Comparing the strength of gravity and electricity

Question

I often see and hear people claiming that "the gravitational force is much weaker than the electromagnetic force". Usually, they justify it by comparing the universal gravity constant to Coulomb's constant. But obviously, such comparison is meaningless, as they differ in dimensions. I'll make myself clear: of course you can say it is true for electron-electron interaction, but I'm talking about whether they can be compared fundamentally somehow in any area of physics.

Answer 1

Yes, they can. Both interactions can be modeled using perturbative quantum field theory, where their strength is parametrized by a dimensionless coupling constant.

Electromagnetic repulsion between two electrons can be written as a power series in $\alpha$, the fine structure constant, which is dimensionless and has a value of roughly 1/137.

Meanwhile, the gravitational attraction between two electrons can be expanded in a similar way in a power series in $\alpha_G$, which is a dimensionless constant with a value of roughly $10^{-45}$.

The precise value of $\alpha_G$ depends somewhat on which particle you're comparing, since ultimately it's the square of the ratio of the particle's mass to the Planck mass. However, for fundamental particles, this ratio does not vary by more than ten orders of magnitude, which still places $\alpha_G$ far smaller than $\alpha$ no matter which fundamental particle you choose to compare.

Answer 2

You may not be able to compare the constants directly, but you can compare the resulting forces. For instance, when you put two electrons $1\mathrm m$ apart, you can calculate their gravitational attraction and their electrostatic repulsion. Guess which of the two forces dwarves the other...

Answer 3

I think the best symbolic comparison is my formula expressing G in terms of the electron mass m_e, crossed Plank constant hbar, speed of light c and the fine structure constant. The formula is:

G = 4/3/2^(1/38) hbar c/m_e^2 alpha^21 or in words that the Gravity constant is equal to 4/3 (the factor of the sphere of radius r volume V=4/3 Pi r^3) divided by small "ellipsoidal" correction the 38-th root of 2 times the crossed Plank constant times speed of light divided by square of the electron mass times the fine structure constant in power 21-st. Or rewriting it as G m_e^2/lambda = 4/3/2^(1/38) 1/4/Pi/epsilon_0 e^2 alpha^20/lambda and implicitly dividing by any distance lambda means that the gravitational energy depletion between two electrons from zero is equal to their Coulomb energy slightly corrected by near-one factor 4/3/2^(1/38) in 20-th power of fine structure constant alpha which is close to 1/137 what clearly means it is small.

Answer 4

Here is how I would compare these two forces in straightforward way:

\begin{equation} \frac{F_{e}}{F_{g}} = \frac{\frac{1}{4 π ε_{0}}\frac{e^2}{r^2}}{G\frac{m_{e}^2}{r^2}} \end{equation}

The terms $r^2$ will cancel out, so it is possible to compute this expression. You get value of approximately $4.1656 \times 10^{42}$.

You can say that: \begin{equation} \frac{F_{e}}{F_{g}} = \frac{\alpha_{e}}{\alpha_{g}} \end{equation} From this you get $\alpha_{g} = 1.7518 \times 10^{-45}$, which agrees with accepted answer.

Answer 5

Welcome to the Physics SE site.

The forces can be compared on a fundamental level with reference to the universal gravitational constant and the Coulomb constant.
These constants have different units indeed but this is only logical as these constants deal with mass ($kg$) and charge combined with mass ($C$ and $kg$, as will become clear). Only the values of these constants are compared.

The universal gravitational constant:

$$\frac{m^3}{s^2kg},$$

And the Coulomb constant:

$$\frac{Nm^2}{C^2}=\frac{m^3kg}{s^2C^2}.$$

Both constants give $F=ma$ if we fill them in in the corresponding formulae of the corresponding forces (the Coulomb force and Newton's law of gravity, taking $C$ for the charges $q_1$ and $q_2$ in the Coulomb force and one kilogram for the masses in Newton's law).

Obviously, these units have a factor of $\frac{m^3}{s^2}$ in common. So what we actually compare are $\frac{1}{kg}$ and $\frac{kg}{C^2}$, which means only the cause of the electric field ($C$) and the cause of the gravitational field ($kg$) are involved in the comparison. So this leads to (setting $\frac{1}{kg}$ equal to $\frac{kg}{C^2}$) $kg^2=C^2.$

Comparing the two (though units differ, we compare the values), and filling in for the Coulomb $6,2\times 10^{18}$ (see here), and for one kilogram, well, 1:

$$1=38,4\times 10^{36}$$

We have to divide the right side still by $4\pi$ which means that the electric force is about $10^{36}$ times as strong as the gravitational force, as it should be.