- #1
MCarroll
- 9
- 0
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= [tex]\frac{2Gm}{s^{3}}[/tex] [tex]\frac{ds}{dt}[/tex]
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 = - [tex]\sqrt{\frac{2}{Gm}}[/tex] * s *[tex]\sqrt{s_{0}-s}[/tex]
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 *[tex]\sqrt{s_{0}-s}[/tex]
I don't know how to solve this for s. Any thoughts?
I am aware of the Gravitational Jerk Equation
J= [tex]\frac{2Gm}{s^{3}}[/tex] [tex]\frac{ds}{dt}[/tex]
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 = - [tex]\sqrt{\frac{2}{Gm}}[/tex] * s *[tex]\sqrt{s_{0}-s}[/tex]
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 *[tex]\sqrt{s_{0}-s}[/tex]
I don't know how to solve this for s. Any thoughts?
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