Newton's Law of Graviation: Why G?

Question

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?

Answer 1

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.