Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs

Dashboard
Notifications
Mark all as read
Q&A

Find jerk of time varying force

+1
−0

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?

Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

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)
Post Feedback (1 comment)

1 answer

+2
−0

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}

Why does this post require moderator attention?
You might want to add some details to your flag.

1 comment thread

Why you wrote acceleration is negative? (3 comments)

Sign up to answer this question »

This community is part of the Codidact network. We have other communities too — take a look!

You can also join us in chat!

Want to advertise this community? Use our templates!