Graavitational Jerk question

1. Sep 27, 2007

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 eqation 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?

Last edited: Sep 27, 2007
2. Sep 27, 2007

ice109

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

although i don't understand how you derived that equation

3. Sep 27, 2007

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.

Last edited: Sep 27, 2007
4. Sep 27, 2007

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 *$$\sqrt{s_{0}-s}$$