I've been wondering, in Coulomb's Law, $k_e = \dfrac{1}{4\pi\epsilon_0}$. Therefore, why do we use $G$ in Newton's Law of Gravitation? What if the constant is more like Coulomb's Law, e.g. $G = \dfrac{1}{4\pi G_0}$ where $G_0$ is some constant.

This would make Newton's Law of Gravitation look like the following: $$\bf{F}_{12} = -\dfrac{m_1 m_2}{4\pi G_0 |\bf{r}_{12}|^2} \bf{\hat{r}}_{12}$$

$GM = \mu$ is used for calculations but so could $\dfrac{M}{4\pi G_0} = \mu$.

If this is not the case, what is the significance of this definition?

  • 1
    $\begingroup$ For all I know, $1/(4 \pi \epsilon)$ is equal to 1 in CGS units. And, $4 \pi$ factor came so that $\epsilon$ could be related to Gauss Law more easily. There is no 'reason'. Most of it is history and empirical. $\endgroup$
    – Cheeku
    Commented Mar 11, 2013 at 0:20
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    $\begingroup$ @Ginger, it makes no difference how you write it. All the physics is still the same. You could write it in your new form in your research if you want (ensuring you define everything in terms of of the conventional units). $\endgroup$
    – Kenshin
    Commented Mar 11, 2013 at 0:23
  • $\begingroup$ The key point is that it's almost never relevant to use Gauss's law in gravity--nearly every mass distribution with a meaningful gravitational field in Newtonian gravitation is either spherically symmetric or a superposition of multiple spherically symmetric sources. $\endgroup$ Commented Mar 11, 2013 at 1:04
  • $\begingroup$ Related question: physics.stackexchange.com/questions/74254/… $\endgroup$
    – Johannes
    Commented Mar 15, 2014 at 15:05
  • $\begingroup$ en.wikipedia.org/wiki/Gravimagnetism $\endgroup$
    – user5402
    Commented Mar 15, 2014 at 16:20

2 Answers 2


Pure convention. There is no reason alternative conventions couldn't be used, apart from the need to avoid confusion. Newton introduced the constant to make the force law simple, whereas the electrostatic definition with the $4\pi$ is designed to make Poisson's equation (one of the equations for the electric field) look simple. You can write a Poisson equation for the gravitational field as well, and it would look simpler in your convention. (Though note that the Poisson equation for gravity gets modified by general relativity, whereas the one for electromagnetism is exact.) The physics is equivalent in both cases.

Note that in high energy physics one often uses the Planck mass which is related to the Newton constant by (up to a normalisation convention)

$$ m_P^2 = \frac{\hbar c}{G_N} \approx (2.2 \times10^{-8}\ \mathrm{kg})^2. $$

So you could write

$$ F = \frac{\hbar c m_1 m_2}{m_P^2 r^2}, $$

which is closer to what you're doing and, with units where $\hbar=c=1$, is the convention in a lot of high energy physics.

  • $\begingroup$ Technically, if I set my new constant $G_0$ = 1, the planck mass would not be the same and therefore, be a new constant ("My Surname" Mass). Therefore, it is possible that I could use this constant to form new units. $\endgroup$
    – Xplane
    Commented Mar 11, 2013 at 18:47
  • $\begingroup$ A rather late nit regarding "Newton introduced the constant to make the force law simple": Newton did not introduce the constant $G$. That first appeared in the late 19th century, a couple centuries after Newton first wrote his Principia. $\endgroup$ Commented Sep 1, 2018 at 12:06

The completely symmetric form of Coulomb's constant, such that $E=H$, is $K_c=c/4\pi$. If you equate the forces of gravity and electricity, you can wrote, eg $M=þQ$, where $þ$ is a constant. Then $G=c/4\pi þ^2$. Stoney's mass is then $eþ$ and planck's mass is $eþ/\sqrt{\alpha}$.

The rationalisation of equations only start when you do, as Heaviside and as Lorentz do, start with maxwell's equations. People are starting to do this with gravity, see, eg gravitomagnetism on the wikipedia.

SI treats rationalisation of quantities in three different ways according to whether it is gravity (unrationalised, no units), or electricity (rationalised, no extra units), or light (unrationalised with units)

It should be noted that gravitational magnetic-theory is somewhat behind the corresponding electric version, because the anticipated size of the field is so small that only now it is possible to try to detect the field.

One should imagine that $G$ is a 'falle-constant'. That is, Newton's equation is not usable as a definition of mass in the way that Coulomb's equation might define charge, so the non-variable constants are lumped together in the manner that $K_c$ defines coulomb's constant. It's only when one has a sufficient theory behind it that one tries to modify the value of $G$. 'Falle-constant' here simply means that the units and value of the constant are 'as they fall'.


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