# Is the curl of the gravitational field required to fully describe Newtonian gravity?

We are familiar with Newton's law of gravitation:

$$\textbf{F} = \frac{-GMm}{r^2} \hat{\textbf{r}},\tag{1}$$

which leads to a gravitational field strength relation:

$$\textbf{g} = \frac{-GM}{r^2} \hat{\textbf{r}}. \tag{2}$$

In terms of vector calculus we can write this in the form:

$$\nabla\cdot\textbf{g} = -4\pi G \rho \tag{3}$$

(where $\rho$ is the mass density) in analogy with Coulombs law and Maxwells first law.

My question is whether this equation is sufficient to fully describe (Newtonian) gravitation, or whether a relation for the curl, $$\nabla \times \textbf{g} = 0,\tag{4}$$ is also required?

If so, it seems unusual to me that Newton's force law requires just one equation, yet a vector calculus approach would require two. (But maybe that's just the way it is!)

Yes, the Newtonian gravitational field ${\bf g}$ is also required to be rotation-free $\nabla \times {\bf g} = 0$. This also follows from the existence of a Newtonian gravitational potential.