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Jerk during Gravitation

  1. Feb 9, 2012 #1
    Lets suppose that we setup two instances involving two masses m1 and m2.

    m1 > m2

    Case 1. The two bodies are separated by a distance r = 20.
    Case 2. The two bodies are separated by a distance r = 10.

    Suppose that in both setups, we were to release all restraining forces on these two bodies at once and let gravity have its way. Both the accelerations will increase over time because r is decreasing but is the jerk of both cases equal?
  2. jcsd
  3. Feb 9, 2012 #2


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    Staff: Mentor

    What "Jerk" are you referring to?
  4. Feb 9, 2012 #3
    it's the rate of change of acceleration, or [itex]\frac{d^{3}\vec{r}}{dt^{3}}[/itex]
  5. Feb 10, 2012 #4
    [tex]Gravitational\,\,acceleration:\quad\vec{a}_1=\frac{Gm_2\vec{r}}{|\vec{r}|^3}\Rightarrow\dot{\vec{a}}_1=\frac{Gm_2\dot{\vec{r}}}{|\vec{r}|^3}-\frac{3Gm_2 (\dot{ \vec{ r}} \cdot \vec{r}) \vec{r}}{| \vec{r}|^5} where\,\,\dot{q}=\frac{\partial{}q}{\partial{}t} for\,\,any \,\,quantity\,\,q[/tex]This equation holds for all types of motion under only (two body) gravitational forces. Suppose the initial relative velocity of the two objects is zero, then near t=0 (when the two objects are released from rest) the motion is well approximated by uniform acceleration, for which the ratio v/r3 is approximately: [tex]|\frac{\dot{\vec{r}}}{|\vec{r}|^3}|\approx\frac{a_ot}{r_o^3}=\frac{Gm_2}{r_o^5}t[/tex]So, the jerk will initially be zero and increase approximately linearly with time (over distances that are small when compared to the initial separation).
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