Amit & Deepali Sinha Foundation

Gravity Loses by Forty Orders

Irodov 3.1DifficultyUniqueness

Problem Irodov 3.1

Two electrons attract each other gravitationally and repel each other electrically. Find the ratio of the electric force to the gravitational force between them, and the same ratio for two protons. Then find the charge-to-mass ratio q/mq/m that two identical particles would need for the two forces between them to have equal magnitudes.

Given
  • two electrons
  • two protons
  • identical particles, charge qq, mass mm
Find
Fe/FgF_e/F_g for each pair; the q/mq/m at which Fe=FgF_e = F_g

Solution

Step 1Both forces fall off the same way

Two identical particles of charge qq and mass mm a distance rr apart feel the Coulomb force and the gravitational force, both along the line joining them:

Fe=14πε0 q2r2Fg=G m2r2(1) \begin{gathered} F_e = \frac{1}{4\pi\varepsilon_0}\,\frac{q^2}{r^2}\\ F_g = G\,\frac{m^2}{r^2} \end{gathered} \tag{1}
Two identical particles a distance r apart. On each, the electric repulsion F_e and the gravitational attraction F_g act along the line joining them. Arrow lengths are schematic.

Two identical particles a distance rr apart. On each, the electric repulsion Fe\vect{F}_e and the gravitational attraction Fg\vect{F}_g act along the line joining them. Arrow lengths are schematic.

Dividing the two forces in (1), the distance cancels:

FeFg=q24πε0 G m2\frac{F_e}{F_g} = \frac{q^2}{4\pi\varepsilon_0\,G\,m^2}
(2)
Step 2Electrons and protons

For electrons the charge is the elementary charge ee and the mass is mem_e. The constants are

e=1.60×10−19 C,me=9.11×10−31 kg,14πε0=9.0×109 m/F,G=6.67×10−11 N m2/kg2\begin{gathered} e = 1.60\times10^{-19}\ \text{C}, \qquad m_e = 9.11\times10^{-31}\ \text{kg},\\ \frac{1}{4\pi\varepsilon_0} = 9.0\times10^{9}\ \text{m/F}, \qquad G = 6.67\times10^{-11}\ \text{N\,m}^2/\text{kg}^2 \end{gathered}

and (2) gives

FeFg=(9.0×109) (1.60×10−19)2(6.67×10−11) (9.11×10−31)2=2.30×10−285.54×10−71=4.2×1042\begin{aligned} \frac{F_e}{F_g} &= \frac{(9.0\times10^{9})\,(1.60\times10^{-19})^2}{(6.67\times10^{-11})\,(9.11\times10^{-31})^2}\\ &= \frac{2.30\times10^{-28}}{5.54\times10^{-71}} = 4.2\times10^{42} \end{aligned}

For protons the charge is the same and the mass is mp=1.672×10−27m_p = 1.672\times10^{-27} kg:

FeFg=(9.0×109) (1.60×10−19)2(6.67×10−11) (1.672×10−27)2=2.30×10−281.86×10−64=1.2×1036\begin{aligned} \frac{F_e}{F_g} &= \frac{(9.0\times10^{9})\,(1.60\times10^{-19})^2}{(6.67\times10^{-11})\,(1.672\times10^{-27})^2}\\ &= \frac{2.30\times10^{-28}}{1.86\times10^{-64}} = 1.2\times10^{36} \end{aligned}
electrons: Fe/Fg≈4×1042protons: Fe/Fg≈1×1036\begin{gathered}\text{electrons: } F_e/F_g \approx 4\times10^{42}\\ \text{protons: } F_e/F_g \approx 1\times10^{36}\end{gathered}
(3)
Step 3The charge per mass that makes them equal

Setting the ratio (2) equal to one and solving for the charge per unit mass:

q2m2=4πε0G=6.67×10−119.0×109=7.4×10−21 C2/kg2\frac{q^2}{m^2} = 4\pi\varepsilon_0 G = \frac{6.67\times10^{-11}}{9.0\times10^{9}} = 7.4\times10^{-21}\ \text{C}^2/\text{kg}^2
qm=4πε0G≈0.86×10−10 C/kg\frac{q}{m} = \sqrt{4\pi\varepsilon_0 G} \approx 0.86\times10^{-10}\ \text{C/kg}
(4)
Check

The critical ratio of Step 3 recovers the electron result of Step 2 by a different route: by (2), Fe/FgF_e/F_g is the square of the particle’s q/mq/m over the critical one, and the electron’s specific charge is e/me=1.76×1011e/m_e = 1.76\times10^{11} C/kg, so

FeFg=(e/meq/m)2=(1.76×10110.86×10−10)2=4.2×1042\frac{F_e}{F_g} = \left(\frac{e/m_e}{q/m}\right)^2 = \left(\frac{1.76\times10^{11}}{0.86\times10^{-10}}\right)^2 = 4.2\times10^{42}

as found. Units: 4πε0G4\pi\varepsilon_0 G is in (C2/N m2)(N m2/kg2)=C2/kg2(\text{C}^2/\text{N\,m}^2)(\text{N\,m}^2/\text{kg}^2) = \text{C}^2/\text{kg}^2, so its root is in C/kg.

Remark

The critical q/mq/m is tiny: an electron carries e/me=1.76×1011e/m_e = 1.76\times10^{11} C/kg, twenty one orders of magnitude more. A body of 1 kg needs an excess of only 0.86×10−100.86\times10^{-10} C, about 5×1085\times10^{8} electrons among its roughly 3×10263\times10^{26}, for its electric repulsion from an identical body to match their gravitational pull. Gravity rules planets and stars only because matter in bulk is neutral to far better than that.