Position as a function of time throught a varying gravitational field

[JJfortherear]

I've been told this is a differential equation, but I have no knowledge of the subject, so I'm hoping it is and that an answer can be found in this forum. The two equations x=1/2at² and a=gm/r², when r is replaced with (r₀-x), with r₀ being the initial radius from the mass from which the gravitational field originates, and x being the distance traveled since t=0, due to the acceleration of the field. Combining these equations gives

x=1/2(gm/(r₀-x)²)t²
I don't know how to solve that equation for x, and I'm hoping this is the place to look for help. Thanks.

 

[Filip Larsen]

The equation x = 1/2at² is the position assuming constant acceleration a, so it does not make much sense to insert a variable acceleration, like from gravity, into it.

The solution you are looking for is a Kepler orbit [1], and if you have the degenerate case of purely radial motion then the solution is a so-called rectilinear Kepler orbit where the eccentricity is 1. For the special case where the initial speed is zero you can also find a relationship between fall time and distance in [2].

[1] http://en.wikipedia.org/wiki/Kepler_orbit
[2] http://en.wikipedia.org/wiki/Free_fall#Inverse-square_law_gravitational_field

 

[JJfortherear]

awesome, exactly what I was looking for. Thanks a lot.