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I read the numerous threads here for example:

Lower limit of the size of the Universe? (WMAP)

Determining the size of the universe to calculate its age

Size of the universe

But I would like to try a novel way to calculate this to illustrate it for middle schoolers.

I would start with the radius of a hydrogen atom calculate its volume then multiply by the number of calculated atoms in the universe.

At this point, the engaged students will point out that the whole universe is not hydrogen. I will point out that 90% of the universe is hydrogen and we are just approximating. (This teachable moment also introduces them to the utility of estimating) Once I have the volume of the universe, I calculate its radius.

I then use the following python code (again a teachable moment about the value of coding)

import math
Hv=math.pi*4/3*(53e-12**3)
print('current volume of hydrogen atom',Hv, 'cubic meters' ) 
print('number of atoms in the universe',1e+83)   
print('current volume of the universe',1e+83*Hv, 'cubic meters' )
print('current radius of the universe',(1e+83*Hv*(3/4)/math.pi) ** (1. / 3), ' meters' )
print('meters in a light year ,',9.461e+15)    
print('current radius of the universe',((1e+83*Hv*(3/4)/math.pi) ** (1. / 3))/9.461e+15, ' light years' )

the output is

current volume of hydrogen atom 6.236145193179834e-31 cubic meters
number of atoms in the universe 1e+83
current volume of the universe 6.236145193179834e+52 cubic meters
current radius of the universe 2.460042081814767e+17  meters
meters in a light year , 9461000000000000.0
current radius of the universe 26.001924551472012  light years

So my rough calculation of 26 million light years is way off. The really engaged students will point out that the hydrogen gas in the universe is not just a bunch of hydrogen atoms next to each other but there is empty space around each one. What they will try to grasp is the density of hydrogen in space. Some of the respondents here suggested using 1 atom/4 cc. So then I can calculate

import math
print (1e+83*4,'cc in the universe')
print ('current volume of the universe',1e+83*4/1e+6,'cubic meters')
print('current radius of the universe',(4e+83/1e+6*(3/4)/math.pi) ** (1. / 3), ' meters' )
print('meters in a light year ,',9.461e+15)    
print('current radius of the universe',((4e+83/1e+6*(3/4)/math.pi) ** (1. / 3))/9.461e+15, ' light years' )
print('current radius of the universe',f'{(1e+83*4/1e+6*(3/4)/math.pi) ** (1. / 3)/9.461e+15:,}', ' light years' )

output

4e+83 cc in the universe
current volume of the universe 4e+77 cubic meters
current radius of the universe 4.570781497340817e+25  meters
meters in a light year , 9461000000000000.0
current radius of the universe 4831182218.941779  light years
current radius of the universe 4,831,182,218.941779  light years

So 4.8 billion light years. Still off by a factor of 10.

EDIT: So sad that I made typos that led to serious errors.

Qmechanic
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aquagremlin
  • 1,729

2 Answers2

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I'm sorry, but your calculation was doomed from the start. To estimate number of atoms in the universe, you start with the assumption that most of the atoms in the universe are in or near galaxies. Then you count galaxies, multiply by an estimate of mass per average galaxy, and divide by hydrogen mass.

As such no amount of manipulation of that estimate can get you an appropriate calculation of the size of all the space that the estimate ignored to begin with.

Your code also has very incorrect values in it. Your estimate for atoms in the universe is under by a factor of $10^{62}$ to $10^{66}$, density of intergalactic medium over by a factor of at least $10^6$. I'm not sure whether your number for universe volume is supposed to be the true value, or the value you're calculating. If it's the former, it is also wrong by a significant factor. If the latter, you should be printing a variable that holds the calculated value, not a hard coded number.

g s
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1

Number of Hydrogen per volume

The critical density of the universe is a handful of hydrogen per cubic meter. The actual number of hydrogen per cubic meter is roughly 0.25. 1 per cubic centimeter is okay for local space, which is much much denser than the Universe on average, because we are in a galaxy, an overdensity of the Universe.

Size of the Universe

The size of the Universe depends on how you define it. As an example, I'll use $R = 4.4\times10^{26}$m (46 billion light-years) corresponding to the observable Universe.

The number of hydrogen atoms in a volume corresponding to that sphere is roughly $9\times10^{79}$.

Hv=math.pi*4/3*(1.2e-11**1.2e-11**1.2e-11)

I believe there is an error, since ** is exponentiation, and you are exponentiating this factor of 1.2e-11 rather than multiplying.

Hv=math.pi*4/3*(1.2e-11)**3

Alwin
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