In Bohr's model, the necessary centripetal force ($F_c$) for the electron's circular motion is provided by the electrostatic Coulomb attraction ($F_e$) between the nucleus and the electron.

1. Formula for centripetal force: $F_c = \frac{mv^2}{r}$
2. Formula for Coulomb force: $F_e = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$ (for Hydrogen, Z=1)
3. Equating the forces: $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$
4. Solving for velocity ($v$): $mv^2 = \frac{e^2}{4\pi\epsilon_0 r} \Rightarrow v^2 = \frac{e^2}{4\pi\epsilon_0 mr}$ $v = \frac{e}{\sqrt{4\pi\epsilon_0 mr}}$

Given that the radius of the ground state orbit is $a_0$, we substitute $r = a_0$ to get: $v = \frac{e}{\sqrt{4\pi\epsilon_0 a_0 m}}$.