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I'm assuming that what's happening is that, for $t<t_0$, there's no gravitational field, then it's mysteriously instantaneously turned on at $t=t_0$.

We can perform a Taylor expansion of $x$ around $t=t_0$ to give \begin{equation}x = x_0 + \dot{x}\left(t-t_0\right) + \frac{1}{2}\ddot{x}\left(t-t_0\right)^2 + \frac{1}{6}\dddot{x}\left(t-t_0\right)^3 + \cdots.\end{equation}

Now, we say that \begin{equation}\dddot{x} = j = \frac{da}{dx}\frac{dx}{dt}.\end{equation} We know $a$ in Newtonian gravity is \begin{equation}a = -\frac{Gm}{x^2},\end{equation} giving \begin{equation}\frac{da}{dx} = \frac{2Gm}{x^3}.\end{equation}

By a similar argument, we also have that \begin{equation}a = \frac{dv}{dx}\frac{dx}{dt} = vv' = \frac{1}{2}\frac{d\left(v^2\right)}{dx}\end{equation} which gives that \begin{equation}v^2 = \frac{2Gm}{x} + C.\end{equation} Assuming that at $x=x_0$, $v=0$ gives that \begin{equation}v^2 = \frac{2Gm}{x} - \frac{2Gm}{x_0}.\end{equation}

This gives us our value of jerk as \begin{equation} j = \frac{2Gm}{x^3}\sqrt{\frac{2Gm}{x} - \frac{2Gm}{x_0}}.\end{equation}