Find jerk of time varying force

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 $$x=x_0+\dot{x}(t-t_0)+\frac{1}{2}\ddot{x}{(t-t_0)}^2+\frac{1}{6}\stackrel{⃛}{x}{(t-t_0)}^3+\cdots .$$

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

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

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

posted over 3 years ago

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Mithrandir24601‭ staff

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