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Find jerk of time varying force

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This gravitational field we move inside has some distance L after which it becomes 0.Before L it is just like any gravitational field. Suppose we move inside that gravitational field.The acceleration we experience depends on the distance from the planet. $$a\sim\frac{1}{r^2}$$

At t=to we enter the gravitational field and assuming the velocity and acceleration was 0 then:

$$x = x_{0}-\frac{1}{6}j(t-t_{0})^3$$

where j is the jerk or

$$\dot{a}\left(t\right)$$

How can we find the jerk?

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3 comment threads

Can't "enter" infinite gravitational field. (4 comments)
I guess I'm a little confused about one of your criteria ("without . . . acceleration"): in this exam... (5 comments)
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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}

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1 comment thread

Why you wrote acceleration is negative? (3 comments)

Comments on Find jerk of time varying force

Why you wrote acceleration is negative?
deleted user wrote about 1 year ago:

Newton said, $g=\frac{Gm}{r^2}$. OK! I am taking $x=r$ and, $a=g$. $a=\frac{Gm}{x^2}$. Although, acceleration is positive. I also know that when we push objects upward then, acceleration becomes negative. But, OP doesn't say we are pushing up or down.

Mithrandir24601‭ wrote about 1 year ago:

In general, you need some kind of reference to distinguish up from down, but lacking that, we normally say 'down' is the direction gravity pulls us. It's an arbitrary choice though, sure. If you swap up and down here, you'll just find a couple of minus sign differences, but as the direction has changed the actual result will be the same

deleted user wrote about 1 year ago:

Gotcha! Thanks.

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