= Solution
Use geometric units $G=c=1$, <metric signature> $(-+++)$, and a <four-velocity> normalized by the physical metric. Write $\phi=1+\Phi$, where $|\Phi|\ll1$. For a pressureless <perfect fluid>, the <dust stress-energy tensor> is $T_{ij}=\rho U_iU_j$ and $g^{ij}U_iU_j=-1$. Since $\eta^{ij}=\phi^2g^{ij}$, its background trace is
$$
\eta^{ij}T_{ij}=-\rho\phi^2.
$$
Consequently the field equation is $(-\partial_t^2+\Delta)\phi=4\pi\rho\phi^3$. Neglecting time derivatives and higher weak-field orders gives the <Newtonian limit of Nordstrom gravity>:
$$
\boxed{\Delta\Phi=4\pi\rho,\qquad\Phi=\phi-1.}
$$
The constant shift between $\phi$ and $\Phi$ does not affect the <Poisson equation> or the force. Here $\rho$ is the physical rest density; if fluid indices are instead defined using the background metric, the intermediate powers of $\phi$ change but the same leading-order equation results.
For the motion, the <Levi-Civita connection> of the <conformally flat metric> is
$$
\Gamma^i{}_{jk}=\delta^i_j\partial_k\log\phi+\delta^i_k\partial_j\log\phi-\eta_{jk}\eta^{i\ell}\partial_\ell\log\phi.
$$
In particular $\Gamma^\alpha{}_{00}=\partial_\alpha\log\phi$. Replacing the <affine parameter> in the <geodesic equation> by coordinate time gives, with $w^i=(1,\boldsymbol v)$ and $\boldsymbol v=d\boldsymbol x/dt$,
$$
\frac{d^2x^\alpha}{dt^2}=-\Gamma^\alpha{}_{ij}w^iw^j+\Gamma^0{}_{ij}w^iw^jv^\alpha
=-(1-|\boldsymbol v|^2)\left(\partial_\alpha\log\phi+v^\alpha\partial_t\log\phi\right).
$$
At small speeds and for a nearly static field, this becomes
$$
\boxed{\frac{d^2\boldsymbol x}{dt^2}=-\boldsymbol\nabla\Phi.}
$$
Thus the <Newtonian gravitational potential> and freely falling trajectories agree to the stated weak-field, slow-motion order. The comparison is an approximation, not an exact equivalence of relativistic trajectories.
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