Question

Check that the ratio $\dfrac{k\,e^{2}}{G\,m_e\,m_p}$ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Hint:

The ratio equals the electric force divided by the gravitational force between an electron and a proton (the \(r^2\) cancels). Units of \(k e^2\) and \(G m_e m_p\) both reduce to N m^2, so it is dimensionless; value \(\approx 2.3\times10^{39}\).

Answer 1

We must show the ratio \(\dfrac{k\,e^2}{G\,m_e\,m_p}\) is dimensionless, evaluate it, and interpret it.

Step 1: Identify the two forces. For an electron and a proton a distance \(r\) apart, the electric (Coulomb) force is \(F_e=k\,e^2/r^2\) and the gravitational force is \(F_g=G\,m_e\,m_p/r^2\). The given ratio is exactly \(F_e/F_g\) because the \(r^2\) factors cancel.

Step 2: Check dimensions. Write the dimensional formulae.
\([k]=[F\,r^2/q^2]=\dfrac{\text{N m}^2}{\text{C}^2}\)
\([G]=[F\,r^2/m^2]=\dfrac{\text{N m}^2}{\text{kg}^2}\)
\([e^2]=\text{C}^2,\qquad [m_e m_p]=\text{kg}^2\).

Step 3: Substitute the units into the ratio.
\[\frac{[k][e^2]}{[G][m_e m_p]}=\frac{(\text{N m}^2\,\text{C}^{-2})(\text{C}^2)}{(\text{N m}^2\,\text{kg}^{-2})(\text{kg}^2)}=\frac{\text{N m}^2}{\text{N m}^2}=1\]
All units cancel, so the ratio is dimensionless.

Step 4: Insert the numerical constants.
\(k=9\times10^{9}\), \(e=1.6\times10^{-19}\ \text{C}\), \(G=6.67\times10^{-11}\), \(m_e=9.11\times10^{-31}\ \text{kg}\), \(m_p=1.67\times10^{-27}\ \text{kg}\).
\[\frac{k e^2}{G m_e m_p}=\frac{(9\times10^{9})(1.6\times10^{-19})^2}{(6.67\times10^{-11})(9.11\times10^{-31})(1.67\times10^{-27})}\]

Step 5: Evaluate.
Numerator: \(9\times10^{9}\times 2.56\times10^{-38}=2.304\times10^{-28}\)
Denominator: \(6.67\times10^{-11}\times 9.11\times10^{-31}\times 1.67\times10^{-27}=1.0147\times10^{-67}\)
\[\frac{k e^2}{G m_e m_p}=\frac{2.304\times10^{-28}}{1.0147\times10^{-67}}\approx 2.3\times10^{39}\]

Step 6: Significance. This number is the ratio of the electrostatic force to the gravitational force between the electron and the proton. It is about \(2.3\times10^{39}\), showing the electric force is enormously, vastly stronger than the gravitational force at the atomic scale, which is why gravity is neglected inside atoms.

\[\boxed{\frac{k e^2}{G m_e m_p}\approx 2.3\times10^{39}\ (\text{dimensionless})}\]

Answer 2

Expert approach: prove dimensionlessness from base SI dimensions, then interpret as a coupling-constant ratio.

Step 1: Reduce each constant to base dimensions \([M],[L],[T],[A]\) (current), since charge \([Q]=[A][T]\).
\([k]=\dfrac{1}{4\pi\epsilon_0}\) has dimensions \(M\,L^3\,T^{-4}\,A^{-2}\).
\([G]=M^{-1}\,L^{3}\,T^{-2}\).
\([e^2]=Q^2=A^2 T^2\).
\([m_e m_p]=M^2\).

Step 2: Assemble the numerator dimension: \([k][e^2]=M L^3 T^{-4} A^{-2}\cdot A^2 T^2=M L^3 T^{-2}\).

Step 3: Assemble the denominator dimension: \([G][m_e m_p]=M^{-1}L^3 T^{-2}\cdot M^2=M L^3 T^{-2}\).

Step 4: The numerator and denominator have identical dimensions \(M L^3 T^{-2}\), so the ratio is a pure number. This is the cleanest possible proof: it never assumed the two forces, only the SI definitions of the constants.

Step 5: Order-of-magnitude value via exponents only. Numerator exponent: \(10^{9}\times(10^{-19})^2=10^{9-38}=10^{-29}\), mantissa \(9\times2.56\approx23\Rightarrow 2.3\times10^{-28}\). Denominator exponent: \(10^{-11}\times10^{-31}\times10^{-27}=10^{-69}\), mantissa \(6.67\times9.11\times1.67\approx101\Rightarrow 1.01\times10^{-67}\). Ratio \(\approx 2.3\times10^{39}\).

Step 6: Physically this is the ratio of the electromagnetic coupling strength to the gravitational coupling strength for the electron-proton pair. Its colossal size, \(\sim10^{39}\), is the reason gravity plays no role in binding atoms.

\[\boxed{\approx 2.3\times10^{39}}\]