Explain cosmic distances to a child I seem to have lost all my sense for simple calculations. I'd like to explain to my son how long cosmic distances are. As an example:
Our Sun has a diameter of $1392684\;\text{km}$. The distance to our next star Alpha Centauri is $4.153\cdot10^{13}\;\text{km}$. Now, if you pretend the Sun to be as large as a pinhead ($1\;\text{mm}$), the distance to Alpha Centauri would be... $6.793\;\text{km}$?
Sorry for this super-trivial question, but I am just not sure enough if my result is correct or a couple of digits wrong. Am I right?
EDIT:
Thank you Claude for noticing my second error - the Sun has a diameter of $1392684\;\text{km}$, not $1392684000\;\text{km}$ as I wrote earlier.
 A: The sun has a diameter of
$$d_{\rm Sun} = 1.39 \times 10^6 \textrm{ km}$$
The distance to Alpha Centauri (in km) is
$$d_{\rm AC} = 4.13 \times 10^{13} \textrm{ km}$$
The ratio of these quantities is
$$\frac{d_{\rm AC}}{d_{\rm Sun}} = \frac{4.13 \times 10^{13}}{1.39\times 10^6} = 2.97 \times 10^7$$
Therefore, if you had a model where the diameter of the sun was 1 mm, the distance to Alpha Centauri in the model would be 2.97 x 10^7 times greater, i.e.
$$2.97 \times 10^7 \textrm{ mm} = 29.7 \textrm{ km}$$
If you took the pinhead representing the Sun and stood next to the Washington Monument in Washington, DC, then Alpha Centauri would be about halfway to Baltimore.

Edit, more appropriate since you live in Austria! If you stood in the center of Vienna, then Alpha Centauri would be about halfway to Bratislava.
The Sun:

Alpha Centauri:

A: Going to powers will simplify. If, as you gave, the diameter of the sun was $1.392684\times 10^9$ km and the distance to Alpha Centauri is $4.153\times 10^{13}$ km, this makes a ratio of $29820$. So, if the diameter of the sun was $1$ mm, the distance to Alpha Centauri would be $29820$ mm, that is to say $29.820$ meters.  
The problem is that the diameter of the sun is $1.392684\times 10^6$ km; the same reasoning will then bring you to $29.820$ kilometers.
