# If the earth was reduced to the size of an atom how big would the galaxy be?

$x = \frac{\left(1 \times {10}^{- 10}\right) \left(1 \times {10}^{21}\right)}{12742000} = \frac{1 \times {10}^{11}}{12742000} \cong 7848 m \approx 8 k m$

#### Explanation:

We have a ratio question - given the size of the Earth and reducing it to the size of an atom, how big will the galaxy be? That looks like:

$\frac{\text{Earth"/"atom"="galaxy}}{x}$

So first off we can solve for x:

x=("atom"xx"galaxy")/"Earth"

And now we need to make a decision - the size of an atom depends on which atom we're talking about. They typically run between $0.1 \to 0.5$ nanometres. I'm going to assume we're using a smaller atom (it'll make the math easier).

And now let's get some numbers:

$\text{Earth} = 12 , 742 k m = 12 , 742 , 000 m$

https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8q=earth+diameter+in+km

$\text{Atom} = 1 \times {10}^{-} 10 m$

$\text{Galaxy} = 1 , 000 , 000 , 000 , 000 , 000 , 000 , 000 = 1 \times {10}^{21} m$

And now the math:

$x = \frac{\left(1 \times {10}^{- 10}\right) \left(1 \times {10}^{21}\right)}{12742000} = \frac{1 \times {10}^{11}}{12742000} \cong 7848 m \approx 8 k m$