Newtonian Gravity involving spheres in a trigonal pyramid

[JennV]

Homework Statement

In deep space, three spherical masses are held in fixed positions (by rods light compared to the masses) at three corners of a square of side length sqrt(18) m as shown. The masses of the three spheres are m1 = m3 = 1.76×10^9 kg, and m2 = 2.75×10^9 kg. A relatively small fourth mass is placed at point A, and released from rest. By symmetry it accelerates straight down.

What is the speed of the fourth mass when it reaches point B, the midpoint between the two 1.76×109 kg spheres? (Note the acceleration is not constant, so you can't use constant acceleration kinematics.)

http://img69.imageshack.us/img69/9571/ch12u.jpg

Homework Equations

GMm / r

The Attempt at a Solution

I found that the fourth small mass falls a distance of approximately 8.999973204m by:
The distance between m1 and m3 is c^2 = (sqrt18)^2 + (sqrt18)^2, which means that the distance between m1 and m3 is 6m.
Then when I know that the distance between m1 and m3 is 6m, I also know that that is the base of the triangle that the fourth small mass fall in between. Since the problem says that the fourth mass falls straight down the middle, so I know that the whole top angle is 36.87 degrees, so one half of that is 18.435 degrees. Now we have a right triangle, so I can use tan to find the distance that the fourth mass falls by 3m/tan18.435 = 8.999973204m. (since the base of 6m, one half of that is 3m).
I found that, but now I'm not sure how to approach by finding the speed of the fourth mass when it reaches point B.

THANK YOU IN ADVANCE.

 

[lanedance]

to find the speed the easiest way would be to compare the gravitational potential of the original configuration to that when the small mass is at B...