Check that the ratio

Solution:

The given ratio is $\frac{k e^{2}}{\mathrm{G} m_{\mathrm{e}} m_{\mathrm{p}}}$.

Where,

G = Gravitational constant

Its unit is $\mathrm{N} \mathrm{m}^{2} \mathrm{~kg}^{-2}$.\

$m_{\mathrm{e}}$ and $m_{\mathrm{p}}=$ Masses of electron and proton.

Their unit is kg.

$e=$ Electric charge.

Its unit is C.

$k=\mathrm{A}$ constant

$=\frac{1}{4 \pi \in_{0}}$

$\epsilon_{0}=$ Permittivity of free space

Its unit is $\mathrm{N} \mathrm{m}^{2} \mathrm{C}^{-2}$.

Therefore, unit of the given ratio $\frac{k e^{2}}{\mathrm{G} m_{\mathrm{e}} m_{\mathrm{p}}}=\frac{\left[\mathrm{Nm}^{2} \mathrm{C}^{-2}\right]\left[\mathrm{C}^{-2}\right]}{\left[\mathrm{Nm}^{2} \mathrm{~kg}^{-2}\right][\mathrm{kg}][\mathrm{kg}]}$

$=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}$

Hence, the given ratio is dimensionless.

$e=1.6 \times 10^{-19} \mathrm{C}$

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

$m_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}$

$m_{\mathrm{p}}=1.66 \times 10^{-27} \mathrm{~kg}$

Hence, the numerical value of the given ratio is

$\frac{k e^{2}}{\mathrm{G} m_{e} m_{p}}=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{6.67 \times 10^{-11} \times 9.1 \times 10^{-3} \times 1.67 \times 10^{-22}} \approx 2.3 \times 10^{39}$

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