# Scale of the universe if a star was a grain of sand

1. Jul 30, 2014

### pelooyen

As a teacher, I am trying to help students understand the scale of the universe.

Using the analogy of sand as stars, and based on the average size of sand as .5mm across, I worked out...

1. The number of stars is greater than the grains of sand on all the beaches of the world
(allowing for the fact that that actual length of coastline is variable depending on how you measure it)

2. The number of stars in our galaxy is about 100 billion stars, and so, in terms of sand, that is about 500 tonne and would fill approx 320 cubic meters, so three large rooms worth. That's considerably smaller than all the beaches of the earth.

3. However, if an average star is about the size of a grain of sand...
- our solar system would be 3.5 m across (give or take where your think the boundary is)
- our galaxy would be still 340,000 km across

My question is:
Am I roughly correct in my estimation? I'd hate to think I am out by many orders of magnitude :D

Paul Looyen

2. Jul 30, 2014

### Staff: Mentor

This interactive video might help with showing the scale of things:

http://htwins.net/scale2/

and then there's the classic Powers of Ten video by Charles Eames:

Last edited by a moderator: Sep 25, 2014
3. Jul 30, 2014

### Matterwave

Let's take your word that sand is .5mm across (pretty fine sand it seems), and do some order of magnitude, back of the envelope calculations then! .5mm across means .25mm in radius. We assume sand is a sphere.

1) There are about 100 billion (dwarf galaxies have fewer) to 1 trillion stars per galaxy (only for the largest elliptical galaxies), and about 200 billion galaxies in our observable universe. Let's just take each galaxy to have approximately 200 billion stars (roughly size of the Milky way). That's 2*10^11 stars per galaxy, and 2*10^11 galaxies for a total of 4*10^22 stars.

Next we have to estimate how much sand is on all the beaches of the world. Certainly not an easy task. The CIA world fact book states that there is about 400,000km of coastline in the World. Now, of course, not all of that coastline is comprised of beaches. But we can use it as an upper bound. If each beach is roughly 50 meters in length, and 10 meters deep (just my personal guesses, if you want to use others, you can do so as well), then that's roughly 4*10^8m*50m*10m=10^11 cubic meters in volume. Looking at problem 2) (I did that one earlier since it's easier) we see that 10 cubic meters of sand contains roughly 10^11 grains, or 1 cubic meter contains 10^10 grains.

As such, 10^11 cubic meters of sand contains roughly 10^11*10^10=10^21 grains of sand. This is comparable to the number of stars within the observable universe, being only 1 order of magnitude lower. We can't conclude for certain, then either way, because we may certainly be off by an order of magnitude or two in either direction. This is probably the most uncertain calculation.

2) I've heard the estimate 200 billion for our Milky way galaxy, but let's use 100 billion here. We said sand was .25mm in radius, that means it's volume is 4/3*pi*(1/4 mm)^3~4*(1/64 mm^3)~1/16 mm^3. Call that 10^-1 mm^3. 10^-1 mm^3 times 10^11 stars equal 10^10 mm^3 which is 10 cubic meters. So we are off from each other by 2 orders of magnitude here. Similarly, I think our mass estimates will be also off from each other. I don't see an error in my calculation, perhaps you can spot one?

3) Our solar system's size is very very variable depending on what you define to be our solar system. If you define our solar system to be to the orbit of Neptune, then the size is ~60AU across. If you define our system to include the Kuiper belt, then the size is ~100AU across (the Heliopause is a bit farther, at 120AU or so). If you include the oort cloud, then the size is ~100,000AU (across).

Stars of course, also vary by size quite a bit, in terms of size. The Sun is ~700,000km in radius, while the largest red giants may have a radius that can be >1AU. Let's just use 1 million km for our calculations, and neglect the massive stars. This is good because there are not all that many massive stars compared with the very tiny ones (the massive stars die very quickly!). So 10^6 km = .25mm. That means 1AU ~150 million km = 150*.25mm~300mm (30 centimeters).

This means the the orbit of Neptune is ~20 meters across, the Kuiper belt and Heliopause makes the solar system ~40 meters across, and the oort cloud makes it a ridiculous 30,000 meters (30 km) across!

Similar concerns happen with our Galaxy. Usually one uses the dimensions of ~50,000 light years in radius, but the halo of stars surrounding our galaxy extends far beyond that. The dark matter distribution extends probably much much farther still. Still, let's take the conventional 10^5 light years across scenario. 10^5 light years, with one light year ~300,000AU means that the galaxy is roughly 3*10^10 AU ~ 10^10 meters (10 million km) across!

It seems here we are off again by a couple of orders of magnitude.

Of course I may be wrong, or off by a couple of orders of magnitude. I did not double/triple check my results.

4. Jul 31, 2014

### davenn

hi ya Paul

welcome to PF

you can add a lot more to your count, here's info from one site and it roughly matches others that I quickly looked at

one quote I saw some long time ago was " there's more stars in the universe than ALL the grains of sand on earth's beaches and deserts

Dave

5. Aug 1, 2014

### Doug Huffman

The Anthropic Cosmological Principle Hardcover – March 6, 1986
by John D. Barrow (Author), Frank J. Tipler (Author), & 1 more
ISBN-13: 978-0198519492 ISBN-10: 0198519494

Would be a good read for you.

6. Aug 1, 2014

### trumpettree

Randall Munroe has tackled the sand:star analogy in his What-If series:
http://what-if.xkcd.com/83/
1) Only for the Milky Way, but a good basis for your average galaxy.
2) "Sand" works for the majority of stars, but the volume of star "gravel" and star "boulders" would dwarf our star beach!