Q:

Newton's law of gravitation in de Sitter space

Given two masses $M$ and $m$ (with $M\gg m$) in a de Sitter background with cosmological constant $\Lambda>0$ and positive spatial curvature ($k=+1$). What is the corresponding (semiclassical "Newtonian") gravitational force between $M$ and $m$?
Form the $g_{00}$ component of the static Schwarzschild-de Sitter solution of the Einstein field equations I would naively guess
$$F\approx -G\frac{Mm}{r^2}+\frac{\Lambda c^2}{3} m \,r,$$
with gravitational constant $G$ and distance $r$. Actually, the second term in this expression is repulsive. Since I have not found any hint in literature I would like to address this question here. 

A:

Your force is correct, that is also the expression for $\ddot{r}$ in the real Schwarzschild De Sitter metric when you set the first proper time derivatives of the spatial coordinates to zero:
The geodesic equation gives the radial component of the 4-acceleration (in natural units):
$$ \ddot{r}= \color{gray}{ \frac{\left(\Lambda  r^3-3\right) \dot{r}^2}{r \left(\Lambda  r^3-3 r+6\right)} } -\frac{\left(\Lambda  r^3-3 r+6\right) \left(\color{gray}{ 3 r^3 \left(\dot{\theta}^2 +\sin ^2 \theta \ \dot{\phi}^2\right) }+\left(\Lambda  r^3-3\right) \dot{t}^2\right)}{9 r^3} $$
where you set $\dot{r}=\dot{\rm \theta}=\dot{\rm \phi}=0$ and plug in
$$ \dot{t}=\sqrt{g^{t t}} \ \color{gray}{ \gamma } = \sqrt{\frac{1 \ / \ (1-2/r-\Lambda r^2/3)}{1-\color{gray}{ v^2 }}}$$
with $v=0$, where $v$ is the velocity measured by local and stationary (relative to the dominant mass) Fidos, then you get
$$\ddot{r} = -\frac{1}{r^2}+\frac{\Lambda r}{3}$$
which is, in natural units, the expression you correctly guessed. The overdot is the differentiation with respect to proper time, but in the newtonian limit the proper and coordinate time are the same.

