How to get space component of weak field (linearized) metric? For Minkowski space with a weak gravitational field the metric takes the form
$$
ds^2 = (1+2\phi/c^2)c^2dt^2 -(1-2\phi/c^2)(dx^2+dy^2+dz^2),
$$
where $\phi$ is the Newtonian gravitational potential.
You can get the $(1+2\phi/c^2)$ factor in front of the $dt^2$ by starting with the geodesic equation and going to the "Newtonian limit" of slow speeds and a slowly changing field.
But is there a way to get the $(1-2\phi/c^2)$ factor for the spatial part of the metric by a similar procedure? What about some clever thought experiment?
 A: *

*The spatial components of the metric 
$$ g_{\mu\nu}~=~\eta_{\mu\nu}+h_{\mu\nu}
~=~\begin{bmatrix}  \mp (1+2\phi/c^2) & 0 \cr 0 &  \pm(1-2\phi/c^2){\bf 1}_{3\times 3}\end{bmatrix}_{4\times 4}
\tag{1}$$ 
can be deduced from the fact that only the temporal component $$\bar{h}_{00}~=~\mp4\phi/c^2\tag{2}$$ 
of the trace-reversed metric perturbation 
$$\bar{h}_{\mu\nu}~:=~h_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}h
\qquad\Leftrightarrow\qquad
h_{\mu\nu}~=~\bar{h}_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}\bar{h}.  \tag{3}$$ 
should survive in the Newtonian limit in order for the linearized EFE to decouple, cf. e.g. this & this related Phys.SE posts. [Here the $\mp$ sign refers to the Minkowski signature convention $(\mp,\pm,\pm,\pm)$.] 

*As for OP's suggestion to use the geodesic equation, note that the spatial components of the metric cannot be reliable extracted in the Newtonian limit as it would be higher order in speed.
A: Start with Einstein's equation
\begin{equation}
 R_{ab}-\frac{1}{2}  g_{ab} R = \kappa\,T_{ab}.
\tag{1}
\end{equation}
Contract both sides of (1) with $g^{ab}$ to get
$$
 R-\frac{N}{2} R = \kappa\,T
\tag{2}
$$
where $N$ is the number of spacetime dimensions and $T$ denotes the trace of $T_{ab}$, just like $R$ denotes the trace of $R_{ab}$. Use (2) in (1) to get
$$
 R_{ab} = \kappa\,\left(T_{ab}+ \frac{g_{ab}}{2-{N}}T\right).
\tag{3}
$$
In the weak-field approximation and with a convenient choice of gauge (as described in many textbooks), we have
$$
    R_{ab}\approx -\frac{1}{2}\partial^2 h_{ab}
\tag{4}
$$
with $h_{ab}\equiv g_{ab}-\eta_{ab}$. Using this in (3) gives
$$
-\frac{1}{2}\partial^2 h_{ab}\approx \kappa\,\left(T_{ab}+ \frac{g_{ab}}{2-{N}}T\right).
\tag{5}
$$
In the usual approximation with $T_{00}$ being the only significant component of $T_{ab}$, we have
$$
T\approx T_{00}.
\tag{6}
$$
Use this in (5) to get
$$
-\frac{1}{2}\partial^2 h_{ab}\approx \kappa\,\left(T_{ab}+ \frac{g_{ab}}{2-{N}}T_{00}\right).
\tag{7}
$$
In particular, equation (7) implies
$$
-\frac{1}{2}\partial^2 h_{00}\approx \kappa\,\frac{3-{N}}{2-{N}}T_{00}
\tag{8}
$$
and
$$
-\frac{1}{2}\partial^2 h_{jk}\approx -\kappa\,\frac{\delta_{jk}}{2-{N}}T_{00}
\tag{9}
$$
for $j,k\neq 0$. For the physically-relevant case $N=4$, equations (8)-(9) are consistent with
$$
h_{jk}\approx \delta_{jk}h_{00}.
\tag{10}
$$
This is consistent with the equation written in the OP, with $2\phi\equiv h_{00}$.
A: The spatial part of the metric is important when test particles have relativistic velocities, like photons. So you need this form of the metric when studying deflection of light by the sun, which was one of the first and significant experimental verifications of GR.
