# Astronomical numbers

1. Feb 2, 2010

### Freeman Dyson

1. The problem statement, all variables and given/known data

Convert astronomical objects into money based on their radius. This is how it works. The earth is worth one penny. $.01 The earth's radius is 6,378 km. So 6,378 km equal one penny in this system. So the sun, which has a radius of 696,000 km. We divide that by 6,378. Which is 109 cents. So the sun is worth 1.09 Solar System is 40 AU. I want this in km. AU is 1.5 x 10^8 km. So do 40 times 1.5 x 10^8 =6,000,000,000km or 6 x 10^9. So then I divide 6 x 10^9 by 6.378 x 10^3. And I get 940,733 cents. Drop 2 decimals to convert to dollars and I get$9,407 for the price of the solar system.

I believe those are right. Anyway, it is the bigger ones that I am having problems with. Next is the Milky Way. Milky way is 50,000 light years away.

2. Relevant equations

3. The attempt at a solution

So this is how I am going about the Milky Way one. Probably wrongly. I want to convert the those 50,000 light years into km. Or do I? Anyway, I'll try it. A light year is 10 trillion km. So I multiply 50,000 by 10 trillion. 5 x 10^4 x 1 x 10^13. So that is 5 x 10^17. Now I divide this by 6.378 x 10^3 and get .78394481 x 10^14. Now what? Should I get rid of some exponents? Maybe 783 x 10^9? Which is 78,300,000,000,000 cents. Take out two points to convert to dollars and you have $783,000,000,000 as the cost of the Milky Way, Is this correct? Is there a better way to do it? 2. Feb 2, 2010 ### Mentallic I guess you're trying to create an analogy of the size of the universe in terms of Earth's radius. First of all, 0.78394481 x 10^14$\approx$784x10^10 so that makes$7.84 trillion.

And what do you mean by the Milky Way is 50,000 light years away? We are IN the milky way and as a note its radius is approximately 50,000 LY.

The way you're doing the calculations is fine. But just to neaten up your work a bit, if you're going to approximate the light year so much so that you say it's 10 trillion km (while 9.46 trillion is a closer approximation) then you should only stick to 2 or 3 significant figures at most.

.78394481 x 10^14 is unnecessary :tongue:

9. Feb 3, 2010

### Mentallic

And does the \$80 billion make the galaxy seem big to you?
I've already accepted that I wouldn't be able to comprehend how big it really is. I can't even come to terms with how big, yet too small, Earth is.

10. Feb 3, 2010

### Freeman Dyson

I already knew the universe was huge. For the Local Group I got 2.5 trillion. And for the visible universe I got 215 quadrillion.

11. Feb 3, 2010

### Mentallic

Yes I know it's huge too, and this is also what anyone would tell you. The only difference is that when I try to comprehend how huge it is, using analogies in a similar fashion as you have still doesn't help me. For e.g. The Earth is already so big, but to take that as being 1 cent and having the visible universe costing 215 quadrillion doesn't help whatsoever since that sum of money is much too big for me to understand.

Another analogy: let the radius of the Earth be the radius of a typical hydrogen atom (10^-10 m) then the galaxy is 800m which is a distance I can come to grips with obviously, but the only problem is that I can't fathom the size of the atom. The visible universe will allow you to circumnavigate the globe 50 times!
I always get either one case where the starting size of say 1km or the globe's diameter is a small but understandable quantity, but then comparing this to the universe the quantity becomes much too big, or, the starting quantity is much too small to comprehend but the outcome I could fathom.