How to show that a radially symmetric central force is conservative?

Question

Let $U\subseteq \mathbb{R}^3$ be open and $f:U\to\mathbb{R}^3$ be a radially symmetric central force, that is, a force field such that $$f(p) = -g(r)u_r$$ where $r=|p|$ and $u_r$ is the unit vector pointing to $p$ for some integrable (this will be important later) function $g:\mathbb{R}_{\ge 0}\to \mathbb{R}$. I want to show that $f$ is a conservative force, and in particular that $$f=-\nabla (G\circ r)$$ where $G' = g$. Put differently, I wish to show that $$\int_Cf \cdot dr = (G\circ r) (t_0) - (G\circ r) (t_1)$$ for some bijective parametrization $r:[t_0,t_1]\to C$ of the smooth curve $C\subseteq U$.

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My interest in the above came after seeing a sort of argument for the above regarding planar curves. In case it is of any help, I include it here. It is from Spivak's Physics for Mathematicians:

Answer 1

Let $\frac{d}{dr} G = g$. Then in spherical coordinates $\vec{r} = r \vec{e_r}$, and $\nabla = \vec{e_r}\frac{\partial}{\partial r} + \vec{e_{\theta}}\frac{\partial}{r\partial \theta} + \vec{e_{\phi}}\frac{\partial}{r\sin(\theta)\partial \phi} $. Then
\begin{equation} -\nabla G(r) = -\vec{e_r}\frac{\partial G}{\partial r} = - g \frac{\vec{r}}{r} \end{equation} since there is no $\theta, \phi$ dependence for $g$ and $G$. By comparison with $f$ the claim follows.

Answer 2

Suppose $U$ a simply connected open subset of $\Bbb R^3$ and $\mathbf F:U\to \Bbb R^3$ a central force, which means that $\mathbf F(\mathbf r)=f(\mathbf r)\mathbf {\hat r}$. If $\mathbf F$ is also spherically symmetric, then $f(\mathbf r)=f(r)$, because $f$ depends only on $|\mathbf r|=r$, thus $\mathbf F(\mathbf r)=f(r)\mathbf {\hat r}$. We calculate $(\nabla \times\mathbf F)(\mathbf r)=\frac {1}{rsinθ} \frac {\partial f(r)}{\partial φ}\mathbf {\hat θ}-\frac 1r \frac {\partial f(r)}{\partial θ}\mathbf {\hat φ}=\mathbf 0$, thus $\mathbf F$ is conservative.