Quote by Nabeshin
I have no idea what it is you're trying to prove here, but whatever it is, you're going about it wrong. First off, you state that you want to prove [itex] V_{esc} \gg v[/itex], what is [itex]v[/itex]? The characteristic velocity of stars in the galaxy? I have absolutely no idea what it is you're trying to derive, it just looks like mindless symbol manipulation and combination of equations which leads me to the second point...
Doing anything on a galactic scale, you cannot use Hubble's velocity relation. It's not merely that the recessional velocities predicted by it will be much smaller than typical velocities of galactic motion (is this what you're trying to prove???), hubble expansion DOESN'T HAPPEN in gravitationally bound clusters. Furthermore, this is completely separate from the cosmological constant! In a traditional LambdaCDM model, the density of dark energy is constant, and gravitationally bound objects (like our galaxy) will stay gravitationally bound indefinitely. (I say traditional because there are some models in which Lambda is somehow a function of time, and grows, leading to 'big rip' scenarios).
As a final point, your equation for escape velocity is incorrect:
[tex]v_{esc}=\sqrt{\frac{2GM}{r}}[/tex]
While
[tex]v_{orbit}=\sqrt{\frac{GM}{r}}[/tex]
Where this [itex]v_{orbit}[/itex] is the velocity of a circular orbit of radius r.

Hi Nabeshin .
I think it would be helpful if I post the actual question than wrongly restating it.
When you say:
the density of dark energy is constant, and gravitationally bound objects (like our galaxy) will stay gravitationally bound indefinitely.

I have few follow up questions to ask..
1) Does that refer to a cluster of galaxies ?
2) if dark energy density stays constant does that mean it's homogeneous , as in not being concentrated at a point ?
Here's the question:
http://i42.tinypic.com/8wzh20.png
P.S : 'v' is the recessional speed of the galaxy.