8. There are 2003 circles each tangent to the two sides of $\angle M O N$, and adjacent circles are externally tangent to each other. Given that $\angle M O N = 60^{\circ}$, and the radius of the smallest circle is 1, then the radius of the 2003rd circle, arranged in increasing order of radius, is $\qquad$
8. Let the radii of the two tangent circles $\odot O_{n}, \odot O_{n+1}$ be $R_{n}, R_{n+1}$, respectively. It is easy to know that $\frac{R_{n+1}}{R_{n}}=\frac{1+\sin 30^{\circ}}{1-\sin 30^{\circ}}=3$, so the sequence $\left\{R_{n}\right\}$, arranged from smallest to largest, is a geometric sequence with the first term 1 and common ratio 3. $R_{2003}=3^{2002}$.