What is the force of the cosmological constant?

Quite accurate on small scales. It's a pseudoforce, like gravity.

Yes. In a Newtonian approximation, you may describe it as the gravitational effect of some uniform "dark energy" with a negative mass density. So, essentially:
[tex]F=4/3 \pi G \rho r m[/tex]
[tex]\rho[/tex] is the double density of dark energy (2*6.8*10^-26 kg/m³).

 

Yes, but gravity is not a force, it's spacetime curvature. You can treat it as a force if you can reasonably define a flat background. If not, the "force-picture" is not really helpful.

Gravity behaves exactly the same way. I'm not considering a point mass here, but uniformly distributed matter. The amount of matter within distance r scales as r³, gravity as r^-2, the net effect scales as r.

You asked for gravitational force, not acceleration. m is the arbitrary mass of the body that experiences the force. The force is proportional to m, the acceleration is independent of it.

Dark Energy has a negative pressure in each space dimension that equals its energy density. As there are three space dimensions, that's three times the energy density. Negative pressure repels the same amount that positive energy density attracts, so there's a net effect of -2 times energy density.

 

Force may not be the best of terms. 'Force' is not a very good way to characterize the effects of gravity, as einstein surmised. I suspect the cosmological constant, as is gravity, may be an artifact of spacetime topology - perhaps different manifestations of the same basic effect. Another way of looking at it - does gravity suck, or space push?

 

Have a look at http://math.ucr.edu/home/baez/einstein/" [Broken] (I'm using the simplified form with Lambda absorbed in the stress-energy-tensor).

Not for a cosmological constant. One might tweak some form of Quintessence to give the desired result (p=-rho/5), but that doesn't seem compatible with observations.

 

Simply integrate and take the negative, that gives
[tex]
\Phi=-2/3 \pi G \rho r^2
[/tex]

Gravitational force increases exactly like "Cosmological constant force". We're not talking about a point mass here, rather a homogeneous sphere. Look at http://www.strw.leidenuniv.nl/~franx/college/sterrenstelsels07/handout2.pdf" [Broken], p. 4/7.

 

Aye, there's the rub. The universe is not a homogenous sphere.

 

Sorry.
I gave you a link with the correct answer: Potential energy increases as r² in a homogenoeus sphere.
You need some handwaving to argue that one may neglegt contributions from outside a sphere in newtonian mechanics. But you can derive the result easily from GR in a Newtonian approximation: from any given center, the attraction towards / repulsion from that center at distance r is caused by the amount of energy and pressure in a sphere with radius r.