.825156=0.34657359\*t 397.679338=t Or about 6.62798897 minutes. That sounds a little better. Maybe. I will probably be back. ---
**Author:** ![samclem](https://avatars.discourse-cdn.com/v4/letter/s/a9a28c/32.png) [@samclem](https://boards.straightdope.com/u/samclem)\ **Post date:** [August 22, 2004, 6:08pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/8 "2004-08-22T18:08:55Z")
[moderator hat on] > [@spingears](#): > > More _homework_ for dopers? :dubious: > Didn’t the instructor furnish the data? This is getting tiresome, **spingears**. If you think that someone is asking homework questions, we’ve devised a method for you to report this. It’s actually less time consuming than writing a snarky reply to the OP. **HIT THE _REPORT BAD POST_** button in the upper right corner of the actual posts. It’s a **!** symbol. Thanks for your future co-operation. [moderator hat off] ---
**Author:** ![ZebraShaSha](https://avatars.discourse-cdn.com/v4/letter/z/eada6e/32.png) [@ZebraShaSha](https://boards.straightdope.com/u/ZebraShaSha)\ **Post date:** [August 22, 2004, 6:14pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/9 "2004-08-22T18:14:05Z")
**Q.E.D.** - Yea, I did not even worry about it being in the form of a gas. What I could of done was taken the molar volume and changed it for when the substance is “right before [its] liquid form” but did not even mess with it. Oh well. ---
**Author:** ![ZebraShaSha](https://avatars.discourse-cdn.com/v4/letter/z/eada6e/32.png) [@ZebraShaSha](https://boards.straightdope.com/u/ZebraShaSha)\ **Post date:** [August 22, 2004, 6:18pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/10 "2004-08-22T18:18:48Z")
> [@ultrafilter](#): > > Is hydrogen diatomic in all situations? I seem to remember reading that nitrogen and oxygen are not diatomic inside stars. 5 replies to one thread. I need to quit. Anyways, yes, you are right. Hydrogen exsists as a single atom in some situations simply because there is no way for it to combine with other hydrogen atoms. I, however, was applying a standard situation to the hydrogen atoms. I shouldn’t of, but I do not think it matters either way. ---
**Author:** ![Aceospades](https://avatars.discourse-cdn.com/v4/letter/a/848f3c/32.png) [@Aceospades](https://boards.straightdope.com/u/Aceospades)\ **Post date:** [August 22, 2004, 6:44pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/11 "2004-08-22T18:44:13Z")
> [@spingears](#): > > Higriger is not listed in the periodic table. > You may need a “Quibic” quantum computer to solve. > More _homework_ for dopers? :dubious: > Didn’t the instructor furnish the data? It is indeed hydrogen and not Higriger (I ran this through the spell checker. I don’t know how this happened). As for this being a homework question, you are wrong. This is a question I thought up last night in bed right before I fell asleep (a time when most of my best questions come to me). Although I agree with the SDMB policy on not answering homework questions, I find your unsupported assumption slightly insulting. Thanks to everyone who contributed to the thread. I would have thought it would have taken a lot longer. Wow, under 10 minutes. I was thinking days. ---
**Author:** ![wolf\_meister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/wolf_meister/32/15202_2.png) [@wolf\_meister](https://boards.straightdope.com/u/wolf_meister)\ **Post date:** [August 22, 2004, 6:44pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/12 "2004-08-22T18:44:43Z")
First, could we settle on the volume of the known Universe? Assuming a Universe of radius 15 billion light years, I get (4/3)_PI_1.5X10^10 Light Years = 1.4137 \* 10[sup]31[/sup] Light Years ---
**Author:** ![wolf\_meister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/wolf_meister/32/15202_2.png) [@wolf\_meister](https://boards.straightdope.com/u/wolf_meister)\ **Post date:** [August 22, 2004, 6:46pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/13 "2004-08-22T18:46:30Z")
Cubic Light Years That Is ---
**Author:** ![ITR\_champion](https://avatars.discourse-cdn.com/v4/letter/i/c67d28/32.png) [@ITR\_champion](https://boards.straightdope.com/u/ITR_champion)\ **Post date:** [August 22, 2004, 6:52pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/14 "2004-08-22T18:52:37Z")
Nothing is diatomic inside the core of stars. The heat required for a fusion reaction to take place is far far greater than what would be needed to tear apart any chemical bond. All figures for the radius of the universe involve a bit of guesswork, and the numbers may change as more powerful telescopes are invented. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [August 22, 2004, 6:53pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/15 "2004-08-22T18:53:32Z")
> [@wolf\_meister](#): > > 1.4137 \* 10[sup]31[/sup] Light Years Try again. First of all, use the correct formula which is 4/3 x pi x r[sup] **2** [/sup]. And check your math. I get 9.42 x 10[sup]20[/sup] LY[sup]3[/sup]. I think you slipped a decimal place or ten someplace. 🙂 ---
**Author:** ![ZebraShaSha](https://avatars.discourse-cdn.com/v4/letter/z/eada6e/32.png) [@ZebraShaSha](https://boards.straightdope.com/u/ZebraShaSha)\ **Post date:** [August 22, 2004, 7:07pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/16 "2004-08-22T19:07:14Z")
> [@Q.E.D.](#): > > Try again. First of all, use the correct formula which is 4/3 x pi x r[sup] **2** [/sup]. And check your math. I get 9.42 x 10[sup]20[/sup] LY[sup]3[/sup]. I think you slipped a decimal place or ten someplace. 🙂 4/3 x pi x r[sup] **3** [/sup], you mean? Anyways, I just realized I messed something ELSE up with my calculation. I am the nation’s official bumbling, awkward, and all around goofy novice mathematician. ln((9460000000000_1.9 × 10^33 cubic light years)/2.5 × 10^-14)=0.34657359_t 137.825156=0.34657359\*t 397.679338=t 1 cubic light year DOES NOT FREAKING EQUAL 1 light year! DAMN ME! 1 light year = 9.46 x 10[sup]12[/sup] km 1 cubic light year = 8.46590536 × 10[sup]38[/sup] cubic km THEREFORE!: ln((8.46590536 × 10[sup]38[/sup] cubic km_1.9 × 10[sup]33[/sup] cubic light years)/2.5 × 10^-14)=0.34657359_t 120.954181=0.34657359\*t 348.999994=t Or 5.81666657 minutes. Yes, I did not do the right volume for Hydrogen gas right before it turns into a liquid. Hopefully I will have this finally resolved (my calculations atleast). Bastard me. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [August 22, 2004, 7:13pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/17 "2004-08-22T19:13:05Z")
> [@ZebraShaSha](#): > > 4/3 x pi x r[sup] **3** [/sup], you mean? Errr, yeah that. Nevermind… ---
**Author:** ![wolf\_meister](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/wolf_meister/32/15202_2.png) [@wolf\_meister](https://boards.straightdope.com/u/wolf_meister)\ **Post date:** [August 22, 2004, 7:58pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/18 "2004-08-22T19:58:57Z")
Okay Zebra, we definitely agree that: 1 cubic light year = 8.46 × 10[sup]38[/sup] cubic km AND Universe Volume = 1.4137e+31 cubic light years and a handy converter to check your work is at: [http://www.1728.com/convert.htm](http://www.1728.com/convert.htm) and volume calculations can be done at: [http://www.1728.com/diam.htm](http://www.1728.com/diam.htm) * * * and **QED** although my number came out correctly, I forgot to put the exponent in the calculation which _should be_: I get (4/3)_PI_(1.5X10[sup]10[/sup])[sup]3[/sup] Light Years Geez, volume calculations can be very tricky. I previewed my calculation, saw it already had ONE exponent but I forgot to show _that_ raised to _another_ exponent. You figure 1 exponent per number always looks right. So, I guess we all are guilty of committing a few mathematical\* faux pas \* in this thread. :smack: (I am SO glad my Ultar Converter was right though). ---
**Author:** ![Squink](https://avatars.discourse-cdn.com/v4/letter/s/b5e925/32.png) [@Squink](https://boards.straightdope.com/u/Squink)\ **Post date:** [August 22, 2004, 8:44pm UTC](https://boards.straightdope.com/t/a-math-question-for-those-who-like-big-numbers/260983/19 "2004-08-22T20:44:13Z")
[1 cubic light year = 1.04008657 X 10[sup]41[/sup] cubic furlongs](http://www.google.com/search?hl=en&lr=&ie=ISO-8859-1&q=1+cubic+light+year+in+cubic+furlongs&btnG=Search)