To add to what Dale was saying, you might use cylindrical coordinates for planetary orbits, where the axial dimension is time. For one orbit of the Earth around the Sun, the equivalent 'distance' covered in time is one light year. It's a very stretched-out spiral.
I'm not sure this really helps, but the Newtonian gravitational potential is expressed in terms of the mass density as
\nabla^{2}\Phi = 4\pi G \rho
\ \rho is mass density.The mass density is the time-time component of the stress energy tensor \ T_{00}.
\nabla^{2}\Phi = 4\pi G T_{00}
Where we're only concerned about the gravitational field of stationary mass, the gravitational potential, in the weak field limit is related to the metric
g_{00} = - \left(1 + 2\Phi \right)
\ \Phi is a small perturbation on the Minkowski metric.
I should add that this is an approximation, not an exact solution, but where \ \Phi << 1
How this squares with what A.T. was saying, I don't know. This is how Sean Carroll presents it.