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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!)

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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.

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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.

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