# How can we compare the ratio of strenghts fundamental forces? [duplicate]

I have read in many books that the ratio of strengths of gravitational force, electromagnetic force, nuclear force is 1:10^36:10^38 (one: 10 raised to thirty six: 10 raised to thirty eight). On what basis are we able to compare this? I mean, there is no connection between any two forces in the above mentioned list of forces. For example, we measure gravitational force by measuring mass, while we measure electrostatic force with charge. So how can we compare those two forces?

Or is it based on SI Units ? If it is, the ratio changes with system of units. Isn't it?

Hope someone could throw some light on this. Thank You.

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If you look at the equations for the EM and gravitational forces in Farcher's answer you'll seek they contain a constant - $\varepsilon_0$ for the EM force and $G$ for gravity. The problem is that these constants have dimensions i.e. their numerical values depend on the units we choose for (in this case) charge and mass. That makes them unsuitable for comparing the fundamental strengths of the forces.

So instead we use a parameter called the coupling constant, and it's the values of these coupling constants that gives us the relative strengths of the forces. The copupling constants are dimensionless so their value is unchanged if we change our definitions of the coulomb, kilogram or whatever.

These coupling constants are approximately:

\begin{align} \text{Strong force}\, \alpha_S &\approx 1 \\ \text{EM force}\, \alpha_{EM} &\approx \tfrac{1}{137} \\ \text{Weak force}\, \alpha_W &\approx 10^{−6} \\ \text{Grav force}\, \alpha_G &\approx 10^{−38} \end{align}

This is the origin of the $1:10^{36}:10^{38}$ ratio that you cite.

Note that the coupling constants $\alpha$ aren't actually constant but change with the interaction energy. However this change is negligable outside special cases like collisions in the Large Hadron Collider.

A ratio is produced by comparing the gravitational force $F_g=\dfrac{Gm^2}{r^2}$ between two protons or electrons of mass $m$ and separation $r$ with the electrostatic force $F_e=\dfrac{e^2}{4\pi\epsilon_o r^2}$ with the same separation.

$$\dfrac{F_e}{F_g}=\dfrac{e^2}{4\pi \epsilon_o Gm^2}$$

The exact ratio you get depends on the charged particles that you choose.
Try putting in the values and see what you get.