# Newton's Law of Graviation: Why G?

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?

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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. –  Cheeku Mar 11 '13 at 0:20
@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). –  Mew Mar 11 '13 at 0:23
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. –  Jerry Schirmer Mar 11 '13 at 1:04

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.

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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. –  Ginger Bill Mar 11 '13 at 18:47