Deriving the Gravitational Jerk Equation and Solving for Time and Position

[MCarroll]

Is anyone aware of a function expressesing position (separation) in terms of time, s=f(x), given only acceleration due to the force of gravity.

I am aware of the Gravitational Jerk Equation

J= \frac{2Gm}{s^{3}} \frac{ds}{dt}

where

m=mass of attracting object
G = gravitational constant

But I can't resolve this for t.

Although I have derived an equation giving dt = f(s) as

t = - \sqrt{\frac{2}{Gm}} * s *\sqrt{s_{0}-s}

given v(o) = 0 (start at rest)

where

s = instantaneous separation
s(o) = orignal separation

which for a given mass of attractor m simplifies to

t = - K * s *\sqrt{s_{0}-s}

I don't know how to solve this for s. Any thoughts?

 

[ice109]

you don't know how to take the square of both sides?

although i don't understand how you derived that equation

 

[MCarroll]

I got some help on the Math forum so I think I am ok but I was not aware of the resolvant quadratic necessary to solve the cubic of s that results from squaring both sides. I should have mentioned my limited math ability in the first place.

I'll show my derivation of t = f(s) when I have some more time, welcoming all criticism.

 

[MCarroll]

EDIT/

here is what I have

s = s(o) - (at^2)/2 where v(0) = 0 and movement is in the negative direction with respect to scalar s(o).

2*(s_{0}-s) = - at^{2}

or

2*(s_{0}-s) = at^{2}

\sqrt{2*(s_{0}-s)} = t * \sqrt{a}

t = \sqrt{2*(s_{0}-s)/a}

but a = - Gm/(s^2), so

t = - \sqrt{\frac{2}{Gm}} * s *\sqrt{s_{0}-s}

please criticize.