What is the magnitude of the gravitational force between two…

What is the magnitude of the gravitational force between two electrons that are $r = 0.2\ \mathrm{nm}$ apart? The mass of an electron is $m = 9 \cdot 10^{-31}\ \mathrm{kg}$, and the gravitational constant is $\gamma = 6.67 \cdot 10^{-11}\ \mathrm{N\,m^2/kg^2}$.

Step 1: Idea

Every two bodies with mass attract each other. For point-like bodies (and electrons certainly are point-like at this distance) the size of the attraction is given by Newton's law of gravitation: it is proportional to the product of the two masses and inversely proportional to the square of their distance.

\[ F_g = \gamma\frac{m_1m_2}{r^2}. \]

Step 2: Substituting the data

Both bodies are electrons, so $m_1 = m_2 = m$ and the product of the masses is $m^2$. The distance must be in metres: $0.2\ \mathrm{nm} = 0.2 \cdot 10^{-9}\ \mathrm{m} = 2 \cdot 10^{-10}\ \mathrm{m}$.

\[ F_g = \gamma\frac{m^2}{r^2} = 6.67 \cdot 10^{-11} \cdot \frac{(9 \cdot 10^{-31})^2}{(2 \cdot 10^{-10})^2}. \]

Work out the fraction first: $(9 \cdot 10^{-31})^2 = 81 \cdot 10^{-62}$ and $(2 \cdot 10^{-10})^2 = 4 \cdot 10^{-20}$, so the fraction is $20.25 \cdot 10^{-42} = 2.025 \cdot 10^{-41}\ \mathrm{kg^2/m^2}$.

\[ F_g = 6.67 \cdot 10^{-11} \cdot 2.025 \cdot 10^{-41}. \]

\[F_g \approx 1.35 \cdot 10^{-51}\ \mathrm{N}\]

Step 3: What the number means

The force is unimaginably small. Two electrons also repel each other electrically, and that force is about $10^{42}$ times larger (see the check below). This is why gravity can be ignored completely in atomic and molecular physics; it only becomes important for very large masses such as planets and stars, which are electrically neutral overall.

Check

Units: $ \mathrm{N\,m^2/kg^2} \cdot \mathrm{kg^2/m^2} $ $= \mathrm{N}\ \checkmark $

Comparison: Coulomb repulsion $ k e^2/r^2 $ $= 9 \cdot 10^9 \cdot (1.6 \cdot 10^{-19})^2 / (4 \cdot 10^{-20}) $ $\approx 5.8 \cdot 10^{-9}\ \mathrm{N} $, about $4 \cdot 10^{42}$ times larger