जाँच द्वारा सुनिश्चित कीजिए कि $k e^{2} / G m_{e} m_{p}$ विमाहीन है। भौतिक नियतांकों की सारणी देखकर इस अनुपात का मान ज्ञात कीजिए। यह अनुपात क्या बताता है?

The given ratio is $\frac{k e^{2}}{G m_{e} m_{p}} .$ Where, $G=$ Gravitational constant. Its unit is $N m ^{2} \,kg ^{-2}$

$m _{ c }$ and $m _{ p }=$ Masses of electron and proton and their unit is kg.

$e =$ Electric charge. Its unit is $C$. $k=\frac{1}{4 \pi \varepsilon_{0}}$ and its unit is $N m ^{2} \,C ^{-2}$

Therefore, unit of the given ratio

$\frac{k e^{2}}{G m_{e} m_{p}}=\frac{\left[N m^{2} C^{-2}\right]\left[C^{-2}\right]}{\left[N\, m^{2}\, k g^{-2}\right][k g][k g]}$$=M^{0} L^{0} T^{0}$

Hence, the given ratio is dimensionless. $e=1.6 \times 10^{-19} \,C$

$G=6.67 \times 10^{-11}\, N m ^{2}\, kg ^{-2}$

$m _{ e }=9.1 \times 10^{-31} \,kg$

$m _{ p }=1.66 \times 10^{-27}\, kg$

Hence, the numerical value of the given ratio is

$\frac{k e^{2}}{G m_{e} m_{p}}=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{6.67 \times 10^{-11} \times 9.1 \times 10^{-31} \times 1.67 \times 10^{-27}}$$\approx 2.3 \times 10^{39}$

This is the ratio of electric force to the gravitational force between a proton and an electron, keeping distance between them constant.