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!)
 A: 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.
A: You can show that if there exists a scalar potential for the field, it will also be curl-free. So the relation is required, but stating it separately is redundant, as it is a consequence of the existence of the potential.
A: Yes, you also need
$$\nabla \times \mathbf{g} = \mathbf{0}$$
This is exactly analogous to the situation in electrostatics, where Gauss's law
$$\nabla \cdot \mathbf{E} = \frac{\rho_q}{\epsilon_0}$$
is insufficient to give the electric field, and we also need
$$\nabla \times \mathbf{E} = \mathbf{0}$$
(together with some assumptions about the underlying "vacuum" [i.e. no radiation is present])
which itself is a special case of Faraday's law
$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$
when the magnetic field remains unchanged with time. In the case of Newtonian gravitation, there is no "gravimagnetic field", as all changes propagate instantaneously, so the right-hand side of the gravitational equation is always zero.
