Move the positive <cosmological constant> to the source side and define
$$
T^{(\Lambda)}_{\alpha\beta}=-\frac{\Lambda}{8\pi}g_{\alpha\beta}.
$$
To first order about <Minkowski spacetime>, $T^{(\Lambda)}=-\Lambda/(2\pi)$. Its contribution to the trace-reversed $00$ source is
$$
T^{(\Lambda)}_{00}-\frac12\eta_{00}T^{(\Lambda)}=-\frac{\Lambda}{8\pi}.
$$
Repeating the <Newtonian limit of general relativity> calculation gives
$$
\boxed{\nabla^2\Phi=4\pi\rho-\Lambda}.
$$
For a point mass, $\nabla^2r^2=6$, so
$$
\Phi=-\frac Mr-Br^2,
\qquad -6B=-\Lambda,
\qquad
\boxed{B=\frac\Lambda6}.
$$
The radial acceleration is
$$
-\frac{d\Phi}{dr}=-\frac M{r^2}+\frac\Lambda3r.
$$
Thus positive $\Lambda$ supplies an outward acceleration and balances the point-mass attraction at the <static radius>
$$
r_{\rm s}=\left(\frac{3M}{\Lambda}\right)^{1/3}.
$$
The <weak-field approximation> requires both $M/r\ll1$ and $\Lambda r^2\ll1$, giving parametrically
$$
\boxed{M\ll r\ll\Lambda^{-1/2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 37D / iv / Solution

{title2=$r_{\rm s}=(3GM/\Lambda)^{1/3}$}

For a point mass in a positive <cosmological constant>, the inward Newtonian acceleration $GM/r^2$ balances the outward cosmological acceleration $\Lambda r/3$ at the static radius $r_{\rm s}=(3GM/\Lambda)^{1/3}$.

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 Incoming links: Static radius