*Originally posted by mtthw*

**Gravitational attraction.
**

Basically, you need to solve this:

d^2x/dt^2 = -GM/4x^2

where x(0) = d/2 and dx/dt(0) = 0

What is t when x = r?

(x being the distance between the mid-point and the centre of one of the spheres)

OK.. so what you are saying is that I have to solve the integral from d to r for t of d^2x/dt^2 = -GM/4x^2 to arrive at the solution?

but that seems the wrong way round... if I solve that I'm solving for x aren't I?

God it is a long time ago.. i used to be really good at this 🙁

d^2x/dt^2 = -GM/4x^2 is d^2x/dt^2 = -GM/4 * (x^-2) because -GM/4 is constant, G being 6.67 * 10^-11 and with M = 1 kg (yes?)

so the only thing to worry about is x^-2

dx/dt = GM/2 * (x^-1) +C

x/t = GM/2 *ln|x| +Cx +D

ehm.. t = GM/2 *xln|x| +Cx^2 +Dx

and x = 45 .. calculator... assume C and D are 0 (what are C and D?)

..i'm lost... I have no idea if this gibberish makes sense.