The gravitational attraction between an electron and a proton in a hydrogen atom is weaker than the Coulomb attraction by a factor of about $10^{-40}$. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction.

(N/A) The radius of the first Bohr orbit is given by the relation:
$r_{1} = \frac{4 \pi \epsilon_{0} (\frac{h}{2 \pi})^{2}}{m_{e} e^{2}} \dots (i)$
Where:
$\epsilon_{0} = 8.854 \times 10^{-12} \, C^{2} N^{-1} m^{-2}$ (Permittivity of free space)
$h = 6.63 \times 10^{-34} \, J s$ (Planck's constant)
$m_{e} = 9.1 \times 10^{-31} \, kg$ (Mass of an electron)
$e = 1.6 \times 10^{-19} \, C$ (Charge of an electron)
$m_{p} = 1.67 \times 10^{-27} \, kg$ (Mass of a proton)
$G = 6.67 \times 10^{-11} \, N m^{2} kg^{-2}$ (Gravitational constant)
Coulomb force: $F_{c} = \frac{e^{2}}{4 \pi \epsilon_{0} r^{2}}$
Gravitational force: $F_{G} = \frac{G m_{p} m_{e}}{r^{2}}$
If the electron and proton were bound by gravitational attraction, we equate the centripetal force to the gravitational force:
$\frac{m_{e} v^{2}}{r} = \frac{G m_{p} m_{e}}{r^{2}}$
Using Bohr's quantization condition $m_{e} v r = \frac{h}{2 \pi}$, we substitute $v = \frac{h}{2 \pi m_{e} r}$ into the force equation:
$r_{1} = \frac{h^{2}}{4 \pi^{2} G m_{p} m_{e}^{2}}$
Substituting the values:
$r_{1} = \frac{(6.63 \times 10^{-34})^{2}}{4 \times (3.14)^{2} \times 6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times (9.1 \times 10^{-31})^{2}}$
$r_{1} \approx 1.21 \times 10^{29} \, m$
Since the observable universe is approximately $1.5 \times 10^{27} \, m$, the calculated radius is much larger than the size of the universe.