BeedS said:
If I take objects for Example the particles of an atomic system and treat them as a gravitational bound “system”… ? ...
You mean the gravitational field of a single atom? It's too tiny for is to know if GR models it correctly. Assuming it does, though, you'd need such incredibly precise measurements to detect the field that it's impossible to say how close you could go and still treat it as a point source.
Note that an atom is not gravitationally bound. Atoms are held together by strong forces in the nucleus and electromagnetic forces between the nucleus and electrons.
BeedS said:
Yes, I was thinking for external objects located outside the system, interacting with the system.
Basically, any system will look like a point source when you're far enough from it. But what "far enough" means in practice depends on how precisely you're measuring. For a simple example, consider the case of the Sun, mass ##M_S## and distance ##R## away from you, and the Earth, mass ##M_E## and distance ##R+r## away from you. We'll have you, the Earth and the Sun in a straight line so we can be lazy about vectors and use Newtonian gravity, because that's valid way before pretending the Earth and Sun are a single point is valid.
The gravitational acceleration you feel is $$\begin{eqnarray*}
g&=&\frac{GM_S}R+\frac{GM_E}{R+r}\\
&=&\frac{GM_S}R+\frac{GM_E}{R}\frac{1}{1+r/R}\\
&=&\frac{GM_S}R+\frac{GM_E}{R}\left(1-\frac{r}{R}+\frac{r^2}{R^2}-\frac{r^3}{R^3}+\ldots\right)\\
&=&\frac{G(M_S+M_E)}R-\frac{GM_E}{R}\left(\frac{r}{R}-\frac{r^2}{R^2}+\frac{r^3}{R^3}-\ldots\right)
\end{eqnarray*}$$Whether you can detect the difference between that and ##G(M_S+M_E)/R^2## depends on whether the experiment you are doing is far enough away that your measurement error is bigger than that last term in the brackets - that is that your distance from the Sun, ##R##, is so much larger than the Earth-Sun distance, ##r##, that all those terms in the brackets are so nearly zero you don't care.
