Gravitational attraction is given by $\frac{GMm}{r^2}$ while attraction due to electric charge is given by $\frac{q_1 q_2}{r^2}$. Why does gravity need a constant while electric charge doesn't? Because we've picked "the right" units for charge so that the constant is 1.

This is my vague understanding of "natural units". My question is: what would natural units for mass look like, and why don't we use them?

  • $\begingroup$ See Plank Mass $\endgroup$
    – user442
    Dec 5, 2010 at 23:53
  • 1
    $\begingroup$ REgarding to your related point, it will probably not last long anymore : Wikipedia (en.wikipedia.org/wiki/…) states "In October 2010, the International Committee for Weights and Measures (known by its French-language initials CIPM) voted to submit a resolution for consideration at the General Conference on Weights and Measures (CGPM), that the kilogram be defined in terms of the Planck constant, h." $\endgroup$ Dec 6, 2010 at 10:25
  • 1
    $\begingroup$ Every time I've practically worked with electric charge, I've always used either $k \frac{q_1 q_2}{r^2}$ or $\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}$. I understand that natural units for electromagnetism exist, but SI units they are not. $\endgroup$
    – Justin L.
    Feb 23, 2011 at 8:49

1 Answer 1


Well, that equation for the force due to electric charges is only true for a very special choice for the unit of the electric charge. Typically, you would write down Coulomb's law as

$k\frac{q_{1}q_{2}}{r^{2}}$, where $k$ is a constant of proportionality chosen to make the units work out. IN the SI system, the unit of charge is the Coulomb (C) and the value of $k$ is approximately $8.99 \cdot 10^{9} \frac{N\cdot m^{2}}{C^{2}}$.

If you choose to express your charge in Gauss units, defining one Gauss to be numerically equal to $\sqrt{8.99\cdot 10^{9}} \,\,C$, then the value of the Coulomb constant becomes 1 in this new system of units.

Now, you could do the same thing for gravity, but your unit of mass would be really small: $\sqrt{6.67 \cdot 10^{-11}} \,\, kg$. It would also mess up the definitions for any unit that had dimensionality of mass, like the Newton, Joule, etc. But you could do that.

Also, there is a scheme called Planck Units that attempts to set as many fundamental constants of nature as possible equal to one.

  • $\begingroup$ Well, the electron volt is a tiny unit, but we just talk in terms of MeV... $\endgroup$
    – Seamus
    Dec 5, 2010 at 12:17
  • $\begingroup$ From the perspective of gravitational physics, where the masses you deal with are things like the sun, it is a tiny unit--eV are a great unit for dealing with atomic physics, because it's a microscopic unit that describes microscopic effects. No one measures the gravity between two neutrons, though, so most people leave Newton's constant in when they do Newtonian gravity problems. That said, most working relativists DO choose units such that $G=1$ in Einstein's equation. $\endgroup$ Dec 5, 2010 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.