<sajid@umerji.fsworld.co.uk> wrote in message 
news:1108502672.596280.145920@c13g2000cwb.googlegroups.com...
> Suppose I drop a ball from a height h, what will the height of the 
> ball
> be at time t ?
>
> I know how to calculate the solution under the assumption that the
> acceleration is constant and equal to g.
>
> But I'm trying to find the solution where the acceleration is equal 
> to:
>
> -GM/x^2
>
> Where x is the height of the ball and M is the mass of the Earth.
>
> Also, to simplify things, I'm assuming the ball does not have any
> gravitational   effect on the Earth.
>
> I can see that:
>
> v^2/2 = GM (1/x - 1/r)
>
> And therefore that:
>                         ___
> (1/x - 1/r)^-1/2 dx = \/2gm t
>
> But I don't know how to calculate the integral on the left hand side
> ...
>
> I would appreciate some help.

 The fundamental theorem of calculus is the statement that the two 
central operations of calculus, differentiation and integration, are 
inverses of each other.
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

Differentiation:

distance h (or x)
time t

speed = derivative of diatance with respect to time = dx/dt  (h/t)

acceleration = the derivative of the speed with respect to time
                   = d(dx/dt) /dt
                   = d^2x / dt^2

Integration:

Reverse the process of differentiation to get back the value of x (or h).

a = g
v = gt
h = (1/2) gt^2

Example
the ball falls for 1 second at 32 fps/s
a = 32
v = 32*1
h = (1/2) 32*1*1
   = 16 ft.

Draw a velocity/time graph. (fixed font needed)
V
32     /
|     /
|    /
|   /
|  /
| /
|/______1_____ t

The area under the line is 1/2 the area of the rectangle 1 wide * 32 high,
because the rectangle is bisected by the line.
So the ball's instantaneous velocity after falling 16 feet for 1 second 
is 32 fps.

The calculation you are attempting, F=GmM/r^2, is for FORCE, not 
acceleration.
F = dp/dt, p = mv, the momentum . This is often written as F = ma.

ma = GmM/r^2
a  = GM/r^2

G remains a universal constant, and the bigger the mass M, the greater
the acceleration.

Hope that helps.

Androcles