Monday, November 28, 2016

newtonian gravity - Newton's Law of Graviation: Why $G$ and not e.g. $dfrac{1}{4pi G_0}$?


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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