2 pt. masses @rest in space. 10m apart. Gravity is only acting force. Each weighs m=1kg. when do they meet

In order to solve this problem you need to know what force is acting upon them... First, let's define some variables.

M1 = Mass #1
M2 = Mass #2
F1 = Force acting upon Mass #1
F1 = Force acting upon Mass #2
d = Distance between masses
G = Gravitational Constant = 6.674 * 10^(-11) N(m/kg)^2

F1 = F2 = G* (M1*M2)/(d^2)

So -

F1 = F2 = (6.674 * 10^(-11) N(m/kg)^2) * (1 kg)(1kg)/(10 m)^2

F1 = F2 = 6.674 * 10^(-13) N

Using this we can determine their acceleration by F= ma -or rather- a = F/M

a1 = a2 = F/M = (6.674 * 10(-13) N)/(1kg) = 6.674 * 10(-13) m/s^2

Since we know they have equal masses and they have equal forces acting upon them, the two will meet in the middle at 5 m. In order to determine how long it takes them to meet, we need to use the following equation...

distance = (1/2) a*t^2

solving for t...

t = (2d/a)^1/2

substituting....

t = ( (2 * 5m)/(6.674 * 10(-13) m/s^2) )^1/2

Answer:

t = 3870854.96 seconds (or about 45 days)

Helicoid and Brian are correct. The above is wrong. The problem is more difficult.... Sorry for my error.

I have a tiny bit of experience with ode's heres an idea
x'' = 2*G / (R - 2*x)^2 from brians answer
Integrate
x' = 2*G(integral(1/((R-2x)^2)dx
Since x'=v
And d=vt or t=d/v
Then t=d/( 2*G(integral(1/((R-2x)^2)dx)
/d=5 R=10
//x =??? I think you do a definite integral over the five meters but ima little confused...

Actually mike is incorrect. You cannot use a constant value for force in this problem. Since force depends on distance, the force between the objects will change as the objects move closer.