## A Universe of Marshmallows

I fear that even contemplating writing the tirade of geekery that’s about to ensue might forever concrete my reputation as a mad scientist. Although I don’t really have a reputation as a mad scientist. Yet. This started out as a slightly bizarre twitter conversation with @Stellar190 and @PenguinGalaxy. About marshmallows. And the Universe. It all got a bit silly and ended up in a rather bizarre piece of recreational mathematics… After all of that, someone had to write a blog entry about it — so here it is! (By the way, if you’re not impressed by geekery for geekery’s sake, you may want to stop reading now.)

Still reading? Ok. Good for you! It all started when I posed the question (for no particular reason):
“If the observable universe is a sphere, 46.5 billion light years in radius, how many marshmallows could you fit into it?”

Well… A few approximations are necessary to figure this out. For a start, marshmallows are basically cylinders, roughly 3cm in diameter and 3cm in depth. For the sake of ease, let’s assume them to be perfect cylinders with a diameter of 1 cm and a depth of 3.085cm. Why 3.085cm? Because that’s one attoparsec!*

Using the dimensions of one marshmallow in parsecs, you can calculate the volume to be 7.85×10-55 cubic parsecs. So, if the observable Universe has a radius of 14 gigaparsecs, then it would have a total volume of 1.149×1031 cubic parsecs. Assuming my calculations are correct, it would, therefore, take 1.465×1085 marshmallows to completely fill the observable Universe. That’s rather a lot of squishy gelatinous goodness.

I’m reminded, however, that 1085 is certainly not 0.85 googol. But it’s still 1 456 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 marshmallows! **

At some point (perhaps inevitably), black holes were mentioned. Making a black hole from marshmallows can be accomplished in one of two ways. Seemingly, both of these are non-trivial tasks. Here’s why.

Firstly, you could conceivably make a black hole from a single marshmallow. Any mass has a schwartzchild radius — a smallest size, below which it would be impossible to prevent gravity from collapsing the object into a singularity. To work that out, you need to know the mass of a single marshmallow. A 175g bag contains about 28 marshmallows, giving them an average mass of 0.00625kg. This would give a marshmallow a schwartzchild radius of 9.28×10-30 metres. Oh dear.

9×10-30 is a small number. A very small number. So small, in fact, that there isn’t really a proper name for it. A bit under one hundred thousandth of a yoctometre (the smallest SI unit of length). This scale is comparable to the amusingly named “shed” — a unit of area so small that it’s only been used a handful of times while studying subatomic particles like neutrinos. The schwartzchild radius for our single marshmallow is, however, still larger than the Planck length (the smallest unit of length with any physical meaning in our Universe!), so it’s still theoretically possible to do this. Theoretically. But somehow, compressing a marshmallow to the size of a subatomic particle doesn’t really seem feasible.

The other option for our marshmallow black hole is to gather enough marshmallows to let gravity do the job for us. The thing is, this isn’t directly possible. A thing about marshmallows is that, excluding the helium, they have a composition not entirely dissimilar to your average red giant star. They’re mostly made of hydrogen, oxygen and carbon, with some traces of things like sodium. Being as they contain hydrogen, if you gather together enough marshmallows, you’d actually form a star. Yes, you read that correctly. A star made from marshmallows.

The minimum mass required for a star to initiate hydrogen fusion in its core (becoming a red dwarf star) isn’t precisely known, but it’s believed to be around 0.075 solar masses. Given that we know the weight of those marshmallows, we can calculate that they have a density of roughly 73 kilograms per metre cubed. This means that 0.075 solar masses of marshmallows would make a sphere with a radius about 1.13 times that of the Sun.*** This sphere would contain approximately 20 nonillion marshmallows.

To be honest, I could probably say more about this. But this is probably the silliest thing I’ve ever written. I think that’s more than enough gratuitous geekery for one evening!

EDIT–So evidently I had a couple of my values wrong in this post. But I’ve fixed them now. I should remember to double check my calculations before posting, the next time I engage in random mathematical silliness, I suppose. Thanks to those who picked me up on my mistakes.

*I knew I’d find some use for that unit of measurement someday…

**Just imagine how many smores you could make!

***Stars are denser than marshmallows. One solar mass of marshmallows would make a sphere with 2.67 times the radius of the Sun. Admittedly though, this is assuming that gravity wouldn’t start to collapse the marshmallows once you gathered enough of them — which it would.

Image credit: “Yella Mella Macra” by flattop341

Molecular astrophysicist, usually found writing frenziedly, staring at the sky, or drinking mojitos.
This entry was posted in Imported from Livejournal, physics and tagged , . Bookmark the permalink.

### 33 Responses to A Universe of Marshmallows

Re: ROTAR : this is your mistakes

2. Anonymous says:

ROTAR : this is your mistakes
2*(pi^sqrt(atan(2418769))-3*E) is not equal with 795.3326

3. moon_razor says:

I initially thought one could assume expansion in the vacuum of space – however a marshmallow filled universe would hardly be considered a vacuum. Then one could assume that surfaces of the marshmallows would touch each other preventing expansion – however in a marshmallow filled universe – gravitational pressure would also have an effect. I hypothesis there must be an equilibrium volume of the marshmallows (which may be different from the standard volume of a marshmallow) which may or may not be stable to perturbations.
I don’t have any desire to do these calculations – so the “ideal marshmallow approximation” works for me :)

Re: I seriously loved this!
Haha… Oh, I’m working on the reputation, evidently. :P
(Although you’re right — some of the blame definitely belongs to the two of you ;)
…so full of goo!
Hehe!

Yeah, I’d considered that. I’m assuming, for the sake of simplicity, that these marshmallows undergo negligible expansion in the vacuum of space. Call it an “ideal marshmallow approximation” if you will. :)
(In the same way, a marshmallow star would be smaller than my estimate, due to marshmallows being condensed under pressure in its core…)

6. Anonymous says:

I seriously loved this!
This was just so fantastic. And if you haven’t got a reputation as a mad scientist yet, why are people so ungrateful? Anyway, if you obtain one, then it’s my fault, and Stellar’s.
Fancy, a marshmallow star. Mmmmm . . . its surface would be so deliciously gooey . . .
Marshmallicitations!
PenguinGalaxy

7. moon_razor says:

Perhaps one should also think about the stability of the marshmallow structure. The given volume of a marshmallow is specific to a marshmallow on Earth because marshmallow is a structure where the internal and external pressures are in equilibrium. Thus, a marshmallows filling the universe will most likely have a different density and volume due to different external pressures (in this case dominated by gravity) and internal pressure (dependent of temperature and also protein structure).

Re: Incorrect calculation
No problem then. :)
And thanks again!

9. Anonymous says:

Re: Incorrect calculation
I apologize if you got the impression that it was patronizing, that was not my intention at all. :)

Hmmm… I must admit that no, I hadn’t heard of those. It sounds like a really interesting technique to learn though — I should read up some more! Thanks! :)

It’s very true that you never quite know where inspiration is going to come from. Plus, it’s good to try and keep your mind active. :)
And thanks!

Absolutely. ;)

Marshmallow fusion. :)

Gigamallow. That’s genius. It sounds a little like a band name…

A multipurpose post, I guess. :P

Easily best when toasted!

Hmmm… I didn’t realise that. So yes, in that case it would likely be a lower mass than 0.075M☉. [Teeth-rotting-sugariness/H] is quite high, after all. :)
Out of interest, does the same apply to brown dwarfs? Or is that only true of hydrogen fusion?

Yes, you’re quite right. Thanks for calling me on such a stupid mistake!
A wektometre? That’s interesting. I hadn’t heard of this system… Although I’d have to wonder how frequently such extremely small prefixes would even be used…
And zepto/yoctomoles? That’s even more interesting. I’m going to be looking for excuses to use those, now. :)

Re: Incorrect calculation
Hmmm… You’re quite right. I have no idea where precisely I went wrong, but I make the new figure to be 1.149x1031pc3. And yes, evidently I took the diameter instead of the radius. Man, that’s a schoolboy error. I need to practice my maths more…
Slightly patronising, but thank you all the same.

21. Anonymous says:

Now I see I go wrong myself… ;)
I took 8.21 x 10^21, not 8.21 x10^20.
A volume of 8.21 x 10^20 cubic parsec corresponds to a radius of 5.8 megaparsec.

22. Anonymous says:

Incorrect calculation
Besides the wrong number for the radius, the calculation is incorrect as well.
“So, if the observable Universe has a radius of 28 gigaparsecs, then it would have a total volume of 8.21×1020 cubic parsecs.”
I really wonder how you came up with this number… You assume a sphere, so the equation to calculate the volume if you know the radius is: V = 4/3 * pi * R^3. For a radius of 28 gigaparsec this is about 9 x 10^31 cubic parsec. But this radius is incorrect, it should be about 14 gigaparsec and that gives a volume of 1 x 10^31 cubic parsec.
A volume of 8.21 x 10^21 cubic parsec mentioned above corresponds to a radius of “only” 12.5 megaparsec.

23. Anonymous says:

“If the observable universe is a sphere, 46.5 billion light years in radius” and: “So, if the observable Universe has a radius of 28 gigaparsecs”.
The last sentence is not correct, 46.5 billion light years is about 14 gigaparsecs, not 28…

24. 6_bleen_7 says:

So sorry, but 1075 is not 0.75 googol. It is 0.0000000000000000000000001 googol. You need to multiply 1075 by 1025 to get 10100 (unless, of course, you’re working on the log scale, and then you’re still really multiplying—it just feels like addition).
A couple of people have proposed extended prefix systems, suggesting that zepto/zetta and yocto/yotta can be continued by working backwards through the Latin alphabet. In this system, 10–30 m is one wektometer.
I’ve seen both zepto- and yocto- used in the biomedical literature, in conjunction with detecting very small quantities of viral RNA. One zeptomole (zmol) is about 602 molecules, and one yoctomole (ymol), about 0.6 molecule. I think I wrote an entry a while back about why it makes sense to use the units zmol and ymol instead of numbers of molecules. (One reason is that it is exceedingly difficult to prepare a solution containing, say, exactly 100 molecules of RNA. Serial dilution of a stock of known concentration will result in a “stochastic” solution in which the actual number of molecules present is a random variable—specifically, a Poisson random variable with mean 100. There are other complicating factors, as well, such as the tendency of nucleic acids to stick to the walls of tubes.)

25. nimblenimbus says:

Delicious. :)

pretty cool stuff!
I don’t think this is gratuitous geekery. I imagine that a lot of great things come out of wild questions such as this, at the very least, you and your friends putting all your knowledge to work.

27. I guess you have heard of Fermi style problems. A question is posed about something (such as your marshmallow question), and you calculate the answer using orders of magnitude, even without a calculator.
We did some of these at the summer school I went to. Some of the questions were:
How much money has been spent so far on cold atom experiments?
By how much would you have to move Earth’s orbit away from the sun to reduce the temperature by 5 degrees? (our group had this one, we worked it out to be 5 light seconds)
What is heavier: the entire atmosphere of the Earth or the Swiss alps?

28. I love this post. That photo of the glowing marshmallows just made my day.

29. Anonymous says:

Whilst working on this yesterday, I started to use yhe measurement GM (GigaMallow)…lol