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Graavitational Jerk question

  1. Sep 27, 2007 #1
    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 eqation 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?
     
    Last edited: Sep 27, 2007
  2. jcsd
  3. Sep 27, 2007 #2
    you don't know how to take the square of both sides? :confused:

    although i don't understand how you derived that equation
     
  4. Sep 27, 2007 #3
    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
  5. Sep 27, 2007 #4
    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*[tex](s_{0}-s)[/tex] = - at[tex]^{2}[/tex]

    or

    2*[tex](s_{0}-s)[/tex] = at[tex]^{2}[/tex]

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

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

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

    t = - [tex]\sqrt{\frac{2}{Gm}}[/tex] * s *[tex]\sqrt{s_{0}-s}[/tex]

    please criticize.
     
    Last edited: Sep 27, 2007
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